Problem 1: Bode magnitude and phase plot of G (jw) -20 -40 -60 -80 -90 -180 -270 -360 101 100 101 Frequency (rad/s) Let P = 0 (i.e., the number of poles of G(s) in the open left-half plane is zero). The Bodé plot of G(jw) is given above. What is the phase crossover frequency wp ? What is the gain crossover frequency wg ? а) 1 b) Choose the approximate Gain Margin GM in dB (i) = 5 dB (ii) = -10 dB (iii) = 20 dB c) (i) = -150° Choose the approximate Phase Margin PM in degrees: (ii) = 30° (iii) a 180° G d) Is the closed-loop system Hyr stable? 1+G Phase (deg) |G (jw) IdB (dB)

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Problem 1:
Bode magnitude and phase plot of G (jw)
-20
-40
-60
-80
-90
-180
-270
-360
101
100
101
Frequency (rad/s)
Let P = 0 (i.e., the number of poles of G(s) in the open left-half plane is zero). The Bodé plot of G(jw)
is given above.
What is the phase crossover frequency wp ?
What is the gain crossover frequency wg ?
а) 1
b)
Choose the approximate Gain Margin GM in dB
(i) = 5 dB
(ii) = -10 dB
(iii) = 20 dB
c)
(i) = -150°
Choose the approximate Phase Margin PM in degrees:
(ii) = 30°
(iii) a 180°
G
d)
Is the closed-loop system Hyr
stable?
1+G
Phase (deg)
|G (jw) IdB (dB)
Transcribed Image Text:Problem 1: Bode magnitude and phase plot of G (jw) -20 -40 -60 -80 -90 -180 -270 -360 101 100 101 Frequency (rad/s) Let P = 0 (i.e., the number of poles of G(s) in the open left-half plane is zero). The Bodé plot of G(jw) is given above. What is the phase crossover frequency wp ? What is the gain crossover frequency wg ? а) 1 b) Choose the approximate Gain Margin GM in dB (i) = 5 dB (ii) = -10 dB (iii) = 20 dB c) (i) = -150° Choose the approximate Phase Margin PM in degrees: (ii) = 30° (iii) a 180° G d) Is the closed-loop system Hyr stable? 1+G Phase (deg) |G (jw) IdB (dB)
ALL PROBLEMS ARE BASED ON THE UNITY-FEEDBACK CONTROL SYSTEM:
R(s)
E(s)
Y(s)
C(s)
+ P(s)
input: r(t) , output: y(t), error: e(t) = r(t) – y(t) ,
G(s) = P(s)C(s)
Y(s)
input-output transfer-function: Hyr(s) =
R(s)
E(s)
R(s)
input-error transfer-function: Her(s)
We denote the unit step function by 1 (t)
The following may be useful:
1
tan 26.5° = 0.5, tan 30° =
V3
tan 14° = 0.25,
tan 45° = 1,
tan 60° =
V3
tan 63.5° = 2,
tan 71.565° = 3,
tan 75.96° = 4,
tan 82.9° = 8 ,
tan 84.3° = 10
log10 2 z 0.3
log10 3 = 0.48,
log1, 5 z 0.7,
log1, 10 = 1, log10 V10 = 0.5, log1o 10° = a
log (a b)
log (a ) + log ( b),
log () = log (a) – log (b)
LABEL axes for all plots.
Transcribed Image Text:ALL PROBLEMS ARE BASED ON THE UNITY-FEEDBACK CONTROL SYSTEM: R(s) E(s) Y(s) C(s) + P(s) input: r(t) , output: y(t), error: e(t) = r(t) – y(t) , G(s) = P(s)C(s) Y(s) input-output transfer-function: Hyr(s) = R(s) E(s) R(s) input-error transfer-function: Her(s) We denote the unit step function by 1 (t) The following may be useful: 1 tan 26.5° = 0.5, tan 30° = V3 tan 14° = 0.25, tan 45° = 1, tan 60° = V3 tan 63.5° = 2, tan 71.565° = 3, tan 75.96° = 4, tan 82.9° = 8 , tan 84.3° = 10 log10 2 z 0.3 log10 3 = 0.48, log1, 5 z 0.7, log1, 10 = 1, log10 V10 = 0.5, log1o 10° = a log (a b) log (a ) + log ( b), log () = log (a) – log (b) LABEL axes for all plots.
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