Automatic Control Systems 9th Edition Solution Manual
So lu t io ns M an ua l
AutomaticControlSystems,9thEdition A
Chapter2Solution ns
Golnarraghi,Kuo
C Chapter 2 2 21(a) Poless:s=0,0,1, 10;
(b) Poles:s=2,,2;
Zeross:s=2,f,f,f.
Zeros:s=0.
Thepoleandzeroats=1ccanceleachotther.
( Poles:s=0,1+j,1j; (c)
(d)Poles:s =0,1,2,f.
Zeross:s=2.
2 2-2)
ሺ௦ାଵሻ
a)
ܩሺݏሻ ൌ
b)
ܩሺݏሻ ൌ ሺ௦ାଵ ሺ ሻሺ௦ାସ
c)
ܩሺݏሻ ൌ
௦ሺ௦ା ௦ ଶሻሺ௦ାଷሻమ ௦మ
ሻ
௦ మ ିଵ ௦ మ ሺ௦ାଷሻሺ௦ାଵሻమ
2 2-3) M MATLABcode e:
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AutomaticControlSystems,9thEdition
Chapter2Solutions
clear all; s = tf('s')
'Generated transfer function:' Ga=10*(s+2)/(s^2*(s+1)*(s+10)) 'Poles:' pole(Ga) 'Zeros:' zero(Ga)
'Generated transfer function:' Gb=10*s*(s+1)/((s+2)*(s^2+3*s+2)) 'Poles:'; pole(Gb) 'Zeros:' zero(Gb)
'Generated transfer function:' Gc=10*(s+2)/(s*(s^2+2*s+2)) 'Poles:'; pole(Gc) 'Zeros:' zero(Gc)
'Generated transfer function:' Gd=pade(exp(-2*s),1)/(10*s*(s+1)*(s+2)) 'Poles:'; pole(Gd) 'Zeros:' zero(Gd)
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AutomaticControlSystems,9thEdition
Chapter2Solutions
Polesandzerosoftheabovefunctions: (a) Poles:00101 Zeros:2 (b) Poles:2.00002.00001.0000 Zeros:01 (c) Poles: 0 1.0000+1.0000i 1.00001.0000i Zeros:2 Generatedtransferfunction: (d)usingfirstorderPadeapproximationforexponentialterm Poles: 0 2.0000 1.0000+0.0000i 1.00000.0000i Zeros: 1
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AutomaticControlSystems,9thEdition
Chapter2Solutions
Golnaraghi,Kuo
2-4) Mathematical representation: In all cases substitute ݏൌ ݆߱ and simplify. The use MATLAB to verify. R
10( jZ 2) 2 I1 Z ( jZ 1)( jZ 10) 10( jZ 2) ( jZ 1)( jZ 10) u 2 Z ( jZ 1)( jZ 10) ( jZ 1)( jZ 10) 10( jZ 2)( jZ 1)( jZ 10) a) I2 Z 2 (Z 2 1)(Z 2 100) jZ 2 jZ 1 jZ 10 R 22 Z 2 1 Z 2 102 Z 2 R(e jI1 e jI2 e jI3 ) I3
10 22 Z 2 1 Z 2 102 Z 2 ; Z 2 (Z 2 1)(Z 2 100)
Z 2 tan 1 2 Z 2
2 2 Z2 2
Z 2 tan 1 1 Z
Z
1 Z2
2 2 tan 1 10 Z
I I1 I2 I3
1
10 102 Z 2
R
10 I1 ( jZ 1) 2 ( jZ 3) 10 ( jZ 1)( jZ 1)( jZ 3) u ( jZ 1)( jZ 1)( jZ 3) ( jZ 1)( jZ 1)( jZ 3) 10( jZ 1)( jZ 1)( jZ 3) b) I2 (Z 2 1) 2 (Z 2 9) jZ 1 jZ 1 jZ 3 R 1 Z2 1 Z2 9 Z2 R (e jI1 e jI2 e jI3 ) I3
10 1 Z 2 9 Z 2 ; (Z 2 1) 2 (Z 2 9) Z 2 tan 1 1 Z
Z 2 tan 1 1 Z
Z 2 tan 1 9 Z
I I1 I2 I3
24
1 1 Z2
1 1 Z2
3 9 Z2
AutomaticControlSystems,9thEdition
Chapter2Solutions
10 jZ ( j 2Z 2 Z 2 ) 10 j (2 Z 2 j 2Z ) u Z ( j 2Z 2 Z 2 ) (2 Z 2 j 2Z ) 10(2Z (2 Z 2 ) j ) Z (4Z 2 (2 Z 2 ) 2 )
c)
R
R
2Z (2 Z 2 ) j 4Z 2 (2 Z 2 ) 2
R (e jI )
10 4Z 2 (2 Z 2 ) 2 Z (4Z 2 (2 Z 2 ) 2 )
10
Z 4Z (2 Z 2 ) 2 2
;
2 Z 2
I
tan 1
4Z (2 Z ) 2
2 2
2Z 4Z 2 (2 Z 2 ) 2 R
e 2 jZ 10 jZ ( jZ 1)( jZ 2) j ( jZ 1)( jZ 2) 2 jZ I1 e 10Z (Z 2 1)(Z 2 2) d) jZ 2 jZ 1 2 jZ jS / 2 R e 22 Z 2 1 Z 2 I2 R (e jI1 e jI2 e jI3 )
1 10Z 2 Z 2 1 Z 2 Z 2
2 2 tan 1 2 Z
2 2 Z2 2
Z 2 tan 1 1 Z
I I1 I2 I3
1 1 Z2
MATLABcode: clear all; s = tf('s')
'Generated transfer function:' Ga=10*(s+2)/(s^2*(s+1)*(s+10)) figure(1)
25
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AutomaticControlSystems,9thEdition
Chapter2Solutions
Nyquist(Ga)
'Generated transfer function:' Gb=10*s*(s+1)/((s+2)*(s^2+3*s+2)) figure(2) Nyquist(Gb)
'Generated transfer function:' Gc=10*(s+2)/(s*(s^2+2*s+2)) figure(3) Nyquist(Gc)
'Generated transfer function:' Gd=pade(exp(-2*s),1)/(10*s*(s+1)*(s+2)) figure(4) Nyquist(Gd)
Nyquistplots(polarplots): Part(a)
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AutomaticControlSystems,9thEdition
Chapter2Solutions
Golnaraghi,Kuo
Nyquist Diagram 15
10
Imaginary Axis
5
0
-5
-10
-15 -300
-250
-200
-150
-100
-50
0
Real Axis
Part(b) Nyquist Diagram 1.5
1
Imaginary Axis
0.5
0
-0.5
-1
-1.5 -1
-0.5
0
0.5
1
1.5
2
2.5
Real Axis
Part(c)
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Chapter2Solutions
Golnaraghi,Kuo
Nyquist Diagram 80
60
40
Imaginary Axis
20
0
-20
-40
-60
-80 -7
-6
-5
-4
-3
-2
-1
0
Real Axis
Part(d)
Nyquist Diagram 2.5 2 1.5
Imaginary Axis
1 0.5 0 -0.5 -1 -1.5 -2 -2.5 -1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
Real Axis
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AutomaticControlSystems,9thEdition
2-5)
Chapter2Solutions
In all cases find the real and imaginary axis intersections.
10( jZ 2) (Z 2 4)
10 ( jZ 2)
G ( jZ )
2
Re ^G ( jZ )` cos I a)
(Z 2 4) Z
Im ^G ( jZ )` sin I
(Z 2 4)
10 (Z 4) 2
, ,
2
I
(Z 2 4)
tan 1
Z (Z 2 4)
10
R
(Z 2 4)
lim Z o0 G ( jZ ) 5; I
90$ 0 lim Z of G ( jZ ) 0; I tan 1 0 180$ 1 Real axis intersection @ jZ 0 tan 1 1
Imaginary axis int er sec tion does not exist. b&c) ݈݅݉ఠ՜ ܩሺ݆߱ሻ = 1 ס0o
݈݅݉ఠ՜ ܩሺ݆߱ሻ = 0 ס-180o ܩሺ݆߱ሻ ൌ
ഘ మ
ଵ ഘ
ଵିቀഘ ቁ ାଶకቀ ഘ ቁ
ൌ
ഘ మ
ഘ
మ ഘ మ
ഘ మ
ቆଵିቀഘ ቁ ିଶకቀ ഘ ቁቇ ൬ଵିቀഘ ቁ ൰ ାସቀഘ ቁ
Therefore: ഘ మ
Re{ G(j) } =
ଵିቀഘ ቁ
మ ഘ మ ഘ మ ൬ଵିቀഘ ቁ ൰ ାସቀഘ ቁ ഘ
Im {G(j)} = െ
ଶకቀ ഘ ቁ
మ ഘ మ ഘ మ ൬ଵିቀഘ ቁ ൰ ାସቀഘ ቁ
29
Golnaraghi,Kuo
2 jZ (Z 2 4)
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AutomaticControlSystems,9thEdition
Chapter2Solutions
If Re{G(j )} = 0
Ö
߱ ൌ ߱୬
If Im{ G(j )} = 0
Ö
߱ൌͲ ൝߱ ՜ Ͳ ߱ ՜
If = n
Ö
ቐ
ܩሺ݆߱ ሻ ൌ ܩסሺ݆߱ ሻ ൌ െͻͲ
If = n and = 1
Ö
ܩሺ݆߱ ሻ ൌ
If = n and ՜ Ͳ
Ö
ܩሺ݆߱ ሻ ՜
If = n and ՜
Ö
ܩሺ݆߱ ሻ ՜ Ͳ
d)
e)
G(j) =
்ఠି ఠሺଵାఠమ ் మ ሻ
Ž‹՜ ሺŒሻ =
ס- 90o
Ž‹՜ ሺŒሻ =
ס-180o
ȁ ܩሺ݆߱ሻȁ ൌ ቚ
షೕഘಽ ଵାఠ்
סG(j) = ס
ቚ ൌ
ଵ
ଵାఠ்
ଵ ξଵାఠమ ் మ
+ ି ݁ סఠ = tan-1 ( T) – L
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AutomaticControlSystems,9thEdition 26
Chapter2Solutions
MATLABcode: clear all; s = tf('s')
%Part(a) Ga=10/(s-2) figure(1) nyquist(Ga)
%Part(b) zeta=0.5;
%asuuming a value for zeta 1
wn=2*pi*10 Gc=1/(1+2*zeta*s/wn+s^2/wn^2) figure(3) nyquist(Gc)
%Part(d) T=3.5 %assuming value for parameter T Gd=1/(s*(s*T+1)) figure(4) nyquist(Gd)
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Chapter2Solutions
Golnaraghi,Kuo
%Part(e) T=3.5 L=0.5 Ge=pade(exp(-1*s*L),2)/(s*T+1) figure(5) hold on; nyquist(Ge)
notes:InordertouseMatlabNyquistcommand,parametersneedstobeassignedwithvalues,andPade approximationneedstobeusedforexponentialterminpart(e). Nyquistdiagramsareasfollows:
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Part(a)
Nyquist Diagram 2.5 2 1.5
Imaginary Axis
1 0.5 0 -0.5 -1 -1.5 -2 -2.5 -5
-4
-3
-2
-1
0
1
Real Axis
Part(b)
Nyquist Diagram 1.5
1
Imaginary Axis
0.5
0
-0.5
-1
-1.5 -1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
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Chapter2Solutions
Golnaraghi,Kuo
Part(c)
Nyquist Diagram 0.8
0.6
0.4
Imaginary Axis
0.2
0
-0.2
-0.4
-0.6
-0.8 -1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
Part(d)
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Chapter2Solutions
Golnaraghi,Kuo
Nyquist Diagram 60
40
Imaginary Axis
20
0
-20
-40
-60 -3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
Real Axis
Part(e)
Nyquist Diagram 0.8
0.6
0.4
Imaginary Axis
0.2
0
-0.2
-0.4
-0.6
-0.8 -1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
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AutomaticControlSystems,9thEdition
2-7)
a)
Chapter2Solutions
ଶሺଶఠାଵሻ
G(j) =
ఠሺǤଵఠାଵሻሺǤଶఠାଵሻ
Steps for plotting |G|: (1) For < 0.1, asymptote is Break point: = 0.5 Slope = -1 or -20 dB/decade (2) For 0.5 < < 10 Break point: = 10 Slope = -1+1 = 0 dB/decade (3) For 10 < < 50: Break point: = 50 Slope = -1 or -20 dB/decade (4) For > 50 Slope = -2 or -40 dB/decade Steps for plotting סG ଶ
(1) ס (2) ס
ଵ ଶఠାଵ
(3) ס
(4) ס
b)
= -90o
ఠ
=ቐ
ଵ Ǥଵఠାଵ
߱ ՜ Ͳǣס
߱ ՜ ס
=ቐ
ଵ Ǥଶఠାଵ
ଵ ଶఠାଵ
߱ ՜ Ͳǣס ߱ ՜ ǣס
=ቐ
՜ െͻͲ ଵ
ଶఠାଵ ଵ
՜ Ͳ
Ǥଵఠାଵ ଵ
Ǥଵఠାଵ ଵ
߱ ՜ Ͳǣס
՜ Ͳ
՜ െͻͲ
Ǥଶఠାଵ ଵ
߱ ՜ ס
՜ െͻͲ
Ǥଶఠାଵ
՜ Ͳ
Let's convert the transfer function to the following form: G(j) =
ଶହ
మǤఱ ഘ ଵఠቀି భబ ఠమ ାభబቁାଵ
Ö G(s) =
ହ
ଵ
ଶ ୱ൬౩మ ାǤଵୱାଵ൰ ర
Steps for plotting |G|: (1) Asymptote: < 1
|G(j)|
2.5 /
Slope: -1 or -20 dB/decade ȁܩሺ݆߱ሻȁఠୀଵ ൌ ʹǤͷ 216
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Chapter2Solutions
Golnaraghi,Kuo
(2) n =2 and = 0.1 for second-order pole break point: = 2 slope: -3 or -60 dB/decade
ȁܩሺ݆߱ሻȁఠୀଶ ൌ
ൌ ͷ
Steps for plotting סG(j): (1) for term 1/s the phase starts at -90o and at = 2 the phase will be -180o (2) for higher frequencies the phase approaches -270o c)
Convert the transfer function to the following form:
for term
ଵ ఠమ
ͲǤͲͳ݆߱ െ ߱ଶ ͳ ܩሺ݆߱ሻ ൌ ߱ଶ െ߱ ଶ ൬ͲǤͲͳ݆߱ െ ͳ൰ ͻ , slope is -2 (-40 dB/decade) and passes through ȁܩሺ݆߱ሻȁఠୀଵ ൌ ͳ
(1) the breakpoint: = 1 and slope is zero (2) the breakpoint: = 2 and slope is -2 or -40 dB/decade |G(j)| = 1 = 2 = 0.01 below the asymptote |G(j)| = 1 =
ଵ ଶ
=
ଵ Ǥଶ
= 50 above the asymptote
Steps for plotting סG:
d)
(1)
phase starts from -180o due to
(2) (3)
סG(j)| =1 = 0 סG(j)| = 2 = -180o
G(j) =
ଵ
మ ଵାଶቀ୨ ቁିቀ ቁ
Steps for plotting the |G|: (1) Asymptote for (2) Breakpoint:
5, the phase remains at -180o.
(2) As is a damping ratio, the phase angles must be obtained for various when 01
28)Usethisparttoconfirmtheresultsfromthepreviouspart. MATLABcode: s = tf('s')
'Generated transfer function:' Ga=2000*(s+0.5)/(s*(s+10)*(s+50)) figure(1) bode(Ga) grid on;
'Generated transfer function:' Gb=25/(s*(s+2.5*s^2+10)) figure(2) bode(Gb) grid on;
'Generated transfer function:' Gc=(s+100*s^2+100)/(s^2*(s+25*s^2+100)) figure(3) bode(Gc) grid on;
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Chapter2Solutions
'Generated transfer function:' zeta = 0.2 wn=8 Gd=1/(1+2*zeta*s/wn+(s/wn)^2) figure(4) bode(Gd) grid on;
'Generated transfer function:' t=0.3 'from pade approzimation:' exp_term=pade(exp(-s*t),1) Ge=0.03*(exp_term+1)^2/((exp_term-1)*(3*exp_term+1)*(exp_term+0.5)) figure(5) bode(Ge) grid on;
Part(a) 219
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Chapter2Solutions
Golnaraghi,Kuo
Bode Diagram 60
Magnitude (dB)
40 20 0 -20 -40 -60 0
Phase (deg)
-45 -90 -135 -180 -2
10
-1
0
10
1
10
2
10
10
3
10
Frequency (rad/sec)
Part(b)
Bode Diagram
Magnitude (dB)
50
0
-50
-100 -90
Phase (deg)
-135 -180 -225 -270 -1
10
0
1
10
10 Frequency (rad/sec)
220
2
10
AutomaticControlSystems,9thEdition
Chapter2Solutions
Golnaraghi,Kuo
Part(c)
Bode Diagram 60
Magnitude (dB)
40 20 0 -20 -40 0
Phase (deg)
-45 -90 -135 -180 -1
0
10
10
Frequency (rad/sec)
Part(d)
221
1
10
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Chapter2Solutions
Golnaraghi,Kuo
Bode Diagram 10
Magnitude (dB)
0 -10 -20 -30 -40 -50 0
Phase (deg)
-45 -90 -135 -180 -1
0
10
1
10
2
10
10
Frequency (rad/sec)
Part(e)
Bode Diagram 0
Magnitude (dB)
-20 -40 -60 -80 -100
Phase (deg)
-120 0
-90
-180
-270 -1
10
0
10
1
10
Frequency (rad/sec)
222
2
10
3
10
AutomaticControlSystems,9thEdition
2-9)
Chapter2Solutions
a)
ª 1 2 0 º « 0 2 3 » A « » ¬« 1 3 1¼»
ª0 0 º « 1 0» B « » ¬« 0 1 »¼
u (t )
ª u1 (t ) º « u (t ) » ¬ 2 ¼
b)
݀ ݔሺݐሻ ۍଵ ې ۑ ݐ݀ ێ െͳ ʹ Ͳ ݔଵ ሺݐሻ ʹ Ͳ ݑሺݐሻ ݔ݀ێଶ ሺݐሻ ۑൌ ʹ ݔ ሺݐሻ Ͳ ͳ൩ ଵ ൨ Ͳ െͳ൩ ଶ ݑଶ ሺݐሻ ۑ ݐ݀ ێ ͵ െͶ െͳ ݔଷ ሺݐሻ Ͳ Ͳ ݔ݀ێଷ ሺݐሻۑ ے ݐ݀ ۏ 2-10) We know that:
ܩ ۓሺݏሻ ൌ න ݃ሺݐሻ݁ ି௦௧ ݀ ݐሺͳሻ ۖ ۖ
ା
۔ ͳ ି௦௧ ۖ ۖ݃ሺݐሻ ൌ ʹߨ݆ න ݃ሺݐሻ݁ ݀ ݐሺʹሻ ە ି Partial integration of equation (1) gives:
݃ሺݐሻ݁ ି௦௧ ͳ ܩሺݏሻ ൌ ቈെ න ݃ ሺݐሻ݁ ି௦௧ ݀ݐ ݏ ݏ
Ö• ሺ•ሻ ൌ ‰ሺͲሻ ࣦሼ‰ ሺ–ሻሽ Ö
՞ • ሺ•ሻȂ ‰ሺͲሻ
Differentiation of both sides of equation (1) with respect to s gives:
ି
ି
݀ܩሺݏሻ ൌ න െሺݐሻ݃ሺݐሻ݁ ି௦௧ ݀ ݐൌ න൫െ݃ݐሺݐሻ൯݁ ି௦௧ ݀ݐ ݀ݏ Comparing with equation (1), we conclude that: 223
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AutomaticControlSystems,9thEdition
Chapter2Solutions
ࣦ ିଵ ቊ
௧
Golnaraghi,Kuo
݀ܩሺݏሻ ቋ ՞ െ݃ݐሺݐሻ ݀ݏ
ௗሺ௧ሻ
2-11) Let g(t) = ݔ ିሺ߬ሻ݀߬ then ݔሺݐሻ ൌ
ௗ௧
Using Laplace transform and differentiation property, we have X(s) = sG(s) Therefore G(s) =
, which means:
ͳ ࣦ ൝ න ݔሺ߬ሻ݀߬ൡ ՞ ܺሺݏሻ ݏ ି
2-12) By Laplace transform definition:
ࣦሼ݃ሺ ݐെ ܶሻݑሺ ݐെ ܶሻሽ ൌ න ݃ሺ ݐെ ܶሻ݁ ି௦௧ ݀ݐ ்
Now, consider = t - T, then:
ࣦሼ݃ሺ ݐെ ܶሻሽ ൌ න ݃ሺ߬ሻ݁ ି௦ሺఛା்ሻ ݀߬ ൌ ݁ ି௦் න ݃ሺ߬ሻ݁ ି௦ఛ ݀߬
Which means:
2-13) Consider: f(t) = g1(t)
ஶ
g2(t) = ିஶ ݃ଵ ሺ߬ሻ݃ଶ ሺ ݐെ ߬ሻ݀߬
By Laplace transform definition: ஶ
ஶ
ܨሺݏሻ ൌ න න ݃ଵ ሺ߬ሻ݃ଶ ሺ ݐെ ߬ሻ݀߬൩ ݁ ି௦௧ ݀ݐ ିஶ ିஶ ஶ
ஶ
ൌ න ݃ଵ ሺ߬ሻ න ݃ଶ ሺ ݐെ ߬ሻ݁ ି௦௧ ݀ݐ൩ ݀߬ ିஶ
ିஶ
By using time shifting theorem, we have: 224
AutomaticControlSystems,9thEdition
Chapter2Solutions ஶ
ܨሺݏሻ ൌ න ݃ଵ ሺ߬ሻሾ݁ ି௦ఛ ܩଶ ሺݏሻሿ݀߬ ିஶ ஶ
ൌ න ݃ଵ ሺ߬ሻ݁ ି௦ఛ ݀߬൩ ܩଶ ሺݏሻ ൌ ܩଵ ሺݏሻ ȉ ܩଶ ሺݏሻ ିஶ
Let's consider g(t) = g1(t) g2(t) ஶ
ܩሺݏሻ ൌ න ݃ଵ ሺݐሻ݃ଶ ሺݐሻ݁ ି௦௧ ݀ݐ
By inverse Laplace Transform definition, we have ାஶ
ͳ ݃ଵ ሺݐሻ ൌ න ܩଵ ሺሻ݁ ௧ ݀ ʹߨ݆ ିஶ
Then ାஶ
ஶ
ܩሺݏሻ ൌ න ܩଵ ሺሻ݀ න ݂ଶ ሺݐሻ݁ ିሺ௦ିሻ௧ ݀ݐ
ିஶ
Where ஶ
ܩଶ ሺ ݏെ ሻ ൌ ݂ଶ ሺݐሻ݁ ିሺ௦ିሻ௧ ݀ݐ therefore: ାஶ
ͳ ܩሺݏሻ ൌ න ܩଵ ሺሻܩଶ ሺ െ ݏሻ݀ ൌ ଵ ሺ•ሻ כଶ ሺ•ሻ ʹߨ݆ ିஶ
2-14) a)
We know that
ࣦቄ
ௗሺ௧ሻ ௗ௧
ஶ ௗሺ௧ሻ
ቅ =
ௗ௧
݁ ି௦௧ ݀ = ݐsG(s) + g(0)
When s Æ , it can be written as:
225
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
ஶ
݈݅݉ න
௦՜ஶ
Chapter2Solutions
݀݃ሺݐሻ ି௦௧ ݁ ݀ ݐൌ ݈݅݉ ሾܩݏሺݏሻ െ ݃ሺͲሻሿ ௦՜ஶ ݀ݐ
As ஶ ௗሺ௧ሻ
݈݅݉௫՜ஶ
ௗ௧
݁ ି௦௧ ݀ ݐൌ Ͳ
Therefore: ݈݅݉௦՜ஶ ܩݏሺݏሻ ൌ ݃ሺͲሻ b)
By Laplace transform differentiation property: ஶ
݈݅݉ න ௦՜
݀݃ሺݐሻ ି௦௧ ݁ ݀ ݐൌ ݈݅݉ሾܩݏሺݏሻ െ ݃ሺͲሻሿ ௦՜ ݀ݐ
As ஶ
ஶ
݀݃ሺݐሻ ି௦௧ ݀݃ሺݐሻ ݈݅݉ න ݁ ݀ ݐൌ න ݀ ݐൌ ݃ሺλሻ െ ݃ሺͲሻ ௦՜ ݀ݐ ݀ݐ
Therefore
݈݅݉ሾܩݏሺݏሻሿ െ ݃ሺͲሻ ൌ ݃ሺλሻ െ ݃ሺͲሻ ௦՜
which means:
݈݅݉ ܩݏሺݏሻ ൌ ݃ሺλሻ ௦՜
215) MATLABcode: clear all; syms t s=tf('s')
f1 = (sin(2*t))^2 L1=laplace(f1)
226
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
Chapter2Solutions
% f2 = (cos(2*t))^2 = 1-(sin(2*t))^2
Golnaraghi,Kuo
===> L(f2)=1/s-L(f1) ===>
L2= 1/s - 8/s/(s^2+16)
f3 = (cos(2*t))^2 L3=laplace(f3)
'verified as L2 equals L3'
^
`
2 MATLABsolutionfor L sin 2t is:
8/s/(s^2+16)
^
`
^
`
2 2 Calculating L cos 2t basedon L sin 2t
^
`
L cos 2 2t =(s^^3+8s)/(s^4+16s^2)
^
`
2 verifying L cos 2t :
(8+s^2)/s/(s^2+16) 216)(a)
G ( s )
(d)
G ( s )
5
s 5 1 2
s 4
2
(b)
G ( s )
(e)
s
4s 2
4
