---
title: "Minimum sampling rate (sampling theorem)"
subject: "Digital Control System"
chapter: 1
key: "semester_6/digital_control_system:minimum-sampling-rate"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/minimum-sampling-rate
site: BCT Engineering
---

# Minimum sampling rate (sampling theorem)

Digital Control System · Chapter 1: Introduction to Discrete Time Control System

- Asked in: 2073 Bhadra, 6 marks
- Video: [SAMPLING THEOREM in digital communication - sampling rate and Nyquist rate](https://www.youtube.com/watch?v=wgS5OglyB80) — Electronics Subjectified
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/minimum-sampling-rate
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-1

## Question

Show that the minimum sampling rate required to recover a signal whose band spectrum is limited to ωₘ is equal to 2ωₘ.

## Short answer

For a signal band-limited to ωₘ, sampling creates spectral replicas at multiples of ωₛ. To avoid overlap (aliasing) the first replica must not overlap the baseband: ωₛ – ωₘ ≥ ωₘ ⇒ ωₛ ≥ 2ωₘ. The minimum sampling rate is thus the Nyquist rate 2ωₘ.

## Answer

**Statement (sampling theorem):** if $x(t)$ has no frequency components above $\omega_m$, it can be recovered completely from its samples only if the sampling frequency satisfies $\omega_s \ge 2\omega_m$ (strictly, $\omega_s > 2\omega_m$). The minimum rate $2\omega_m$ is the **Nyquist rate**.

#### Proof

The ideal sampler output is

$$
x^*(t) = x(t)\sum_{k=-\infty}^{\infty}\delta(t-kT)
$$

The impulse train is periodic with period $T$, so expand it as a Fourier series ($\omega_s = 2\pi/T$):

$$
\sum_{k}\delta(t-kT) = \frac{1}{T}\sum_{n=-\infty}^{\infty} e^{jn\omega_s t}
$$

Then

$$
x^*(t) = \frac{1}{T}\sum_{n=-\infty}^{\infty} x(t)\,e^{jn\omega_s t}
$$

Using the frequency-shift property $\mathcal{F}[x(t)e^{jn\omega_s t}] = X(j(\omega - n\omega_s))$:

$$
X^*(j\omega) = \frac{1}{T}\sum_{n=-\infty}^{\infty} X(j(\omega + n\omega_s))
$$

So the spectrum of the sampled signal is the original spectrum (scaled by $1/T$) repeated at every multiple of $\omega_s$.

#### Condition for no overlap

```text
 ws > 2wm: no overlap
   /\        /\        /\
 -/--\------/--\------/--\--> w
 -ws   -wm  0  wm   ws
          |<-LPF->|

 ws < 2wm: overlap (aliasing)
   /\  /\  /\
 -/--\/--\/--\--> w
```

The primary band occupies $-\omega_m$ to $\omega_m$. The first copy occupies $\omega_s-\omega_m$ to $\omega_s+\omega_m$. They do not overlap if

$$
\omega_s - \omega_m \ge \omega_m \quad\Rightarrow\quad \omega_s \ge 2\omega_m
$$

Under this condition, an ideal low-pass filter with gain $T$ and cut-off $\omega_s/2$ removes all copies and returns $X(j\omega)$, i.e. $x(t)$ is recovered exactly. If $\omega_s < 2\omega_m$, the bands overlap (**aliasing**): high-frequency components appear as low frequencies, and no filter can separate them.

Hence the minimum sampling rate is $\omega_s = 2\omega_m$.

**Example:** a signal limited to 50 Hz must be sampled at more than 100 Hz. In practice control systems sample much faster (often 10 times the closed-loop bandwidth).

## Check yourself

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

1. The minimum sampling rate for a signal band-limited to wm is:
   A. wm
   B. 2 wm
   C. wm/2
   D. 4 wm
   Answer: B. This is the Nyquist rate. (Key point: Nyquist rate)

2. Multiplying x(t) by an impulse train repeats the spectrum:
   A. Only at ws
   B. Scaled by T at wm
   C. At every multiple of ws, scaled by 1/T
   D. Only at 2 wm
   Answer: C. Fourier series of the train gives X*(jw) = (1/T) sum X(j(w + n ws)). (Key point: Sampled spectrum)

3. The primary band and first copy do not overlap if:
   A. ws + wm >= wm
   B. ws <= wm
   C. ws - wm <= wm
   D. ws - wm >= wm
   Answer: D. This gives ws >= 2 wm. (Key point: No-overlap condition)

4. Recovery is by a low-pass filter with:
   A. Gain T and cut-off ws/2
   B. Gain 1/T and cut-off wm/2
   C. Gain T and cut-off 2 ws
   D. Gain 1 and cut-off 0
   Answer: A. It removes all copies and returns X(jw). (Key point: Recovery filter)

## Related in Digital Control System

- [Types of sampling operations compared](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Acompare-sampling-operations) — Digital Control System, chapter 1, asked 3×
- [Sampling operations and DCS advantages](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Asampling-operations-advantages) — Digital Control System, chapter 1
- [Z-transform of ramp-and-hold signal (T = 1 s)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Awaveform-ramp-to-1-at-4) — Digital Control System, chapter 2

## Related in other subjects

- [Band-limited signals; aliasing; Nyquist rate](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Abandlimited-aliasing-nyquist) — Signals and Systems, chapter 4
- [Aliasing in a band-limited signal](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aaliasing-bandlimited-w) — Signals and Systems, chapter 4
- [Why band-limit a signal before sampling](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Asampling-bandlimit) — Data Communication, chapter 5
- [TDM of 24 voice signals: PAM bandwidth and PCM sampling rate](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Atdm-24-voice-pam-pcm) — Communication Systems, chapter 5