Why do the same signals in the continuous-time and discrete-time differ in their frequency?

For example, if $x(t)$ is a continuous-time signal taking values only at $t= 1,2,3,4$ and $x(n)$ takes the same values as $x(t)$ at $n =1,2,3,4$.

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    $\begingroup$ A couple of points: First, the function x(t) will have values for the whole domain, not just the integer points, otherwise it can't be continuous. Second, in Mathematics, a function on the domain of the integers is known as a sequence and is denoted with a subscript $x_n$. The convention here seems to be to use brackets instead, like $ x[n] $. $\endgroup$ Jan 25, 2018 at 15:41
  • $\begingroup$ As @CedronDawg points out, if $x(t)$ takes values only at discrete time instants, then it's a discrete-time signal... $\endgroup$
    – MBaz
    Jan 25, 2018 at 15:52

2 Answers 2


I disagree with the comments on the question.

A continuous-time signal $x(t)$ is one for which the value of $x(t)$ is defined for all real numbers $t$ (or for all real numbers $t$ in some interval of the real line, e.g. $[0,\infty)$ or $[0,2\pi)$ or $(-3,+5)$, whatever). There is no requirement that the values of $x(t)$ be nonzero, or that $x(t)$ be continuous function of time (on the real line, or the specified interval as the case may be) or anything like that. Now, the question the OP is asking is about frequency content of $x(t)$ and for this, we need to put more restrictions on $x(t)$ so that $x(t)$ can enjoy having a Fourier transform, e.g. $x(t)$ has finite energy (that is, $x(t)$ is square-integrable). Now, the OP's $x(t)$ which has nonzero value only at $t=1,2,3,4$ and is $0$ otherwise is indeed a continuous-time function but it is a zero-energy signal and its Fourier transform $X(f)$ has value $0$ for all $f$. "But wait a minute," you say, "whatever happened to $$x(t) = \int_{-\infty}^\infty X(f) \exp(j2\pi ft) \,\mathrm df \implies x(1) = \int_{-\infty}^\infty X(f) \exp(j2\pi f) \,\mathrm df??$$ I know that $x(1)$ is nonzero but that integral has value $0$ since you say that $X(f)$ is $0$ for all $f$. Something is wrong!" Well, the Fourier integral is not guaranteed to converge to $x(t)$ for each $t$ but only to some $\tilde{x}(t)$ such that $x(t)-\tilde{x}(t)$ has zero energy; $\tilde{x}(t) = 0$ for all $t$ in this case, and the Fourier integral does give the correct value $(0)$ of $x(t)$ for all $t$ except $t = 1,2,3,4$.

Work in progress

Turning to the discrete-time signal $x[n]$ is obtained from $x(t)$ via $x[n] = x(nT)$ for some given $T$, its Discrete-Time Fourier Transform (DTFT)

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    $\begingroup$ Since we're quibbling, my comment makes sense whether you consider the continuous applying to the domain or the function. The OP said the function only took values at those specific points. Assuming that not having values in between implies a value of zero, is an assumption on the OP's intent, not his statement. I was trying to point out that his statement was poorly worded, and technically self-contradictory. I haven't a clue if the OP's intent about "the frequency" is what you are addressing. Also, notice I said "whole domain", not "for all reals". $\endgroup$ Jan 25, 2018 at 17:30
  • $\begingroup$ when progress is complete, i will remove down vote $\endgroup$
    – user28715
    Jan 29, 2018 at 14:20

A continuous time signal can be evaluated arbitrarily at any time,at least in theory, and can take on values at arbitrary points between any evaluation point. The Fourier transform, while not always well defined, can take on values at frequencies over $-\infty$ to $\infty$ and that value can be zero, and the signal in continuous time also takes on values over the infinite interval and those values can also be zero.

A discrete time signal only takes on values at discrete time intervals. It is only meaningful to consider values intermediate in time, intermediate to those samples, if for example the discrete time signal was sampled from a continuous signal, or if you have a reason (justification). In general, the values intermediate to the samples are not defined. You can consider the price of a stock or pork bellies on days where there are no trades. No one is buying or selling. The price isn’t the last traded price and it isn’t zero. It does have a Fourier Transform, computed as an infinite summation, and the frequency is uniquely (continuously defined) limited over $-\pi$ to $\pi$.

So there is only a subset of discrete time signals that we can meaningfully compare their Fourier Transforms, and those are signals that are Nyqvist Sampled. Band limiting and periodic sampling alter the original continuous time signal, which alters the frequency spectrum, which almost answers your question.

The question of whether something is meaningful at intermediate points between samples in both time and frequency is more subtle.

I can have a periodic discrete time sequence and the frequencies in the Fourier transform are likewise values that are discrete in frequency, but are the intermediate values in frequency zero?

The most used DSP text book is Oppenheim and Schaefer and the most recent edition has 3 interpretations of the DFT which I interpret as yes , no, and maybe.

I can also interpolate pork belly prices, but is it meaningful?

I think the answer lays with how you want to represent something.

If you want to describe a discrete time signal in the continuous time domain, a sequence of Dirac Delta functions is suitable where intermediate values are zero is OK. In other contexts, it may be meaningless.

Another example worth considering is a realization of a Poisson process. You have a set of "arrival times" that you can store on a computer, limited by the amount of memory and numerical precision available on the machine. It is a list of numbers and the intermediate values denote no "arrivals" at those intermediate times. The Poisson process is continuous time. One typically doesn't perform Fourier analysis on a Poisson Process. There is some discretization in the computer representation, so one can quibble that it isn't strictly continuous but it is a practical representation of a continuous signal. This is also an example of a continuous signal where Nyquist sampling is irrelevant.


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