f
(c)
G ( s)
¦e
1 s2
1 e
k 0
227
G ( s )
1
kT ( s 5 )
T ( s 5 )
4 2
s 4s 8
AutomaticControlSystems,9thEdition Chapter2Solutions Golnaraghi,Kuo 217)Note:%section (e) requires assignment of T and a numerical loop calculation MATLABcode: clear all; syms t u
f1 = 5*t*exp(-5*t) L1=laplace(f1)
f2 = t*sin(2*t)+exp(-2*t) L2=laplace(f2)
f3 = 2*exp(-2*t)*sin(2*t) L3=laplace(f3)
f4 = sin(2*t)*cos(2*t) L4=laplace(f4)
%section (e) requires assignment of T and a numerical loop calculation
(a) g (t )
5te 5t u s (t )
Answer:5/(s+5)^2 (b) g ( t ) ( t sin 2t e 2t )us ( t ) Answer:4*s/(s^2+4)^2+1/(s+2) (c) g ( t ) 2e 2t sin 2t u s (t ) Answer:4/(s^2+4*s+8)
228
AutomaticControlSystems,9thEdition (d) g ( t ) sin 2t cos 2t u s (t )
Chapter2Solutions
Golnaraghi,Kuo
Answer:2/(s^2+16) f
(e) g ( t )
¦e
5 kT
G ( t kT )
whereG(t)=unitimpulsefunction
k 0
%section (e) requires assignment of T and a numerical loop calculation
218(a) u s (t ) 2u s (t 1) 2u s ( t 2) 2u s (t 3)
g (t )
1
G(s)
s
gT (t ) GT ( s )
1 e
1 2e s 2e 2 s 2e 3s
s 1 e
u s (t ) 2u s (t 1) u s (t 2) 1 s
1 2e s e 2 s
1 s
g ( t )
1 e s
¦
2
f
gT ( t 2 k )u s ( t 2 k )
s
0dtd2
f
s
1
¦s
G( s)
k 0
(1 e
s 2 2 ks
) e
1 e
s
s(1 e
k 0
s
)
(b)
g (t )
G ( s )
2tu s (t ) 4(t 0.5)u s (t 0.5) 4(t 1)u s (t 1) 4(t 1.5)u s (t 1.5)
2 s
2
1 2e
0.5 s
2e
s
2e
1.5 s
0.5 s 0.5 s 2 s 1 e
2 1 e
gT ( t ) 2tu s ( t ) 4( t 0.5)u s ( t 0.5) 2( t 1)u s ( t 1)
GT ( s )
2 s
2
1 2e0.5 s e s
2 s
2
1 e0.5 s
f
g (t )
¦ g T (t k )u s (t k ) k 0
2
f
G ( s)
0 d t d 1
¦ s2 2
1 e
k 0
0.5 s 2
e
ks
0.5s 2 0.5 s s 1 e 2 1 e
219)
g ( t )
( t 1)u s ( t ) ( t 1)u s ( t 1) 2u s ( t 1) ( t 2 )u s ( t 2) ( t 3)u s ( t 3) u s ( t 3)
229
AutomaticControlSystems,9thEdition
Chapter2Solutions
Golnaraghi,Kuo
1
G ( s )
s
2
1 e s e 2 s e 3s s 1 2e s e 3s 1
2-20) ்
ࣦሼ݂ሺݐሻሽ ൌ න ݂ሺݐሻ݁
ି௦௧
் ଶ
்
݀ ݐൌ න ݁ ି௦௧ ݀ ݐ න ሺെͳሻ݁ ି௦௧ ݀ݐ
்௦
்௦
் ଶ
்௦ ଶ ͳ െ ݁ ି ଶ ݁ ି்௦ െ ݁ ି ଶ ͳ ൌ ൌ ͳ െ ݁ ି ଶ ൨ ݏ ݏ ݏ
2-21) ஶ
ࣦሼ݂ሺݐሻሽ ൌ න ݂ሺݐሻ݁
ି௦௧
ଶ ݁ ି௦௧ ݁ ି௦௧ ݀ ݐൌ න ଶ ݀ ݐെ න ଶ ݀ݐ ܮ ܮ ଶ
݁ ି௦௧ ݁ ି௦௧ ͳ െ ݁ ି௧ ݁ ିଶ௧ െ ݁ ି௧ ൌ ቈെ ଶ ቈ ଶ ൌ ܮݏ ܮݏ ܮݏଶ ܮݏଶ ͳ ൌ ଶ ሺͳ െ ݁ ି௧ ሻଶ ܮݏ
2-22) ࣦ ቄ ࣦቄ ࣦቄ
ௗ య ௬ሺ௧ሻ ௗ௧ మ ௗ మ ௬ሺ௧ሻ ௗ௧ మ ௗ௬ሺ௧ሻ ௗ௧ మ
ቅ ൌ ݏଷ ܻሺݏሻ െ ݏଶ ݕᇱᇱ ሺͲሻ െ ݕݏᇱ ሺͲሻ െ ݕሺͲሻ ቅ ൌ ݏଶ ܻሺݏሻ െ ݕݏᇱ ሺͲሻ െ ݕሺͲሻ
ቅ ൌ ܻݏሺݏሻ െ ݕሺͲሻ
ࣦሼെ݁ ି௧ ݑ௦ ሺݐሻሽ ൌ െ
ଵ ௦ାଵ
Ö ݏଷ ܻሺݏሻ ݏെ ݏ ʹ ݏଶ ܻሺݏሻ ʹ ݏെ ʹ െ ܻݏሺݏሻ ʹܻሺݏሻ ൌ െ
230
ଵ ௦ାଵ
AutomaticControlSystems,9thEdition
Chapter2Solutions
Ö ሺ ݏଷ ʹ ݏଶ െ ݏ ʹሻܻሺݏሻ ʹ ݏെ ʹ ൌ െ
ଵ ௦ାଵ
ଶ௦ మ ିଷ
Ö ܻሺݏሻ ൌ ሺ௦ାଶሻሺ௦మ
ାଵሻሺ௦ାଵሻ
223
MATLABcode: clear all; syms t u s x1 x2 Fs
f1 = exp(-2*t) L1=laplace(f1)/(s^2+5*s+4);
Eq2=solve('s*x1=1+x2','s*x2=-2*x1-3*x2+1','x1','x2') f2_x1=Eq2.x1 f2_x2=Eq2.x2
f3=solve('(s^3-s+2*s^2+s+2)*Fs=-1+2-(1/(1+s))','Fs')
HereisthesolutionprovidedbyMATLAB: Part(a):F(s)=1/(s+2)/(s^2+5*s+4) Part(b):X1(s)=(4+s)/(2+3*s+s^2) X2(s)=(s2)/(2+3*s+s^2) Part(c):F(s)=s/(1+s)/(s^3+2*s^2+2)
231
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition 224)
Chapter2Solutions
MATLABcode: clear all; syms s Fs f3=solve('s^2*Fs-Fs=1/(s-1)','Fs') Answer from MATLAB: Y(s)=1/(s-1)/(s^2-1)
225) MATLABcode: clear all; syms s CA1 CA2 CA3 v1=1000; v2=1500; v3=100; k1=0.1 k2=0.2 k3=0.4
f1='s*CA1=1/v1*(1000+100*CA2-1100*CA1-k1*v1*CA1)' f2='s*CA2=1/v2*(1100*CA1-1100*CA2-k2*v2*CA2)' f3='s*CA3=1/v3*(1000*CA2-1000*CA3-k3*v3*CA3)' Sol=solve(f1,f2,f3,'CA1','CA2','CA3') CA1=Sol.CA1 CA3=Sol.CA2 CA4=Sol.CA3
232
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition SolutionfromMATLAB:
Chapter2Solutions
Golnaraghi,Kuo
CA1(s)= 1000*(s*v2+1100+k2*v2)/(1100000+s^2*v1*v2+1100*s*v1+s*v1*k2*v2+1100*s*v2+1100*k2*v2+k1*v1*s*v2+ 1100*k1*v1+k1*v1*k2*v2) CA3(s)= 1100000/(1100000+s^2*v1*v2+1100*s*v1+s*v1*k2*v2+1100*s*v2+1100*k2*v2+k1*v1*s*v2+1100*k1*v1+k1* v1*k2*v2) CA4(s)= 1100000000/(1100000000+1100000*s*v3+1000*s*v1*k2*v2+1100000*s*v1+1000*k1*v1*s*v2+1000*k1*v1*k 2*v2+1100*s*v1*k3*v3+1100*s*v2*k3*v3+1100*k2*v2*s*v3+1100*k2*v2*k3*v3+1100*k1*v1*s*v3+1100*k1 *v1*k3*v3+1100000*k1*v1+1000*s^2*v1*v2+1100000*s*v2+1100000*k2*v2+1100000*k3*v3+s^3*v1*v2*v3+ 1100*s^2*v1*v3+1100*s^2*v2*v3+s^2*v1*v2*k3*v3+s^2*v1*k2*v2*v3+s*v1*k2*v2*k3*v3+k1*v1*s^2*v2*v3 +k1*v1*s*v2*k3*v3+k1*v1*k2*v2*s*v3+k1*v1*k2*v2*k3*v3) 2-26) (a)
1
G( s)
3s
(b)
G( s)
1
2( s 2)
1
1
1 2t 1 3t e e 3 2 3
g (t )
3( s 3)
t t 0
2.5
5
s 1 ( s 1)
2
2.5
g (t )
s3
2.5e
t
5te
t
2.5e
3t
t t 0
(c)
G(s)
50 s
20
s 1
30 s 20 2
s 4
e
s
>50 20e (t 1) 30 cos 2(t 1) 5 sin 2(t 1) @ us (t 1)
g (t )
(d)
G( s)
1 s
s 1
1
2
s s2
g (t )
1 1.069e
(e)
g (t )
0.5t
s
1 2
s s2
s 2
s s2
TakingtheinverseLaplacetransform,
>sin 1.323t sin 1.323t 69.3o @ 2 t
0.5t e
1 e
t t 0
(f)Try using MATLAB >> b=num*2
233
0.5t
1.447 sin 1.323t cos1.323t t t 0
AutomaticControlSystems,9thEdition
Chapter2Solutions
b= 2
2
2
>>num = 1
1
1
>> denom1=[1 1] denom1 = 1
1
>> denom2=[1 5 5] denom2 = 1
5
5
>> num*2 ans = 2
2
2
>> denom=conv([1 0],conv(denom1,denom2)) denom = 1
6
10
5
0
5
0
>> b=num*2 b= 2
2
2
>> a=denom a= 1
6
10
>> [r, p, k] = residue(b,a) r= -0.9889 234
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
Chapter2Solutions
2.5889 -2.0000 0.4000 p= -3.6180 -1.3820 -1.0000 0 k=[] If there are no multiple roots, then
The number of poles n is r r1 r 2 ... n k s p1 s p2 s pn
b a
In this case, p1 and k are zero. Hence, 0.4 0.9889 2.5889 2 s s 3.6180 s 1.3820 s 1
G ( s)
0.4 0.9889e 3.618t 1.3820e 2.5889t 2e t
g (t ) (g) ܩሺݏሻ ൌ ሺ
ൌ
ଶ ௦ାଵሻሺ௦ାଶሻ
ଶ ௦ାଵ
െ
ଶ ௦ାଶ
ଶ షೞ ௦ାଵ
ଶ షೞ ௦ାଵ
Ö ࣦ ିଵ ሼ ሺ•ሻሽ ൌ ʹ‡ି୲ െ ʹ‡ିଶ୲ ʹ‡ିሺ୲ିଵሻ ݑሺ ݐെ ͳሻ (h)
ܩሺݏሻ ൌ
ଶ௦ାଵ ሺ௦ାଵሻሺ௦ାଶሻሺ௦ାଷሻ
ൌെ
భ మ
௦ାଵ
ଷ ௦ାଶ
ଵ
ହ
ଶ
ଶ
െ
Ö ࣦ ିଵ ሼܩሺݏሻሽ ൌ െ ݁ ି௧ ͵݁ ିଶ௧ െ ݁ ିଷ௧
235
ହ ଶሺ௦ାଷሻ
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
ܩሺݏሻ ൌ
(i)
Ö
ଷ௦ య ାଵ௦ మ ା଼௦ାହ ௦ యర ାହ௦ య ା௦ మ ାହ௦ା
Chapter2Solutions
ൌ
ଵ ௦ାଶ
ଵ ௦ାଷ
Golnaraghi,Kuo
௦ ௦ మ ାଵ
ࣦ ିଵ ሼܩሺݏሻሽ ൌ ݁ ିଶ௧ ݁ ିଷ௧ ܿݐݏ
227 MATLABcode: clear all; syms s
f1=1/(s*(s+2)*(s+3)) F1=ilaplace(f1)
f2=10/((s+1)^2*(s+3)) F2=ilaplace(f2)
f3=10*(s+2)/(s*(s^2+4)*(s+1))*exp(-s) F3=ilaplace(f3)
f4=2*(s+1)/(s*(s^2+s+2)) F4=ilaplace(f4)
f5=1/(s+1)^3 F5=ilaplace(f5)
f6=2*(s^2+s+1)/(s*(s+1.5)*(s^2+5*s+5)) F6=ilaplace(f6)
s=tf('s') f7=(2+2*s*pade(exp(-1*s),1)+4*pade(exp(-2*s),1))/(s^2+3*s+2) %using Pade command for exponential term
236
AutomaticControlSystems,9thEdition
Chapter2Solutions
Golnaraghi,Kuo
[num,den]=tfdata(f7,'v') %extracting the polynomial values syms s f7n=(-2*s^3+6*s+12)/(s^4+6*s^3+13*s^2+12*s+4) %generating sumbolic function for ilaplace F7=ilaplace(f7n)
f8=(2*s+1)/(s^3+6*s^2+11*s+6) F8=ilaplace(f8)
f9=(3*s^3+10^s^2+8*s+5)/(s^4+5*s^3+7*s^2+5*s+6) F9=ilaplace(f9)
SolutionfromMATLABfortheInverseLaplacetransforms: Part(a):
G ( s)
1 s( s 2)( s 3)
G(t)=1/2*exp(2*t)+1/3*exp(3*t)+1/6
Tosimplify:
symst
digits(3)
vpa(1/2*exp(2*t)+1/3*exp(3*t)+1/6)
ans=.500*exp(2.*t)+.333*exp(3.*t)+.167
Part(b):
G ( s)
10 ( s 1) 2 ( s 3)
G(t)=5/2*exp(3*t)+5/2*exp(t)*(1+2*t) 237
AutomaticControlSystems,9thEdition Part(c):
G( s)
100( s 2) s( s 2 4 )( s 1)
Chapter2Solutions
Golnaraghi,Kuo
e s
G(t)=Step(t1)*(4*cos(t1)^2+2*sin(t1)*cos(t1)+4*exp(1/2*t+1/2)*cosh(1/2*t1/2)4*exp(t+1)cos(2*t2) 2*sin(2*t2)+5)
Part(d):
G ( s)
2( s 1) s( s 2 s 2 )
G(t)=1+1/7*exp(1/2*t)*(7*cos(1/2*7^(1/2)*t)+3*7^(1/2)*sin(1/2*7^(1/2)*t))
Tosimplify:
symst
digits(3)
vpa(1+1/7*exp(1/2*t)*(7*cos(1/2*7^(1/2)*t)+3*7^(1/2)*sin(1/2*7^(1/2)*t)))
ans=1.+.143*exp(.500*t)*(7.*cos(1.32*t)+7.95*sin(1.32*t))
Part(e):
G ( s)
1 ( s 1) 3
G(t)=1/2*t^2*exp(t)
Part(f):
G( s)
2( s 2 s 1) s( s 15 . )( s 2 5s 5)
G(t)=4/15+28/3*exp(3/2*t)16/5*exp(5/2*t)*(3*cosh(1/2*t*5^(1/2))+5^(1/2)*sinh(1/2*t*5^(1/2)))
Part(g):G ( s )
2 2 se s 4e 2 s s 2 3s 2
238
AutomaticControlSystems,9thEdition G(t)=2*exp(2*t)*(7+8*t)+8*exp(t)*(2+t)
Part(h):
G ( s)
Chapter2Solutions
Golnaraghi,Kuo
2s 1 s 6 s 2 11s 6 3
G(t)=1/2*exp(t)+3*exp(2*t)5/2*exp(3*t)
Part(i):
G ( s)
3s 3 10 s 2 8s 5 s 4 5 s 3 7 s 2 5s 6
G(t)= 7*exp(2*t)+10*exp(3*t) 1/10*ilaplace(10^(2*s)/(s^2+1)*s,s,t)+1/10*ilaplace(10^(2*s)/(s^2+1),s,t)+1/10*sin(t)*(10+dirac(t)*(exp( 3*t)+2*exp(2*t)))
239
AutomaticControlSystems,9thEdition
2-28)
ୢ୶ሺ୲ሻ ୢ୲
Chapter2Solutions
Golnaraghi,Kuo
ൌ šሺ–ሻ —ሺ–ሻ
a)
݀ ݔሺݐሻ ۓଵ ൌ െݔଶ ሺݐሻ ʹݔଷ ሺݐሻ െ ݑଶ ሺݐሻ ۖ ݀ݐ ݀ݔଶ ሺݐሻ ൌ ݔଵ ሺݐሻ ݔଷ ሺݐሻ ݑଵ ሺݐሻ ݀ݐ ۔ ۖ݀ݔଷ ሺݐሻ ሺݐሻ െ ʹݔଶ ሺݐሻ ݔଷ ሺݐሻ ݐ݀ ەൌ െݔଵ b)
݀ ݔሺݐሻ ۓଵ ൌ ͵ݔଵ ሺݐሻ ݔଶ ሺݐሻ െ ʹݔଷ ሺݐሻ െ ݑሺݐሻ ۖ ݀ݐ ݀ݔଶ ሺݐሻ ൌ െݔଵ ሺݐሻ ʹݔଶ ሺݐሻ ʹݔଷ ሺݐሻ ݀ݐ ۔ ݀ݔଷ ሺݐሻ ۖ ൌ ݔଷ ሺݐሻ ʹݑሺݐሻ ە ݀ݐ
2-29) (a)
3s 1
Y (s) 3
s 2 s 5s 6
R( s)
(c)
e)
2
(b)
R( s)
Y (s)
s ( s 2)
R( s)
s 10 s 2 s s 2
4
3
Y (s)
2
5 4
2
s 10 s s 5
(d)
Y (s) R( s)
1 2e 2
s
2s s 5
ݔሺݐሻ ൌ ݕሺ ݐ ͳሻ Ö
ௗ మ ௫ሺ௧ሻ ௗ௧ మ
ସௗ௫ሺ௧ሻ ௗ௧
ͷݔሺݐሻ ൌ
ௗሺ௧ሻ ௗ௧
௧
ʹݎሺݐሻ ʹ ିஶ ݎሺ߬ሻ݀ ߬
By using Laplace transform, we have:
ݏଶ ܺሺݏሻ Ͷܺݏሺݏሻ ͷܺሺݏሻ ൌ ܴݏሺݏሻ ʹܴሺݏሻ As ሺ•ሻ ൌ ‡ିୱ ሺ•ሻ, then
240
ܴሺݏሻ ݏ
AutomaticControlSystems,9thEdition
Chapter2Solutions
ሺ ݏଶ Ͷ ݏ ݐሻ݁ ି௦ ܻሺݏሻ ൌ
Golnaraghi,Kuo
ݏଶ ʹ ݏ ͳ ܴሺݏሻ ݏ
Then: ሺ௦ሻ ோሺ௦ሻ
f)
ൌ
ሺ௦ାଵሻమ ೞ ௦ሺ௦ మ ାସ௦ା௦ሻ
By using Laplace transform we have:
ʹ ൬ ݏଷ ʹ ݏଶ ݏ ʹ ൰ ܻሺݏሻ ൌ ି ݁ݏ௦ ܴሺݏሻ ʹ݁ ି௦ ܴሺݏሻ ݏ As a result: ሺ௦ሻ ோሺ௦ሻ
ൌ
௦ሺ௦ାଶሻ షೞ ௦ ర ାଶ௦ య ା௦ మ ାଶ௦ାଶ
230) AftertakingtheLaplacetransform,theequationwassolvedintermsofY(s),andconsecutivelywasdividedby inputR(s)toobtainY(s)/R(s):
MATLABcode: clear all; syms Ys Rs s
sol1=solve('s^3*Ys+2*s^2*Ys+5*s*Ys+6*Ys=3*s*Rs+Rs','Ys') Ys_Rs1=sol1/Rs
sol2=solve('s^4*Ys+10*s^2*Ys+s*Ys+5*Ys=5*Rs','Ys') Ys_Rs2=sol2/Rs
sol3=solve('s^3*Ys+10*s^2*Ys+2*s*Ys+2*Ys/s=s*Rs+2*Rs','Ys') Ys_Rs3=sol3/Rs
sol4=solve('2*s^2*Ys+s*Ys+5*Ys=2*Rs*exp(-1*s)','Ys') Ys_Rs4=sol4/Rs
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Golnaraghi,Kuo
%Note: Parts E&F are too complicated with MATLAB, Laplace of integral is not executable in MATLAB.....skipped
MATLABAnswers: Part(a):
Y(s)/R(s)=(3*s+1)/(5*s+6+s^3+2*s^2);
Part(b):
Y(s)/R(s)=5/(10*s^2+s+5+s^4)
Part(c):
Y(s)/R(s)=(s+2)*s/(2*s^2+2+s^4+10*s^3)
Part(d):
Y(s)/R(s)=2*exp(s)/(2*s^2+s+5)
%Note: Parts E&F are too complicated with MATLAB, Laplace of integral is not executable in MATLAB.....skipped
231 MATLABcode: clear all; s=tf('s')
%Part a Eq=10*(s+1)/(s^2*(s+4)*(s+6)); [num,den]=tfdata(Eq,'v'); [r,p] = residue(num,den)
%Part b Eq=(s+1)/(s*(s+2)*(s^2+2*s+2)); [num,den]=tfdata(Eq,'v'); [r,p] = residue(num,den)
%Part c Eq=5*(s+2)/(s^2*(s+1)*(s+5)); [num,den]=tfdata(Eq,'v');
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[r,p] = residue(num,den)
%Part d Eq=5*(pade(exp(-2*s),1))/(s^2+s+1); %Pade approximation oreder 1 used [num,den]=tfdata(Eq,'v'); [r,p] = residue(num,den)
%Part e Eq=100*(s^2+s+3)/(s*(s^2+5*s+3)); [num,den]=tfdata(Eq,'v'); [r,p] = residue(num,den)
%Part f Eq=1/(s*(s^2+1)*(s+0.5)^2); [num,den]=tfdata(Eq,'v'); [r,p] = residue(num,den)
%Part g Eq=(2*s^3+s^2+8*s+6)/((s^2+4)*(s^2+2*s+2)); [num,den]=tfdata(Eq,'v'); [r,p] = residue(num,den)
%Part h Eq=(2*s^4+9*s^3+15*s^2+s+2)/(s^2*(s+2)*(s+1)^2); [num,den]=tfdata(Eq,'v'); [r,p] = residue(num,den)
Thesolutionsarepresentedintheformoftwovectors,randp,whereforeachcase,thepartialfraction expansionisequalto: 243
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b( s ) a( s)
Chapter2Solutions
r r1 r 2 ... n s p1 s p 2 s pn
Followingarerandpvectorsforeachpart: Part(a): r=0.6944 0.9375 0.2431 0.4167 p=6.0000 4.0000 0 0 Part(b): r=0.2500 0.25000.0000i 0.2500+0.0000i 0.2500 p=2.0000 1.0000+1.0000i 244
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Chapter2Solutions
0 Part(c): r=0.1500 1.2500 1.4000 2.0000 p=5 1 0 0 Part(d): r=10.0000 5.00000.0000i 5.0000+0.0000i p=1.0000 0.5000+0.8660i 0.50000.8660i 245
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Chapter2Solutions
Part(e): r=110.9400 110.9400 100.0000 p=4.3028 0.6972 0 Part(f): r=0.2400+0.3200i 0.24000.3200i 4.4800 1.6000 4.0000 p=0.0000+1.0000i 0.00001.0000i 0.5000 0.5000 0 246
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Chapter2Solutions
r=0.1000+0.0500i 0.10000.0500i 1.1000+0.3000i 1.10000.3000i p=0.0000+2.0000i 0.00002.0000i 1.0000+1.0000i 1.00001.0000i Part(h): r=5.0000 1.0000 9.0000 2.0000 1.0000 p=2.0000 1.0000 1.0000 0 0 247
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AutomaticControlSystems,9thEdition 232)
Chapter2Solutions
MATLABcode: clear all; syms s
%Part a Eq=10*(s+1)/(s^2*(s+4)*(s+6)); ilaplace(Eq)
%Part b Eq=(s+1)/(s*(s+2)*(s^2+2*s+2)); ilaplace(Eq)
%Part c Eq=5*(s+2)/(s^2*(s+1)*(s+5)); ilaplace(Eq)
%Part d exp_term=(-s+1)/(s+1) %pade approcimation Eq=5*exp_term/((s+1)*(s^2+s+1)); ilaplace(Eq)
%Part e Eq=100*(s^2+s+3)/(s*(s^2+5*s+3)); ilaplace(Eq)
%Part f Eq=1/(s*(s^2+1)*(s+0.5)^2); ilaplace(Eq)
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%Part g Eq=(2*s^3+s^2+8*s+6)/((s^2+4)*(s^2+2*s+2)); ilaplace(Eq)
%Part h Eq=(2*s^4+9*s^3+15*s^2+s+2)/(s^2*(s+2)*(s+1)^2); ilaplace(Eq)
MATLABAnswers: Part(a): G(t)=15/16*exp(4*t)+25/36*exp(6*t)+35/144+5/12*t Tosimplify:
symst
digits(3)
vpa(15/16*exp(4*t)+25/36*exp(6*t)+35/144+5/12*t) ans=.938*exp(4.*t)+.694*exp(6.*t)+.243+.417*tPart(b): G(t)=1/4*exp(2*t)+1/41/2*exp(t)*cos(t) Part(c): G(t)=5/4*exp(t)7/5+3/20*exp(5*t)+2*t
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Chapter2Solutions
Part(d): G(t)=5*exp(1/2*t)*(cos(1/2*3^(1/2)*t)+3^(1/2)*sin(1/2*3^(1/2)*t))+5*(1+2*t)*exp(t) Part(e): G(t)=100800/13*exp(5/2*t)*13^(1/2)*sinh(1/2*t*13^(1/2)) Part(f): G(t)=4+12/25*cos(t)16/25*sin(t)8/25*exp(1/2*t)*(5*t+14) Part(g): G(t)=1/5*cos(2*t)1/10*sin(2*t)+1/5*(11*cos(t)3*sin(t))*exp(t) Part(h): G(t)=2+t+5*exp(2*t)+(1+9*t)*exp(t)
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Chapter2Solutions
2-33)(a)Polesareat s 0, 15 . j16583 . , 15 . j16583 .
