written 5.8 years ago by
teamques10
★ 68k
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•
modified 5.8 years ago
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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.
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teamques10
★ 68k