Question

7. Consider the system with transfer function 100 G(s) = (s + 202 (a) Sketch the bode plot and Nyquist diagrams and determine the range of proportional closed loop gain K for stability. (b) What positive gain K will yield a phase margin of 30 degrees ?

Answer 1

Here we have given

.

100 G(S) = ss + 20)

a) For plotting Bode plot we have to write equation in time constant form as,

$G(s)= \frac{100}{20^2s(\frac{s}{20}+1)^2}$

0.25 G(S) = 5 + 1)2

Now for bode plot magnitude in dB as,

starting gain,

M= 20log(0.25)= -12dB.

starting slope is -20dB/dec

Now again slope changes at w=20 rad/sec to -60dB/dec because there is two pole at w=20.

and for phase plot starting phase is due to pole at 0 is -90degree.

after that it changes with -180degree due to two poles at w=20.

Now drawing plot we have,

M (dB) 10 w 20 - 20dBldec -35 do- -35 -60 dB/dec Bode magnitude plot Olindegree). 20 -go 200 W - 180 -270 Phose plot

Now for sketching nyquist plot, we have to firstly draw the polar plot for this,

we have G(s) as

100 G(S) = ss + 20)

Now replacing s with jw we have,

G(jw) = w 100 +20)

Now taking magnitude and phase of G(jw) we have,

$M= |G(jw)|= \frac{100}{w(w^2+400)}$

Q = LG(jw) = -90 - 2 tan-1

Now writting the table for magnitude and phase for different w.

w	M	$\Phi$
0	$\infty$	$-90^{\circ}$
$\infty$	0	$-270^{\circ}$

Now plotting polar plot we have,

366 -21გ° Q-ო | |-180

Now taking mirror image and joining it we can able to plot nyquist plot.

Hence we have nyquist plot is,

wzo Co -270 -270 Avant Nyaunt plot wyt iso

Now we have to determine the range of propotional K for the stability of the system.

Adding K system becomes,

$G(s)= \frac{100K}{s(s+20)^2}$

Now taking 1+G(s)H(s)= 0.

we have characteristic equation like,

Now using routh array table,

$s^3$	1	400
$s^2$	40	100K
$s^1$	400- 2.5K	0
	100K	0

We know that for the stability no sign change should happen in 1st column. Hence sign is possitive,

so,

and

100K > 0

so K > 0.

So we have range of K for stability is .

(b) Now we have to find the value of K so that phase margin is 30 degree.

We have,

$G(s)= \frac{100K}{s(s+20)^2}$

We know that at the point of phase margin gain of the system is unity ie |G(jw)|=1 at w= wpc,

Since phase margin is,

$PM= 180^{\circ}+\phi$

$30= 180^{\circ}+\phi$

Hence $\phi = -150^{\circ}$ .

Since

Q = LG(jw) = -90 - 2 tan-1

$\tan^{-1}\frac{w}{20}=-30^{\circ}$.

$\because w= \frac{20}{\sqrt 3}$

putting this value in,

$M= |G(jw)|= \frac{100K}{w(w^2+400)}$

we have M=1. then,

$100K= \frac{20}{\sqrt3}(\frac{400}{3}+400)$

solving this we can get,

$K=\frac{320}{3\sqrt3}= 61.6$

Hence for K=61.6 we have phase margin as 30 degree in the system.