Onepolesats=0.Marginallystable. Twopoleson jZ axis.Marginallystable.
(b)Polesareat s
5, j 2 , j 2
(c)Polesareat s
0.8688, 0.4344 j 2.3593, 0.4344 j 2.3593TwopolesinRHP.Unstable.
(d)Polesareat s
5, 1 j , 1 j
(e)Polesareat
s
(f)Polesareat s
22.8487 r j 22.6376, 213487 . r j 22.6023
AllpolesintheLHP.Stable.
13387 . , 16634 . j 2.164, 16634 . j 2.164 TwopolesinRHP.Unstable.
TwopolesinRHP.Unstable.
2-34) Find the Characteristic equations and then use the roots command. (a) p= [ 1 3 5 0] sr = roots(p)
p=
1
3
5
0
sr = 0 -1.5000 + 1.6583i -1.5000 - 1.6583i
p=conv([1 5],[1 0 2])
(b)
sr = roots(p) p= 1
5
2
10
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Chapter2Solutions
sr = -5.0000 0.0000 + 1.4142i 0.0000 - 1.4142i
(c) >> roots([1 5 5])
ans =
-3.6180 -1.3820
(d) roots(conv([1 5],[1 2 2])) ans =
-5.0000 -1.0000 + 1.0000i -1.0000 - 1.0000i (e) roots([1 -2 3 10]) ans =
1.6694 + 2.1640i 1.6694 - 2.1640i -1.3387
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Chapter2Solutions
(f) roots([1 3 50 1 10^6]) -22.8487 +22.6376i -22.8487 -22.6376i 21.3487 +22.6023i 21.3487 -22.6023i
Alternatively Problem234 MATLABcode: % Question 2-34, clear all; s=tf('s')
%Part a Eq=10*(s+2)/(s^3+3*s^2+5*s); [num,den]=tfdata(Eq,'v'); roots(den)
%Part b Eq=(s-1)/((s+5)*(s^2+2)); [num,den]=tfdata(Eq,'v'); roots(den)
%Part c Eq=1/(s^3+5*s+5); [num,den]=tfdata(Eq,'v'); roots(den)
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Chapter2Solutions
%Part d Eq=100*(s-1)/((s+5)*(s^2+2*s+2)); [num,den]=tfdata(Eq,'v'); roots(den)
%Part e Eq=100/(s^3-2*s^2+3*s+10); [num,den]=tfdata(Eq,'v'); roots(den)
%Part f Eq=10*(s+12.5)/(s^4+3*s^3+50*s^2+s+10^6); [num,den]=tfdata(Eq,'v'); roots(den)
MATLABanswer: Part(a) 0 1.5000+1.6583i 1.50001.6583i Part(b) 5.0000 254
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AutomaticControlSystems,9thEdition 0.0000+1.4142i
Chapter2Solutions
0.00001.4142i Part(c) 0.4344+2.3593i 0.43442.3593i 0.8688 Part(d) 5.0000 1.0000+1.0000i 1.00001.0000i Part(e) 1.6694+2.1640i 1.66942.1640i 1.3387 Part(f)
255
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Chapter2Solutions
Golnaraghi,Kuo
22.848722.6376i 21.3487+22.6023i 21.348722.6023i
2-35) (a) s 3 25 s 2 10 s 450 0
Roots: 25.31, 0.1537 j 4.214, 0.1537 4.214
RouthTabulation: s
3
1
10
s
2
25
450
s
1
s
0
250 450 25
8
Twosignchangesinthefirstcolumn.TworootsinRHP.
0
450
3
2
(b) s 25 s 10 s 50
Roots: 24.6769, 0.1616 j1.4142, 0.1616 j1.4142
0
RouthTabulation: s
3
1
10
s
2
25
50
s
1
s
0
250 50
8
Nosignchangesinthefirstcolumn.NorootsinRHP.
0
25 50
(c) s 3 25 s 2 250 s 10 0
RouthTabulation:
Roots: 0.0402, 12.48 j 9.6566, j 9.6566
256
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s
3
1
250
s
2
25
10
s
1
s
0
6250 10
249.6
Chapter2Solutions
Golnaraghi,Kuo
Nosignchangesinthefirstcolumn.NorootsinRHP.
0
25 10
4
3
2
(d) 2 s 10 s 5.5 s 5.5 s 10
RouthTabulation:
s
4
2
5.5
s
3
10
5.5
s
2
55 11
4.4
0 Roots: 4.466, 1116 . , 0.2888 j 0.9611, 0.2888 j 0.9611
10
10
10 s
4.4
s
24.2 100
1
0
75.8
10
Twosignchangesinthefirstcolumn.TworootsinRHP.
(e) s 6 2 s 5 8 s 4 15 s 3 20 s 2 16 s 16 0 Roots: 1.222 r j 0.8169, 0.0447 r j1153 . , 0.1776 r j 2.352
RouthTabulation:
s
6
1
8
20
s
5
2
15
16
s
4
s
3
16 15
0.5
40 16
2
2
33
48
16
12
257
AutomaticControlSystems,9thEdition
s
2
s
1
396 24
11.27
33 5411 . 528
. 116
11.27 s
0
Chapter2Solutions
Golnaraghi,Kuo
16 0
0
Foursignchangesinthefirstcolumn.FourrootsinRHP.
4
3
2
(f) s 2 s 10 s 20 s 5
RouthTabulation:
s
4
1
10
s
3
2
20
s
2
20 20
0
Roots: 0.29, 1788 . , 0.039 j 3105 . , 0.039 j 3105 .
5
0
5
2 s
2
s
1
s
0
H
Replace 0 in last row by H
5
20H 10
H
#
10
H
Twosignchangesinfirstcolumn.TworootsinRHP.
5
(g) s8
1
8
20
16
0
s7
2
12
16
0
0
s6
2
12
16
0
0
s5
0
0
0
0
0
ܣሺݏሻ ൌ ʹ ݏ ͳʹ ݏହ ͳ ݏସ
݀ܣሺݏሻ ൌ ͳʹ ݏହ Ͳ ݏସ Ͷ ݏଷ ݀ݏ 258
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Chapter2Solutions
s5
12
60
64
0
s4
2
ͳ ͵
0
0
s3
28
64
0
0
s2
0.759
0
0
0
s1
28
0
s0
0
2-36) Use MATLAB roots command a) roots([1 25 10 450]) ans =
-25.3075 0.1537 + 4.2140i 0.1537 - 4.2140i b) roots([1 25 10 50]) ans =
-24.6769 -0.1616 + 1.4142i -0.1616 - 1.4142i c) roots([1 25 250 10]) ans = -12.4799 + 9.6566i -12.4799 - 9.6566i
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Chapter2Solutions
-0.0402 d) roots([2 10 5.5 5.5 10]) ans =
-4.4660 -1.1116 0.2888 + 0.9611i 0.2888 - 0.9611i e) roots([1 2 8 15 20 16 16]) ans =
0.1776 + 2.3520i 0.1776 - 2.3520i -1.2224 + 0.8169i -1.2224 - 0.8169i 0.0447 + 1.1526i 0.0447 - 1.1526i f) roots([1 2 10 20 5]) ans =
0.0390 + 3.1052i 0.0390 - 3.1052i -1.7881 -0.2900 g) roots([1 2 8 12 20 16 16]) 260
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Chapter2Solutions
ans =
0.0000 + 2.0000i 0.0000 - 2.0000i -1.0000 + 1.0000i -1.0000 - 1.0000i 0.0000 + 1.4142i 0.0000 - 1.4142i
AlternativelyusetheapproachinthisChapter'sSection214: 1. ActivateMATLAB 2. GotothedirectorycontainingtheACSYSsoftware. 3. Typein Acsys
4. Thenpressthe"transferfunctionSymbolicbutton 261
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Chapter2Solutions
Golnaraghi,Kuo
5. Enterthecharacteristicequationinthedenominatorandpressthe"RouthHurwitz"push button. RH =
[ 1, 10] [ 25, 450] [ -8, 0] [ 450, 0] Two sign changes in the first column. Two roots in RHP=> UNSTABLE
2-37) Use the MATLAB "roots" command same as in the previous problem.
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Chapter2Solutions
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2-38) To solve using MATLAB, set the value of K in an iterative process and find the roots such that at least one root changes sign from negative to positive. Then increase resolution if desired. Example: in this case 0Norighthandsidepole Part(c): 274
AutomaticControlSystems,9thEdition RHchart:
Chapter2Solutions
[1,250] [25,10] [1248/5,0] [10,0] Stablesystem>>Norighthandsidepole Part(d): RHchart: [2,11/2,10] [10,11/2,0] [22/5,10,0] [379/22,0,0] [10,0,0] Unstablesystemdueto379/22onthe4throw. 2complexconjugatepolesonrighthandside.Allthepolesare: 4.4660 1.1116 0.2888+0.9611i 0.28880.9611i
275
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Chapter2Solutions
Part(e): RHchart: [1,8,20,16] [2,15,16,0] [1/2,12,16,0] [33,48,0,0] [124/11,16,0,0] [36/31,0,0,0] [16,0,0,0] Unstablesystemdueto33and36/31onthe4thand6throw. 4complexconjugatepolesonrighthandside.Allthepolesare: 0.1776+2.3520i 0.17762.3520i 1.2224+0.8169i 1.22240.8169i 0.0447+1.1526i 0.04471.1526i Part(f): RHchart: [1,10,5] [2,20,0]
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AutomaticControlSystems,9thEdition [eps,5,0]
Chapter2Solutions
[(10+20*eps)/eps,0,0] [5,0,0] Unstablesystemdueto((10+20*eps)/eps)onthe4th. 2complexconjugatepolesslightlyonrighthandside.Allthepolesare: 0.0390+3.1052i 0.03903.1052i 1.7881 0.2900 Part(g): RHchart: [1,8,20,16,0] [2,12,16,0,0] [2,12,16,0,0] [12,48,32,0,0] [4,32/3,0,0,0] [16,32,0,0,0] [8/3,0,0,0,0] [32,0,0,0,0] [0,0,0,0,0] Stablesystem>>Norighthandsidepole
277
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AutomaticControlSystems,9thEdition 6poleswtzerorealpart:
Chapter2Solutions
0 0 0.0000+2.0000i 0.00002.0000i 1.0000+1.0000i 1.00001.0000i 0.0000+1.4142i 0.00001.4142i
278
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Chapter2Solutions
Golnaraghi,Kuo
(a) s 4 25 s 3 15 s 2 20 s K 0
RouthTabulation: s
4
s
3
s
2
1
K
15 25
20
375 20
K
14.2
25 s
1
284 25 K 14.2
s
0
. K 20 176
. K ! 0 or K 1136 . 20 176
K!0
K
Thus,thesystemisstablefor01,and 9 K 1 ! 0 .Since K isalwayspositive,the
lastconditioncannotbemetbyanyrealvalueofK.Thus,thesystemisunstableforallvaluesofK.
2
279
2
AutomaticControlSystems,9thEdition
Chapter2Solutions
Golnaraghi,Kuo
3
2
(c) s ( K 2 ) s 2 Ks 10
0
RouthTabulation: s
3
1
2K
s
2
K2
10
K ! 2
2
s
1
s
0
2 K 4 K 10
2
K 2K 5 ! 0
K2 10
Theconditionsforstabilityare:K>2and K 2 K 5 ! 0 or(K+3.4495)(K1.4495)>0,
orK>1.4495.Thus,theconditionforstabilityisK>1.4495.WhenK=1.4495thesystemis
marginallystable.Theauxiliaryequationis A( s )
Thefrequencyofoscillationis1.7026rad/sec.
2
3
2
(d) s 20 s 5 s 10 K
2
3.4495 s 10
0. Thesolutionis s
2
2.899 .
0
RouthTabulation: s
3
1
5
s
2
20
10 K
s
1
100 10 K 20
s
0
5 0.5K
5 0.5K ! 0 or K 10
K!0
10 K
Theconditionsforstabilityare:K>0andK>symsk >>kval=solve(5*k10+k^3,k); >>eval(kval) kval= 1.4233 0.7117+2.5533i 0.71172.5533i SoK>1.4233.
Thus,theconditionsforstabilityis:K>2
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AutomaticControlSystems,9thEdition 4
3
2
(f) s 12.5 s s 5 s K
Chapter2Solutions
Golnaraghi,Kuo
0
RouthTabulation: s
4
1
1
s
3
12.5
5
s
2
12.5 5
K
K
0.6
12.5
s
1
3 12.5 K 0.6
s
0
5 20.83 K
5 20.83 K ! 0 or K 0.24
K!0
K
Theconditionforstabilityis00,and K
282
4T 2 3T 1
.Theregionsofstabilityinthe
AutomaticControlSystems,9thEdition
Chapter2Solutions
Golnaraghi,Kuo
TversusKparameterplaneisshownbelow.
240UsetheapproachinthisChapter'sSection214: 1. ActivateMATLAB 2. GotothedirectorycontainingtheACSYSsoftware. 3. Typein Acsys
283
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Chapter2Solutions
4. Thenpressthe"transferfunctionSymbolicbutton."
284
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Chapter2Solutions
Golnaraghi,Kuo
5. Enterthecharacteristicequationinthedenominatorandpressthe"RouthHurwitz"push button. RH= [1,50000,24*k] [600,k,80*k] [1/600*k+50000,358/15*k,0] [(35680*k1/600*k^2)/(1/600*k+50000),80*k,0]
[24*k*(k^221622400*k+5000000000000)/(k30000000)/(35680*k1/600*k^2)*(1/600*k+50000), 0,0] [80*k,0,0] 6. FindthevaluesofKtomakethesystemunstablefollowingthenextsteps. (a)Characteristicequation: s 5 600 s 4 50000 s 3 Ks 2 24 Ks 80 K 0
RouthTabulation:
s
5
1
s
4
600
s
3
K
7
2
s
3 u 10 K
14320 K
600
600
21408000 K K
80 K
1
s
80 K 11
7.2 u 10 3113256 u 10 K 14400 K .
2
7
12
K 2.162 u 10 K 5 u 10 K!0
80 K
285
K 21408000
2
600(21408000 K ) 0
7
2
7
s
K 3 u 10
3 u 10 K 16
24 K
50000
0
AutomaticControlSystems,9thEdition
Conditionsforstability:
Fromthe s row:
Fromthe s row:
Chapter2Solutions
3
K 3 u 10
2
K 2.1408 u 10
Fromthe s row:
1
K 2.162 u 10 K 5 u 10
Thus,
2.34 u 10 K 2.1386 u 10
Fromthe s row:
0
K>0
Golnaraghi,Kuo
7
7
2
7
12
5
7
0 or ( K 2.34 u 10 )( K 2.1386 u 10 ) 0
5
7
5
7
2.34 u 10 K 2.1386 u 10
Thus,thefinalconditionforstabilityis:
5
When K
2.34 u 10
Z
10.6 rad/sec.
When K
2.1386 u 10
Z
188.59 rad/sec.
7
(b)Characteristicequation: s 3 ( K 2 ) s 2 30 Ks 200 K 0
Routhtabulation:
s
3
1
30 K
s
2
K2
200 K
K ! 2
2
s
1
s
0
30 K 140 K
K ! 4.6667
K2
K!0
200 K
StabilityCondition:K>4.6667
WhenK=4.6667,theauxiliaryequationis A( s )
Thefrequencyofoscillationis11.832rad/sec.
2
6.6667 s 933.333
3
2
(c)Characteristicequation: s 30 s 200 s K
0
286
0 .Thesolutionis s
2
140.
AutomaticControlSystems,9thEdition
Chapter2Solutions
Golnaraghi,Kuo
Routhtabulation: s
3
1
200
s
2
30
K
s
1
6000 K
K 6000
30 s
0
K!0
K
0 K 6000
StabililtyCondition:
WhenK=6000,theauxiliaryequationis A( s )
Thefrequencyofoscillationis14.142rad/sec.
2
30 s 6000
0. Thesolutionis s
2
200.
3
2
(d)Characteristicequation: s 2 s ( K 3) s K 1 0
Routhtabulation: s
3
1
K3
s
2
2
K +1
s
1
s
0
K 5
K ! 5
30
K ! 1
K +1
K>1.WhenK=1thezeroelementoccursinthefirstelementofthe
Stabilitycondition:
s row.Thus,thereisnoauxiliaryequation.WhenK=1,thesystemismarginallystable,andone
ofthethreecharacteristicequationrootsisats=0.Thereisnooscillation.Thesystemresponse
wouldincreasemonotonically.
0
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242Stateequation:
Openloopsystem:
A
Closedloopsystem:
A BK
x ( t )
ª 1 2 º «10 0 » ¬ ¼
x ( t )
2 º
k 2 »¼
Ax ( t ) Bu ( t )
ª0 º «1 » ¬ ¼
B
ª 1 «10 k ¬ 1
Golnaraghi,Kuo
( A BK )x ( t )
Characteristicequationoftheclosedloopsystem:
sI A BK
s 1
2
10 k1
s k2
s k 2 1 s 20 2k1 k 2 2
0
Stabilityrequirements:
k 2 1 ! 0 or k 2 ! 1
20 2k1 k 2 ! 0 or k 2 20 2k1
Parameterplane:
243)Characteristicequationofclosedloopsystem:
sI A BK
s
1
0
0
s
1
k1
k2 4
s k3 3
s k 3 3 s k 2 4 s k1 3
2
288
0
AutomaticControlSystems,9thEdition
Chapter2Solutions
Golnaraghi,Kuo
RouthTabulation:
s
3
s
2
s
1
s
k2 4
1 k3 3
k
3
k 3 +3>0 or k 3 ! 3
k1
3 k 2 4 k1
k
k3 3 0
k
3
3 k 4 k ! 0 2
1
k !0 1
1
StabilityRequirements:
k3 ! 3,
k1 ! 0,
k
3
3 k 2 4 k1 ! 0
244(a)SinceAisadiagonalmatrixwithdistincteigenvalues,thestatesaredecoupledfromeachother.The
secondrowofBiszero;thus,thesecondstatevariable, x 2 isuncontrollable.Sincetheuncontrollable
statehastheeigenvalueat3whichisstable,andtheunstablestate x3 withtheeigenvalueat2is
controllable,thesystemisstabilizable.
(b)Sincetheuncontrollablestate x1hasanunstableeigenvalueat1,thesystemisnostabilizable.
2-45) a) ܩሺݏሻ ൌ If
ௗమ௬ ௗ௧
ܻሺݏሻ ܨሺݏሻ
ሺ௦ሻ
௦మି
െ ݕൌ ݖ, then ݏଶ ܻሺݏሻ െ ܻሺݏሻ ൌ ܼሺݏሻ or ܻሺݏሻ ൌ
If ݂ሺݐሻ ൌ
ఛௗఛ ௗ௧
ݖ, then ܨሺݏሻ ൌ ሺ߬ ݏ ͳሻܼሺݏሻ. As a result: ܼሺݏሻ ݃ ݏଶ െ ͳ ݈ ൌ ܩሺݏሻ ൌ ሺ߬ ݏ ͳሻܼሺݏሻ ቀ ݏଶ െ ݃ቁ ሺ߬ ݏ ͳሻ ݈ 289
AutomaticControlSystems,9thEdition
bሻ ቊ
Chapter2Solutions
Golnaraghi,Kuo
ܨሺݏሻ ൌ ሺ߬ ݏ ͳሻܼሺݏሻ ା ሺ௦ሻ Ö ܼሺݏሻ ൌ ܧሺݏሻ ఛ௦ାଵ ܨሺݏሻ ൌ ൫ܭ ܭௗ ݏ൯ܧሺݏሻ
As a result:
ܭ ܭௗ ݏ ܻሺݏሻ ൌ ܩሺݏሻܪሺݏሻ ൌ ݃ ܧሺݏሻ ሺ߬ ݏ ͳሻ ቀ ݏଶ െ ቁ ݈ ܭ ܭௗ ݏ ܻሺݏሻ ܩሺݏሻܪሺݏሻ ൌ ൌ ܺሺݏሻ ͳ ܩሺݏሻܪሺݏሻ ሺ߬ ݏ ͳሻ ቀ ݏଶ െ ݃ቁ ܭݏ ܭ ௗ ݈ Y ( s) X (s)
G ( s) H (s) (1 G ( s ) H ( s ))
( K p K d s) ((W s 1)( s 2 g / l ) K p K d s )
( K p K d s) (W s (W ( g / l ) 1) s 2 K d s g / l K p ) 3
c) lets choose
ൌ ͳͲܽ݊݀߬ ൌ ͲǤͳ.
