Question

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

• A. $\left(\dfrac{15W}{4}\right)\cos(10.5Wt)$
• B. $\left(\dfrac{15W}{2}\right)\cos(10.5Wt)$
• C. $\left(\dfrac{15W}{8}\right)\cos(10.5Wt)$
• D. $(15W)\cos(10.5Wt)$

Hint:

$\big(\tfrac{\sin Wt}{\pi t}\big)^2$ has a {triangular} spectrum of base $4W$ and peak $W/\pi$ (with the $1/2\pi$-FT convention). Multiplying by $\cos\omega_c t$ shifts the spectrum to $\omega=\pm\omega_c$; then evaluate $H(\omega)$ at the input tone.

Answer 1

The Correct Option is A

Step 1: Recall Fourier transform pair
With definitions \(X(\omega) = \int x(t)e^{-j\omega t}dt,\; x(t) = \frac{1}{2\pi}\int X(\omega)e^{j\omega t} d\omega\), we know: \[ \frac{\sin(Wt)}{\pi t} \;\;\longleftrightarrow\;\; \mathrm{rect}\!\left(\frac{\omega}{2W}\right), \] which means the spectrum has unit gain for \(|\omega| < W\).

Step 2: Square in time domain
Now consider \[ \left(\frac{\sin(Wt)}{\pi t}\right)^{2}. \] By the convolution property in frequency domain: \[ \left(\frac{\sin Wt}{\pi t}\right)^{2} \;\longleftrightarrow\; \frac{1}{2\pi}\big(\mathrm{rect} * \mathrm{rect}\big)(\omega). \] The convolution of two rectangular functions gives a triangular function: \[ \frac{1}{2\pi}(2W - |\omega|), \quad |\omega| \le 2W,\quad 0 \text{ otherwise}. \]

Step 3: Multiply by factor \(\pi\) in time
Multiplying the time function by \(\pi\) scales the frequency spectrum by the same factor \(\pi\): \[ H_0(\omega) = \frac{\pi}{2\pi}(2W - |\omega|) = W - \frac{|\omega|}{2}, \quad |\omega| \le 2W. \] Outside this range, \(H_0(\omega)=0\).

Step 4: Account for cosine modulation
The given impulse response has an additional \(\cos(10Wt)\) multiplier. In frequency domain this produces shifts: \[ H(\omega) = \tfrac{1}{2}\Big[ H_0(\omega - 10W) + H_0(\omega + 10W)\Big]. \]

Step 5: Evaluate response at input frequency
Input tone frequency: \(\omega_0 = 10.5W\). \[ H_0(\omega_0 - 10W) = H_0(0.5W) = W - \tfrac{0.5W}{2} = \tfrac{3W}{4}. \] \[ H_0(\omega_0 + 10W) = H_0(20.5W) = 0 \quad (\text{outside support}). \] So, \[ H(\omega_0) = \tfrac{1}{2}\cdot \tfrac{3W}{4} = \tfrac{3W}{8}. \]

Step 6: Form the output signal
For input \(x(t) = 10\cos(\omega_0 t)\), the output is amplitude-scaled: \[ y(t) = 10 \cdot H(\omega_0) \cos(\omega_0 t) = 10 \cdot \tfrac{3W}{8} \cos(10.5Wt). \] Simplify: \[ y(t) = \left(\tfrac{15W}{4}\right)\cos(10.5Wt). \]

Final Answer:
\[ \boxed{y(t) = \left(\tfrac{15W}{4}\right)\cos(10.5Wt)} \]