Question: Construct the bode plot for the open loop transfer function. Comment on stability. G(s) = $\frac{288(s+4)}{5(s+1)(s^2 +48s + 144)}$ .
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H(s) = 1

Mumbai University > Electronics Engineering > Sem 4 > Linear Control Systems

Topic: Stability Analysis in Frequency Domain

Difficulty : High

Marks : 10M

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modified 4 days ago by gravatar for Mayank Aggarwal Mayank Aggarwal ♦♦ 0 written 27 days ago by gravatar for abhishektiwari1712 abhishektiwari1712 ♦♦ 30
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Given => $\frac{288(s+4)}{s(s+1)(s^2 +4.8s +144)}$

step I => Bring equation in the standard fims constant form

G(s)H(s) = $\frac{288*4*(\frac{s}{4} +1)}{144s(s+1)(\frac{s^2}{144} + \frac{4.8s}{144}+1)}$

Step 2 => Get freq domain transfer function s=jw,

GH(jw) = $\frac{288*4*(\frac{jw}{4} +1)}{144jw(jw+1)(1+ j0.033w-\frac{w^2}{144}}$

comparing the quadrate pole with standard equation,

1 + 2jw$\frac{w}{w_n}$ - $\frac{w^2}{w_{n^2}}$ = 1 + j0.033w - $\frac{w^2}{144}$

=> $w^2_n$ = 144

w= 12

$w_c - w_n$ =12

similarly $\frac{2E_1}{w_n}$ = 0.033

$E_1$ = 0.2

The correction magnitude at $w_n$ = 12 will be -20log$(4E^2_1)^\frac{1}{2}$ = 7.45 dB

step 3 => i) constant k=8 i.e. 20 logk = 18.06 dB

ii) pole at origin $\frac{1}{jw}$

iii) first order pole $\frac{1}{1+jw}$

iv) first order pole (1+$\frac{1}{1+jw}$)

v) fsecond order pole $\frac{1}{1+0.033jw- \frac{w^2}{144}}$

Step 4 =>

SR No. Factors Magnitude Phase
1 k=8 18.06 dB straight line $\phi$=0
2 $\frac{1}{jw}$ straight line of slope -20 dB/sec phassing through, w=1,0dB point $\phi$ = -90 for all value of w
3 $\frac{1}{1+jw}$ line slopes are i) 0dB/dec for w<1 ii) -20dB/dec for w>1 $\phi$=ta$n^{-}$(w)
4 (1+$\frac{jw}{4}$) lines slops are i) 0dB/dec for w<4 ii) +20 dB/dec for w>4 $\phi$=ta$n^{-}(\frac{w}{4})$
5 $\frac{1}{1+0.033jw -\frac{w^2}{144}}$ i) 0 dB/dec for w<12 ii) -40 db/dec for w>12 $\phi$=ta$n^{-}(\frac{0.033w}{1-\frac{w^2}{144}})$

Step 5=> Magnitude plot

SR No. Factors Resultant slop start point End point
1 k 18.06 dB straight line 0.1 $\infty$
2 $\frac{1}{jw}$ -20 dB/dec 0.1 1
3 $\frac{1}{1+jw}$ -20+(-20) = -40 dB/dec 1 4
4 (1+$\frac{jw}{4}$) -40+(20) = -20 dB/dec 4 12
5 $\frac{1}{1+0.033jw -\frac{w^2}{144}}$ -20 +(-40) = -60 dB/dec 12 $\infty$

Step 6 => Phase angle equation

i) For k=8,$\phi$ = 0

ii) For $\frac{1}{jw}$ $\phi$ = -90

iii) For $\frac{1}{1+jw}$ $\phi= -tan^- (w)$

iv) For (1+$\frac{jw}{4}$), $\phi= -tan^- (\frac{w}{4})$

v) For $\frac{1}{1+0.033jw -\frac{w^2}{144}}$ $\phi= -tan^- (\frac{0.033w}{1-\frac{w^2}{144}})$

Step 7 => Phase angle

w $\frac{1}{jw}$ $-tan^- (w)$ $tan^- (\frac{w}{4})$ $tan^- (\frac{0.33w}{1-\frac{w^2}{144}})$ $\phi_{12}$
0.1 -90 -3.71 1.43 -0.19 -94.47
0.5 -90 -26.57 7.13 -0.95 -110.39
1 -90 -45 14.04 -1.9 -122.86
4 -90 -75.96 45 -8.45 -129.41
5 -90 -78.69 51.34 -11.29 -128.41
8 -90 -82.87 63.43 -25.42 -134.86
9 -90 -83.66 66.04 -34.17 -141.79
10 -90 -84.89 68.2 -47.2 -153.29
11 -90 -84.81 70.02 -66.25 -171.04
15 -90 -86.19 75.05 41.35-180 -239.77
50 -90 -88.85 85.43 -174.24 -268.96
100 -90 -89.71 87.71 -177.24 -268.96
500 -90 -89.89 89.54 -179.46 -269.81
1000 -90 -89.94 89.77 -179.73 -269.9

Step 8 => draw the bode plot

GM = + 8dB,

PM = 48

$w_{gc}$ = 3rad/sec

$w_{pc}$ = 12 rad/sec

The given system is stable.

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modified 4 days ago  • written 4 days ago by gravatar for Mayank Aggarwal Mayank Aggarwal ♦♦ 0
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