UsetheapproachinthisChapter'sSection214: 1. ActivateMATLAB 2. GotothedirectorycontainingtheACSYSsoftware. 3. Typein Acsys
290
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Chapter2Solutions
4. Thenpressthe"transferfunctionSymbolicbutton."
291
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Chapter2Solutions
Golnaraghi,Kuo
5. Enterthecharacteristicequationinthedenominatorandpressthe"RouthHurwitz"push button. RH =
[
1/10,
kd]
[
eps,
kp-10]
[ (-1/10*kp+1+kd*eps)/eps, [
kp-10,
0] 0]
For the choice of g/l or W the system will be unstable. The quantity W g/l must be >1. Increase W g/l to 1.1 and repeat the process. 292
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Chapter2Solutions
Golnaraghi,Kuo
d) Use the ACSYS toolbox as in section 2-14 to find the inverse Laplace transform. Then plot the time response by selecting the parameter values. Or use toolbox 2-6-1.
UsetheapproachinthisChapter'sSection214: 1. ActivateMATLAB 2. GotothedirectorycontainingtheACSYSsoftware. 3. Typein Acsys
4. Thenpressthe"transferfunctionSymbolicbutton."
293
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5. Enterthecharacteristicequationinthedenominatorandpressthe"InverseLaplaceTransform" pushbutton. ---------------------------------------------------------------Inverse Laplace Transform ----------------------------------------------------------------
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Golnaraghi,Kuo
G(s) =
[
kd
kp
[-----------------------[
]
------------------------------]
3
3
[1/10 s + s kd + kp - 10
]
1/10 s + s kd + kp - 10]
G(s) factored:
[
kd
[10 -------------------------[
kp
]
10 --------------------------]
3
[ s + 10 s kd + 10 kp - 100
3
]
s + 10 s kd + 10 kp - 100]
Inverse Laplace Transform: g(t) = matrix([[10*kd*sum(1/(3*_alpha^2+10*kd)*exp(_alpha*t),_alpha=RootOf(_Z^3+10*_Z*kd +10*kp100)),10*kp*sum(1/(3*_alpha^2+10*kd)*exp(_alpha*t),_alpha=RootOf(_Z^3+10*_Z*kd+1 0*kp-100))]]) While MATLAB is having a hard time with this problem, it is easy to see the solution will be unstable for all values of Kp and Kd. Stability of a linear system is independent of its initial conditions. For different values of g/l and , you may solve the problem similarly – assign all values (including Kp and Kd) and then find the inverse Laplace transform of the system. Find the time response and apply the initial conditions.
Lets chose g/l=1 and keep =0.1, take Kd=1 and Kp=10.
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Y (s) X (s)
G (s) H (s) (1 G ( s ) H ( s ))
Chapter2Solutions
( K p K d s) ((W s 1)( s 2 g / l ) K p K d s )
(10 s ) (0.1s (0.1(1) 1) s 2 s 1 10) 3
(10 s ) (0.1s 0.9 s 2 s 9) 3
Using ACSYS: RH =
[ 1/10,
1]
[ 9/10,
9]
[ 9/5,
0]
[
0]
9,
Hence the system is stable ---------------------------------------------------------------Inverse Laplace Transform ---------------------------------------------------------------G(s) =
s + 10 ------------------------3
2
1/10 s + 9/10 s + s + 9
296
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Chapter2Solutions
Golnaraghi,Kuo
G factored:
Zero/pole/gain: 10 (s+10) ----------------(s+9) (s^2 + 10)
Inverse Laplace Transform: g(t) = -10989/100000*exp(2251801791980457/40564819207303340847894502572032*t)*cos(79057/25000*t)+868757373/25000 0000*exp(2251801791980457/40564819207303340847894502572032*t)*sin(79057/25000*t)+10989/100000*ex p(-9*t)
Use this MATLAB code to plot the time response: fori=1:1000 t=0.1*i; tf(i)=10989/100000*exp( 2251801791980457/40564819207303340847894502572032*t)*cos(79057/25000*t)+868757373/250 000000*exp( 2251801791980457/40564819207303340847894502572032*t)*sin(79057/25000*t)+10989/100000*e xp(9*t); end figure(3) plot(1:1000,tf)
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252)USEMATLAB symst f=5+2*exp(2*t)*sin(2*t+pi/4)4*exp(2*t)*cos(2*tpi/2)+3*exp(4*t) F=laplace(f) cltF=F/(1+F) f= 5+2*exp(2*t)*sin(2*t+1/4*pi)4*exp(2*t)*sin(2*t)+3*exp(4*t) F= (8*s^3+44*s^2+112*s+160+8*2^(1/2)*s^2+16*2^(1/2)*s+2^(1/2)*s^3)/s/(s^2+4*s+8)/(s+4) cltF=
298
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Chapter2Solutions
Golnaraghi,Kuo
(8*s^3+44*s^2+112*s+160+8*2^(1/2)*s^2+16*2^(1/2)*s+2^(1/2)*s^3)/s/(s^2+4*s+8)/(s+4)/(1+(8*s^3 +44*s^2+112*s+160+8*2^(1/2)*s^2+16*2^(1/2)*s+2^(1/2)*s^3)/s/(s^2+4*s+8)/(s+4)) symss cltFsimp=simplify(cltF) NexttypethedenominatorintoACSYSRouthHurwitzprogram. char=collect(s^4+16*s^3+68*s^2+144*s+160+8*2^(1/2)*s^2+16*2^(1/2)*s+2^(1/2)*s^3) char= 160+s^4+(16+2^(1/2))*s^3+(8*2^(1/2)+68)*s^2+(16*2^(1/2)+144)*s >>eval(char) ans= 160+s^4+4901665356903357/281474976710656*s^3+2790603031573437/35184372088832*s^2+293 1340519928765/17592186044416*s >>sym2poly(ans) ans= 1.000017.414279.3137166.6274160.0000 HencetheCharacteristicequationis: '
s 4 17.4142 s 3 79.3137 s 2 166.6274 s 160
USEACSYSRouthHurwitztoolasdescribedinpreviousproblemsandthisChapter'ssection214. RH= [1,5581205465083989*2^(46),160] [87071/5000,5862680441794645*2^(45),0] [427334336632381556219/6127076924293382144,160,0]
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[238083438912827127943602680401244833403/1879436288300987963959490983755776000, 0,0] [160,0,0] Thefirstcolumnisallpositive,andthesystemisSTABLE. Fortheothersection symss G=(s+1)/(s*(s+2)*(s^2+2*s+2)) g=ilaplace(G) G= (s+1)/s/(s+2)/(s^2+2*s+2) g= 1/41/2*exp(t)*cos(t)+1/4*exp(2*t) cltG=G/(1+G) cltG= (s+1)/s/(s+2)/(s^2+2*s+2)/(1+(s+1)/s/(s+2)/(s^2+2*s+2)) cltGsimp=simplify(cltG) cltGsimp= (s+1)/(s^4+4*s^3+6*s^2+5*s+1) NexttypethedenominatorintoACSYSRouthHurwitzprogram.
2100
AutomaticControlSystems,9thEdition
Chapter2Solutions
RH= [1,6,1] [4,5,0] [19/4,1,0] [79/19,0,0] [1,0,0] STABLE
2101
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AutomaticControlSystems,9thEdition
Chapter3Solutions
Golnaraghi,Kuo
Chapter 3__________________________________________________________________________ 3-1)
a)
ܩሺݏሻܪሺݏሻ ൌ ቂ
b)
ܩሺݏሻ ൌ
c) d) e)
ாሺ௦ሻ ோሺ௦ሻ
ൌ
௦ሺ௦ାሻ
௦ሺ௦ାሻ
ሺ௦ሻ
ൌ
ವ ௦ା
ଵ ଵିீሺ௦ሻுሺ௦ሻ
Feedback ratio =
ሺ௦ሻ
ቃ ܭ ݏൌ
ீሺ௦ሻ ଵିீሺ௦ሻுሺ௦ሻ
ൌ
௦ା ௦ାି ವ
ୋሺୱሻୌሺୱሻ ଵିீሺ௦ሻுሺ௦ሻ
ൌ
ൌ
ವ ௦ାି ವ
௦൫௦ାିವ ൯
3-2)
Characteristic equation: ݏሺ ݏ ͳሻሺ ݏ ʹሻ ͳ ൌ Ͳ Ö ݏଷ ͵ ݏଶ ʹ ݏ ͳ ൌ Ͳ
31
AutomaticControlSystems,9thEdition
Chapter3Solutions
Golnaraghi,Kuo
3-3)
G
1
1 G1 H 1
H2 G2 G1 1 G1 H 1
G1G2 1 G1 H 1
G3
H2 G2
ܩଵ ܩଶ ܻሺݏሻ ܩଵ ܩଶ ܩଷ ܩଵ ܪଶ ܪଶ ͳ െ ܩଵ ܪଵ ൌ ൬ܩଷ ൰ ൌ ܩଶ ͳ െ ܩଵ ܪଵ ܩଵ ܩଶ ܪଷ ܺሺݏሻ ͳ ܩଵ ܩଶ ܪଷ ͳ െ ܩଵ ܪଵ
32
AutomaticControlSystems,9thEdition
Chapter3Solutions
Golnaraghi,Kuo
3-4)
G2 1 G 2 G3 H 3
X
+
G1 -
G2 1 G2G3 H 3 G2 H 2
H1
33
G3
Y
AutomaticControlSystems,9thEdition A
X
+ -
Chapter3Solution ns
G1G2 G3 1 G 2 G3 H 3 G 2 H 2
H1 G3
ܻሺݏሻ ܩଵ ܩଶ ܩଷ ൌ ܺሺݏሻ ͳ ܩଵ ܩଶ ܪଵ ܩଶ ܪଶ ܩଶ ܩଷ ܪଷ
3 3-5)
34
Golnarraghi,Kuo
Y
AutomaticControlSystems,9thEdition
Chapter3Solutions
3-6) MATLAB symss G=[2/(s*(s+2)),10;5/s,1/(s+1)] H=[1,0;0,1] A=eye(2)+G*H B=inv(A) Clp=simplify(B*G) G= [ 2/s/(s+2), 10] [ 5/s, 1/(s+1)] H= 1 0
0 1
A= [ 1+2/s/(s+2), 10] [ 5/s, 1+1/(s+1)] B= [ [
s*(s+2)/(s^2-48*s-48), -10/(s^2-48*s-48)*(s+1)*s] -5/(s^2-48*s-48)*(s+1), (s^2+2*s+2)*(s+1)/(s+2)/(s^2-48*s-48)]
Clp = [ [
-2*(24+25*s)/(s^2-48*s-48), 10/(s^2-48*s-48)*(s+1)*s] 5/(s^2-48*s-48)*(s+1), -(49*s^2+148*s+98)/(s+2)/(s^2-48*s-48)]
35
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition A
Chapter3Solution ns
3-7)
3-8)
36
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition A
Chapter3Solution ns
3-9)
3 3-10)
3 3-11)
37
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition A
Chapter3Solution ns
3-12)
3-13) symst f=100*(10.3 3*exp(6*t)0 0.7*exp(10*t)) F=laplace(f) symss F=eval(F) Gc=F*s M=30000 symsK Olp=simplify(K K*Gc/M/s) Kt=0.15 Clp=simplify(Olp/(1+Olp*K Kt)) s=0 Ess=eval(Clp) f= 10030*exp(6 6*t)70*exp(10*t) F= 80*(11*s+75)/s/(s+6)/(s+1 10) ans= 38
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition A Chapter3Solution ns (880*s+6000)/s/(s+6)/(s+1 10) Gc= (880*s+6000)/(s+6)/(s+10 0) M= 30000 Olp= 1/375*K*(11**s+75)/s/(s+6 6)/(s+10) Kt= 0.1500 Clp= 2 20/3*K*(11*s s+75)/(2500*s^3+40000*ss^2+150000*ss+11*K*s+75*K) s s= 0 E Ess= 2 20/3
3 3-14)
39
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition
Chapter3Solutions
3-15) Note: If
1
G(s)=g(t),then
1
{easG(s)}=u(ta)•g(ta)
symsts f=100*(10.3*exp(6*(t0.5))) F=laplace(f)*exp(0.5*s) F=eval(F) Gc=F*s M=30000 symsK Olp=simplify(K*Gc/M/s) Kt=0.15 Clp=simplify(Olp/(1+Olp*Kt)) s=0 Ess=eval(Clp) digits(2) Fsimp=simplify(expand(vpa(F))) Gcsimp=simplify(expand(vpa(Gc))) Olpsimp=simplify(expand(vpa(Olp))) Clpsimp=simplify(expand(vpa(Clp))) f= 10030*exp(6*t+3) F= (100/s30*exp(3)/(s+6))*exp(1/2*s) F= (100/s2650113767660283/4398046511104/(s+6))*exp(1/2*s) Gc= (100/s2650113767660283/4398046511104/(s+6))*exp(1/2*s)*s M= 30000 Olp= 1/131941395333120000*K*(2210309116549883*s2638827906662400)/s/(s+6)*exp(1/2*s) Kt= 0.1500 Clp=
310
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition A Chapter3Solution ns Golnarraghi,Kuo 2 20/3*K*(2210 03091165498 883*s263882 27906662400 0)*exp(1/2*s)/(87960930 02220800000 0*s^2 5 52776558133 324800000*s+ +2210309116 6549883*K*eexp(1/2*s)*s2638827906 6662400*K*exxp(1/2*s)) s= 0 Ess= 2 20/3 Fsimp= 50*s)*(5.*s6 6.)/s/(s+6.) .10e3*exp(.5 Gcsimp= .10e3*exp(.5 50*s)*(5.*s6 6.)/(s+6.) Olpsimp= .10e2*K*exp p(.50*s)*(17 7.*s20.)/s/(s++6.) Clpsimp= 5 5.*K*exp(.50 0*s)*(15.*s1 17.)/(.44e4*ss^2.26e5*s+1 11.*K*exp(.5 50*s)*s13.*K K*exp(.50*s)))
3-16)
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3-17)
3-18) ͳ ܺଵ ሺݏሻ ൌ ሾെͷܺଵ ሺݏሻ െ ܺଶ ሺݏሻ ͵ܺଷ ሺݏሻ ͲǤͷܷଵ ሺݏሻሿ ݏ ͳ ܺଶ ሺݏሻ ൌ ሾܺଵ ሺݏሻ െ ܺଷ ሺݏሻ ͲǤͷܷଶ ሺݏሻሿ ݏ ۔ ͳ ۖ ሺݏሻ ൌ ሾെͲǤͷܺଵ ሺݏሻ ͳǤͷܺଶ ሺݏሻ ͲǤͷܺଷ ሺݏሻ ͲǤͷܷଵ ሺݏሻ ͲǤͷܷଶ ሺݏሻሿ ܺەଷ ݏ ۓ ۖ
൜
ܼଵ ሺݏሻ ൌ ͲǤͷܺଵ ሺݏሻ ͲǤͷܺଶ ሺݏሻ ܼଶ ሺݏሻ ൌ ͲǤͷܺଵ ሺݏሻ ͲǤͷܺଷ ሺݏሻ
0.5
u1
0.5
z1
1
3 0.5 1/s 0.5
0.5 u2
0.5
-1
1/s
-6 x2 1
x3
x1
0.5 0.5
-5
1
1.5
0.5 -0.5
312
1/s
z2
AutomaticControlSystems,9thEdition
Chapter3Solutions
Golnaraghi,Kuo
3-19)
ܩሺݏሻ ൌ
ܤଵ ݏ ܤ ܻሺݏሻ ൌ ଶ ܷሺݏሻ ݏ ܣଵ ݏ ܣ
Ö
ሺ ݏ ܣଵ ݏ ܣ ሻܻሺݏሻ ൌ ሺܤଵ ݏ ܤ ሻܷሺݏሻ
Ö
ቀ ݏ ܣଵ
Ö
ቊ
Ö Ö
బ ௦
ቁ ܻሺݏሻ ൌ ܤଵ ܷሺݏሻ
బ ௦
ܷሺݏሻ
ܻݏሺݏሻ ൌ െܣଵ ܻሺݏሻ ܺሺݏሻ ܤଵ ܷሺݏሻ ܺሺݏሻ ൌ െ
బ ௦
ܻሺݏሻ
బ ௦
ܷሺݏሻ
ܻݏሺݏሻ ൌ െܣଵ ܻሺݏሻ ܺሺݏሻ ܤଵ ܷሺݏሻ ൜ ܺݏሺݏሻ ൌ െܣ ܻሺݏሻ ܤ ܷሺݏሻ ݕሶ ൌ െ ݕܣ ݔ ܤଵ ݑሺݐሻ ൜ ݔሶ ൌ െܣ ݕ ܤ ݑሺݐሻ B1 u
B0
1/s
1 x
313
y -A1
-A0 3-20)
1/s
AutomaticControlSystems,9thEdition A
Chapter3Solution ns
3 3-21)
3 3-22)
314
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition A
Chapter3Solution ns
3-23)
315
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition A
Chapter3Solution ns
3-24)
3 3-25)
3 3-26)
3 3-27)
316
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition A
Chapter3Solution ns
3-28)
317
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition A
Chapter3Solution ns
3-29)
318
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition A
Chapter3Solution ns
3-30) Use Mason'ss formula:
3-31) MATLAB symssK G=100//(s+1)/(s+5) g=ilaplace(G/s) H=K/s YN=sim mplify(G/(1+G*H)) Yn=ilap place(YN/s) G= 100/(ss+1)/(s+5) g= p(5*t)+20 25*exxp(t)+5*exp H= K/s 319
Golnarraghi,Kuo
AutomaticControlSystems,9thEdition
Chapter3Solutions
YN= 100*s/(s^3+6*s^2+5*s+100*K) ApplyRouthHurwitzwithinSymbolictoolofACSYS(seechapter3)
RH= [1,5] [6,100*k] [50/3*k+5,0] [100*k,0] Stabilityrequires:0 K M 1 M 2 B1 B2 B2 B3 B1 B3 @ s B1 B2 K
^
3
2
B3 s K
s M 1 M 2 s > B1 B3 M 2 B2 B3 M 1 @ s > K M 1 M 2 B1 B2 B2 B3 B1 B3 @ s B1 B2 K 3
2
(b)Forceequations:
46
` `
AutomaticControlSystems,9thEdition
2
d y1 dt
2
B
B2 dy1
1
M
dt
Chapter4Solutions
B2 dy 2 M dt
1
f
Golnaraghi,Kuo
dy2
dy1
dt
dt
M
K B2
y2
(i)Statediagram:
Definetheoutputs oftheintegrators asstatevariables, dy1 x1 y 2 , x 2 dt .
Stateequations:
dx1 dt
K B2
dx1
K B2
x1 x3
Transferfunctions:
Y1 ( s) F (s)
dx 2
K M
x1
B1 M
y2 , x2 dx3
x3
dt
dt
K M
B2 s K
x1
1
f
M
B1 M
dy1
.
dt x3
1
f
M
Y2 ( s )
B2
F (s)
MB2 s ( B1 B2 KM ) s ( B1 B2 ) K
s ¬ª MB2 s B1 B2 KM s B1 B2 K ¼º
2
(c)Forceequations:
dy1
dy2 dt
1 B1
2
f
d y2 dt
2
B
1
B2 dy2 M
dt
(i)Statediagram:
47
x2
y1 , x3
2
dt
dt
(ii)Stateequations:Statevariables: x1
dt
dx 2
x1 x 2
B1 dy2 M dt
B1 dy1 M dt
K M
y2
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
Stateequatio ons:Definetheeoutputsofintegratorsassttatevariables.
dx1
x2
dt
d dt
K M
(ii)Stateequations:statevariaables: x1
dx1 dt
dxx 2
x3
1
dx 2
f
x3
dt
B1
x1
B2 M
x2
y1 , x 2 dxx3
f
M y 2 , x3
d dt
1
K
x2
M
B2 M
dy 2
.
dt x3
1
f
M
Statediagram m:
Transferfuncctions: Y1 ( s )
Ms B1 B2 s K
F (s)
B1 s Ms B2 s K
2
2
Y2 ( s )
1
F (s)
Ms B2 s K
4-55)
48
2
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
(a)Forceequaations:
y1
1 K2
2
d y2
( f Mg ) y 2
dt
2
B dy 2 M dt
K1 K 2 M
y2
K2
y1
M
Staatediagram:
Staateequations: Deffinethestate varriablesas: dy 2 x1 y 2 , x 2 d dt . dx1
dtt
d 2 dx
x2
dt
K1 M
x1
B M
Transferfunction ns:
dy1 dt
1 B1
> f (t ) Mg @
( f Mg )
M
2
Y2 ( s )
2
K 2 ( Ms Bs K1 )
F ( s)
(b)Forcceequations:
1
s Bs K1 K 2
Y1 ( s )
x2
dy 2 dt
K1 B1
y
y2 1
F ( s)
2
d y2 dt
2
1 2
Mss Bs K1
§ dy1 dy2 · K1 y y B2 y y B2 dy2 1 2 1 2 ¨ ¸ M M © dt dt ¹ M M dt B1
Stattediagram:(W Withminimumnumberofinttegrators)
49
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
ToobtainthetranssferfunctionsY1 ( s ) / F ( s ) andY2 ( s ) / F ( s ), weneedtoredefinethesttatevariablesaas:
x1
Stattediagram:
y2 , x2
dy 2 / dtt , and x3
y1 .
