Question

A unit step input is applied to a system with impulse response \( H(s) = \dfrac{1-s/\omega_z}{1+s/\omega_p} \) at \( t = 0 \). The output of the system \( y(t) \) at \( t = 0^+ \) is
 

• A. 1
• B. \( -\omega_z/\omega_p \)
• C. \( -\omega_p/\omega_z \)
• D. 0

Hint:

For systems with a transfer function and a step input, evaluate the system at \( s = 0 \) to find the output at \( t = 0^+ \).

Answer 1

The Correct Option is C

Given the transfer function \( H(s) = \frac{1-s/\omega_z}{1+s/\omega_p} \), we need to determine the output \( y(t) \) when a unit step input is applied at \( t = 0^+ \). The Laplace transform of a unit step input is \( \frac{1}{s} \). Thus, the output \( Y(s) \) is given by: \[ Y(s) = H(s) \cdot \frac{1}{s}. \] Substitute the expression for \( H(s) \) into the above equation: \[ Y(s) = \frac{1-s/\omega_z}{s(1+s/\omega_p)}. \] Now, we take the inverse Laplace transform to find \( y(t) \). Since we are interested in the value of \( y(t) \) at \( t = 0^+ \), we can evaluate \( Y(s) \) at \( s = 0 \) for this step input, leading to the value of the output at \( t = 0^+ \). After evaluating, the correct value of \( y(t) \) at \( t = 0^+ \) is: \[ y(0^+) = -\frac{\omega_z}{\omega_p}. \] Thus, the correct answer is Option (A): \( -\omega_z/\omega_p \).