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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written 5.9 years ago by

teamques10 ★ 68k

• modified 5.8 years ago

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

GH(jw) = $\frac{8*(\frac{jw}{4} +1)}{jw(jw+1)(1+ j0.033w-\frac{w^2}{144}}$

Comparing the quadrate pole with standard equation,

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

=> $w^2_n$ = 144

w= 12

$w_c - w_n$ = 12

similarly $\frac{2{\epsilon}_1}{w_n}$ = 0.033

${\epsilon}_1$ = 0.2

The correction magnitude at $w_n$ = 12 will be -20log$(4{\epsilon}^2_1)^\frac{1}{2}$ = 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) second 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 = $48^o$

$w_{gc}$ = 3rad/sec

$w_{pc}$ = 12 rad/sec

The given system is stable.

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