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Homework Help: Matching Discrete Fourier Transform (DFT) Pairs

  1. Jun 26, 2017 #1
    1. The problem statement, all variables and given/known data

    I am trying to match each of the following 28-point discrete-time signals with its DFT:

    Set #1:


    Set #2:


    2. Relevant equations

    3. The attempt at a solution

    Set #1
    We have already established (here) that:

    ##Signal 1 \leftrightarrow DFT3##
    ##Signal 4 \leftrightarrow DFT2##

    Now, Signal 3 looks like Signal 4. The only difference is that the temporal sample spacing has been increased. So, instead of having a single central peak in the DFT, we now have two peaks (at 7 and 21). Why?

    Likewise, Signal 2 looks like Signal 1, except it was sampled at twice the sampling interval. So, why does increasing the sample spacing create an additional frequency peak in each case? :confused:

    Set #2:
    These signals look like rectangular pulses (but none of them are a full period of a periodic rectangular signal). The DFT of a rectangular pulse is samples of the sinc function. DFT1 & 3 look like sincs, but I am not sure how to interpret DFT2.

    Clearly, each signal has a different average value. For instance, Signal 3 should have the highest DC value because it has more samples at 1 than the other two signals. But the axes of the DFTs are not labeled. So, how else can I match these?

    Any suggestions would be greatly appreciated.
  2. jcsd
  3. Jun 26, 2017 #2


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    Staff: Mentor

    Is there really an additional frequency? Think about it. What is special about the sampling rate for signals 1 and 4?

    Going back to the previous thread, think about how these signals fit in with respect to signals 1 and 6 in that problem.
  4. Jun 26, 2017 #3
    In direct space, the period is 2. For instance, we can see that in signal 4, it is the largest frequency to represent a cosine with. Signal 3 looks like another cosine oscillating at a slower frequency. Where does the additional frequency come from?

    Could you please explain more? I don't see how it relates to this problem.

    Signal 1 & 6 in that problem show that a constant function corresponds to a Dirac-##\delta## spectrum, and conversely a ##\delta## impulse corresponds to a constant.

    I also know this relationship between sample spacing and span in each domain:

    & \text{Time} & \text{Freq}\\
    \hline \text{Spacing} & \Delta T & 1/N\Delta T\\
    \text{Span} & N\Delta T & 1/\Delta T
  5. Jun 26, 2017 #4


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    Staff: Mentor

    I'll ask again: are there really two frequencies? (Think about the ordering of the DFT, as per the previous thread.)

    What is the relation between the length of a signal in time and the width of its frequency spectrum?
  6. Jun 26, 2017 #5
    Thank you. So the highest frequency is in the center, so in DFT2 and DFT3 the two DFT coefficients overlap.

    They are inversely proportional. If the length of the temporal signal is ##\text{NT}## (where ##\text{T}## is the intersample spacing, and ##\text{N}## is the number of samples), then the length of the frequency spectrum is ##\text{1/T}##.

    So basically more of the overall sinc function is contained in the DFT of the signal with a smaller period? Signals 1, 2, and 3 correspond to DFTs 3, 1, and 2 respectively?

    But this can't be right because from my notes, for a full period of a rectangular pulse we have this pair:


    In our problem, Signal 3 looks most like the signal shown above. So its DFT should also look more like that (i.e. ##\text{Signal3} \leftrightarrow \text{DFT3}##).
  7. Jun 27, 2017 #6


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    Staff: Mentor

    Correct. Since signals 2 and 3 correspond to half the frequency, the positive and negative components of the same absolute frequency now appear as separated.

    That's not correct, and as you indicate, is not compatible with the other signal you have in your notes. The conclusion is that the shorter the duration of the rectangular pulse, the broader its frequency range is. Using that heuristic, can you now assign the DFT for signals 1, 2 and 3?
  8. Jun 27, 2017 #7
    DFT2 has 1 zero, DFT1 has 3 zeros, DFT3 has 7 zeros. So, DFT2 is the broadest because there are more frequency components present in its spectrum?

    In summary:
    ##\text{Signal 1} \leftrightarrow \text{DFT 2}##
    ##\text{Signal 2} \leftrightarrow \text{DFT 1}##
    ##\text{Signal 3} \leftrightarrow \text{DFT 3}##

    Is that correct?
  9. Jun 27, 2017 #8


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    Staff: Mentor

  10. Jun 27, 2017 #9
    Thank you so much for your help. It makes perfect sense now.
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