Let x(t)=10(10.5Wt) be passed through an LTI system having impulse response h(t)=( Wt t)210Wt. The output of the system is

$\left(\dfrac{15W}{2}\right)\cos(10.5Wt)$

$\left(\dfrac{15W}{8}\right)\cos(10.5Wt)$

GATE 2023 EC – Question 49

Networks, Signals and Systems · Continuous-time Signals · 2 marks · Multiple choice

Let $x(t)=10\cos(10.5Wt)$ be passed through an LTI system having impulse response $h(t)=\pi\left(\dfrac{\sin Wt}{\pi t}\right)^2\cos10Wt$. The output of the system is

$\left(\dfrac{15W}{4}\right)\cos(10.5Wt)$
$\left(\dfrac{15W}{2}\right)\cos(10.5Wt)$
$\left(\dfrac{15W}{8}\right)\cos(10.5Wt)$
$(15W)\cos(10.5Wt)$

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Show answer and explanation

Correct answer: (A) $\left(\dfrac{15W}{4}\right)\cos(10.5Wt)$

Explanation

$\dfrac{\sin Wt}{\pi t}$ has a rectangular spectrum of height 1 on $[-W,W]$. Its square has the spectrum $\frac1{2\pi}(\text{rect}*\text{rect})$, a triangle on $[-2W,2W]$ with peak $\dfrac{2W}{2\pi}=\dfrac W\pi$; multiplying by $\pi$ gives the triangle $T(\omega)=W\left(1-\dfrac{|\omega|}{2W}\right)$. Multiplying by $\cos10Wt$ shifts it to $\pm10W$ with gain $\tfrac12$: $H(\omega)=\tfrac12\left[T(\omega-10W)+T(\omega+10W)\right]$. At $\omega=10.5W$: $H=\tfrac12W\left(1-\dfrac{0.5W}{2W}\right)=\tfrac12W(0.75)=\dfrac{3W}8$. The output amplitude is $10\times\dfrac{3W}{8}=\dfrac{15W}{4}$.