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

# Prove sampling theorem; Nyquist of product

Signals and Systems · Chapter 4: Sampling

- Asked in: 2082 Kartik, 3+3 marks
- Video: [Sampling Theorem | Proof | aliasing | Module 3 | S&S Lect 42](https://www.youtube.com/watch?v=EaO3p8qWHQs) — IMPLearn
- 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-nyquist
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

State and prove the sampling theorem. Find the Nyquist rate and Nyquist interval for the signal x(t) = (1/2π) cos(4000πt) cos(1000πt)

## Short answer

Sampling theorem: if X(jω)=0 for |ω|>ωm, samples at ωs ≥ 2ωm determine x(t); proof uses impulse-train sampling, repeated spectrum with spacing ωs, and ideal low-pass recovery when copies do not overlap. Convert product to (1/4π)[cos5000πt + cos3000πt]; fm = 2500 Hz. Nyquist rate = 5000 samples/s, interval = 0.2 ms.

## Answer

#### Sampling theorem

**Statement:** A 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 $\omega_s = \frac{2\pi}{T} \ge 2\omega_m$.

**Proof:** Model sampling as multiplication by an impulse train:

$$
x_p(t) = x(t)\,p(t), \qquad p(t) = \sum_{n=-\infty}^{\infty}\delta(t - nT)
$$

The FT of the periodic impulse train is $P(j\omega) = \frac{2\pi}{T}\sum_k\delta(\omega - k\omega_s)$. Multiplication in time is convolution in frequency:

$$
\begin{aligned}
X_p(j\omega) &= \frac{1}{2\pi}X(j\omega) * P(j\omega) \\
&= \frac{1}{T}\sum_{k=-\infty}^{\infty}X(j(\omega - k\omega_s))
\end{aligned}
$$

So the sampled spectrum is the original spectrum repeated every $\omega_s$. The copy at $k=0$ spans $-\omega_m$ to $\omega_m$, and the next starts at $\omega_s - \omega_m$. They do not overlap if $\omega_s - \omega_m \ge \omega_m$, i.e. $\omega_s \ge 2\omega_m$. Then an ideal low-pass filter with gain $T$ and cutoff $\omega_m \le \omega_c \le \omega_s - \omega_m$ passes only the $k=0$ copy, giving back $X(j\omega)$ and hence $x(t)$ exactly. If $\omega_s < 2\omega_m$, the copies overlap (aliasing) and recovery is impossible. Hence proved.

#### Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(4000\pi t)\cos(1000\pi t)$

$$
x(t) = \frac{1}{4\pi}\left[\cos(5000\pi t) + \cos(3000\pi t)\right]
$$

Frequencies: $\frac{5000\pi}{2\pi} = 2500$ Hz and $\frac{3000\pi}{2\pi} = 1500$ Hz, so $f_m = 2500$ Hz.

$$
f_N = 2f_m = 5000\ \text{Hz}, \qquad T_N = \frac{1}{5000} = 0.2\ \text{ms}
$$

**Answer:** Nyquist rate $= 5000$ samples/s (5 kHz); Nyquist interval $= 0.2$ ms.

## Check yourself

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

1. In the proof, sampling is modelled as
   A. Convolving x(t) with an impulse train
   B. Integrating x(t)
   C. Multiplying x(t) by an impulse train
   D. Adding an impulse train to x(t)
   Answer: C. x_p(t) = x(t) p(t). (Key point: Sampling model)

2. Multiplication in time becomes in frequency
   A. Multiplication, giving one copy of X
   B. Convolution, giving copies scaled by T
   C. Division by P(jω)
   D. Convolution, giving copies of X repeated every ωs scaled by 1/T
   Answer: D. X_p = (1/2π) X * P = (1/T) sum X(j(ω-kωs)). (Key point: Spectrum result)

3. Which filter recovers x(t) when ωs >= 2ωm?
   A. Ideal LPF with gain T and cutoff between ωm and ωs - ωm
   B. Ideal HPF with gain 1
   C. Ideal LPF with gain 1/T
   D. Ideal LPF with cutoff above ωs
   Answer: A. It passes only the k = 0 copy, restoring X(jω). (Key point: Recovery filter)

4. For x(t) = (1/2π)cos(4000πt)cos(1000πt), what is fm?
   A. 2000 Hz
   B. 2500 Hz
   C. 1500 Hz
   D. 4000 Hz
   Answer: B. Sum frequency 5000π gives 2500 Hz. (Key point: Product signal fm)

## Close variants set in other exams

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