Question

The corner frequencies of a system having transfer function \[ G(s)=\frac{100(1+0.2s)}{1+0.5s} \] are

• A. \(\omega_z=2,\; \omega_p=5\)
• B. \(\omega_z=0,\; \omega_p=0.5\)
• C. \(\omega_z=0.5,\; \omega_p=0.2\)
• D. \(\omega_z=5,\; \omega_p=2\)

Hint:

For a transfer function term \[ \boxed{ 1+\tau s, } \] the corresponding corner (break) frequency is \[ \boxed{ \omega=\frac{1}{\tau}\ \text{rad/s}. } \]

Answer 1

The Correct Option is D

Step 1: Recall the corner frequency formula. For a factor of the form \[ 1+\tau s, \] the corner frequency is \[ \omega=\frac{1}{\tau}. \]

Step 2: Determine the zero and pole frequencies. The zero term is \[ 1+0.2s, \] so \[ \omega_z=\frac{1}{0.2}=5\ \text{rad/s}. \] The pole term is \[ 1+0.5s, \] so \[ \omega_p=\frac{1}{0.5}=2\ \text{rad/s}. \] Hence, \[ \boxed{\omega_z=5,\qquad \omega_p=2.} \] Therefore, \[ \boxed{(D)} \] is the correct answer.