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Please, check the following discrete periodic sequence when the period $N=2$.

$x[k]=\exp(j\frac{2\pi}{N}k), N=\text{period}$

$..., x[0]= 1, x[1]= -1, x[2]= 1, x[3]= -1, ... , N=2$

According to my DTFS calculation, DTFS coefficient of above sequence is as follows.

$..., D_{0}=0, D_{1}=1, D_{2}=0, D_{3}=1, ... , N=2 $

If I shift above DTFS by 1 in the frequency domain like this,

$..., D_{0}=1, D_{1}=0, D_{2}=1, D_{3}=0, ...$

Inverse DTFS of above coefficient is as follows according to my calculation.

$..., x[0]=1, x[1]=1, x[2]=1, x[3]=1, ... N=2$

Actually, if DTFS of $x[k]$ is $D_{n}$, then DFTS of $\exp(ja\frac{2\pi}{N}k)x[k]$

will be $D_{n-a}$ by DTFS frequency shifting definition.

So, DFTS of $\exp(j(-1)\frac{2\pi}{N}k)x[k]$ will be $D_{n+1}$.

In this case, $\exp(j(-1)\frac{2\pi}{2}k)x[k]$ is $\exp(-j\frac{2\pi}{2}k)$ $\exp(j\frac{2\pi}{2}k)=1$.

Shifting by $-1$ shows the same result in this case because $N=2$.

So, I think above result is correct.

I know DTFS of $x[k]=1$ with $N=1$ is as follows.

$..., x[0]=1, x[1]=1, x[2]=1, x[3]=1, ... N=1$

My conclusion is as follows.

$x[k]=..., 1, -1, 1, -1,... \longleftrightarrow D_{n}=..., 0, 1, 0, 1,... N=2$ case 1

$x[k]=..., 1, 1, 1, 1,... \longleftrightarrow D_{n}=..., 1, 0, 1, 0,... N=2$ case 2

$x[k]=..., 1, 1, 1, 1,... \longleftrightarrow D_{n}=..., 1, 1, 1, 1,... N=1$ case 3

I can't understand the last 2 lines. I think case 2 and case 3 should have

different sequences in the time domian. What did I miss?

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    $\begingroup$ hay kappy, you need to learn how to use $\LaTeX$ math. and you need to define what you mean by "DTFS". $\endgroup$ – robert bristow-johnson Sep 27 '20 at 17:42
  • $\begingroup$ now what is the "Discrete Time Fourier Series"? is that the same thing as the "Discrete Fourier Series"? $\endgroup$ – robert bristow-johnson Sep 30 '20 at 1:28
  • $\begingroup$ I think most people use the word "DTFS" instead of "DFS". Please, check the chapter 11 and 12 of the following pdf file. cfile24.uf.tistory.com/attach/173B49494EC1B160298290 DTFS, DTFT, Z transform, DFT and FFT are not the same things. $\endgroup$ – kappy super Oct 7 '20 at 15:17
  • $\begingroup$ Oppenheim and Schafer call the Discrete Fourier Series the "DFS" and it is precisely the same thing that the Discrete Fourier Transform (DFT) is but is not the same as the Discrete-Time Fourier Transform (DTFT) nor the Z Transform. "DTFS" is not a term I have ever seen in my 40+ years in electrical engineering and signal processing. $\endgroup$ – robert bristow-johnson Oct 7 '20 at 16:58
  • $\begingroup$ If you search the "DTFS" in the ieeexplore.ieee.org, you can find several papers about "DTFS". Please, check the following link. ieeexplore.ieee.org/search/… $\endgroup$ – kappy super Oct 13 '20 at 3:18

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