We know that the magnitude spectrum of a continuous time fourier series representation of a real periodic signal is an even function (i.e. symmetric about y axis). Does this hold true for discrete time periodic signals as well ? Can anyone give me references as well for this.

I found two questions where the above does not hold true.

  1. x[n] = {0,1,2,3,0,1,2,3,...}.

  2. Compute the DTFSC of a periodic discrete signal that repeats with period = 4 and has two impulses of amplitude 2 and 1,as shown in Fig.3.24(a).graphical representation of the signal

I know the function hear appears as an even function , but if you look at the period from -3 to 3 , it does not look so.

please excuse me , if i am asking a very silly question . I am a complete noob here . I am learning this stuff on my own. I dont have anyone to clarify my doubts . that is the reason why i am asking.

  • 2
    $\begingroup$ If you know the proof for the continuous Fourier series, why don't you just adapt it to the discrete-time case? It's absolutely basic. $\endgroup$
    – Matt L.
    Apr 15 at 7:59
  • $\begingroup$ Write down the definition for the DFT, assume a real signal, replace $n$ by $-n$, see what happens. $\endgroup$
    – Hilmar
    Apr 15 at 12:12
  • $\begingroup$ I don't see how this property doesn't hold in your examples. Please show us your computation of the Fourier coefficients. $\endgroup$
    – Matt L.
    Apr 15 at 13:44
  • $\begingroup$ When solving for Continuous time fourier series of a real periodic signal , we get the coefficients in the form of positive and negative complex exponentials.But when we solve for Discrete time fourier series of a discrete time periodic signal ,(in some examples in textbooks ,I have seen that ) we get either only positive complex exponential coefficients or both positive and negative complex exponential coefficients. $\endgroup$
    – amoghfyi
    Apr 15 at 13:58
  • $\begingroup$ @MattL.thank you for answering. But if you see in fig(c). they have shaded the portion where k = 2 is not part of the fundamental coefficients. $\endgroup$
    – amoghfyi
    Apr 15 at 14:01

1 Answer 1


Just to get this out of the way:

The DFT of a time sequence $x[n]$ is defined as

$$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j2\pi\frac{kn}{N}} \tag{1}$$

For the negative frequencies we get

$$X[-k] = \sum_{n=0}^{N-1}x[n]e^{j2\pi\frac{kn}{N}} \tag{2}$$

If $x[n]$ is real, than equation 2 is simply the complex conjugate if equation one and we get

$$X[-k] = X^*[k], x[n] \in \mathbb{R} \tag{3}$$

and from that follow directly that

$$|X[-k]| = |X[k]|, \angle X[-k] = - \angle X[k] \tag{4}$$

So the magnitude is even symmetric and the phase is odd symmetric. I think the confusion comes from the definition of "fundamental period". A DFT of length 4 has a period of 4. I.e. the fundamental period can NOT be even magnitude symmetric unless $X[2] = 0$.

However the DFT is a periodic function and the whole function definitely has even magnitude symmetry. In other words there is "circular symmetry".

The fact that there's a value at $k=−2$ but not at $k=2$ in figure (b) is a printing error;

I don't think so. The fundamental period has by definition only 4 values, so can either pick $k=2$ or $k=-2$ but not both. Otherwise your period is 5 samples long.

  • $\begingroup$ I think you're right about their intention of showing just one period. I just found it strange that they show values of $k$ outside that period without the corresponding Fourier coefficients, because by definition the Fourier coefficients are periodic too. $\endgroup$
    – Matt L.
    Apr 15 at 16:09
  • $\begingroup$ I love that you're using the notation "$X[-k]$" instead of saying "$X[N-k]$". $\endgroup$ Apr 15 at 21:33
  • $\begingroup$ Thanks for answering ;but still why this coefficients do not come in nicely symmetric form as in Continuous time fourier series ,I mean if you see in the above example .if you calculate coefficients over any interval of 4 samples , you get the above fourier coefficients which do not appear symmetric about y axis . why is that so ? sorry if this question is silly . I am just curious. $\endgroup$
    – amoghfyi
    Apr 17 at 20:10
  • $\begingroup$ The coefficients ARE symmetric, Why do you think they are not? We have $X[-1]=X[+1]$, $X[-7]=X[+7]$,$X[-2]=X[+2]$, etc. That's the very definition of symmetry. The "fundamental period" isn't symmetric simply because an even length sequence CAN'T be symmetric. But that's irrelevant: the fundamental period (however you choose to define it) is just a part of the whole thing and the whole thing definitely IS symmetric. $\endgroup$
    – Hilmar
    Apr 18 at 23:43
  • $\begingroup$ @amoghfyi did Hilmar answer your question? If so, please select it so we can close this out, or comment on what is still confusing you. Thanks! $\endgroup$ Apr 20 at 20:00

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