Question

In the system shown below, $x(t)=\sin(t)u(t)$. In steady-state, the response $y(t)$ will be 

• A. $\dfrac{1}{\sqrt{2}}\sin(t-\pi/4)$
• B. $\dfrac{1}{\sqrt{2}}\sin(t+\pi/4)$
• C. $\dfrac{1}{\sqrt{2}}e^{-t}\sin(t)$
• D. $\sin(t)-\cos(t)$

Hint:

Steady-state sinusoidal response depends only on system frequency response at that frequency.

Answer 1

The Correct Option is A

Step 1: From the block diagram, the system transfer function is \( H(s) = \dfrac{1}{s+1} \), and the sinusoidal input has angular frequency \( \omega = 1 \).
Step 2: Evaluate the frequency response at \( s = j\omega = j1 \): \( H(j1) = \dfrac{1}{1+j} \), which has magnitude \( |H(j1)| = \dfrac{1}{\sqrt{2}} \) and phase \( \angle H(j1) = -\tan^{-1}(1) = -\dfrac{\pi}{4} \).
Step 3: The steady-state response to a sinusoidal input is the same frequency sinusoid scaled by this magnitude and shifted by this phase. \[ \boxed{y(t) = \dfrac{1}{\sqrt{2}}\sin\!\left(t-\dfrac{\pi}{4}\right)} \]

Answer 2

An alternative way to reach the steady-state output is to represent the input sinusoid as a complex exponential, pass it through the system's frequency response, and then extract the real (sinusoidal) part at the end, checking each option along the way.

1. Option \( \dfrac{1}{\sqrt{2}}\sin(t-\pi/4) \): Writing \( \sin(t) = \text{Im}\{e^{jt}\} \), the steady-state output to \( e^{jt} \) is \( H(j1)e^{jt} = \dfrac{1}{1+j}e^{jt} \). Converting \( \dfrac{1}{1+j} \) to polar form gives magnitude \( \dfrac{1}{\sqrt{2}} \) and angle \( -\dfrac{\pi}{4} \), so \( H(j1)e^{jt} = \dfrac{1}{\sqrt{2}}e^{j(t-\pi/4)} \); taking the imaginary part recovers \( \dfrac{1}{\sqrt{2}}\sin\!\left(t-\dfrac{\pi}{4}\right) \), matching this option exactly.
2. Option \( \dfrac{1}{\sqrt{2}}\sin(t+\pi/4) \): This would require the angle of \( \dfrac{1}{1+j} \) to be \( +\pi/4 \), but converting \( 1+j \) to polar form gives an angle of \( +\pi/4 \) for the denominator, meaning the angle of its reciprocal is \( -\pi/4 \), not \( +\pi/4 \); this option has the sign flipped.
3. Option \( \dfrac{1}{\sqrt{2}}e^{-t}\sin(t) \): A term like \( e^{-t} \) arises from the system's natural (homogeneous) response due to its pole at \( s=-1 \), which decays away over time and is therefore part of the transient, not the steady-state, response that persists as \( t\to\infty \).
4. Option \( \sin(t) - \cos(t) \): Converting this to phasor form, its amplitude is \( \sqrt{1^2+1^2} = \sqrt{2} \), which is the reciprocal of the correct magnitude \( 1/\sqrt{2} \); this mismatch in magnitude rules it out.

Working through the complex-exponential (phasor) method and converting back to a real sinusoid reproduces the same magnitude and phase found by the direct frequency-response evaluation.

So the correct answer is \( \dfrac{1}{\sqrt{2}}\sin(t-\pi/4) \).