Question

The Bode magnitude plot of
$H\left(j\omega \right)=\frac{{10}^4(1+j\omega )}{(10+j\omega )(100+j\omega {)}^2}is$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$H\left(j\omega \right)=\frac{{10}^4(1+j\omega )}{(10+j\omega )(100+j\omega {)}^2}$

$=\frac{(1+j\omega )}{10\times (1+\frac{j\omega }{10})\times {(1+\frac{j\omega }{100})}^2}$

$H\left(j\omega \right)=\frac{k(1+j\omega )}{(1+\frac{j\omega }{10})\times {(1+\frac{j\omega }{100})}^2}$

where,
$k=\frac{1}{10}=0.1$

Corner frequencies
${\omega }_1=1rad/sec;{\omega }_2=10rad/secand{\omega }_3=100rad/sec$

Gain of the system is constant as there is no pole at origin.

Gain=20 log k =20 log 0.1=-20 dB

$At\omega ={\omega }_1=1rad/secorlog{\omega }_1=log=0$

There is zero , so system gain increases with slope +20 dB/decade and system gain becomes 0 dB at $\omega =10rad/secorlog\omega {|}_{\omega =10}=log10=1$

$At{\omega }_2=10\left(or\right)log\omega {|}_{{\omega }_2=10}=1og10=1$

There is a pole, so slope is -20 dB/decade.

$Overallslope{\omega }_2<\omega <{\omega }_3$

=20 dB/decade-20 dB/decade

=0 dB/decade

So, gain remains constant between
${\omega }_2<\omega <{\omega }_3$

$or1<log\omega <2$

$At\omega ={\omega }_3=100rad/secorlog\omega {|}_{{\omega }_3=100}=log100=2$

Double pole are present.

So, system gain decrease with -40 dB/decade.