---
title: "Frequency of a signal sampled at 300 Hz"
subject: "Signals and Systems"
chapter: 4
key: "semester_4/signals_and_systems:sampled-signal-frequency-300"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/sampled-signal-frequency-300
site: BCT Engineering
---

# Frequency of a signal sampled at 300 Hz

Signals and Systems · Chapter 4: Sampling

- Asked in: 2072 Magh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/sampled-signal-frequency-300
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

The signal is sampled at the rate Fs = 300 samples/sec. Determine the frequency of the discrete-time signal x[n] = xₐ(nT), T = 1/Fs [The analog signal xₐ(t) is not given on the paper.]

## Short answer

Assuming the standard signal x_a(t)=3cos50πt+10sin300πt−cos100πt, analog frequencies are 25, 150, 50 Hz. With F_s=300 Hz, discrete frequencies f = F/F_s are 1/12, 1/2, 1/6 cycles/sample. The sine term vanishes (sin(πn)=0), so x[n]=3cos(πn/6)−cos(πn/3).

## Answer

The analog signal is not printed in the question. This answer assumes the standard textbook signal (Proakis) whose Nyquist rate is exactly 300 Hz:
$$
x_a(t) = 3\cos 50\pi t + 10\sin 300\pi t - \cos 100\pi t
$$
The same method works for any given $x_a(t)$.

#### Method
Sampling at $t = nT = n/F_s$ turns an analog frequency $F$ into a discrete-time frequency
$$
f = \frac{F}{F_s}\ \text{cycles/sample}, \qquad \omega = 2\pi\frac{F}{F_s}\ \text{rad/sample}
$$

#### Frequencies of each term

| Term | $F$ (Hz) | $f = F/300$ | $\omega = 2\pi f$ |
|---|---|---|---|
| $3\cos 50\pi t$ | $25$ | $1/12$ | $\pi/6$ |
| $10\sin 300\pi t$ | $150$ | $1/2$ | $\pi$ |
| $-\cos 100\pi t$ | $50$ | $1/6$ | $\pi/3$ |

#### Discrete-time signal
$$
\begin{aligned}
x[n] &= x_a\!\left(\frac{n}{300}\right) \\
&= 3\cos\frac{\pi n}{6} + 10\sin(\pi n) - \cos\frac{\pi n}{3}
\end{aligned}
$$
Since $\sin(\pi n) = 0$ for every integer $n$, the 150 Hz term gives all-zero samples:
$$
x[n] = 3\cos\frac{\pi n}{6} - \cos\frac{\pi n}{3}
$$
**Answer: the discrete-time frequencies are $f = 1/12,\ 1/2,\ 1/6$ cycles/sample ($\omega = \pi/6,\ \pi,\ \pi/3$ rad/sample).** The highest frequency is 150 Hz, so the Nyquist rate is 300 Hz. Sampling at exactly 300 Hz loses the sine term, which shows why $F_s$ must be strictly greater than $2F_{max}$.

## Check yourself

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

1. Analog frequency F maps to discrete frequency
   A. f = F Fs
   B. f = Fs/F
   C. f = F/2Fs
   D. f = F/Fs cycles/sample
   Answer: D. Put t = n/Fs. (Key point: Mapping)

2. The 150 Hz term at Fs = 300 Hz has ω =
   A. π rad/sample
   B. π/2
   C. 2π
   D. π/6
   Answer: A. 150/300 = 1/2, ω = 2π x 1/2 = π. (Key point: 150 Hz at 300 Hz)

3. 10sin(πn) is
   A. 10 for all n
   B. Zero for every integer n
   C. Alternating ±10
   D. Aliased to DC with amplitude 10
   Answer: B. sin(πn) = 0. (Key point: Lost term)

4. Sampling exactly at 2Fmax shows that
   A. Fs = 2Fmax is always safe
   B. Fs can be less than Fmax
   C. Fs must be strictly greater than 2Fmax
   D. Sine terms are always preserved
   Answer: C. A sine at Fs/2 gives all-zero samples. (Key point: Lesson)

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