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

Topic: Stability Analysis in Frequency Domain

Difficulty : Medium

Marks : 5M

Given: $G(s)=\cfrac{12}{s(1+s)}$

Step 1

$\Longrightarrow $ We convert the transfer function to the frequency domain by replacing $s$ by $j\omega.$ $\therefore G(j\omega )=\cfrac { 12 }{ j\omega (1+j\omega ) }$ ------(1) Step 2 $\Longrightarrow $ Calculate magnitude and phase from equation (1). **Magnitude** $\Longrightarrow $ $|G(j\omega)|=\cfrac { 12 }{ \omega \sqrt{1+\omega^{2} } }$ At $\omega=0,$ $|G(j\omega)|=\cfrac { 12 }{0 \sqrt{1+0^{2} } }= \infty$ At $\omega=\infty,$ $|G(j\omega)|=\cfrac { 12 }{\infty \sqrt{1+\infty^{2} } }= 0$ At $\omega=1,$ $|G(j\omega)|=\cfrac { 12 }{1 \sqrt{1+1^{2} } }=\cfrac{12}{\sqrt{2}}= 8.48$ **Phase**$\Longrightarrow $ $\angle G(j\omega )=\cfrac { \tan ^{ -1 }{ \left( \cfrac { 0 }{ 12 } \right) } }{ \tan ^{ -1 }{ \left( \cfrac { \omega }{ 0 } \right) } +\tan ^{ -1 }{ \left( \cfrac { \omega }{ 1 } \right) } } $ $\angle G(j\omega )=\cfrac { 0 }{ { 90 }^{ \circ }+\tan ^{ -1 }{ \left( \omega \right) } } $ $\angle G(j\omega )=-{ 90 }^{ \circ }-\tan ^{ -1 }{ \left( \omega \right) } $ At $\omega=0,$ $\angle G(j\omega )=-{ 90 }^{ \circ }-\tan ^{ -1 }{ \left( 0 \right) } =-{ 90 }^{ \circ }$

At $\omega=\infty,$ $\angle G(j\omega )=-{ 90 }^{ \circ }-\tan ^{ -1 }{ \left( \infty \right) } =-{ 90 }^{ \circ }-{ 90 }^{ \circ }=-{ 180 }^{ \circ }$

At $\omega=1,$ $\angle G(j\omega )=-{ 90 }^{ \circ }-\tan ^{ -1 }{ \left( 1 \right) } =-{ 90 }^{ \circ }-{ 45 }^{ \circ }=-{ 135 }^{ \circ }$