Parseval's theorem; output PSD of LTI system

Parseval's theorem for power signals

For a periodic power signal $x(t)$ with period $T_0$ and complex Fourier series coefficients $c_n$, the average power computed in the time domain equals the sum of powers of its harmonics:

$$P = \frac{1}{T_0}\int_{T_0} |x(t)|^2\,dt = \sum_{n=-\infty}^{\infty} |c_n|^2$$

In general (any power signal), $P = \int_{-\infty}^{\infty} S_x(f)\,df$, where $S_x(f)$ is the power spectral density. So power can be found either in the time domain or in the frequency domain.

Proof: $S_y(f) = |H(f)|^2 S_x(f)$

Let a power signal $x(t)$ be applied to an LTI system with impulse response $h(t)$ (real) and transfer function $H(f)$. The output is

$$y(t) = \int_{-\infty}^{\infty} h(\alpha)\,x(t-\alpha)\,d\alpha$$

Step 1: autocorrelation of the output. With $R_x(\tau) = \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2} x(t)x(t-\tau)\,dt$,

$$\begin{aligned} R_y(\tau) &= \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2} y(t)\,y(t-\tau)\,dt \\ &= \int\!\!\int h(\alpha)h(\beta)\left[\lim_{T\to\infty}\frac1T\int x(t-\alpha)x(t-\tau-\beta)\,dt\right]d\alpha\,d\beta \\ &= \int\!\!\int h(\alpha)h(\beta)\,R_x(\tau - \alpha + \beta)\,d\alpha\,d\beta \end{aligned}$$

(the bracket is $R_x$ at the lag $(t-\alpha) - (t-\tau-\beta) = \tau - \alpha + \beta$).

Step 2: take the Fourier transform with respect to $\tau$. By the Wiener–Khinchine theorem $S_y(f) = \mathcal{F}\{R_y(\tau)\}$:

$$\begin{aligned} S_y(f) &= \int\!\!\int h(\alpha)h(\beta)\left[\int R_x(\tau-\alpha+\beta)e^{-j2\pi f\tau}d\tau\right]d\alpha\,d\beta \end{aligned}$$

Put $\lambda = \tau - \alpha + \beta$, so $\tau = \lambda + \alpha - \beta$:

$$\begin{aligned} S_y(f) &= \int h(\alpha)e^{-j2\pi f\alpha}d\alpha \int h(\beta)e^{+j2\pi f\beta}d\beta \int R_x(\lambda)e^{-j2\pi f\lambda}d\lambda \\ &= H(f)\,H^*(f)\,S_x(f) \end{aligned}$$

since for real $h(t)$, $\int h(\beta)e^{j2\pi f\beta}d\beta = H(-f) = H^*(f)$.

Result:

$$S_y(f) = |H(f)|^2\,S_x(f)$$

The output PSD equals the input PSD multiplied by the squared amplitude response; the phase response has no effect on the PSD. The output power is then $P_y = \int |H(f)|^2 S_x(f)\,df$.

Example: white noise of PSD $N_0/2$ through an ideal LPF of bandwidth $B$ and gain 1 gives $S_y = N_0/2$ for $|f| < B$ and output power $N_0B$.