Question

A system has the transfer function $\dfrac{Y(s)}{X(s)}=\dfrac{s-\pi}{s+\pi}$. Let $u(t)$ be the unit–step function. The input $x(t)$ that results in a steady–state output $y(t)=\sin(\pi t)$ is _____.

• A. $x(t)=\sin(\pi t)\,u(t)$
• B. $x(t)=\sin\!\left(\pi t+\dfrac{\pi}{2}\right) u(t)$
• C. $x(t)=\sin\!\left(\pi t-\dfrac{\pi}{2}\right) u(t)$
• D. $x(t)=\cos\!\left(\pi t+\dfrac{\pi}{4}\right) u(t)$

Hint:

For LTI systems with sinusoidal input: $y(t)=|H(j\omega)|\sin\!\big(\omega t+\phi_{\text{in}}+\angle H(j\omega)\big)$. Choose input phase to offset $\angle H(j\omega)$.

Answer 1

The Correct Option is C

For sinusoidal steady state at $\omega=\pi$, the frequency response is \[ H(j\omega)=\frac{j\omega-\pi}{j\omega+\pi} \Rightarrow H(j\pi)=\frac{j\pi-\pi}{j\pi+\pi}. \] Magnitude: $|H(j\pi)|=\dfrac{\sqrt{\pi^2+\pi^2}}{\sqrt{\pi^2+\pi^2}}=1$ (no gain change).
Phase: $\angle(j\pi-\pi)=135^\circ$, $\angle(j\pi+\pi)=45^\circ$ $\Rightarrow$ $\angle H(j\pi)=90^\circ=\dfrac{\pi}{2}$.
Thus the output lags the input by $-\,\dfrac{\pi}{2}$? (Equivalently, the output phase $=\text{input phase}+\dfrac{\pi}{2}$.) To obtain $y(t)=\sin(\pi t)$ (zero phase), the input must be \[ x(t)=\sin\!\big(\pi t-\tfrac{\pi}{2}\big)u(t), \] which is option \fbox{(C)}.