Transferfuncctions:
F (s)
4-66)
Ms B1 B2 s K1 2
Y1 ( s ) s
2
M s B B > MB 1
1
2
MK1 @
Bs K1
Y2 ( s ) F (s)
s
2
M s B B > MB 1
1
2
MK1 @
a) y1
y2
m
K ( y1 y 2 )
K ( y1 y 2 )
M
Pmgy 2 b)
PMgy1
From Newtton's Law: ݕܯሷଵ ൌ ܨെ ܭሺݕଵ െ ݕଶ ሻ െ ߤݕ݃ܯଵሶ ݉ݕሷ ଶ ൌ ܭሺݕ ݕଵ െ ݕଶ ሻ െ ߤ ߤ݉݃ݕଶሶ If y1 and y2 are considerred as a posiition and v1 and a v2 as vellocity variabbles ݕଵሶ ൌ ݒଵ ݕଶሶ ൌ ݒଶ Then: ൞ ݒଵሶ ൌ ܨെ ܭሺሺݕଵ െ ݕଶ ሻ െ ߤݒ݃ܯଵ ݒܯሶ ݉ݒሶ ݒଶሶ ൌ ܨെ ܭሺሺݕଵ െ ݕଶ ሻ െ ߤ݉݃ݒଶ The output equation cann be the veloocity of the engine, e whicch means ݖൌ ݒଶ 410
F
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
c) ݏܯଶ ܻଵ ሺݏሻ ൌ ܨെ ܭ൫ܻଵ ሺݏሻ െ ܻଶ ሺݏሻ൯ െ ߤܻݏ݃ܯଵ ሺݏሻ ቐ ݉ ݏଶ ܻଶ ሺݏሻ ൌ ܭ൫ܻଵ ሺݏሻ െ ܻଶ ሺݏሻ൯ െ ߤܻ݉݃ݏଶ ሺݏሻ ܼሺݏሻ ൌ ܸଶ ሺݏሻ ൌ ܻݏሺݏሻ Obtaining
ሺ௦ሻ ிሺ௦ሻ
requires solving above equation with respect to Y2(s)
From the first equation: ሺ ݏܯଶ ܭ ߤݏ݃ܯሻܻଵ ሺݏሻ ൌ ܨ ܻܭଶ ሺݏሻ ܻଵ ሺݏሻ ൌ
ܨ ܻܭଶ ሺݏሻ ߤ ݏ݃ܯ ܭ
ݏܯଶ
Substituting into the second equation: ݉ ݏଶ ܻଶ ሺݏሻ ൌ
ܨܭ ܭଶ ܻଶ ሺݏሻ െ ܻܭଶ ሺݏሻ െ ߤܻ݉݃ݏଶ ሺݏሻ ݏܯଶ ߤ ݏ݃ܯ ܭ
By solving above equation: ܼሺݏሻ ܻݏଶ ሺݏሻ ݉ ݏଶ ݉ߤ݃ ݏ ͳ ൌ ൌ ݏ݉ܯଷ ሺʹ݃ߤ݉ܯሻ ݏଶ ሺ ݇ܯ ݉ܯሺߤ݃ሻଶ ݉݇ሻ ݏ ݃ߤܭሺ ܯ ݉ሻ ܨሺݏሻ ܨሺݏሻ d) Ͳ Ͳ ۍ ݕଵሶ ܭ ێ ݕଶሶ ൦ ൪ ൌ ێെ ݕଷሶ ݉ ێ ݕସሶ ܭ ێ ݉ ۏ
Ͳ Ͳ ܭ ܯ ܭ െ ܯ
ͳ Ͳ െߤ݃ Ͳ
Ͳ Ͳ ͳ ݕ ېଵ ېͲۍ ݕ ۑ ۑ ێ Ͳ ۑ൦ݒଶ ൪ ܨ ۑ ͳ ێ ۑଵ ۑ ܯێ ݒ ۑଶ ۏ Ͳے െߤ݃ے
ݕଵ ݕଶ ܼ ൌ ሾͲͲͲͳሿ ൦ ݒ൪ ሾͲሿܨ ଵ ݒଶ 4-7)(a)Forceequations:
f (t )
§ dy1 dy2 · ¸ © dt dt ¹
2 § dy1 dy2 · M d y2 B dy2 ¸ t 2 dt dt © dt dt ¹
K h y1 y2 Bh ¨
(b)Statevariables: x1
y1 y 2 , x 2
K h y1 y 2 Bh ¨
dy 2
dt
411
AutomaticControlSystems,9thEdition
Stateequations:
dx1 dt
Kh Bh
x1
1
Chapter4Solutions
dx 2
f (t )
dt
Bh
Bt M
x2
1
Golnaraghi,Kuo
f (t )
M
4-8)
D2
D1
For the left pendulum: ܶ௧ ൌ ݈݉ ଶ ߙଶሷ ܷ ൌ െ݈݉݃ •‹ ߙଶ
଼
଼
ܶ ൌ ܭቀ ݈ቁ ሺ•‹ ߙଶ െ •‹ ߙଵ ሻ …'• ߙଶ ቀ ݈ቁ ൌ ܭ Ö ܶ௧ ܷ ܶ ൌ Ͳ Ö ݈݉ ଶ ߙଶሷ ݈݉݃ •‹ ߙଶ
ସଽ
ସଽ ଶ ݈ ሺ•‹ ߙଶ ସ
െ •‹ ߙଵ ሻ …'• ߙଶ
݈ܭଶ ሺ•‹ ߙଶ െ •‹ ߙଵ ሻ …'• ߙଶ
ସ
For the right pendulum, we can write the same equation: ݈݉ଶ ߙଵሷ ݈݉݃ •‹ ߙଵ
ସଽ ସ
݈ܭଶ ሺ•‹ ߙଵ െ •‹ ߙଶ ሻ …'• ߙଵ
since the angles are small: •‹ ߙଶ ؆ ߙଶ …'• ߙଶ ؆ ͳ Ö ൞ •‹ ߙଵ ൌ ߙଵ …'• ߙଵ ൌ ͳ
4-9)
Ö
ቐ
݈݉ߙଵሷ ݉݃ߙଵ ݈݉ߙଶሷ ݉݃ߙଶ
a)
412
ସଽ ସ ସଽ ସ
݈ܭሺߙଵ െ ߙଶ ሻ ൌ Ͳ ݈ܭሺߙଶ െ ߙଵ ሻ ൌ Ͳ
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
Fy
T
x(t)
m
Fx
mg M
f
T
b)
If we consider the coordinate of centre of gravity of mass m as (xg, yg), Then ݔ ൌ ݔ ݈ •‹ ߠ and ݕ ൌ ݈ …'• ߠ From force balance, we have: ݔܯሷ ݉ݔሷ ൌ ݂ Ö ݔܯሷ ݉ݔሷ ݉൫ߠሷ …'• ߠ െ ߠሶ ଶ •‹ ߠ൯ ൌ ݂ From the torque balance, we have: ሺܨ௫ …'• ߠሻ݈ െ ൫ܨ௬ •‹ ߠ ൯݈ ൌ ሺ݉݃ •‹ ߠሻ݈ Where: ܨ௫ ൌ ݉ݔሷ ൌ ݉൫ݔሷ െ ݈ߠሶ ଶ ߠ݊݅ݏ ݈ߠሷ …'• ߠ൯ ቊ ܨ௬ ൌ ݉ݕሷ ൌ െ݉൫݈ߠሶ ଶ …'• ߠ ݈ߠሷ •‹ ߠ൯ Substituting these equation: ݉ݔሷ …'• ߠ ݈݉ߠሷ ൌ ݉݃ •‹ ߠ
4-10) a)
413
AutomaticControlSystems,9thEdition
Chapter4Solutions
Fy1
Golnaraghi,Kuo
Fy 2 Fx 2
Fx1 m1 g
m2 g
T1
b)
T2
Kinetic energy (i) For lower pendulum: ଶ ଶ ͳ ͳ ݀ ݀ ଶ ሶ ܶଵ ൌ ܬଵ ߠଵ ݉ଵ ቊ ሺ݈ଵ •‹ ߠଵ ሻ൨ ሺ݈ଵ …'• ߠଵ ሻ൨ ቋ ʹ ʹ ݀ݐ ݀ݐ
For upper pendulum: ଶ ଶ ͳ ͳ ݀ ݀ ܶଶ ൌ ܬଶ ߠሶଶଶ ݉ଶ ቊ ሺ݈ଶ •‹ ߠଶ ሻ൨ ሺ݈ଶ …'• ߠଶ ሻ൨ ቋ ʹ ʹ ݀ݐ ݀ݐ ଵ
For the cart: ܶଷ ൌ ݔܯሶ ଶ ଶ
(ii) Potential energy: For lower pendulum:
ܷଵ ൌ ݉ଵ ݈݃ଵ …'• ߠଶ
For upper pendulum:
ܷଶ ൌ ݉ଶ ݈݃ଶ …'• ߠଶ
For the cart:
ܷଷ ൌ Ͳ
(iii) Total kinetic energy:
ܶଵ ൌ ܶଵ ܶଶ ܶଷ
Total potential energy: ܷ ൌ ܷଵ ܷଶ ܷଷ 414
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
The Lagrangian equation of motion is: ݀ ߲ܶ ߲ܶ ߲ܷ ۓ ൌ݂ ൬ ൰െ ݀ݔ߲ ݐሶ ߲ݔ߲ ݔ ۖ ۖ ݀ ߲ܶ ߲ܶ ߲ܷ ቆ ቇ െ ൌͲ ሶ ݀ݐ ߲ߠ ߲ߠ ߲ߠ ଵ ଵ ଵ ۔ ۖ ۖ ݀ ቆ ߲ܶ ቇ െ ߲ܶ ߲ܷ ൌ Ͳ ߲ߠଶ ߲ߠଶ ߠ߲ ݐ݀ەଶሶ Substituting T and U into the Lagrangian equation of motion gives: ଶ ଶ ሺ݉ଵ ݉ଶ ܯሻݔሷ ݉ଵ ݈ଵ ߠଵ …'• ߠଵ ݉ଶ ݈ଶ ߠଶ …'• ߠଶ ൌ ݉ଵ ݈ଵ ߠଵሶ •‹ ߠଵ ݉ଶ ݈ଶ ߠଶሶ •‹ ߠଶ ݂ ൞ ݉ଵ ݈ଵ ݔሷ …'• ߠଵ ሺܬଵ ݉ଵ ݈ଵଶ ሻߠଵሷ ൌ ݉ଵ ݈ଵ ݃ •‹ ߠଵ ݉ଶ ݈ଶ ݔሷ …'• ߠଶ ሺܬଶ ݉ଶ ݈ଶଶ ሻߠଶሷ ൌ ݉ଶ ݈ଶ ݃ •‹ ߠଶ
4-11) a)
From the Lagrangian equation of motion: ൬
b)
ܬ ݉൰ ሷ ݉݃ •‹ ߙ െ ݉ߙሶ ଶ ൌ Ͳ ݎଶ
As: ߙൌ
݀ ߠ ܮ
Then ൬
ܬ ݀ߠሶ ݀ ݉൰ ሷ ݉݃ •‹ ቆ ቇ െ ݉ߠ ሶ ଶ ൌ Ͳ ଶ ݎ ܮ ܮ
If we linearize the equation about beam angle = 0, then sin and sin Then: ܬ ݀ ݉൰ ሷ ൌ െ݉݃ ߠ ଶ ݎ ࣦ ܬ ݉݃݀ ߠሺݏሻ ൬ ଶ ݉൰ ݏଶ ܲሺݏሻ ൌ െ ݎ ܮ ܲሺݏሻ ݉݃݀ ൌ ߠሺݏሻ ݏଶ ܮቀ ܬ ݉ቁ ݎଶ ൬
415
AutomaticControlSystems,9thEdition
c)
Chapter4Solutions
Considering ൜
ሶ ൌ ݍ ሷ ൌ ݍሶ
Then the state-space equation is described as:
ሶ Ͳ ൨ൌቂ ݍሶ Ͳ
d)
G ( s)
ͳ ቃቂ ቃ ൦ Ͳ ݍ
mgd ( s L( J / r 2 m)) 2
clear all % select values of m, d, r, and J %Step input g=10; J=10; M=1; D=0.5; R=1; L=5; G=tf([M*g*D],[L*(J/R^2+M) 0 0]) step(G,10) xlabel( 'Time(sec)'); ylabel('Amplitude'); Transfer function: 5 -----55 s^2
416
Ͳ ݉݃݀ ൪ߠ ܬ ܮቀ ଶ ݉ቁ ݎ
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
4-12) If the aircraft is at a constant altitude and velocity, and also the change in pitch angle does not change the speed, then from longitudinal equation, the motion in vertical plane can be written as: ݔ ݑ ۓሶ ൌ ݉ െ ݃ •‹ ߠ െ ߱ݍ ۖ ۖ߱ሶ ൌ ݖെ ݃ …'• ߠ ݑݍ ݉ ܯ ۔ ݍሶ ൌ ۖ ܫ௬௬ ۖ ߠሶ ൌ ݍ ە Where u is axial velocity, is vertical velocity, q is pitch rate, and is pitch angle. Converting the Cartesian components with polar inertial components and replace x, y, z by T, D, and L. Then we have: ͳ ܸ ۓሶ ൌ ሾܶ …'• ߙ െ ܦെ ݉݃ •‹ ߛሿ ݉ ۖ ͳ ۖ ߛሶ ൌ ሾܶ •‹ ߙ ܮെ ݉݃ …'• ߛሿ ܸ݉ ۔ ܯ ݍሶ ൌ ۖ ܫ௬௬ ۖ ߠሶ ൌ ݍ ە Where = – is an attack angle, V is velocity, and is flight path angle. 417
AutomaticControlSystems,9thEdition
Chapter4Solutions
It should be mentioned that T, D, L and M are function of variables and V. Refer to the aircraft dynamics textbooks, the state equations can be written as: ߙሶ ൌ ܣଵ ߙ ܤଵ ݍ ܥଵ ߛ ቐݍሶ ൌ ܣଶ ߙ ܤଶ ݍ ܥଶ ߛ ߠሶ ൌ ܣଷ ݍ b)
The Laplace transform of the system is: ܩሺݏሻ ൌ
ߠሺݏሻ ߛሺݏሻ
By using Laplace transform, we have: ߙݏሺݏሻ ൌ ܣଵ ߙሺݏሻ ܤଵ ݍሺݏሻ ܥଵ ߛሺݏሻሺͳሻ ݍݏሺݏሻ ൌ ܣଶ ߙሺݏሻ ܤଶ ݍሺݏሻ ܥଶ ߛሺݏሻሺʹሻ ߠݏሺݏሻ ൌ ܣଷ ݍሺݏሻሺ͵ሻ From equation (1): ߙሺݏሻ ൌ
ܥଵ ܤଵ ݍሺݏሻ ߛሺݏሻ ݏെ ܣଵ ݏെ ܣଵ
Substituting in equation (2) and solving for q(s): ݍሺݏሻ ൌ
ܥଷ ሺ ݏെ ܣଵ ሻ ܣଶ ܥଵ ߛሺݏሻ ݏሺ ݏെ ܣଵ ሻ െ ܤଶ ሺ ݏെ ܣଵ ሻ െ ܣଶ ܤଵ
Substituting above expression in equation (3) gives: ሺܥଶ ݏ ܣଶ ܥଵ െ ܥଶ ܣଵ ሻܣଷ ߠሺݏሻ ൌ ଶ ߛ ሺݏሻ ݏሾ ݏെ ሺܣଵ ܤଶ ሻ ݏെ ሺܤଶ ܣଵ ܣଶ ܤଵ ሻሿ If we consider ݑൌ ߱ଶ •‹ ߱ݐ, then ݕܯሷ ݕܤሶ ݕܭൌ ݈݉ݑ By using Laplace transform: ሺ ݏܯଶ ݏܤ ܭሻܻሺݏሻ ൌ ݈ܷ݉ሺݏሻሺͶሻ
Which gives: ܻሺݏሻ ݈݉ ൌ ܷሺݏሻ ݏܯଶ ݏܤ ܭ 418
Golnaraghi,Kuo
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
For plotting g state flow diagram, equation u (4) muust be rewritten as: ൬ ݏ
ܭ ݈݉ ܤ ܷሺݏሻ ൰ ܻሺݏሻ ൌ ܵܯ ܵܯ ܯ
or ܻݏሺݏሻ ൌ െ
൞
ܺሺݏݏሻ ܤ ܤ ܻሺݏሻ ܺሺݏሻ ՜ ܻ ܻሺݏሻ ൌ െ ܻሺݏሻ ܯ ܯ ݏ ܭ ݈݉ ܺሺݏሻ ൌ െ ܻሺݏሻ ܷሺݏሻ ܵܯ ܵܯ
S the state flow diagram So, m will plotteed as: ml
u
1/Ms
1 x
1/s
y
-B/M -K
Also look at seection 4-11 4-113)(a)Torqueequation: 2
d T dt
2
B dT J dt
1
Stateequations:
dt
Stated diagram:
T (t )
dx 2
x2
J
dx1
dt
Transsferfunction:
4( s )
1
T ( s)
s( Js B )
B J
x2
1
T
J
(b)Torqueequations: d T1 2
dt
2
K J
T
1
T2
1 J
T
K T 1 T 2
B
dT 2 dt
Statediagram:(minimumnumbeerofintegratorrs)
419
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
Stateequations:
dx1 dt
K B
dx 2
x1 x 2
dt
Stateequations:Let x1
dx1 dt
K B
x1
K B
x2
Statediagram:
Tran nsferfunctionss:
41 ( s )
T 2 , x2 dx 2 dt
Bs B K
s BJs JKs BK
T (s)
2
(c)Torqqueequationss:
T (t )
d T1 2
J1
dt
2
K T 1 T 2
K J
x1
1
T
J dT 1
T 1, and x3 x3
dx3
K
dt
J
42 ( s)
x1
K J
x2
1
T
J
K
s BJs JK Ks BK
T (s)
K T 1 T 2
2
d T2 2
J2
Stattediagram:
420
.
dt
dt
2
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
State eequations:sttatevariables:: x1
dx1 dt
dx 2
x2
dt
K J2
x1
K J2
x3
dt
d 3 dx dt
41 ( s )
J2s K
T (s)
s ª¬ J 1 J 2 s K J 1 J 2 º¼
2
2
2
(d)Torqueequationss:
T (t )
d Tm 2
x4
dT 1
T 1, x 4 d 4 dx
K
dt
J1
.
d dt
x1
K J1
x3
1
T
J1
Jm
dt
2
K1 T m T 1 K 2 T m T 2
42 ( s)
K
T (s)
s ª¬ J 1 J 2 s K J 1 J 2 º¼ 2
K1 T m T 1
421
, x3
nsferfunctionss: Tran
dT 2
T 2 , x2
2
d T1 2
J1
dt
2
K 2 T m T 2
d T2 2
J2
dt
2
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
Statediiagram:
Stateequatio ons: x1
dx1
x 2 x3
dt
dx 2
K1
dt
J1
dx3
x1
dt
dT 1 dtt
K1 Jm
x1
dt K2
, x4
x4
Jm
1
dT 2
T m T 2 , x5
T
Jm
dx 4 dt
.
dt
x3 x5
dx5
K2
dtt
J2
41 ( s )
K1 ( J 2 s K 2 )
x4
T (s)
s ª¬ s K1 J 2 J m K 2 J 1 J m K1 J 1 J 2 K 2 J 1 J 2 s K1 K 2 J m J 1 J 2 º¼
42 (s)
K 2 ( J 1 s K1 )
T (s)
s ª¬ s K1 J 2 J m K 2 J 1 J m K1 J 1 J 2 K 2 J 1 J 2 s K1 K 2 J m J 1 J 2 º¼
2
2
4
2
4
d 2T m dt
2
K1 Jm
2
2
(e)Torqqueequationss:
T m T1
2
K2 Jm
T m T 2
1 Jm
d T1 2
T
d dt
2
K1 J1
Statediagram m:
422
dT m
, x3
Transferfuncctions:
T m T 1, x2
T
m
T1
B1 dT 1 J 1 dt
d T2 2
dt
2
K2 J2
T
m
T1
B2 dT 2 J 2 dt
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
dx1 dt
dT 1
T m T 1, x2
SStatevariabless: x1
Stateequatio ons: dx 2 K1 B x 2 x3 x1 1 x 2 J1 dt J1
dt
dx3
dt
K1 Jm
dT m
, x3
x1
dt
K2 Jm
x4
T m T 2 , x5
, x4
1
dx4
T
Jm
dt
x3 x 5
dT 2
.
dt
dx5
K2
dt
J2
x4
B2 J2
x5
Transferfuncctions: K1 J 2 s B2 s K 2
T (s)
'(s)
2
42 (s)
K 2 J 1 s B1 s K1
T (s)
'(s)
2
s { J 1 J 2 J m s J m B1 B2 s > K1 J 2 K 2 J 1 J m K1 K 2 J 1 J 2 B1 B2 J m @ s
'( s)
41 ( s )
2
4
3
> B1 K 2 B2 K1 J m B1 K 2 J 2 B2 K1 J 1 @ s K1 K 2 J m J 1 J 2 }
2
4-114) d T1 2
Tm (t )
(a))
Jm
dt
2
T1
N1
T1
T3
N2 N4
T1
T2
N3 N4
T3
N3
d T3
N2
N1 N 3
T2
T4
N4
N3 N4
dt
2
d T1 2
Tm
Jm
423
JL
T4
2
JL
d T3 2
T4
dt
2
dt
T2
2
N1 N 3 N2 N4
T4
T3
T2
N1 N2
T1
2 ª ª N1 N 3 º º d 2T 1 «Jm « » JL » 2 ¬ N 2 N 4 ¼ »¼ dt ¬«
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
(b) d T1
d T2
2
Tm
Jm
T2
N1
dt
N2
2
T1
2
T1
T2
Tm (t ) J m
d T1 dt
2
2
dt
T1
N2 N4
T2
N1 §
d T2
N2 ©
dt
2
¨ J2
T4
J
d T2
N3
T3 2
N1 N 3
T3
2
J2
2
N3
N4
J2
J
dt
J4
3
2
d T3 2
JL
3
dt
d T2 2
T4
N4
J2
dt
N1
T1
2
2
N3
N4
N2
J
3
T2
N3
T3
JL
N4
T4
d T3 2
dt
2
2 2 ª º d 2T1 § N1 · § N1 N 3 · J J J J « m ¨ ¸ 2 ¨ ¸ 3 L » dt 2 2 ¸ dt ¹ «¬ © N2 ¹ © N2 N4 ¹ »¼
d T3 · 2
4-15)(a) 2
Tm
Jm
Tm
Jm
2
d Tm dt
T1
2
d Tm
T2 d TL
2
Set
dt
wD L
2
JL
dt
dt
nTL
2
T
2
TL
N1
T1
N2
T2
2
nTm n TL 2
Thus, D L
2 nTL J m n J L 2 nJ L nTm n J L
m
T m N1 T L N 2
nT2
§ J m nJ · D nT ¨ L ¸ L L © n ¹
2
nJ L
0.
wn
d TL
2
Jm n JL 2
2
0 Or, n
J mTL J LTm
n
Jm
0
JL
Optimalgearratio:
n
J mTL 2 J LTm
2
2
2
J mTL 4 J m J LTm
wherethe+signhasbeenchosen.
2 J LTm
(b)When TL
0 ,theoptimalgearratiois
n
Jm / J L
4-16)(a)Torqueequationaboutthemotorshaft: 2
Tm
Jm
d Tm dt
2
2
Mr
2
d Tm dt
2
Bm
dT m
Relationbetweenlinearandrotationaldisplacements:
y
dt
rT m
(b)TakingtheLaplacetransformoftheequationsinpart(a),withzeroinitialconditions,wehave
Tm ( s )
J
m
Mr
2
s 4 2
m
( s ) Bm s4 m ( s )
424
Y (s)
r4m (s)
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
Transfferfunction:
Y ( s)
r
s ª¬ J m Mr
Tm ( s )
r
s B
m
º¼
4-117)(a) d Tm 2
Tm
Jm
dt
2
r T1 T2
2
T1 T2
M
d y dt
2
K 2 rT m rT p
T1 d Tm 2
Thu us, Tm
Jm
dtt
2
K 2 rT m y
r K1 K 2 rT m y
T2 2
M
d y dt
2
K
K1 y rT m
1
K 2 rT m y
(c)Stateequations:
dx1
rx3 x2
dt
dx2
K1 K 2
dt
M
x1
dx3
r K1 K 2
dt
Jm
(d)Transferfunction:
r K1 K 2
Y ( s) Tm ( s )
s ¬ª J m Ms K1 K 2 J m rM º¼ 2
2
(e)Charaacteristicequaation:
s
2
ª¬ J m Ms 2 K1 K 2 J m rM º¼ 0
4-118) (a)Systeemequations: 425
x1
1 Jm
Tm
AutomaticConttrolSystems,9thEdition
Tm
J
K i ia
TD
d
m
JL
e
(sec)
V
dZ m dt
r b
Chapteer4Solutionss
BmZ m
b
Ra ia La
ea
Ks y
Ea ( s )
dia dt
Golnarraghi,Kuo
K bZ m
y
nT m
y
y t TD
KGc ( s ) E ( s )
Blockkdiagram:
(b)Forw wardpathtransferfunction n:
Y (s) E (s)
KK i nGc ( s )e
TD s
s ^ Ra La s > J m J L s Bm @ K b K i `
Clossedlooptranssferfunction:
Y ( s) R(s)
KK K i nGc ( s )e
TD s
s Ra La s > J m J L s Bm @ K b K i s KGc ( s ) K i ne
TD s
4-119) (a)Torquueequations: d Tm 2
Tm (t )
Jm
dt
2
Bm
dT m dt
K T m T L
K T m T L
Statediagram:
426
d TL 2
JL
dt
2
BL
dT L dt
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
(b)Transsferfunctions:
4 L ( s)
K
4m (s)
J L s BL s K
Tm ( s )
'(s)
Tm ( s )
'( s)
2
(c)Charaacteristicequaation: '( s )
'(s)
s ª¬ J m J L s Bm J L BL J m s KJ m KJ L Bm BL s Bm K º¼ 3
2
0
dystateperforrmance: Tm ( t ) (d)Stead
Tm
constant. t Tm ( s )
Tm
.
s
J L s BL s K
1
J m J L s Bm J L BL J m s KJ K m KJ L Bm BL s Bm K
Bm
2
lim Z m (t )
lim s: m ( s )
t of
s o0
lim s o0
3
2
Thuss,inthesteadyystate,Z m
Z L .
(e)TheesteadystatevvaluesofZ m and a Z L donotdependon J m and J L .
4-220)(a)Torqqueequation:(AbouttheceenterofgravityyC) 2
J
d T dt
2
Ts d 2 sin G Fd d1
JD D 1
Fa d1
2
Thuss,
J
d T dt
2
2
(b) Js 4 ( s ) K F d14 ( s )
K F d1T
sin G # G
2
Ts d 2G K F d1T
J
d T dt
2
K F d1T
Ts d 2G
Ts d 2 ' ( s )
(c)With hCandPintercchanged,thetorqueequationaboutCis:
Ts d1 d 2 G FD d 2
dT 2
J
dt
2
Ts d1 d 2 G K F d 2T
427
dT 2
J
dt
2
AutomaticControlSystems,9thEdition
Chapter4Solutions
Ts d1 d 2 ' ( s )
Js 4 ( s ) K F d 2 4 ( s ) 2
Golnaraghi,Kuo
4( s )
Ts d1 d 2
'(s)
Js K F d 2 2
4-21)(a)Nonlineardifferentialequations:
dx ( t )
dv ( t )
v(t )
dt
With Ra
0 ,I ( t )
dt e( t )
K f i f (t )
Kb v ( t )
k ( v ) g ( x ) f (t )
Bv ( t ) f ( t )
K f i f (t )
Then, ia ( t )
2
KiI ( t )ia ( t )
f ( t )
Ki e ( t ) 2 Kb K f
dv ( t )
Thus,
.
