---
title: "PSDF of an arbitrary signal"
subject: "Communication Systems"
chapter: 1
key: "semester_6/communication_systems:psdf-arbitrary-signal"
source: https://www.bctengineer.com/q/semester_6/communication_systems/psdf-arbitrary-signal
site: BCT Engineering
---

# PSDF of an arbitrary signal

Communication Systems · Chapter 1: Introduction

- Asked in: 2073 Magh (CS I), 4+6 marks
- Video: [What is Power Spectral Density (PSD)?](https://www.youtube.com/watch?v=DoSLMEEo1Y0) — Iain Explains Signals, Systems, and Digital Comms
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/psdf-arbitrary-signal
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-1

## Question

What is power spectral density function (PSDF)? Derive an expression for PSDF of an arbitrary signal X(t).

## Short answer

Power spectral density gives distribution of average power over frequency, with area equal to average power. Truncate the signal to x_T(t), compute energy via Rayleigh, divide by T, and take T to infinity to obtain S_x(f)=lim |X_T(f)|^2/T. It is the Fourier transform of autocorrelation.

## Answer

#### Power spectral density function (PSDF)
The **power spectral density** $S_x(f)$ of a power signal describes how its average power is distributed over frequency. Its unit is W/Hz, and the area under it equals the average power:

$$
P = \lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt = \int_{-\infty}^{\infty} S_x(f)\,df
$$

Properties: $S_x(f) \ge 0$; even in $f$ for real signals; $S_x(f)$ is the Fourier transform of the autocorrelation $R_x(\tau)$ (Wiener–Khinchine theorem); for an LTI system $S_y(f) = |H(f)|^2S_x(f)$.

#### Derivation of PSDF for an arbitrary signal $x(t)$
A power signal (e.g. a non-periodic random-like waveform) has infinite energy, so its Fourier transform may not exist. We therefore use a **truncated** version:

$$
x_T(t) = \begin{cases} x(t), & -T/2 \le t \le T/2 \\ 0, & \text{elsewhere} \end{cases}
$$

$x_T(t)$ has finite energy, so it has a Fourier transform $X_T(f) = \int_{-T/2}^{T/2} x(t)e^{-j2\pi ft}dt$.

**Step 1: energy of the truncated signal** (Rayleigh's theorem):

$$
E_T = \int_{-\infty}^{\infty}|x_T(t)|^2dt = \int_{-T/2}^{T/2}|x(t)|^2dt = \int_{-\infty}^{\infty}|X_T(f)|^2df
$$

**Step 2: average power over the interval $T$:**

$$
P_T = \frac{E_T}{T} = \frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt = \int_{-\infty}^{\infty}\frac{|X_T(f)|^2}{T}\,df
$$

**Step 3: let $T \to \infty$.** The left side becomes the average power of $x(t)$:

$$
P = \lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt = \int_{-\infty}^{\infty}\left[\lim_{T\to\infty}\frac{|X_T(f)|^2}{T}\right]df
$$

**Step 4: identify the PSD.** Since $P = \int S_x(f)\,df$, the integrand is the power spectral density:

$$
S_x(f) = \lim_{T\to\infty}\frac{|X_T(f)|^2}{T}
$$

The quantity $|X_T(f)|^2/T$ is called the periodogram; its limit is the PSD.

**Special case, periodic signal:** with Fourier coefficients $c_n$ and fundamental $f_0$, the PSD is a set of impulses,

$$
S_x(f) = \sum_{n=-\infty}^{\infty}|c_n|^2\,\delta(f - nf_0)
$$

**Example:** for $x(t) = A\cos 2\pi f_0 t$, $c_{\pm1} = A/2$, so $S_x(f) = \frac{A^2}{4}[\delta(f-f_0) + \delta(f+f_0)]$ and $P = A^2/4 + A^2/4 = A^2/2$, as expected.

**Relation to autocorrelation:** it can also be shown that $S_x(f) = \int R_x(\tau)e^{-j2\pi f\tau}d\tau$ where $R_x(\tau) = \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)x^*(t-\tau)dt$.

## Check yourself

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

1. The PSD S_x(f) of a power signal gives:
   A. How average power is distributed over frequency, in W/Hz
   B. The phase of each component
   C. The bandwidth only
   D. Total energy in joules
   Answer: A. The area under S_x(f) equals average power. (Key point: PSD meaning)

2. Why is the signal truncated to x_T(t) in the derivation?
   A. To make it periodic
   B. To remove noise
   C. To lower power
   D. A power signal has infinite energy so its Fourier transform may not exist; the truncated one has finite energy
   Answer: D. x_T has finite energy and therefore a Fourier transform. (Key point: Why truncate)

3. The derived PSD expression is:
   A. S_x(f) = lim |X_T(f)|^2 / T
   B. S_x(f) = lim T |X_T(f)|
   C. S_x(f) = lim X_T(f)/T
   D. S_x(f) = |X_T(f)|^2 T
   Answer: A. Dividing Rayleigh's energy by T and letting T go to infinity gives the periodogram limit. (Key point: Result)

4. For x(t) = A cos 2 pi f0 t, the total power from the PSD is:
   A. A/2
   B. A^2/4
   C. A^2
   D. A^2/2
   Answer: D. Two impulses of A^2/4 each sum to A^2/2. (Key point: Periodic example)

## Close variants set in other exams

- [ESD vs PSD; PSD of an arbitrary signal](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aesd-vs-psd-arbitrary-signal) — Communication Systems, chapter 1, asked 2×
- [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×

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