I have just started DSP self-learning. I am a little confused by this 'end of the chapter' exercise question from Chapter 2 of "Understanding Digital signal processing" 3rd edition by Richard G. Lyons.

Consider a continuous time-domain sine wave defined by

$$x(t)=\cos(4000\pi t)$$

that was sampled to produce the discrete sine wave sequence defined by

$$x(n)=\cos(n \pi/2)$$

What is the sample rate ($f_s$ measured in $\textrm{Hz}$) that would result in the sequence $x(n)$?

There are a few things that I have not understood about this question:

  1. Why is the argument of the cosine function so different after sampling?

  2. I am not sure why the argument of the cosine after sampling does not include $t$ so that we can get $nT$ where $T$ is the sample time.

  3. What determined the argument of the cosine after sampling?

  4. Can I determine the frequency of the sampled signal from the argument of the cosine in the $x(n)$ equation?

  5. And of course a cheeky one, the answer to the book question please :)


1 Answer 1


The sequence $x[n]$ equals the continuous-time signal $x(t)$ sampled at $t=nT$, where $T=1/f_s$ is the sampling period:

$$x[n]=x(nT)=\cos(4000\pi nT)=\cos(4000\pi n/f_s)\tag{1}$$

So if $x[n]=\cos(n\pi/2)$ you just have to compare this argument to the argument in $(1)$ to figure out what $f_s$ is.

The normalized frequency of the discrete-time signal $x[n]$ is $\omega_0=\pi/2$ (in radians). It is related to the frequency $f$ in Hertz by

$$\omega_0=\frac{2\pi f}{f_s}$$

  • $\begingroup$ I got sampling rate = 8000Hz. There are other questions that need the book reader to understands the concept of nomalised frequency. The book makes it hard for a self learner and a DSP new comer by omitting such a fundamental concept of normalised frequency in my opinion. I hope nothing else is omitted. $\endgroup$
    – Chika
    Jun 29, 2016 at 5:20

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