2
K f ia ( t )
v (t )
dt
Bv ( t )
Ki 2 Kb K f
(b)Stateequations: ia ( t ) asinput.
dx ( t )
dv ( t )
v (t )
dt
dt
2
Bv ( t ) Ki K f ia ( t )
(c)Stateequations:I ( t ) asinput. f (t )
2
K i K f ia ( t )
ia ( t )
dx ( t ) dt
v (t )
dv ( t ) dt
Bv ( t )
428
i f (t )
I (t ) Kf Ki Kf
2
I (t )
e( t 0 Kb K f v ( t ) 2
2
v (t )
e (t )
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
4-22) Define as the angle between mass m and the horizontal axis (positive in c..c.w. direction): m =t
ym
Use Newton's second law: m( y ym ) Fm y Fm By Ky ( M m)
ym
eZ 2 sin Zt
Ÿ My By Ky
meZ 2 sin Zt
Where M is the Mass of the overall block system. M-m is the mass of the block alone.
Y meZ 2 R Ms 2 Bs K Zero i.c. and input r (t ) sin Zt
G ( s)
Note T Zt . So in case of a step response as asked in the question, is a step input and angle increases with time – i.e. it is a ramp function. Hence, ym is a sinusoidal function, where the Laplace transform of a Z sine function is sin(Zt ) 2 s Z2 Pick values of the parameters and run MATLAB. See toolbox 5-8-2 clear all m=20.5 %kg M=60 %kg K=100000 %N/m Om=157 %rad/s B=60 %N-m/s e=0.15 %m G=tf([m*e*Om^2],[M B K]) t=0:0.01:1; u=1*sin(Om*t); lsim(G,u,t) xlabel( 'Time(sec)'); ylabel('Amplitude');
429
AutomaticControlSystems,9thEdition
Chapter4Solutions
m = 20.5000 M = 60 K = 100000 Om = 157 B = 60 e = 0.1500 Transfer function: 7.58e004 ---------------------60 s^2 + 60 s + 100000
430
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
4-23) a)
Chapter4Solutions
summation of vertical forces gives: ൜
ݕܯሷ ሺʹ ܭ ݇ሻ ݕെ ݇ ݔൌ ܨሺͳሻ ݉ݔሷ െ ݕܭ ݇ ݔൌ Ͳሺʹሻ
If we consider ݕሶ ൌ ݍand ݔሶ ൌ ݎ, then: ݍܯሶ ሺʹ ܭ ݇ሻ ݕെ ݇ ݔൌ ܨ ൜ ݉ݎሶ െ ݕܭ ݇ ݔൌ Ͳ The state-space model is: Ͳ ۍ Ͳ ݕሶ ێെʹ ܭെ ݇ ൦ݔሶ ൪ ൌ ێ ݍሶ ܯ ێ ܭ ݎሶ ێ ݉ ۏ ܩሺݏሻ ൌ
b)
Ͳ Ͳ
ͳ Ͳ
݇
Ͳ
݇
Ͳ
Ͳ Ͳ ͳݕ ې ېͲۍ ۑ ۑ ێ Ͳ ۑ ݔ ܨۑͳێ ݍ ۑ ۑ ܯێ ݎ ۑ ۏ Ͳے Ͳے
ሺ௦ሻ ሺ௦ሻ
By applying Laplace transform for equations (1) and (2), we obtain: ሾ ݏܯଶ ሺʹ ܭ ݇ሻሿܻሺݏሻ െ ݇ܺሺݏሻ ൌ ܨሺݏሻ ൜ ሺ݉ ݏଶ ݇ሻܺሺݏሻ ൌ ܻ݇ሺݏሻ Which gives: ܺሺݏሻ ൌ
݇ ܻሺݏሻ ݇
݉ ݏ௦
and ቈ ݏܯଶ ሺʹ ܭ ݇ሻ െ
݇ଶ ܻሺݏሻ ൌ ܨሺݏሻ ݉ ݏଶ ݇
Therefore: ܻሺݏሻ ݉ ݏଶ ݇ ൌ ܨሺݏሻ ݏ݉ܯସ ሺ ݇ܯ ʹ ݉ܭ ݉݇ሻ ݏଶ ʹ݇ܭ
4-24) a) Summation of vertical forces gives: ൜
ݕܯሷ ሺ ܤ ܾሻݕሶ െ ܾݔሶ ሺ ܭ ݇ሻ ݕെ ݇ ݔൌ ܨ ݉ݔሷ െ ܾݕሶ ܾݔሶ െ ݇ ݕെ ݇ ݔൌ Ͳ
Consider ݕሶ ൌ ݍand ݔሶ ൌ ݎ, then 431
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
ݍܯሶ ሺ ܤ ܾሻ ݍെ ܾ ݎ ሺ ܭ ݇ሻ ݕെ ݇ ݔൌ ܨ ൜ ݉ݎሶ െ ܾ ݍ ܾ ݎെ ݇ ݕെ ݇ ݔൌ Ͳ So, the state-space model of the system is: Ͳ Ͳ ۍ Ͳ Ͳ ݕሶ ێെሺ ܭ ݇ሻ ݇ܭ ൦ݔሶ ൪ ൌ ێ ݍሶ ܯ ܯ ێ ݇ ݇ ݎሶ ێ ۏ ܯ ݉
ͳ Ͳ ܤܾ െ ܯ ܾ ݉
Ͳ Ͳ ͳ ݕ ې ېͲۍ ۑ ܾ ۑ ݔ ܨ ۑۑ ͳ ێێ ݍ ۑ ܯ ۑ ܯێ ܾݎ ۑ ےͲۏ െ ے ݉
b) The Laplace transform of the system is defined by: ܻሺݏሻ ܩሺݏሻ ൌ ܺሺݏሻ where ൫ ݏܯଶ ሺ ܤ ܾሻ ݏ ሺ ܭ ݇ሻ൯ܻሺݏሻ െ ሺܾ ݏ ܭሻܺሺݏሻ ൌ ܨሺݏሻ ቊ ሺ݉ ݏଶ ܾ ݏെ ݇ሻܺሺݏሻ ൌ ሺܾ ݏ ݇ሻܻሺݏሻ as a result: ܺሺݏሻ ൌ
ܾ ݏ ݇ ܻሺݏሻ ݉ ݏଶ ܾ ݏെ ݇
Substituting into above equation: ൣ൫ ݏܯଶ ሺ ܤ ܾሻ ݏ ሺ ܭ ݇ሻ൯ሺ݉ ݏଶ ܾ ݏ ݇ሻ െ ሺܾ ݏ ݇ሻଶ ൧ܻሺݏሻ ൌ ሺ݉ ݏଶ ܾ ݏെ ݇ሻܨሺݏሻ ܻሺݏሻ ݉ ݏଶ ܾ ݏെ ݇ ൌ ܺሺݏሻ ሾ ݏܯଶ ሺ ܤ ܾሻ ݏ ሺ ܭ ݇ሻሿሾ݉ ݏଶ ܾ ݏെ ݇ሿ െ ሺܾ ݏ ݇ሻଶ
4-25) a) According to the circuit: ݒ െ ݒଵ ݀ ݒ௨௧ െ ݒଵ ݒ ܥଵ ൌͲ ݀ݐ ʹܴ ʹܴ ݀ ܥ ݒଶ ݀ ܥ ሺݒ െ ݒଶ ሻ െ ሺ ݒെ ݒଶ ሻ ൌ Ͳ ܴ ʹ ݀ ݐ௨௧ ݐ݀ ʹ ۔ ݀ ܥ ݒଵ െ ݒ௨௧ ۖ ሺݒଶ െ ݒ௨௧ ሻ ൌͲ ە ʹ ݀ݐ ʹܴ ۓ ۖ
By using Laplace transform we have:
432
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
୧୬ ሺ•ሻ െ ଵ ሺ•ሻ ୭୳୲ ሺ•ሻ െ ଵ ሺ•ሻ •ଵ ሺ•ሻ ൌͲ ʹ ʹ • ଶ ሺ•ሻ • ൫୧୬ ሺ•ሻ െ ଶ ሺ•ሻ൯ െ ൫୭୳୲ ሺ•ሻ െ ଶ ሺ•ሻ൯ ൌ Ͳ ʹ ʹ۔ ଵ ሺ•ሻ െ ୭୳୲ ሺ•ሻ • ۖ ൫ ሺ•ሻ െ ୭୳୲ ሺ•ሻ൯ ൌͲ ە ʹ ଶ ʹ ۓ ۖ
From above equations: ͳ ൫ ሺ•ሻ ୭୳୲ ሺ•ሻ൯ ʹሺ ͳሻ ୧୬ ۔ሺ•ሻ ൌ ൫ ሺ•ሻ ୭୳୲ ሺ•ሻ൯ ଶ ʹሺ ͳሻ ୧୬ ە ۓଵ ሺ•ሻ ൌ
Substituting V1(s) and V2(s) into preceding equations, we obtain: ܸ௨௧ ሺݏሻ ܴଶ ܥଶ ݏଶ ͳ ൌ ଶ ଶ ଶ ܴ ݏ ܥ Ͷܴ ݏܥ ͳ ܸ ሺݏሻ b)
Measuring Vout requires a load resistor, which means:
i1 Vin
i2 VC1
Then we have: ݀ ܮۓଵ ݅ଵ ൌ ݒ െ ݅ݎଵ െ ݒଵ ۖ ݀ݐ ݀ ۖ ۖ ܥଵ ݒଵ ൌ ݅ଵ െ ݅ଶ ݀ݐ ݀ ܮ ۔ ݅ ൌ ݒଵ െ ݒଶ ଶ ݀ ݐଶ ۖ ۖ ۖ ݒ ݀ ܥൌ ݅ െ ݒଶ ଶ ەଶ ݀ ݐଶ ܴ ܴ When ݒ௨௧ ൌ
ܴ ݒ ܴ ܴ ଶ
If RL >>R, then ݒ௨௧ ൌ ݒଶ 433
RL VC 2
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
By using Laplace transform we have: ܮଵ ܫݏଵ ሺݏሻ ൌ ܸ ሺݏሻ െ ܫݎଵ ሺݏሻ െ ܸଵ ሺݏሻ ۓ ܥଵ ܸݏଵ ሺݏሻ ൌ ܫଵ ሺݏሻ െ ܫଶ ሺݏሻ ۖ ܮଶ ܫݏଶ ሺݏሻ ൌ ܸଵ ሺݏሻ െ ܸଶ ሺݏሻ ۔ ܸଶ ሺݏሻ ۖ ܥଶ ܸݏଶ ሺݏሻ ൌ ܫଶ ሺݏሻ െ ە ܴ ܴ Therefore: ܫଶ ሺݏሻ ൌ ܸଵ ሺݏሻ ൌ
ܥଶ ሺܴ ܴ ሻ ͳ ܸଶ ሺݏሻ ܴ ܴ
ܮଶ ܥଶ ݏሺܴ ܴ ሻ ݏ ሺܴ ܴ ሻ ܸଶ ሺݏሻ ܴ ܴ
ܮଶ ܥଶ ܥଵ ݏଶ ሺܴ ܴ ሻ ܥଵ ݏଶ ܥଵ ݏሺܴ ܴ ሻ ܥଶ ሺܴ ܴ ሻ ͳ ܫଵ ሺݏሻ ൌ ܸଶ ܴ ܴ మ ሺ௦ሻ ሺ௦ሻ
can be obtained by substituting above expressions into the first equation of the state variables
of the system. 4-26) a) The charge q is related to the voltage across the plate:
ݍൌ ܥሺ݀ሻݒ
The force fv produced by electric field is: ݂௩ ൌ
ݍଶ ʹߝܣ
Since the electric force is opposes the motion of the plates, then the equation of the motion is written as: ݀ܯሷ ݀ܤሶ ݀ܭ ൣ݊݃ݏ൫݀ሶ൯൧݂௩ ൌ ݂ሺݐሻ The equations for the electric circuit are: ݀ ݒൌ ܴ݅ ݅ ܮ ݒ ݀ݐ ൞ ݀ݒ ܥ ൌ݅ ݀ݐ As we know, ݅ ൌ
ௗ ௗ௧
ݍൌ ݍሶ and ݍൌ ݒܥ , then:
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ݍ ݒൌ ܴݍሶ ݍܮሷ ܥ ൞ ݀ݒ ܥ ൌ݅ ݀ݐ Since ܥሺ݀ሻ ൌ
ఌ ௗ
, then : ݍଶ ݀ܯۓሷ ݀ܤሶ ݀ܭ ൣ݊݃ݏ൫݀ሶ൯൧ ൌ ݂ሺݐሻ ʹߝܣ ۖ ݀ݍ ݒ ൌ ܴݍሶ ݍܮሷ ߝܣ ۔ ݀ݒ ۖ ܥ ൌ ݍሶ ە ݀ݐ
b) As ݍൌ ݒܥ then ݍଶ ൌ ݒݍܥ If ݊݃ݏ൫݀ሶ ൯ ൌ ͳ ݒݍܥ ൌ ݂ሺݐሻ ݀ܯۓሷ ݀ܤሶ ݀ܭ ʹߝܣ ۖ ݀ݍ ݒ ൌ ܴݍሶ ݍܮሷ ߝܣ ۔ ݀ݒ ۖ ܥ ൌ ݍሶ ە ݀ݐ Then the transfer function is:
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ܳܥሺݏሻ ܸ כ ሺݏሻ ൌ ܨሺݏሻ ʹߝܣ ܦሺݏሻ ܳ כሺݏሻ ܸ ሺݏሻ ൌ ሺ ݏܮଶ ܴݏሻܳሺݏሻ ߝܣ ܸܥ ሺݏሻ ൌ ܳሺݏሻ
ۓሺ ݏܯଶ ݏܤ ܭሻܦሺݏሻ ۖ ۔ ۖ ە
If ݊݃ݏ൫݀ሶ ൯ ൌ െͳ ݒݍܥ ൌ ݂ሺݐሻ ݀ܯۓሷ ݀ܤሶ ݀ܭെ ʹߝܣ ۖ ݀ݍ ݒ ൌ ܴݍሶ ݍܮሷ ߝܣ ۔ ݀ݒ ۖ ܥ ൌ ݍሶ ە ݀ݐ Then the transfer function is: ܳܥሺݏሻ ܸ כ ሺݏሻ ൌ ܨሺݏሻ ʹߝܣ ܦሺݏሻ ܳ כሺݏሻ ܸ ሺݏሻ ൌ ሺ ݏܮଶ ܴݏሻܳሺݏሻ ߝܣ ܸܥ ሺݏሻ ൌ ܳሺݏሻ
ۓሺ ݏܯଶ ݏܤ ܭሻܦሺݏሻ െ ۖ ۔ ۖ ە
4-27) a) The free body diagram is:
where F is required force for holding the core in the equilibrium point against magnetic field b)
The current of inductor, i, and the force, F, are function of flux, , and displacement, x. Also, we know that ݅ൌ
ߔ ܮሺݔሻ
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The total magnetic field is: ః
ܹሺߔǡ ݔሻ ൌ න
ߔଶ ߔ ݀ߔ ൌ ʹܮሺݔሻ ܮሺݔሻ
where W is a function of electrical and mechanical power exerted to the inductor, so: μ Ȱ ൌ μȰ ሺšሻ μ Ȱଶ ݀ܮሺݔሻ ۔ ൌ ൌ െ ۖ μš ʹଶ ሺšሻ ݀ݔ ە ۓ ۖ
‹ ൌ
As v = ߔሶ, then: ݀݅ ݀ܮሺݔሻ ݅ݔሶ ݀ݐ ݀ݐ ͳ ݀ܮሺݔሻ ଶ ܨൌെ ݅ ʹ ݀ݔ
ݒൌ ܮሺݔሻ
൞
c)
Changing the flux requires a sinusoidal movement, and then we can conclude that: ݔൌ ݐ߱ ݊݅ݏ ܣ if the inductance is changing relatively, then L(x) = Lx, where L is constant. Also, the current is changing with the rate of changes in displacement. It means: ‹ ൌ െšሶ So: ݅ ൌ െݐ߱ ݏܿ ߱ܤܣ ܮሺݔሻ ൌ ݐ߱݊݅ݏܣܮ Substituting these equations into the state-space equations gives: Ö ݒൌ ݐ߱ ݊݅ݏ ܣܮሺ߱ ܤܣଶ ݐ߱ ݊݅ݏሻ ܮሺെݐ ߱ݏܿ ߱ܤܣሻሺݐ߱ ݏܿ ߱ܣሻ ൌ ܣܤܮଶ ߱ଶ ሺ݊݅ݏଶ ߱ ݐെ ܿ ݏଶ ߱ݐሻ ͳ ܨൌ െ ܮሺݐ߱ ݏܿ ߱ܤܣሻଶ ʹ Therefore: ሺ•ሻ ʹ ൌ ࣦ ൜െ ሾ–ƒଶ ɘ ݐെ ͳሿൠ ሺ•ሻ 437
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4-28) a) The free body diagram is:
where F is the external force required for holding the plate in the equilibrium point against the electrical field. b) The voltage of capacitors, ݒ, and the force, ܨ, are function of charge, ݍ, and displacement, ݔ. Also, we know ݒൌ
ݍ ܥሺݔሻ
The total electrical force between plates is:
ܹሺݍǡ ݔሻ ൌ න
ݍଶ ݍ ݀ ݍൌ ʹܥሺݔሻ ܥሺݔሻ
Where W is a function of electrical and mechanical power exerted to the capacitor, so: ߲ܹ ݍ ൌ ߲ݍ ܥሺݔሻ ߲ܹ ݍଶ ݀ܥሺݔሻ ۔ ܨ ൌ ൌ െ ۖ ߲ݔ ʹ ܥଶ ሺݔሻ ݀ݔ ە ۓ ۖ
As ݅ ൌ
ௗ ௗ௧
ݒൌ
, then:
൞
c)
݀ܥ݀ ݒሺݔሻ ݔݒሶ ݀ݐ ݀ݐ ͳ ݀ܥሺݔሻ ଶ ݒ ܨൌെ ʹ ݀ݔ
݅ ൌ ܥሺݔሻ
The same as Problem 4.28, ݔൌ ݐ ߱݊݅ݏ ܣ Consider: ൝ ܥሺݔሻ ൌ ݔܥ ݒൌ െݔܤሶ Then solve the equations.
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4-29) According to the circuit: ݒ െ ̴ݒ ̴ݒെ ݒ௨௧ ൌ ݎ ܴ or ̴ݒൌ
ܴ ݎ ݒ ݒ ܴݎ ܴ ݎ௨௧
As an op-amp is modeled with the following equation: ܣ ሾ ݒെ ̴ݒሿ ݒ௨௧ ൌ ݏͳ ା Then: ܣ ܴ ܴݎ ݒ െ ݒ െ ݒ൨ ݏͳ ܴݎ ܴ ݎ௨௧ ݎ ݎ ܣ ݒ െ ൌ ݒቃ ቂ ܴ ݎ௨௧ ݏͳ ܴݎ ݎܣ ሺ ݒെ ݒ௨௧ ሻ ൌ ሺ ݏ ͳሻሺܴ ݎሻ
ݒ௨௧ ൌ
ݎܣ ݒ௨௧ ݎܣ ሺ ݏ ͳሻሺܴ ݎሻ ൌ ൌ ݎܣ ሺ ݏ ͳሻሺܴ ݎሻ ݎܣ ݒ ͳ ሺ ݏ ͳሻሺܴ ݎሻ
4-30) a) Positive feedback ratio: ܨ ൌ
ݎ ݎ ܴ
b) Negative feedback ratio: ܨே ൌ
ܴ ܴ ܴ
c) According to the circuit: ݒ െ ିݒ ିݒെ ݒ௨௧ ൌ ܴ ܴ ൞ ݒെݒ ݒା ௨௧ ା ൌ ܴ ݎ Therefore: 439
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ିݒۓൌ ۔ ە
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ܴ ܴ ݒ ݒ ൌ ሺͳ െ ܨே ሻݒ ܨே ݒ௨௧ ܴ ܴ ܴ ܴ ௨௧ ݎ ݒା ൌ ݒ ൌ ܨ ݒ௨௧ ݎ ܴ ௨௧
As ݒ௨௧ ൌ
ͳͲ ሾ ݒെ ିݒሿ ݏͳ ା
then: ݒ௨௧
ͳͲ ൌ ൣ ݒ ܨെ ሺͳ െ ܨே ሻݒ െ ܨே ݒ௨௧ ൧ ݏ ͳ ௨௧
which gives: ݒ௨௧ ͳ െ ܨே ൌ ݒ ͳͲ ൣ ݏ ͳ െ ͳͲ ൫ܨ ܨே ൯൧
It is stable when ͳ െ ͳͲ ሺܨ ܨே ሻ Ͳ which means: ܨ ܨே ͳͲି 4-31) a) If the drop voltage of Rin is called v1 Then: ሺݒ െ ݒሻ ݒଵ ݀ െ െ ݒ ܥଵ ൌ Ͳ ݀ݐ ܴ ܴ Also:
ݒ െ ݒଵ ݒ௨௧ ൌͲ ܴ ܴ
Then: ݒଵ ൌ ݒ
ܴ ݒ ܴ ௨௧
Substituting this expression into the above equation gives: ܴ ܴݒ݀ ܥ௨௧ ݒ ݀ݒ ͳ ܥ ൭൬ͳ ൌͲ ൰ ݒ௨௧ ൱ ܴ ܴ ݀ݐ ܴ ܴ ݀ݐ As a result:
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ͳ ͳ ܴ ൬ ݏܥ൰ ݒ ሺݏሻ ൌ െ ൬ͳ ൰ ܴݏܥ൨ ݒ௨௧ ሺݏሻ ܴ ܴ ܴ Or ܴ ሺܴ ݏܥ ͳሻ ݒ௨௧ ൌെ ݒ ܴܴ ݏܥ ሺܴ ܴሻ b) If the dropped voltage across resistor Rf is called vf, then ݒ ݒ ൌͲ ۓ ܴ ܴ ݒ۔ െ ݒ௨௧ ݀ ܥ൫ ݒെ ݒ൯ െ ݒ ൌ Ͳ ௨௧ ݀ ݐ ܴ ܴ ە As a result: ܴ ݒ ൌ െ ݒ ܴ Substituting into the second equation gives: െ
ܴ ݀ݒ ܴ ݒ௨௧ ݀ ݒ ݒ െ െܥ െ ݒ ܥ௨௧ െ ൌͲ ݀ݐ ܴܴ ܴ ܴ ݀ݐ ܴ
or െ൬
ܴ ܴ ܴܥ ݀ݒ ݀ݒ௨௧ ݒ௨௧ ܥ ݒ ൰ൌ ܴ ݀ݐ ܴ ݀ݐ ܴܴ
As a result: ݒ௨௧ ͳ ܴܴ ݏܥ ܴ ܴ ൌെ ܴ ݒ ܴ ݏܥ ͳ 4-32) The heat flow-in changes with respect to the electric power as: ݍሶ
ݒଶ ൌܭ ܴ
where R is the resistor of the heater. The heat flow-out can be defined as: ݍሶ ௨௧ ൌ
ܶଵ െ ܶଶ ܭ
where Kf is the heat flow coefficient between actuator and air, T1 and T2 are temperature of actuator and ambient.
