Construct the bode plot for the following Transfer function. Comment on stability. G(s).H(s) = $\frac{10}{s(s+1)(s+5)}$ .

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written 6.1 years ago by

teamques10 ★ 69k

• modified 6.1 years ago

Given $\Longrightarrow G(s)\cdot H(s)=\cfrac { 10 }{ s(s+1)(s+5) }$

Step 1 $\Longrightarrow$ First bring the given G(s) into standard time constant form.

$G(s)\cdot H(s)=\cfrac { 10 }{ s(s+1)5\left( 1+\cfrac { s }{ 5 } \right) }$

$G(s)\cdot H(s)=\cfrac { 2 }{ s(s+1)\left( 1+\cfrac { s }{ 5 } \right) }$

Step 2 $\Longrightarrow$ To convert it into frequency domain replace 's' by $j\omega$ .

$G(j\omega )\cdot H(j\omega )=\cfrac { 2 }{ j\omega (j\omega +1)\left( 1+\cfrac { j\omega }{ 5 } \right) }$

Step 3 $\Longrightarrow$ In the given transfer function following factors are present,

i) Constant, k=2

$\therefore 20 \log{k}=20 \log{2}=6.02dB$

ii) Pole at origin $\Longrightarrow \cfrac{1}{j \omega}$

iii) First order pole $\Longrightarrow \cfrac{1}{1+j \omega}$

$\therefore \omega_{c1}=1$

iv) First order pole $\Longrightarrow \cfrac{1}{1+\cfrac{j \omega}{5}}$

$\therefore \omega_{c2}=5$

Step 4 $\Longrightarrow$

Serial No.	Factor	Magnitude curve	Phase curve
1	k=2	straight line at 6.02 dB	$\phi=0^{\circ}$
2	$\cfrac{1}{j\omega}$	straight line of slope -20dB/dec passing through $\omega=1,0$ dB point	$\phi=90^{\circ}$
3	$\cfrac{1}{1+j\omega}$	Line slopes are: 1) 0dB/dec for $\omega \leq 1$ 2) -20dB/dec for $\omega \gt 1$	$\phi=\tan^{-1}{(\omega)}$ for all values of $\omega$
4	$\cfrac{1}{1+\cfrac{j\omega}{5}}$	Line slopes are: 1) 0dB/dec for $\omega \leq 5$ 2) -20dB/dec for $\omega \gt 5$	$\phi=\tan^{-1}{(\omega)}$ for all values of $\omega$

Step 5 $\Longrightarrow$ Magnitude plot

Serial No.	Factor	Resultant slope	Start point ( $\omega$ )	End point ( $\omega$ )
1	k=2	straight line at 6.02 dB	0.1	$\infty$
2	$\cfrac{1}{j\omega}$	-20dB/dec	0.1	1
3	$\cfrac{1}{1+j\omega}$	-20dB/dec+(-20dB/dec)=-40dB/dec	1	5
4	$\cfrac{1}{1+\cfrac{j\omega}{5}}$	-40dB/dec+(-20dB/dec)=-60dB/dec	5	$\infty$

Step 6 $\Longrightarrow$ Phanse angle

$\phi(\omega)=-90+(-\tan^{-1}{\omega})+(-\tan^{-1}\cfrac{\omega}{5})$

$\omega$	$\cfrac{1}{j\omega}$	$-\tan^{-1}{\omega}$	$-\tan^{-1}{\cfrac{\omega}{5}}$	$\phi(\omega)$
0.1	$-90^{\circ}$	$-5.71^{\circ}$	$-1.145^{\circ}$	$-96.85^{\circ}$
0.5	$-90^{\circ}$	$-26.56^{\circ}$	$-5.71^{\circ}$	$-121.66^{\circ}$
1	$-90^{\circ}$	$-45^{\circ}$	$-11.3^{\circ}$	$-146.3^{\circ}$
10	$-90^{\circ}$	$-84.28^{\circ}$	$-63.43^{\circ}$	$-237.71^{\circ}$
100	$-90^{\circ}$	$-89.42^{\circ}$	$-87.13^{\circ}$	$-266.55^{\circ}$
500	$-90^{\circ}$	$-89.88^{\circ}$	$-89.42^{\circ}$	$-269.3^{\circ}$
1000	$-90^{\circ}$	$-89.94^{\circ}$	$-89.71^{\circ}$	$-269.65^{\circ}$

Step 7 $\Longrightarrow$ Bode plot is as shown in figure. From the bode plot

i) Gain margin=12dB

ii) Phase margin=15 $^{\circ}$

iii) Gain cross over frequency= $\omega_{gc}=1.4 rad/sec$

iv) Phase cross over frequency= $\omega_{pc}=2.7 rad/sec$

Since the gain margin and the phase margin are both positive, so the given system is stable.

enter image description here

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