I started studying DSP and the first things that came out were the differences between continuous and discrete time signals. So, I was wondering if I understood well these concepts before I keep going into more difficult topics.

The first thing I realized is that discrete-time signals replace the continuous parameter $t$ (time) for $n$ (sample) where $n$ is always an integer, leading this to two main differences:

Complex exponential sequences $Ae^{j\omega n}$ with frequencies $\omega + 2\pi k$ where $k$ is an integer are indistinguishable from one another since $e^{j2\pi n} = 1$ always. This is not true for continuous time signals since $e^{j2\pi t}$ can possibly be different from $1$.

Concerning periodicity, in the discrete time case, the fundamental period $N$ is always an integer equal to $\frac{k}{F}$ where $k$ is an integer, meaning that $N$ is not necessarily equal to $\frac{1}{F}$.

In the continuous case, this is not true since $T = \frac{1}{F}$.

So, my questions are:

  1. Am I correct about this two differences? If this is not the case, please explain it to me better.
  2. Are these two the more important to take into account, or am I missing something (else) big?

2 Answers 2


In general I'd say that you understand the differences pretty well.

  1. You nailed the first difference- continuous time vs. discrete time (samples).
  2. You were also correct about there being sets of indistinguishable frequencies in the discrete realm. A corollary of this is that all discrete signals are band-limited, whereas continuous signals do not have to be. This concept is most often referred to as the Nyquist frequency.
  3. I don't entirely agree with your point about periodicity, though I believe I understand what you're getting at. If by periodicity you mean having the exact same values repeated, then yes, a discrete period would have to be an integer number of samples. On the other hand, if you take periodic to mean that the "underlying" signal repeats, then it is possible to have the samples not be the exact same if the period is not an integer multiple of samples. You could still detect the periodicity with an autocorrelation though. Anyway, if this point is confusing, don't worry about it- it's not terribly important.

It is, by the way, possible to view discrete signals as a special case of continuous signals. You can model discrete signals as continuous signals that have been multiplied by an infinite set of evenly spaced delta functions. If the delta functions are separated by $T_s$ seconds, then in the frequency domain they can be represented by another set of delta functions that are separated by $\frac{1}{T_s}$ Hz. The multiplication in the time domain is equivalent to convolution in the frequency domain, so copies of the signal are reproduced in the frequency domain at each of the delta functions.

Though the mathematicians in the audience will probably scream at me for saying this, I think that you can likewise look at continuous signals as a special case of discrete signals, where we take the limit of $T_s$ as it approaches zero.


Jim Clay's answer is a good one, but I thought I'd expand a little on point 3 about periodicity.

A signal, $x\{s\}$, is said to be periodic with period $P$ if $$ x\{s\} = x\{s + P\} $$

For the continuous-time case, $s = t \in \mathbb{R} $ (real numbers) and we get: $$ x(t) = x(t + P) $$ with $P \in \mathbb{R}$.

For the discrete-time case, $s = n \in \mathbb{Z}$ (integers) and we get $$ x[n] = x[n + P]. $$ Because $x[t]$ with $t\in \mathbb{R}$ makes no sense for $t$ not also an integer, we must have the condition $P \in \mathbb{Z}$ here.

I think that's what you mean when you say

$\ldots N$ is not necessarily equal to $\frac{1}{f}$

because, in the discrete-time case "periodicity" is a much more strict condition (periodic with a period that is an integer number of samples long).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.