GATE EC 2014 Set 3 — Question 60

The power spectral density (PSD) of the random process $X(t)$ is given by:$$S_X(f) = \begin{cases} 10^{-6}(3000-|f|) & \text{for } |f| \leq 3 \text{ kHz} \\ 0 & \text{otherwise} \end{cases}$$This is a triangular PSD centered at $f=0$. The peak value at $f=0$ is $10^{-6}(3000-0) = 3 \times 10^{-3}$ Watts/Hz. The base extends from -3 kHz to 3 kHz.The signal $X(t)$ modulates a carrier $\cos(16000 \pi t)$. The carrier frequency is $f_c = \frac{16000 \pi}{2 \pi} = 8000 \text{ Hz} = 8 \text{ kHz}$.
Assuming DSB-SC modulation, the PSD of the modulated signal $Y(t)$ is given by:$$S_Y(f) = \frac{1}{4} [S_X(f-f_c) + S_X(f+f_c)]$$$S_X(f-f_c)$ is a triangular PSD centered at $f_c = 8 \text{ kHz}$, extending from $8-3=5 \text{ kHz}$ to $8+3=11 \text{ kHz}$.
$S_X(f+f_c)$ is a triangular PSD centered at $-f_c = -8 \text{ kHz}$, extending from $-8-3=-11 \text{ kHz}$ to $-8+3=-5 \text{ kHz}$.The resultant signal is passed through an ideal band-pass filter with a centre frequency of 8 kHz and a bandwidth of 2 kHz.
This means the filter passes frequencies in the range $[f_c - BW/2, f_c + BW/2]$ and $[-f_c - BW/2, -f_c + BW/2]$.
Passband for positive frequencies: $[8 \text{ kHz} - 2/2 \text{ kHz}, 8 \text{ kHz} + 2/2 \text{ kHz}] = [7 \text{ kHz}, 9 \text{ kHz}]$.
Passband for negative frequencies: $[-9 \text{ kHz}, -7 \text{ kHz}]$.The output power $P_{out}$ is the integral of $S_Y(f)$ over the filter's passband:$$P_{out} = \int_{-9000}^{-7000} S_Y(f) df + \int_{7000}^{9000} S_Y(f) df$$Substitute $S_Y(f)$:$$P_{out} = \int_{-9000}^{-7000} \frac{1}{4} [S_X(f-f_c) + S_X(f+f_c)] df + \int_{7000}^{9000} \frac{1}{4} [S_X(f-f_c) + S_X(f+f_c)] df$$Consider the positive frequency integral:$$\int_{7000}^{9000} \frac{1}{4} S_X(f-f_c) df + \int_{7000}^{9000} \frac{1}{4} S_X(f+f_c) df$$The second term is zero because $S_X(f+f_c)$ is centered at -8 kHz and has no overlap with $[7 \text{ kHz}, 9 \text{ kHz}]$.
So, the positive frequency contribution is $\frac{1}{4} \int_{7000}^{9000} S_X(f-f_c) df$.
Let $u = f-f_c = f-8000$. When $f=7000, u=-1000$. When $f=9000, u=1000$.
This integral becomes $\frac{1}{4} \int_{-1000}^{1000} S_X(u) du$.Consider the negative frequency integral:$$\int_{-9000}^{-7000} \frac{1}{4} S_X(f-f_c) df + \int_{-9000}^{-7000} \frac{1}{4} S_X(f+f_c) df$$The first term is zero because $S_X(f-f_c)$ is centered at 8 kHz and has no overlap with $[-9 \text{ kHz}, -7 \text{ kHz}]$.
So, the negative frequency contribution is $\frac{1}{4} \int_{-9000}^{-7000} S_X(f+f_c) df$.
Let $v = f+f_c = f+8000$. When $f=-9000, v=-1000$. When $f=-7000, v=1000$.
This integral becomes $\frac{1}{4} \int_{-1000}^{1000} S_X(v) dv$.Combining these, the total output power is:$$P_{out} = \frac{1}{4} \int_{-1000}^{1000} S_X(u) du + \frac{1}{4} \int_{-1000}^{1000} S_X(v) dv = \frac{1}{2} \int_{-1000}^{1000} S_X(f) df$$Now, we need to calculate the integral of $S_X(f)$ from -1 kHz to 1 kHz.
Since $S_X(f)$ is an even function, we can write:$$\int_{-1000}^{1000} S_X(f) df = 2 \int_{0}^{1000} S_X(f) df$$For $f \in [0, 1000]$, $S_X(f) = 10^{-6}(3000-f)$.$$\int_{0}^{1000} 10^{-6}(3000-f) df = 10^{-6} \left[3000f - \frac{f^2}{2}\right]_{0}^{1000}$$$$= 10^{-6} \left[(3000 \times 1000) - \frac{1000^2}{2}\right] - 0$$$$= 10^{-6} [3 \times 10^6 - 0.5 \times 10^6] = 10^{-6} [2.5 \times 10^6] = 2.5$$So, $\int_{-1000}^{1000} S_X(f) df = 2 \times 2.5 = 5$.Finally, the output power is:$$P_{out} = \frac{1}{2} \times 5 = 2.5 \text{ Watts}$$The final answer is $\boxed{2.5}$.