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Is the Nyquist Sampling Rate Satisfied? (Homework Problem)
March 2, 2022

Table of Contents

Introduction

In this blog I answer a question I received about how to apply Nyquist’s sampling rate mathematically and thought it’s worth sharing. One of the difficult parts about getting started in DSP is the concepts are not intuitive which is further compounded by the requirement that the early work must done mathematically, rather than by building or simulating. Take heart if you feel this way, you are not alone! I hope this blog is useful in helping to understand the mathematics of DSP. Please leave a comment below with other questions you have.

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Homework Question

A continuous-time signal x_{0}(t) = \sin\left( 2\pi f_{0} t) is modulated by a second continuous-time signal x_{1}(t) = \sin\left( 2\pi f_{1} t\right) such that

(1)   \begin{equation*}y(t) = x_{0}(t) \cdot x_{1}(t).\end{equation*}

The resulting signal y(t) is sampled with period T_s.

Questions:

For f_0 = 5000 Hz, f_1 = 1000 Hz, and T_s = 0.00025 seconds,

  1. Write y(t) as the sum of two sines or cosines
  2. Sample the continuous-time signal y(t) to produce the discrete-time signal y[n]
  3. What is the Nyquist sampling rate needed to sample y(t)?
  4. What is the sampling frequency being applied to y(t)?
  5. Is the signal y(t) sampled below or above the Nyquist sampling rate?

Question 1: Modulated Sinusoids

A trigonometric identity can be used on the modulated, or multiplied, sinusoids to represent them as the sum of two cosines:

(2)   \begin{equation*}\text{sin}\left( \alpha \right) \text{sin} \left( \beta \right) = \frac{1}{2} \left( \text{cos}\left( \alpha - \beta) - \text{cos}\left( \alpha + \beta \right) \right).\end{equation*}

Substituting \alpha = 2\pi f_{0} t and \beta = 2\pi f_{1} t the signal y(t) can be written as

(3)   \begin{equation*}\begin{split}y(t) & = \frac{1}{2} \left( \text{cos} \left( 2\pi f_{0}t - 2\pi_f{1}t \right) - \text{cos} \left( 2\pi f_{0}t + 2\pi f_{1}t \right) \right) \\& = \frac{1}{2} \left( \text{cos} \left( 2\pi \left( f_{0} - f_{1} \right) t \right) - \text{cos} \left( 2\pi \left( f_{0} + f_{1} \right) t \right) \right)\end{split}\end{equation*}

The multiplication of the two sines of frequency f_{0} and f_{1} resulted in the addition of two cosines, with frequency f_{0} - f_{1} and f_{0} + f_{1}.

Question 2. Sampling a Continuous-Time Signal

The continuous-time t is converted into discrete-time by n T_s where n = 0, 1, 2, 3, \dots~. The quantity n T_s therefore represents the n^{th} sampling instance at time t = 0, T_s, 2T_s, 3T_s, \dots ~.

Discrete-time is represented by the time index n such that the discrete-time signal is y[n]. The continuous-time signal y(t) is sampled by indexing with n T_s

(4)   \begin{equation*}\begin{split}y[n] & = y\left( n T_s \right) \\& = \frac{1}{2} \left[ \text{cos} \left( 2\pi \left( f_{0} - f_{1} \right) t \right) - \text{cos} \left( 2\pi \left( f_{0} + f_{1} \right) t \right) \right] \\& = \frac{1}{2} \left[ \text{cos} \left( 2\pi \left( f_{0} - f_{1} \right) n T_s \right) - \text{cos} \left( 2\pi \left( f_{0} + f_{1} \right) n T_s \right] \right)\end{split}\end{equation*}

Question 3: Finding the Nyquist Sampling Rate

Nyquist’s theorem states that to avoid distortion the sampling frequency f_{Nyq} must be greater than twice the largest frequency f_{max} of the signal to be sampled, f_{Nyq} > 2 f_{max}. The sampled signal (4) has two frequencies,

(5)   \begin{equation*}f_{0} - f_{1} = 5000 - 1000 = 4000~\text{Hz}\end{equation*}

(6)   \begin{equation*}f_{0} + f_{1} = 5000 + 1000 = 6000~\text{Hz}.\end{equation*}

The larger of the frequencies is f_{max} = 6000 and the minimum sampling frequency f_{Nyq} needed to sample the signal without distortion is therefore

(7)   \begin{equation*}f_{Nyq} > 2 f_{max} = 2 \cdot 6000 = 12,000~\text{Hz}.\end{equation*}

Question 4: Calculating Sampling Frequency

Any time period and frequency are inverses of one another,

(8)   \begin{equation*}T = \frac{1}{f}\end{equation*}

and equivalently,

(9)   \begin{equation*}f = \frac{1}{T}.\end{equation*}

The sampling period T_s is related to the sampling frequency

(10)   \begin{equation*}f_s = \frac{1}{T_s}\end{equation*}

therefore the sampling frequency is

(11)   \begin{equation*}f_s = \frac{1}{0.00025~\text{seconds}} = 4000~\text{Hz}.\end{equation*}

Question 5: Is Nyquist's Sampling Theorem Satisfied?

The sampling rate f_{Nyq} > 12,000 Hz is the minimum sampling rate needed to sample y(t) without distortion. However, the signal has been sampled by frequency f_s = 4000 Hz. Since f_s < f_{Nyq}, the signal is undersampled and Nyquist’s sampling theorem is not satisfied.

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