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Since the temperature changes with the differences in heat flows: ݀ܶଵ ͳ ݒଶ ͳ ͳ ൌ ሺݍሶ െ ݍሶ ௨௧ ሻ ൌ ൭ ܭെ ሺܶଵ െ ܶଶ ሻ൱ ܥ ܥ ݀ݐ ܴ ܭ where C is the thermal capacitor. The displacement of actuator is changing proportionally with the temperature differences: ݔൌ ܣሺܶଵ െ ܶଶ ሻ If we consider the T2 is a constant for using inside a room, then ܶଵ ൌ
ݔ ܶଶ ܣ
Therefore: ݀ܶଵ ͳ ݀ݔ ൌ ݀ݐ ݐ݀ ܣ ͳ ݀ݔ ݔ ܭଶ ቆ െ ቇൌ ݒ ܭ ݐ݀ ܣ ܶܥ By linearizing the right hand side of the equation around point ݒൌ ݒ ͳ ݀ݔ ݔ ݒܭ ሺʹݒଶ െ ͳሻ ቆ െ ቇൌ ܴ ݐ݀ ܣ ܴܥ Or ݔ ݔ݀ ܴܥ ቆ െ ቇ ൌ ʹ ݒെ ͳ ݒܣܭ ݀ܭ ݐ If we consider the right hand side of the above equation as two inputs to the system as: ݑଵ ሺݐሻ ൌ ʹݒ and ݑଶ ሺݐሻ ൌ ͳ or ݑଶ ሺݐሻ ൌ ݑ௦ ሺݐሻ, then: ቈ
ܺሺݏሻ ܸሺݏሻ ௨
ൌ మ ሺ௧ሻୀ
ʹݒܣܭ ܴܥሺ ݏെ ͳሻ
4-33) Due to insulation, there is no heat flow through the walls. The heat flow through the sides is:
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ʹߨܭ௩ ܪ ݍۓଵǡଶ ൌ ݎሺܶଵ െ ܶଶ ሻሺͳሻ ۖ Ž ቀ ଶ ቁ ݎଵ ʹߨܭ ܪ ݍ۔ൌ ଶǡ ݎଷ ሺܶଶ െ ܶ ሻሺʹሻ ۖ Ž ቀ ቁ ە ݎଶ Where T1 and T2 are the temperature at the surface of each cylinder. Asݍଵǡଶ ൌ ݍଶǡ , then from equation (1) and (2), we obtain: ܶଶ ൌ
ݎ Ž ቀݎଷ ቁ ଶ
ʹߨܭ ܪ
ݍଵǡଶ ܶ ሺ͵ሻ
The conduction or convection at: ଶ ݈݄݅݁ݐ݂݂݁ܿܽݎݑݏ݄݁ݐۓǣݍ ൌ ܥ ሺߨݎଵ ሻሺܶଵ െ ܶ ሻሺͶሻ ۖ ݃݊݅݃ݎ݂݂݂݄݁ܿܽ݁ݐǣ ݍൌ ܣ ܥ൫ܶ െ ܶ ൯ሺͷሻ ଵ ܭ ۔ ௩ ሺߨݎଵଶ ሻሺܶଵ െ ܶ ሻሺሻ ۖݐܽݒ݄݁ݐݐܽ݉ݐݐܾ݄݁ݐǣݍ௩ ൌ ݄ ە
The thermal capacitance dynamics gives:
൞
݀ ܶ ൌ ݍ െ ݍଵǡଶ െ ݍ௩ െ ݍ ሺሻ ݀ ݐଵ ݀ ݉ܶ ܥ ൌ െݍ ሺͺሻ ݀ݐ
݉ ܥ
Where ݉ ൌ ߨݎଵଶ ݀ܪ According to the equation (7) and (8), T1 and Tf are state variables. Substituting equation (3), (4), (5) and (6) into equation (7) and (8) gives the model of the system. 4-34) As heat transfer from power supply to enclosure by radiation and conduction, then: ܥ ݍ ൌ
݀ ܶ ൌ ݍ െ ݍ െ ݍ ሺͳሻ ݀ ݐ
ߪ൫ܶସ െ ܶସ ൯ ߪ൫ܶସ െ ܶସ ൯ ൌ ሺʹሻ ͳ ͳ െ ߝଵ ͳ െ ߝଶ ͳ ൨ ܴ ܨ ܣ ܴ ܣ ܨ ߝଵ ܣ ߝଶ ܣ ݍ ൌ ൬
ܶ െ ܶ௦ ܭଵ ܣଵ ሺ͵ሻ ൰ ൫ܶ െ ܶ ൯ ൌ ߂ݔ ܴா
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Also the enclosure loses heat to the air through its top. So: ܥ
݀ ܶ ൌ ݍ ݍ െ ݍ െ ܥ௧ ܣ௧ ሺܶ െ ܶ ሻሺͶሻ ݀ ݐ
Where ݍ ൌ ൬
ܶ െ ܶ௦ ܭଶ ܣଶ ሺͷሻ ൰ ሺܶ െ ܶ௦ ሻ ൌ ߂ݔ ܴ௦
And Ct is the convective heat transfer coefficient and At is the surface area of the enclosure. The changes if the temperature of heat sink is supposed to be zero, then: ܥ
݀ ܶ ൌ ݍ െ ݍ௦ ൌ Ͳ ݀ ݐ௦
Therefore ݍ ൌ ݍ௦ where ݍ௦ ൌ ܥ௦ ܣ௦ ሺܶ௦ െ ܶ ሻ, as a result: ܶ െ ܶ௦ ൌ ܥ௦ ܣ௦ ሺܶ௦ െ ܶ ሻሺሻ ܴ௦ According to the equations (1) and (4), Tp and Te are state variables. The state model of the system is given by substituting equations (2), (3), and (6) into these equations give.
4-35) If the temperature of fluid B and A at the entrance and exit are supposed to be ܶே and ܶ , and TAN and TAX, respectively. Then: ൜
ݍ ൌ ݉ሶ ܥ ሺܶ െ ܶே ሻሺͳሻ ݍ ൌ ݉ሶ ܥ ሺܶ െ ܶே ሻሺʹሻ
The thermal fluid capacitance gives: ݀ ܶ௫ ൌ െݍ െ ݍି ሺ͵ሻ ݀ܶ ൞ ݀ ܶ ൌ െݍ ݍି ሺͶሻ ܥ ݀ܶ ௫ ܥ
From thermal conductivity: ݍି ൌ
ܶ௫ െ ܶ௫ ሺͷሻ ܴ Ž ቀ ቁ ܴ ͳ ͳ ܥ ܣ ʹߨܥ ܮܭ ܣ
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Where Ci and Co arre convectivve heat transffer coefficiennt of the inner and outerr tube; Ai andd Ao are t the surfface of innerr and outer tuube; Ri and Ro are the raadius of inneer and outer tube. Substittuting equations (1), (2), and (5) intoo equations (3) and (4) giives the statee model of thhe system m.
4-336)(a)Blockkdiagram:
(b)Transferfunction:
W D s
:( s )
K1 K 4 e
D ( s)
Js JK L B s K 2 B K 3 K 4 e 2
W D s
(c)Characteristicequattion:
Js JK L B s K 2 B K 3 K 4 e 2
W D s
0
(d)Transferfunction:
:( s )
Charaacteristicequaation:
D (s)
#
K1 K 4 2 W D s '( s)
' ( s ) # J W D s 2 J JK 2W D BW D s 2 JK 2 2 B W D K 2 B W D K 3 K 4 s 2 K 2 B K 3 K 4 3
2
4-337) The tottal potential energy is: ͳ ͳ ܷ ൌ ߤ ݕܣଶ െ ൬െ ߤܣ ݕܣଶ ൰ ൌ ߤݕܣ ݕଶ ʹ ʹ The tottal kinetic en nergy is:
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ߤܮܣଶ ݕሶ ʹ݃
Therefore: ߤܮܣଶ ݕሶ ൌ ߤ ݕܣଶ ʹ݃ ܮଶ ݕሶ ൌ ݕଶ ʹ݃ As a result: ݕሶ ൌ ඨ
ʹ݃ ݕ ܮ
So, the natural frequency of the system is calculated by:
߱ ൌට
Also, by assuming ݕሺݐሻ ൌ ܻ •‹ሺ߱ ݐ ߠሻ and substituting into
ଶ௬
ଶ
ݕሶ ଶ ൌ ݕଶ yields the same result
when calculated for maximum displacement.
4-38) If the height of the reservoir, the surge tank and the storage tank are assumed to be H, h1 and h2, then potential energy of reservoir and storage tank are: ܲ ൌ ߩ݃ܪ ൜ ଵ ܲ௧ ൌ ߩ݄݃ଶ For the pipeline we have: ݈ܲ
݀ ܳ ൌ ܣሺܲଵ െ ܲଶ ሻ ߩ݃ܣሺݖଵ െ ݖଶ ሻ െ ܨ ݀ݐ
The surge tank dynamics can be written as: ܲ௦ ൌ ߩ݄݃ଵ ݀ ܣ௦ ݄ଵ ൌ ܳଶି௦ ሺܾ݁݇݊ܽݐ݁݃ݎݑݏ݀݊ܽʹݐ݊݅ݐܽ݁݅݊݁݁ݓݐሻ ݀ݐ At the turbine generator, we have: ൫ܲ௧ െ ܲ௧ ൯ܳଶି௩ ൌ ܫ where I is a known input and Q2-v is the fluid flow transfer between point 2 and valve. The behaviour of the valve in this system can be written as: 446
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo ଵ
ቐ
ܳଶି௦ ൌ ܥ௦ ݊݃ݏሺܲଶ െ ܲ௦ ሻሺȁܲଶ െ ܲ௦ ȁሻఈೞ
ଵ
ܳଶି௩ ൌ ܥ௩ ݊݃ݏ൫ܲ௩ െ ܲ௧ ൯൫หܲ௩ െ ܲ௧ ห൯ఈೡ
Regarding Newton's Law: ൜
ܲଶ ൌ ܲ௩ ܳ ൌ ܳଶି௩ ܳଶି௦
According to above equations, it is concluded that Q and h1 are state variables of the system. The state equations can be rewritten by substituting P2, Pv, Ps and Q2-v from other equations.
4-39)
D
If the beam rotate around small angle of ሺ…'• ߙ ؆ ͳሻ, then ݀ ߱ ൌ ܶ െ ݀݃ܯെ ܮܨ ݀ݐ ܧܣሺ ߙܮെ ݕሻ ۔ ܨൌ ܪെݕ ە ܬۓ
where A and E are cross sectional area and elasticity of the cable; H is the distance between point O and the bottom of well, and y is the displacement. On the other hand, Newton's Law gives: ݉
݀ ݒൌ ܲ ܣ௪ ܨെ ܲ ܣ௪ െ ݒܤଶ െ ݉݃ ݀ݐ
447
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
Pa AZ
Bv 2
Pb AZ
where B is the viscous friction coefficient, Aw is the cross sectional area of the well; P1 and P2 are pressures above and below the mass m. The dynamic for the well can be written as two pipes separating by mass m: Pb AZ Ps AZ
Q Q
y Ff
P1 AZ
Pa AZ
b)
a
ܨ ݀ܳ ܣ௪ ܣ௪ ݃ ሺܲଵ െ ܲ ሻ ሺͲ െ ݕሻ െ ൌ ݀ݐ ݕ ߩݕ ߩݕ ܨଵ ݀ܳ ݃ ͺܣ ܣ ௪ ௪ ۔ଵൌ ሺܲ െ ܲ௦ ሻ ሺͲ െ ݕሻ െ ݕ ߩሾ ܪെ ܦെ ݕሿ ߩݕ ݐ݀ ە ۓ
Where D is the distance between point O and ground, Ps is the pressure at the surface and known. If the diameter of the well is assumed to be r, the Ff for the laminar flow is ܨ ൌ ͵ʹ Therefore:
448
ߤܳݕ ݎଶ
AutomaticControlSystems,9thEdition
Chapter4Solutions
݀ܳ ܣ௪ ߤܳ ሺܲଵ െ ܲ ሻ െ ܣ௪ ݃ െ ͵ʹ ଶ ൌ ݀ݐ ߩݎ ߩݕ ݀ܳ ܣ ߤܳଵ ௪ ۔ଵൌ ሺܲ െ ܲ௦ ሻ െ ܣ௪ ݃ െ ͵ʹ ଶ ߩሾ ܪെ ܦെ ݕሿ ߩݎ ݐ݀ ە ۓ
The state variables of the system are , v, y, Q, Q1. 4-40) For the hydraulic amplifier, we have: ܳ ൌ ܰݕ௩ ݀ݕ ൝ ܳൌܣ ݀ݐ As a result ݀ ܰ ݕ ൌ ݕ௩ ݀ݐ ܣ where N is a constant and A is the cross sectional area. For the walking beam: ݕ௩ ൌ For the spring:
݈ଵ ݕଶ െ ݈ଶ ݕ ݈ଵ ݈ଶ
ܨൌ ܭሺݕଵ െ ݕଶ ሻ
The angular velocity of the lever is assumed as: ߱௫ ൌ Ͳ ՜ ߗ௫ ൌ Ͳ ߱ ቐ ௬ ൌ ߱ ՜ ߗ௬ ൌ ߱ ߱௭ ൌ ߙሶ ՜ ߗ௭ ൌ Ͳ The moments of inertia of the lever are calculated as: ܬ௫௫ ൌ ݉ܮଶ •‹ଶ ߙ ܬ௫௬ ൌ ݉ሺ ߙ •'… ܮ ݎሻሺߙ ‹• ܮሻ ܬ௬௬ ൌ ݉ሺ ߙ •'… ܮ ݎሻଶ ۔ ܬ௬௭ ൌ ܬ௭௫ ൌ Ͳ ۖ ܬە௭௭ ൌ ݉ሾሺ ߙ •'… ܮ ݎሻଶ ܮଶ •‹ଶ ߙሿ ۓ ۖ
where L is the length of lever and r is the offset from the center of rotation. According to the equation of angular motion:
449
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
ሶ ߱ െ ܬ௫௬ ߱ሶ ܬ௭௭ ߱௭ ߱ ܶ௫ ൌ െܬ௫௬ ሶ ߱ ܬ௬௬ ߱ሶ ܶ௬ ൌ ܬ௬௬ ൞ ሶ ߱௭ ܬ௭௭ ߱ሶ ௭ ܬ௫௬ ߱ଶ ܶ௭ ൌ ܬ௭௭
Also: ܶ௭ ൌ ݂௬ ݀
ܨ ሺ ݎെ ݀ •‹ ߙሻ െ ݉݃ሺ ߙ •'… ܮ ݎሻ ʹ
Due to force balance, we can write: ݂௬ െ
ܨ െ ݉݃ ൌ ݉ሺ߱ ߙ ‹• ܮሶ ଶ െ ߱ ߙ ‹• ܮሶ ଶଶ ሻ ʹ
Therefore ߱ሶ ଶ can be calculated form above equations. On the other hand, െݕଶ ൌ ߙ ‹• ݎ, and
ௗ௬మ ௗ௧
ൌ ݒଶ and
ௗ ௗ௧
ߙ ൌ ߱ଶ , the dynamic of the system is:
݀ ݊ଵ ܳ ߱ൌ െ ߱ܤെ ܶ ቐ ݀ݐ ߱ ܳ ൌ ݊ଶ ݕሶ ܬ
where B is the viscous friction coefficient, and n1 and n2 are constant. The state variables of the systems are ,yp, and 2.
450
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Chapter4Solutions
Golnaraghi,Kuo
4-41) If the capacitances of the tanks are assumed to be C1 and C2 respectively, then ݄݀ଵ ܥ ۓଵ ൌ ሺݍଵ െ ͳݍሻ ݀ݐ ۖ ۖ ݄݀ଶ ۖܥଶ ൌ ሺݍଵ ݍଶ െ ݍ ሻ ݀ݐ ݄ଵ െ ݄ଶ ۔ ݍଵ ൌ ܴଵ ۖ ۖ ݄ଶ ۖ ݍ ൌ ە ܴଶ Therefore: ݄݀ଵ ͳ ݄ଵ െ ݄ଶ ൌ ൬ݍଵ െ ൰ ܥଵ ݀ݐ ܴଵ ݄݀۔ଶ ൌ ͳ ൬݄ଵ െ ݄ଶ ݍെ ݄ଶ ൰ ଶ ݐ݀ ە ܥଶ ܴଵ ܴଶ ۓ
Asa result: ͳ ݄݀ଵ ۍെ ܴଵ ܥଵ ൦ ݀ ݐ൪ ൌ ێ ݄݀ଶ ͳ ێ ܴ ۏଵ ܥଶ ݀ݐ
ͳ ͳ ې ۍ ܴଵ ܥଵ ݄ ۑଵ ܥ ൨ ێଵ ܴଵ ܴଶ ݄ ۑଶ Ͳێ െ ۏ ܴଵ ܴଶ ܥଶ ے
4-42) The equation of motion is: ݔܯሷ ܤሺݔሶ െ ݕሶ ሻ ܭሺ ݔെ ݕሻ ൌ Ͳ Considering ݖൌ ݔെ ݕgives: ܯሺݖሶ െ ݕሷ ሻ ݖܤሶ ݖܭൌ Ͳ or ݖሷ
ܭ ܤ ݖሶ ݖൌ ݕሷ ܯ ܯ
4-43)(a)Blockdiagram:
451
Ͳې ݍ ۑቂ ଵ ቃ ݍ ͳ ۑଶ ܥଶ ے
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
(b)Transfferfunction:
TAO ( s )
Tr ( s )
KM KR
3.51
1 W s 1 W s K s
c
m
KR
20 s 122 s 4.51 2
4-444) Sysstemequations: d Tm 2
Tm (t )
Jm
dt
2
Bm
dT m
ET L
uation: eo Outputequ
Statediiagram:
Transferr
dt
20S
K T m T L
d TL 2
K T m T L
JL
dt
2
Bp
dT L
d dt
fun nction:
4L (s) Tm ( s ) Eo ( s ) Tm ( s )
K s ª¬ J m J L s Bm J L B p J m s J m K J L K Bm B p s Bm K º¼ 3
2
KE / 20S
s ª¬ J m J L s Bm J L B p J m s J m K J L K Bm B p s Bm K º¼ 3
2
4-445)(a)Statteequations:
dT L dt
ZL
dZ L
K2
dtt
JL
Tm
K2 JL
TL
dT t dt
Zt
452
dZ t
K1
dtt
Jt
Tm
K1 Jt
Tt
AutomaticConttrolSystems,9thEdition
dT m
dZ m
Zm
dt
dtt
Bm Jm
Chapteer4Solutionss
Zm
K
1
K2 Jm
Tm
K1 Jm
Tt
K2 Jm
TL
Golnarraghi,Kuo
1 Jm
Tm
(b)Stateddiagram:
(c)Transfferfunctions:
4 L (s)
K 2 J t s K1
Tm ( s )
'(s)
2
'( s)
4t ( s)
K1 J L s K 2
Tm ( s )
'(s)
2
4m (s)
J t J L s K1 J L K 2 J t s K1 K 2
Tm ( s )
'(s)
4
2
s[ J m J L s Bm J L J t s K1 J L J t K 2 J L J t K1 J m J L K 2 J m J t s 5
4
Bm J L K1 K 2 s K1 K 2 J L J t J m s Bm K1 K 2 ] 2
(d)Charaacteristicequattion:
' s) '(
3
0
0 .
4-446) (a)Transfferfunction:
G(s)
Ec ( s )
1 R2 Cs
E (s)
1 R1 R2 Cs
(b)Blockdiagram:
453
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
(c)Forwardpathtransferfunction:
K 1 R2 Cs
:m (s)
>1 R
E (s)
1
R2 Cs @ K b K i Ra J L s
(d)Closedlooptransferfunction:
:m (s)
Fr ( s )
(e)
Gc ( s )
K I K 1 R2 Cs
>1 R
R2 Cs @ K b K i Ra J L s K I KK e N 1 R2 Cs
1
Ec ( s )
1 R Cs
E (s)
R1Cs
2
Forwardpathtransferfunction:
:m (s)
E (s)
Closedlooptransferfunction:
Fr ( s )
R1Cs K b K i Ra J L s K I KK e N 1 R2 Cs
(f) f r
120 pulses / sec
K I K 1 R2 Cs
36 pulses / rev
NK eZ m
R1Cs K b K i Ra J L s
:m (s)
Ke
fZ
K 1 R2 Cs
36 / 2S pulses / rad
Zm
120 pulses / sec
5.73 pulses / rad.
200 RPM
200(2S / 60 ) rad / sec
N ( 36 / 2S ) 200(2S / 60 ) 120 N pulses / sec
Thus,N=1.ForZ m 1800 RPM, 120 N ( 36 / 2S )1800(2S / 60) 1080 N . Thus, N 9.
4-47) If the incremental encoder provides a pulse at every edge transition in the two signals of channels A and B, then the output frequency is increased to four times of input frequency. 4-48)(a) 1 §
:m (s) TL ( s )
K H (s) · K1 § H i (s) · 1 K1 H e ( s ) 1 i H e (s) ¨ ¸ ¨ ¸ B Js © Ra La s ¹ B Js © Ra La s ¹ # 0 Zr 0
'( s)
'(s)
Thus,
454
AutomaticControlSystems,9thEdition
H e (s)
Chapter4Solutions
H i (s)
H i (s)
Ra La s
H e (s)
Golnaraghi,Kuo
Ra La s
K1 K i (b)
'(s)
a
:r (s)
1 K1 H e ( s )
a
K1 K b
R
La s B Js
:m (s)
dt
R
a
TL 0
d Tm
Ra La
2
2
dt
d To 2
dt
ia
1 La
Bm dT m J m dt
KL
2
JL
T
2
e
Te
Stateequations:
dt
1 J
R
a
dt nK L Jm
Tm
nK L Jm
nT
m
e
a
La s ( B Js
#
1
Kb H e (s)
K sT e
ea
Ke
Tm
T2
nT m
K i ia
To
T2
KL JL
2
x1
n KL Jm
x3
x1 Bm Jm
Z o , x3
nK L JL x4
T m , x4
n
Z m , x5
dx3
x3
ia
x4
dt
Ki Jm
x5
dx5 dt
(b)Statediagram:
455
R
La s B Js
T r To Tm
T o , x2
dx 2
x2
dx 4
Ra La s
K1 K i K b H e ( s )
To
eb
a
Statevariables: x1
dt
La s B Js K i K b K1 K i K b H e ( s )
dx1
K1 H i ( s )
K1 K i
:r ( s)
dia
K1 K i
La s B Js
4-49)(a)Causeandeffectequations:
'( s)
K1 K b
R
La s B Js
TL 0
a
1
R
:m ( s)
KK s La
x1
Kb La
x4
Ra La
x5
KK s La
Tr
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
(c)Forw wardpathtran nsferfunction: 4o ( s)
4 e ( s )
KK s K i nK L
n R K J
3
2
L
a
L
2
2
Ra K L J m Bm K L La s K i K b K L Ra Bm K L º¼
Clossedlooptranssferfunction: 4o ( s)
s ª¬ J m J L La s J L Ra J m Bm J m Bm La s n K L La J L K L J m La Bm Ra J L s 4
4 r ( s )
KK s K i nK L 4
n R K J 2
a
(d) K L
J m J L La s J L Ra J m Bm J m Bm La s n K L La J L K L J m La Bm Ra J L s 5
f, T o
T2
L
L
2
3
Ra K L J m Bm K L La s K i K b K L Ra Bm K L s nKK s K i K L 2
nT m . J L is reflected to motor side so J T
2
J m n J L .
Staateequations::
dZ m dtt
Bm JT
Zm
Ki JT
ia
dT m dt
Zm
dia dt
Statediagram:
456
Ra La
ia
KK s La
Tr
KK s La
nT m
Kb La
Zm
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
Forrwardpathtraansferfunction n:
Closedlooptranssferfunction:
4o ( s )
KK s K i n
4e ( s)
s ª¬ J T La s Ra J T Bm La s Ra Bm K i K b º¼ 2
4o ( s)
KK s K i n
4r (s)
J T La s Ra J T Bm La s Ra Bm K i K b s KK s K i n 3
f, all the terms withoutt K L in 4 o ( s ) / 4 e ( s ) and 4 o ( s ) / 4 r ( s ) cann be neglectedd.
Frompaart(c),when K L
2
Thesam meresultsasab boveareobtained.
