---
title: "Parseval's theorem; output PSD of LTI system"
subject: "Communication Systems"
chapter: 1
key: "semester_6/communication_systems:parseval-power-lti-psd"
source: https://www.bctengineer.com/q/semester_6/communication_systems/parseval-power-lti-psd
site: BCT Engineering
---

# Parseval's theorem; output PSD of LTI system

Communication Systems · Chapter 1: Introduction

- Asked in: 2077 Chaitra (CS I), 3+5 marks
- Video: [Relation Between Input Energy Spectral Density & Output Energy Spectral Density](https://www.youtube.com/watch?v=BuhpOnfv9Bk) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/parseval-power-lti-psd
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-1

## Question

Define Parseval's theorem for power signal. Prove that in a LTI system when a power signal x(t) is applied the PSD of its output is equal to the PSD of its input multiplied by the squared amplitude response of the system.

## Short answer

Parseval's theorem for power signals states average power equals sum of squared magnitude of Fourier series coefficients or integral of PSD. For an LTI system, output autocorrelation R_y(τ) derived, its Fourier transform yields S_y(f)=|H(f)|^2 S_x(f); output PSD equals input PSD times squared amplitude response.

## Answer

#### 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 $N_0B$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a periodic power signal with Fourier coefficients c_n, Parseval's theorem states the average power equals:
   A. The sum over n of |c_n|^2
   B. The sum of c_n
   C. |c_0|^2 only
   D. The integral of x(t) over one period
   Answer: A. P = (1/T0) integral |x|^2 dt = sum |c_n|^2: total power is the sum of harmonic powers. (Key point: Parseval for periodic power signal)

2. In the proof, the output autocorrelation is expressed as:
   A. A double integral of h(alpha) h(beta) R_x(tau - alpha + beta)
   B. The square of R_x(tau)
   C. R_x(tau) + h(tau)
   D. R_y(tau) = h(tau) R_x(tau)
   Answer: A. Substituting y = h*x gives R_y = double integral h(a)h(b)R_x(tau-a+b)da db. (Key point: Output autocorrelation)

3. Taking the Fourier transform of R_y and substituting lambda = tau - alpha + beta yields:
   A. S_y(f) = |H(f)| S_x(f)^2
   B. S_y(f) = H(f) S_x(f)
   C. S_y(f) = |H(f)|^2 S_x(f)
   D. S_y(f) = S_x(f) / |H(f)|
   Answer: C. The integrals produce H(f) H*(f) S_x(f) = |H(f)|^2 S_x(f). (Key point: Wiener-Khinchine and result)

4. What effect does the system's phase response have on the output PSD?
   A. It makes the PSD negative
   B. None, only the squared amplitude response matters
   C. It scales PSD by theta(f)
   D. It rotates the PSD
   Answer: B. H H* = |H|^2 removes phase, so phase does not affect PSD. (Key point: Role of phase response)

## Related in Communication Systems

- [Parseval's theorem; PSD–autocorrelation relation for white noise](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aparseval-psd-autocorrelation-white-noise) — Communication Systems, chapter 9
- [LTI output of a WSS input is WSS](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Alti-wssp-output) — Communication Systems, chapter 9
- [Random process; LTI output is WSS if input is WSS](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Arandom-process-lti-wssp) — Communication Systems, chapter 9
- [PSD of a power signal and its interpretation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apsdf-power-signal-interpretation) — Communication Systems, chapter 1, asked 2×
- [LTI system output in time domain](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Alti-output-time-domain) — Communication Systems, chapter 1
- [Periodic and non-periodic signals; LTI output is convolution](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aperiodic-nonperiodic-convolution) — Communication Systems, chapter 1

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