---
title: "Prove sampling theorem; aliasing effect"
subject: "Signals and Systems"
chapter: 4
key: "semester_4/signals_and_systems:prove-sampling-theorem-aliasing"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/prove-sampling-theorem-aliasing
site: BCT Engineering
---

# Prove sampling theorem; aliasing effect

Signals and Systems · Chapter 4: Sampling

- Asked in: 2082 Bhadra (new course), 3+2 marks
- Video: [Sampling Theorem](https://www.youtube.com/watch?v=XyZerq7-IvA) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/prove-sampling-theorem-aliasing
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

State and prove the sampling theorem. What is aliasing effect? Explain.

## Short answer

Sampling theorem: a band-limited signal with max frequency ω_M can be exactly reconstructed from its samples if ω_s > 2ω_M, as impulse-train sampling produces non-overlapping spectral replicas. Aliasing is the overlapping when ω_s < 2ω_M, causing high frequencies to fold back (e.g., 700 Hz at 1 kHz appears as 300 Hz).

## Answer

#### Statement
A continuous-time signal $x(t)$ band-limited to $\omega_M$ ($X(j\omega) = 0$ for $|\omega| > \omega_M$) is uniquely determined by its samples $x(nT)$ if the sampling frequency $\omega_s = 2\pi/T > 2\omega_M$.

**Proof.** Let $x(t)$ be band-limited: $X(j\omega) = 0$ for $|\omega| > \omega_M$.

**Step 1 – sampling model.** Sample with the impulse train $p(t) = \sum_n \delta(t-nT)$:
$$
x_p(t) = x(t)p(t) = \sum_{n=-\infty}^{\infty} x(nT)\,\delta(t-nT)
$$
**Step 2 – transform of $p(t)$.** $p(t)$ is periodic with period $T$; its Fourier series coefficients are all $1/T$, so
$$
P(j\omega) = \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}\delta(\omega - k\omega_s), \qquad \omega_s = \frac{2\pi}{T}
$$
**Step 3 – multiplication property.**
$$
\begin{aligned}
X_p(j\omega) &= \frac{1}{2\pi}X(j\omega) * P(j\omega) \\
&= \frac{1}{T}\sum_{k=-\infty}^{\infty} X\big(j(\omega - k\omega_s)\big)
\end{aligned}
$$
**Step 4 – no-overlap condition.** The $k=0$ copy lies in $[-\omega_M, \omega_M]$ and the $k=1$ copy starts at $\omega_s - \omega_M$. They do not overlap if
$$
\omega_s - \omega_M > \omega_M \;\Rightarrow\; \omega_s > 2\omega_M
$$
**Step 5 – recovery.** Pass $x_p(t)$ through an ideal LPF $H(j\omega) = T$ for $|\omega| < \omega_c$, 0 otherwise, with $\omega_M < \omega_c < \omega_s - \omega_M$. Only the $k=0$ term survives:
$$
X_r(j\omega) = H(j\omega)X_p(j\omega) = T\cdot\frac{1}{T}X(j\omega) = X(j\omega)
$$
Hence $x_r(t) = x(t)$: the samples determine $x(t)$ uniquely. $\blacksquare$

#### Aliasing effect
If $\omega_s < 2\omega_M$, the $k=1$ copy in $X_p(j\omega)$ starts at $\omega_s - \omega_M$, which is less than $\omega_M$, so neighbouring copies **overlap**. In the overlap region the spectral values add, and a component at $\omega_0 > \omega_s/2$ appears at $\omega_s - \omega_0$. The LPF then outputs a distorted signal. This is aliasing.

Example: 700 Hz sampled at 1000 Hz gives the same samples as 300 Hz.

It is avoided by sampling above the Nyquist rate and using an anti-aliasing LPF before sampling.

## Check yourself

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

1. What condition makes the k=0 and k=1 copies not overlap?
   A. ωs < ωM
   B. ωs = 0
   C. ωs - ωM > ωM, i.e. ωs > 2ωM
   D. ωs - ωM < ωM
   Answer: C. The k=1 copy starts at ωs - ωM, which must exceed ωM. (Key point: Sampling condition)

2. Why does Xp(jω) = (1/2π) X(jω) * P(jω) become a sum of shifted copies?
   A. Convolving X with an impulse train places a copy of X at each impulse
   B. Convolution multiplies X by ωs
   C. Impulses cancel X
   D. P(jω) is constant
   Answer: A. Convolution with δ(ω - kωs) shifts X by kωs. (Key point: Multiplication property)

3. Which LPF recovers x(t)?
   A. Gain 1 with cutoff ωs
   B. Gain T with cutoff ωM < ωc < ωs - ωM
   C. Gain 1/T with any cutoff
   D. Gain T with cutoff below ωM
   Answer: B. Gain T cancels 1/T and the cutoff passes only the baseband copy. (Key point: Recovery)

4. In aliasing, where does a component ω0 > ωs/2 appear?
   A. ω0 - ωs/2
   B. ω0
   C. 2ωs
   D. ωs - ω0
   Answer: D. Overlap folds the component to ωs - ω0. (Key point: Aliasing)

## Related in Signals and Systems

- [Sampling theorem; aliasing cause; Nyquist](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampling-aliasing-cause-nyquist) — Signals and Systems, chapter 4
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- [Aliasing via impulse-train sampling](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aaliasing-impulse-train-frequency) — Signals and Systems, chapter 4
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