4-550)(a)Systemequations: f
K i ia
MT
dv dt
BT v
ea
Ra ia La Las
dia dt
Las
dis dt
eb
0
Rs is Ls Las
dis dt
Las
dia
dt
othsidesoftheelastthreeequ uations,withzeroinitialcond ditions,wehavve (b)TaketheLaplacetraansformonbo
Ki I a ( s) 0
s BT V ( s )
>R L
Ea ( s )
a
a
Ki M T s BT
Y (s)
I a (s)
1 Ra La Las s
>E
a
V (s)
Ki
s
s M T s BT
( s ) Las sI s ( s ) K bV ( s ) @
Blockd diagram:
457
Las s @ I a ( s ) Las sI s ( s ) K bV ( s )
Rearrrangingtheseequations,weeget
I a (s)
T
Las sI a ( s ) > Rs s Ls Las @ I s ( s )
V (s)
M
I a (s)
I s (s)
Las s Ra La Las s
I a (s)
AutomaticConttrolSystems,9thEdition
Chapteer4Solutionss
Golnarraghi,Kuo
(c)Trannsferfunction:: K i > Rs Ls Las s @
Y (s)
s > Ra La Las s @ > Rs Ls Las s @ M T s BT K i K b > Rs La Las s @ Las s 2
Ea ( s )
2
M
T
s BT
4-551) (a)Causeeandeffecteqquations:
Te
Tr TL
Tm
K i ia
e
dZ m dt d
Kb
Statteequations:
dT L dt
ZL
K sT e 1 Jm
Ks Tm
Bm Jm
15.5 V / KRPM
dZ L
KL
dt
JL
Tm
KL JL
TL
Z
ea
1 V/rrad KL Jm
T
m
TL
15.5
dZ m
Zm
dt
ea eb
ia
dZ L
KL
dt
JL
T
Ra m
TL
eb
K bZ m
0.148 V / rad / sec
1000 u 2S / 60
dT m
Ke
dt
Bm Jm
Zm
KL Jm
TL
1 Ki J m Ra
KK T s
e
K bZ m
(b)Stattediagram:
(c)Forw wardpathtran nsferfunction:
G(ss )
K i KK Ks KL s ª¬ J m J L Ra s Bm Ra K i K b J L s Ra K L J L J m s K L Bm Ra K i K b º¼ 3
2
458
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
0.03 u 115 . u 0.05
J m Ra J L
10 u 115 . u 0.05
Bm Ra J L
0.001725
115 . u 50000 u 0.05
Ra K L J L
K L Bm Ra K i K b
(d)Closedlooptransferfunction:
115 . u 50000 u 0.03 1725
Ra K L J m
2875
50000(10 u 1.15 21 u 0.148)
21 u 0.148 u 0.05
Ki Kb J L
0.575
0.1554
21 u 1 u 50000 K
Ki KK s K L
1050000 K
730400
608.7 u 10 K 6
G(s)
M ( s )
s s 423.42 s 2.6667 u 10 s 4.2342 u 10 3
2
6
8
4L (s)
G(s)
K i KK s K L
4r ( s)
1 G(s)
J m J L Ra s Bm Ra K i K b J L s Ra K L J L J m s K L Bm Ra K i K b s K i KK s K L 4
3
2
8
M ( s)
6.087 u 10 K
Characteristicequationroots:
4
3
6 2
8
8
s 423.42 s 2.6667 u 10 s 4.2342 u 10 s 6.087 u 10 K
K
K
K
1
s
1.45
s
r j1000
s
405 r j1223.4
s
159.88
s
2117 . r j1273.5
s
617.22 r j1275
s
13105 . r j1614.6
2738
5476
4-52)(a)Nonlineardifferentialequations:
dx ( t )
dv ( t )
v(t )
dt
With Ra
0 ,I ( t )
dt e( t )
K f i f (t )
Kb v ( t )
k ( v ) g ( x ) f (t )
Bv ( t ) f ( t )
K f i f (t )
Then, ia ( t )
2
KiI ( t )ia ( t )
f ( t )
Ki e ( t ) 2 Kb K f
.
2
K f ia ( t )
dv ( t )
Thus,
v (t )
dt
Bv ( t )
Ki 2 Kb K f
(b)Stateequations: ia ( t ) asinput.
dx ( t )
v (t )
dt
dv ( t ) dt
(c)Stateequations:I ( t ) asinput.
459
2
Bv ( t ) Ki K f ia ( t )
e( t 0 Kb K f v ( t ) 2
2
v (t )
e (t )
AutomaticConttrolSystems,9thEdition
2
f (t )
Chapteer4Solutionss
K i K f ia ( t )
ia ( t )
dx ( t )
dv ( t )
v (t )
dt
dt
I (t )
i f (t )
Bv ( t )
Golnarraghi,Kuo
Kf Ki Kf
2
I (t )
4-553)(a)Diffeerentialequations: d Tm 2
K i ia
Jm
dt
2
Bm
dT m dt
§ dT m dT L d dt © dt
K T m T L B ¨
2 § dT m dT L · § J d T L B dT L ¸ ¨ L 2 L dt ¹ © dt dt © dt
K T m T L B ¨
· ¸ ¹
· ¸ TL ¹
(b)TaketheLaplacetraansformoftheedifferentialeq quationswithzeroinitialcon nditions,wegeet
Ki I a ( s)
Bs K 4
4 L (s)
m
s Bm s Bs K 4 m ( s ) Bs K 4 L ( s ) 2
m
( s ) Bs K 4 L ( s )
J s L
2
s BL s4 L ( s ) TL ( s)
a 4 L ( s ) fromthelasttwo oequations,weehave Solvingfor 4 m ( s ) and 4m (s)
J
Ki J m s Bm B s K 2
Bss K J L s BL B s K 2
I a (s) 4m (s)
Bs K J m s Bm B s K 2
4 L (s)
TL ( s ) J L s BL B s K 2
Signaalflowgraph:
460
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
(c)Transfermatrix: ª K i ª¬ J L s 2 BL B s K º¼ « 'o (s) ¬ K i Bs K
ª4m (s) º « 4 (s) » ¬ L ¼
' o ( s )
1
º ª I a (s) º »« » J m s Bm B s K ¼ ¬ TL ( s ) ¼ Bs K
2
J L J m s > J L Bm B J m BL B @ s > BL Bm BL BM B J m J L K @ s K BL B s 3
3
2
4-54) As ݁ ି் ௦ can be estimated by: ݁
ି் ௦
ܶௗ ݏ ቁ ʹ ؆ ʹ െ ܶௗ ݏ ؆ ܶ ݏ ͳ ቀ ௗ ቁ ʹ ܶௗ ݏ ʹ ͳെቀ
Therefore: ܩሺݏሻ ൌ
ܭሺʹ െ ܶௗ ݏሻ ሺʹ ܶௗ ݏሻሺ߬ଵ ݏ ͳሻሺ߬ଶ ݏ ͳሻ
As a result: Poles: െ zeros:
ଶ
ଵ ఛభ
ǡെ
ଵ ఛమ
ǡെ
ଶ ்
்
4-55) By approximating݁ ି௦் : ݁
ି்௦
ܶݏ ʹ ൌ ܶݏ ͳ ʹ ͳെ
a) ܩሺݏሻ ൌ
ͳെ
ݏܮ ʹ
ሺܶ ݏ ͳሻ ቀͳ
ݏܮ ቁ ʹ
Therefore: ܩሺ݆߱ሻ ൌ
ͳെ
ሺ݆ܶ߱ ͳሻ ቀͳ
461
݆߱ܮ ʹ
݆߱ܮ ቁ ʹ
AutomaticControlSystems,9thEdition
Chapter4Solutions
b) ݏ ͳെݏ ʹ ʹ ʹݏ ݏ Ͷͳ ݏ ͳ ʹ ܩሺݏሻ ൌ ଶ ݏ ͵ ݏ ʹ ʹሺʹ ݏሻሺͳ ݏሻ ݏሺʹ െ ݏሻሺͳ ݏሻ Ͷሺͳ െ ݏሻሺʹ ݏሻ ൌ ሺ ݏ ʹሻଶ ሺ ݏ ͳሻଶ െ ݏଷ െ ݏଶ Ͷ ݏ Ͷ ൌ ሺ ݏ ʹሻଶ ሺ ݏ ͳሻଶ ͳെ
4-56) MATLAB clearall L=1 T=0.1 G1=tf([1/21],conv([0.11],[1/21])) figure(1) step(G1) G2=tf([1144],conv(conv([12],[12]),conv([11],[11]))) figure(2) step(G2) L= 1 T= 0.1000 Transferfunction: 0.5s+1 0.05s^2+0.6s+1 Transferfunction: s^3s^2+4s+4 s^4+6s^3+13s^2+12s+4
462
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
Chapter4Solutions
ª 4-57) 423(a)Differentialequations: « L ( y ) ¬
Lº
e ( t )
Ri ( t )
d L ( y )i ( t ) dt
Ri ( t ) i ( t )
y »¼
My ( t )
Mg
Ki ( t ) 2
y (t )
dL( y ) dy ( t ) dy
2
At equilibrium,
dt
di ( t )
0,
dt
L di ( t ) y dt
dy ( t ) dt
463
Golnaraghi,Kuo
Ri ( t )
L y
2
i(t )
dt
2
0,
d y(t ) dt
2
dy ( t )
0
L di ( t ) y dt
AutomaticControlSystems,9thEdition
Thus, ieq
Eeq
dy eq
R
dt
Chapter4Solutions
yeq
0
(b)Definethestatevariablesas x1
Eeq
x1eq
Then,
i, x 2
K
R
Mg dy
y, and x3
x 2eq
R
Eeq
Eeq
K
R
Mg
Golnaraghi,Kuo
.
dt x3eq
0
Thedifferentialequationsarewritteninstateequationform:
dx1
dt
R L
x1 x 2
x1 x3 x2
x2
e
dx 2
f1
L
x3
dt
2
dx3
f2
K x1
g
dt
f3
2
M x2
(c)Linearization:
wf1
wx1
R L
x 2eq
x3eq
x 2eq
wf1
x 2 eq
1
K Eeq
wf 2
we
L
L
Mg R
wx1
wf 3
wx1
2 K x1eq M
2 x 2 eq
2 Rg Eeq
Eeq
wf1
K
wx 2
L
Mg
0
wf 2
2
wf 3
2 K x1eq
wx 2
3 x 2eq
R L
wf 2
0
wx 2
M
wx 3
x1 x3
x1eq
2 x2
wf 2
1
Mg
wf 3
Eeq
K
we
L
A
Mg º
» K » 0 » » » 0 » ¼
0 0 2 Rg
Mg
Eeq
K
B
(a)Differentialequations: 2
M1
d y1 ( t ) dt
2
M1 g B
dy1 ( t ) dt
2
Ki ( t ) 2 y1 ( t )
464
2
Ki ( t )
x1eq
Mg
wx 3
x 2eq
K
1 y 2 ( t ) y1 ( t )
2
A 'x B 'e
ª Eeq K º « » « RL Mg » « » 0 » « 0 « » « » ¬ ¼
4-58)
wf1
0
Thelinearizedstateequationsabouttheequilibriumpointarewrittenas: 'x
ª Eeq K « « L Mg « 0 « 2 Rg « « E eq ¬
0
0
we
2 Rg
Eeq
AutomaticControlSystems,9thEdition
Chapter4Solutions
2
M2
d y2 (t ) dt
2
Definethestatevariablesas x1
Thestateequationsare:
dx1
x2
dt
dx2
M1
dt
Atequilibrium,
dx1
dx 2
0,
dt
2
2
X1
SolvingforI,with X 1
dt
Ki
x
3
dx3
0,
KI
X
3
y 2 ( t ) y1 ( t )
, x3
dx3
x1
2
M2g
0
2
Y2
X3
§ M M2 · 1 ¨ 1 ¸ © M2 ¹
(b)Nonlinearstateequations: dx1 dt
dx2
x2
dt
(c)Linearization: w f1
w x1
B
w x1
M 1 x1
w f2
2 KI
2
3
K
2 KI
2
3
w x1
M 2 X 3 X1
w f4
2 3
w x2
Linearizedstateequations: M1
dx3
X1
3
x1
2
0.
0
2
w f1
0
M1
w x3
M1 X 3 X1
w f3
w x1
w x2
w f4
2 KI
w x3
M 2 X 3 X1
1, g
w f4
2
32.2, B
3
0.1, K
w x4 1.
2
M 2 x3 x1
2
0
w x4
3
w f3
0
w x3
Ki
w f2
2
w f3
0
M2
x4
0
wi 2 KI
0
B
g
dt
w f2
w f3
2, M 2
dx4
B
465
x
2
2
x4
dt
w x4
w x2
0
2
w f1
0
w f2
2
2
M 1 x3 x1
w x3
§ 1 · 1 ¨ 2 2 ¸ M1 © X1 X 3 X1 ¹ 2 KI
dt
Ki
1/ 2
Ki
i
w f1
M1 ( X 3 X1 )
w f4
M 2 g Bx4
§ M1 M 2 g · ¨ ¸ K © ¹
I
2
M 1 x1
0
w x2
2 KI
x2
M1
w f1
0
w f2
wi
g
3
dx4
M2
0 and x 4 eq
KI
X
.
dt
1,wehave
dy 2
0. Thus, x 2eq
dt
2
X1
x4
dt
1/ 2
2
y2 , x4
2
dx 4
0,
dt
2
KI
dy1
2
x1
dt
M1g
Ki
Ki ( t )
dt
y1 , x 2
M 1 g Bx2
2
dy 2 ( t )
2
M2 g B
Golnaraghi,Kuo
w x4
w f3
1
wi
0
B
w f4
2 KI
M2
wi
M 2 X 3 X1
2
AutomaticControlSystems,9thEdition
Chapter4Solutions
1/ 2
§ 32.2(1 2) · X ¨ ¸ 1 1 © ¹
I
1
X 3
A
1 2 X1
96.6 X 1
2.732 X 1
Y2
0 ª « 2 · 1 « 2 KI §¨ 1 3 ¸ 3 « M1 © X1 X 3 X1 ¹ « 0 « 2 « 2 KI « 3 M 2 X 3 X1 «¬
B
9.8285 X 1
0 ª º « » ·» 1 « 2 KI §¨ 1 « M 1 © X 12 X 3 X 1 2 ¹¸ » « » 0 « » « » 2 KI « » 2 M 2 X 3 X1 «¬ »¼
X1 X 3 X1
2.732
0
B
2 KI
M1
M1 X 3 X1
0
0 2 KI
1
1
9.8285 1.732
0 º
1
0
Golnaraghi,Kuo
»
2 3
2
M 2 X 3 X1
3
0 »
» » 1 » B » » M 2 »¼
1 0 0 º ª 0 « 115.2 0.05 18.59 0 » « » 0 0 1 » « 0 « » 0 37.18 0.1¼ ¬ 37.18
ª 0 º « 6.552 » « » « 0 » « » ¬ 6.552 ¼
4-59) a)
Z
F3
T
F1
b)
F2
The equation of the translational motion is:
466
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
ݒ݀ܯ ۓ ൌ ݃ܯെ ܨଵ ܨଶ െ ܨଷ ሺͳሻ ۖ ݀ݐ ܨଵ ൌ ݕܭ ݀ݕ ۔ ൌݒ ۖ ݀ݐ ە ܨଶ ൌ െݒܤ The equation of rotational motion is: ݀ ߱ ൌ ܨଷ ݎ ܬ ൞ ݀ݐ ݀ߠ ൌ߱ ݀ݐ ଵ
where ܬൌ ݉ ݎଶ ଶ
Also, the relation between rotational and translational motion defines: ݒൌ ߱ݎ ቄ ݕൌ ߠݎ Therefore, substituting above expression into the first equation gives: ܨଷ ൌ ቀ
݉ ቁ ሺ ݃ܯെ ݕܭെ ݒܤሻ ʹ ܯ ݉
The resulted state space equations are: ݀ ʹ ݃ܯെ ߠݎܭെ ߱ݎܤ ߱ ۓൌ ൬ ൰൬ ൰ ݀ݐ ʹߤ ݉ ݎ ݀ ۔ ߠൌ߱ ە ݀ݐ c) According to generalized elements: 1) Viscous friction can be replaced by a resistor where R = B 2) Spring can be replaced by a capacitor where ܥൌ
ଵ
3) Mass M and m can be replaced by two inductors where ܮଵ ൌ ܯand ܮଶ ൌ ݉. Then the angular velocity is measured as a voltage of the inductor L2 4) The gear will be replaced by a transformer with the ratio of ܰ ൌ 5) The term Mg is also replaced by an input voltage of ܸ ൌ ݃ܯ
467
ଵ
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
4-60) As the base is not moving then the model can be reduced to:
Therefore: 1) As ݉
ௗ௩ ௗ௧
ൌ ܨ, they can be replaced by a inductor with L = m
2) Friction B can be replaced by a resistor where R = B 3) Spring can be replaced by a capacitor where ܥൌ
ଵ
4) The force F is replaced by a current source where Is = F
4-61)
V2
ଵ
ܴ ൌ ݂ோ ሺܳሻ ൌ ܥሺȁܲ െ ܲଶ ȁሻቀఈቁ ൌ
ɏ‰
ܸ ൌ ߩ݄݃
468
AutomaticControlSystems,9thEdition
Chapter4Solutions
4-62) Recall Eq. (4-324) Z (s) Y( s )
1 s 2]Z n s Z n 2 2
Z 1, Set Y( s ) impulse , pick n for simplicity. 9 1 clear all G=tf([-1],[1 2 1]) figure(1) impulse(G) Transfer function: -1 ------------s^2 + 2 s + 1
469
Golnaraghi,Kuo
AutomaticControlSystems,9thEdition
Chapter4Solutions
Golnaraghi,Kuo
4-63) Use Eq. (4-329).
Z (s)
Kmr Ra
§ La · 2 K m Kb s ¨ s 1¸ ( Js Bs K ) Ra © Ra ¹ § La · ¨ s 1¸ r © Ra ¹
§ La · 2 K m Kb s ¨ s 1¸ ( Js Bs K ) Ra © Ra ¹
For
Va ( s )
mrY( s )
La =0 (very small) the format of the equation is similar to Eq. (4-324), and we expect the same Ra
response for the disturbance input. Except, Z ( s )
Kmr Ra K K ( Js Bs K ) m b s Ra 2
effects of disturbance. See Chapter 6.
470
Va ( s ) can be used to reduce the
AutomaticControlSystems,9thEdition
Chapter5Solutions
Golnaraghi,Kuo
Chapter 5 51(a)] t 0.707 Z n t 2 rad / sec
(b) 0 d ] d 0.707 Z n d 2 rad / sec
(c)] d 0.5 1 d Z n d 5 rad / sec
(d) 0.5 d ] d 0.707 Z n d 0.5 rad / sec
52(a)Type0 (b)Type0(c)Type1(d)Type2(e)Type3(f)Type3 (g)
type 2
53(a) K p
lim G ( s )
lim G ( s )
(h)
type 1
1000
Kv
lim sG ( s )
0
Ka
lim s G ( s )
2
0
f
Kv
lim sG ( s ) 1
Ka
lim s G ( s )
2
0
so 0
so 0
so 0
(b) K p
so 0
so 0
51
so 0
AutomaticControlSystems,9thEdition
Chapter5Solutions
Golnaraghi,Kuo
2
lim G ( s )
f
Kv
lim sG ( s )
K
Ka
lim s G ( s )
lim G ( s )
f
Kv
lim sG ( s )
f
Ka
lim s G ( s ) 1
(e) K p
lim G ( s )
f
Kv
lim sG ( s ) 1
Ka
lim s G ( s )
2
0
(f) K p
lim G ( s )
f
Kv
lim sG ( s )
Ka
lim s G ( s )
2
K
ErrorConstants
(c) K p
so 0
so 0
so 0
0
(d) K p
so 0
so 0
2
so 0
so 0
so 0
so 0
so 0
f
so 0
so 0
54(a)Input
SteadystateError
________________________________________________________________________________
u s ( t )
K p
1000
1 1001
tu s ( t )
K v
0
f
t us (t ) / 2
K a
0
f
Input
ErrorConstants
2
(b)
SteadystateError
________________________________________________________________________________
u s ( t )
K p
f
tu s ( t )
K v
1
1
t us (t ) / 2
K a
0
SteadystateError
2
0
f
(c)Input
ErrorConstants
________________________________________________________________________________
u s ( t )
K p
f
52
0
AutomaticControlSystems,9thEdition
Chapter5Solutions
Golnaraghi,Kuo
tu s ( t )
t us (t ) / 2
TheaboveresultsarevalidifthevalueofKcorrespondstoastableclosedloopsystem.
2
K v
K
1/ K
K a
0
f
(d)Theclosedloopsystemisunstable.Itismeaninglesstoconductasteadystateerroranalysis. (e)
Input
ErrorConstants
SteadystateError
________________________________________________________________________________
u s ( t )
Kp
f
0
tu s ( t )
Kv
1
1
t us (t ) / 2
Ka
0
f
Input
SteadystateError
2
(f)
ErrorConstants
________________________________________________________________________________
u s ( t )
Kp
f
0
tu s ( t )
K v
f
0
t us (t ) / 2
K a
K
1/ K
TheclosedloopsystemisstableforallpositivevaluesofK.Thustheaboveresultsarevalid.
2
55(a) K H
H ( 0 ) 1
UnitstepInput:
s 1
G( s)
M ( s)
1 G( s) H ( s)
a0
3, a1
ess
§ b0 K H · ¨1 a ¸ KH © ¹ 0 1
3, a2
3
2, b0
2
3
53
2
s 2 s 3s 3 1, b1
1.
AutomaticControlSystems,9thEdition
Unitrampinput:
Chapter5Solutions
a0 b0 K H
3 1 2 z 0. Thus e ss
a0 b0 K H
2 z 0 and a1 b1 K H
Golnaraghi,Kuo
f.
UnitparabolicInput:
1 z 0. Thus e ss
f.
(b) K H
H ( 0)
5
G( s)
UnitstepInput:
UnitrampInput:
i
e ss
M ( s)
UnitparabolicInput:
1
1 G( s) H ( s)
ess
e ss
2
s 5s 5
a0
5, a1
5, b0
1, b1
0.
§ b0 K H · 1 § 5 · ¨1 a ¸ 5 ¨1 5 ¸ 0 KH © © ¹ ¹ 0 1
0: a0 b0 K H
i
0
a1 b1 K H
5
1
a0 K H
25
5
1: a1 b1 K H
5 z 0
f
(c) K H
H ( 0 ) 1 / 5
M ( s )
UnitstepInput:
UnitrampInput:
s5
G( s) 1 G( s) H ( s) a0
ess
1, a1
4
3
1, a2
50, a3
Thesystemisstable.
15, b0
§ b0 K H · § 5 / 5 · ¨1 a ¸ 5 ¨1 1 ¸ 0 KH © ¹ ¹ © 0 1
54
2
s 15s 50 s s 1
5, b1
1
AutomaticControlSystems,9thEdition
0: a0 b0 K H
i
e ss
UnitparabolicInput:
e ss
Chapter5Solutions 1: a1 b1 K H
i
0
a1 b1 K H
1 1/ 5
a0 K H
1/ 5
Golnaraghi,Kuo 4 / 5 z 0
4
f
(d) K H
H ( 0 ) 10
G( s)
M ( s )
UnitstepInput:
UnitrampInput:
i
e ss
ess
UnitparabolicInput:
1 G( s) H ( s) a0
e ss
10, a1
4
3
2
UnitstepInput:
Unitrampinput:
a0
ess
i
4, a1
s 12 s 5 s 10
5, a2
12, b0
0: a0 b0 K H
0 i
Thesystemisstable.
1, b1
a1 b1 K H
5
a0 K H
100
1: a1 b1 K H
0, b2
0
5 z 0
0.05
f
K H 1
4, a2
Thesystemisstable.
48, a3
16, b0
4, b1
1, b2
§ b0 K H · § 4 · ¨1 a ¸ ¨1 4 ¸ 0 KH © ¹ ¹ © 0 1
0: a0 b0 K H
0
i
1: a1 b1 K H
55
§ b0 K H · 1 § 10 · ¨ 1 a ¸ 10 ¨ 1 10 ¸ 0 KH © © ¹ ¹ 0
s 16 s 48 s 4 s 4
2
1
s4
56(a) M ( s )
1 3
4 1 3 z 0
0, b3
0
AutomaticControlSystems,9thEdition
e ss
UnitparabolicInput:
a1 b1 K H
4 1
3
a0 K H
4
4
e ss
Chapter5Solutions
Golnaraghi,Kuo
f
K ( s 3)
(b) M ( s )
3
2
s 3s ( K 2) s 3K a0
UnitstepInput:
UnitrampInput:
i
e ss
3 K , a1
KH K 2, a2
1
Thesystemisstablefor K ! 0.
3, b0
3 K , b1
K
§ b0 K H · § 3K · ¨ 1 a ¸ ¨ 1 3K ¸ 0 KH © ¹ ¹ © 0 1
ess
0: a0 b0 K H
i
0
a1 b1 K H
K 2 K
2
a0 K H
3K
3K
UnitparabolicInput:
TheaboveresultsarevalidforK>0.
e ss
1: a1 b1 K H
K 2 K
f
(c) M ( s )
s5 4
3
2
s 15 s 50 s 10 s
UnitstepInput:
UnitrampInput:
a0
ess
0, a1
H ( s) 10, a2
10 s s5 50, a3
KH 15, b0
so 0
H ( s)
2
s
5, b1
§ a2 b1 K H · 1 § 50 1 u 2 · ¨ a ¸ 2 ¨© 10 ¸¹ 2.4 KH © ¹ 1 1
56
lim
1
2 z 0
AutomaticControlSystems,9thEdition
UnitparabolicInput:
e ss
f
e ss
f
Chapter5Solutions
Golnaraghi,Kuo
K ( s 5)
(d) M ( s )
4
3
1
KH
2
s 17 s 60 s 5 Ks 5 K
Thesystemisstablefor0
Automatic Control Systems 9th Edition Solution Manual
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