Question

Let X(t) be a white Gaussian noise with power spectral density $\frac{3}{2}$ W/Hz. If X(t) is input to an LTI system with impulse response $e^{-t}u(t)$. The average power of the system output is ___________ W (rounded off to two decimal places).

Hint:

The average power of a signal can also be found by integrating the squared magnitude of its impulse response if the input is white noise with unity PSD. For a general white noise input with PSD $N_0/2$, the output power is $P_Y = (N_0/2) \int_{-\infty}^{\infty} |h(t)|^2 dt$. This avoids Fourier transforms. Here, $\int_0^\infty (e^{-t})^2 dt = 1/2$, so $P_Y = (3/2) \times (1/2) = 3/4 = 0.75$.

Answer 1

Correct Answer: 0.24

The average power of the output signal $Y(t)$ is related to its power spectral density (PSD), $S_Y(f)$, by the formula: $P_Y = \int_{-\infty}^{\infty} S_Y(f) df$.
The output PSD is related to the input PSD, $S_X(f)$, and the system's transfer function, $H(f)$, by: $S_Y(f) = S_X(f) |H(f)|^2$.
We are given the input PSD as a constant value, $S_X(f) = \frac{3}{2}$ W/Hz. This is the two-sided PSD, $N_0/2$.
The impulse response is $h(t) = e^{-t}u(t)$.
Step 1: Find the transfer function $H(\omega)$ or $H(f)$. Let's use angular frequency $\omega$. $H(\omega) = \mathcal{F}\{h(t)\} = \int_{-\infty}^{\infty} e^{-t}u(t) e^{-j\omega t} dt = \int_{0}^{\infty} e^{-(1+j\omega)t} dt = \frac{1}{1+j\omega}$.
$|H(\omega)|^2 = \frac{1}{|1+j\omega|^2} = \frac{1}{1^2 + \omega^2} = \frac{1}{1+\omega^2}$.
Step 2: Calculate the output power using integration in the $\omega$ domain. $P_Y = \frac{1}{2\pi} \int_{-\infty}^{\infty} S_Y(\omega) d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} S_X(\omega) |H(\omega)|^2 d\omega$.
Since $S_X(f) = 3/2$ W/Hz, the PSD in the $\omega$ domain is $S_X(\omega) = S_X(f) = 3/2$ for all $\omega$.
$P_Y = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{3}{2} \cdot \frac{1}{1+\omega^2} d\omega = \frac{3}{4\pi} \int_{-\infty}^{\infty} \frac{1}{1+\omega^2} d\omega$.
Step 3: Evaluate the integral. $\int_{-\infty}^{\infty} \frac{1}{1+\omega^2} d\omega = [\arctan(\omega)]_{-\infty}^{\infty} = \arctan(\infty) - \arctan(-\infty) = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.
Step 4: Calculate the final power. $P_Y = \frac{3}{4\pi} \times \pi = \frac{3}{4} = 0.75$ W.
The average power of the system output is 0.75 W.