Steady-state output of a first-order LTI system
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A causal LTI system has impulse response $h(t) = e^{-3t}u(t)$. The input $x(t) = 2\cos(3t)$ is applied. The steady-state output $y(t)$ is:
Show the step-by-step derivation
Answer
(A) 2 3 ! (3t - 45^ )
Step-by-step derivation
- For an LTI system driven by a sinusoid, the steady-state output is the same sinusoid scaled by $|H(j\omega)|$ and shifted by $\angle H(j\omega)$. So find the frequency response first.
- Take the Fourier transform of the impulse response: $$H(j\omega) = \int_0^{\infty} e^{-3t}e^{-j\omega t}\,dt = \left[\frac{e^{-(3+j\omega)t}}{-(3+j\omega)}\right]_0^{\infty} = \frac{1}{3 + j\omega}.$$
- The input frequency is $\omega = 3$ rad/s. Substituting: $$H(j3) = \frac{1}{3 + j3}.$$
- Magnitude: $$|H(j3)| = \frac{1}{|3+j3|} = \frac{1}{\sqrt{3^2+3^2}} = \frac{1}{\sqrt{18}} = \frac{1}{3\sqrt{2}}.$$
- Phase: $$\angle H(j3) = -\arctan\!\left(\frac{3}{3}\right) = -\arctan(1) = -45^\circ.$$ The sign is negative because the $j$ term is in the denominator.
- Apply both to the input amplitude of $2$: $$y(t) = 2 \cdot \frac{1}{3\sqrt{2}} \cos\!\left(3t - 45^\circ\right) = \frac{2}{3\sqrt{2}}\cos\!\left(3t - 45^\circ\right).$$
- Rationalise the amplitude: $$\frac{2}{3\sqrt{2}} = \frac{2\sqrt{2}}{3 \cdot 2} = \frac{\sqrt{2}}{3} \approx 0.471.$$ So $y(t) = \dfrac{\sqrt{2}}{3}\cos(3t - 45^\circ)$, option (A).
The trap this question is built around
Forgetting that the phase is negative. A pole contributes lagging phase, so a first-order low-pass output always trails its input. Option (C) is the same magnitude with the sign flipped, and it is the most-picked wrong answer.