Question

Let \(X(t)\) be a white Gaussian noise with power spectral density \(\dfrac{1}{2}\) 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:

For white input with two-sided PSD \(S_X(f)=N_0/2\), the output power is \(\displaystyle \int S_X(f)|H(j2\pi f)|^2 df\). For \(h(t)=e^{-t}u(t)\), the integral collapses to \(\frac{1}{2}.\frac{1}{2}=0.25\).

Answer 1

Step 1: System frequency response.
For \(h(t)=e^{-t}u(t)\), \[ H(j\omega)=\int_{0}^{\infty} e^{-t}e^{-j\omega t}\,dt=\frac{1}{1+j\omega},\qquad |H(j\omega)|^{2}=\frac{1}{1+\omega^{2}}. \] Step 2: Output power using PSD.
With two-sided PSD \(S_X(f)=\tfrac{1}{2}\) W/Hz (white), the output average power is \[ P_Y=\int_{-\infty}^{\infty} S_X(f)\,|H(j2\pi f)|^{2}\,df = \frac{1}{2}\int_{-\infty}^{\infty}\frac{df}{1+(2\pi f)^2}. \] Use \(\displaystyle \int_{-\infty}^{\infty}\frac{df}{1+(af)^2}=\frac{\pi}{a}\) with \(a=2\pi\): \[ P_Y=\frac{1}{2}. \frac{\pi}{2\pi}=\frac{1}{4}=0.25\ \text{W}. \] \[ \boxed{0.25} \]