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I am learning DSP by myself and I encountered a problem that bewilders me. If I have a sequence of length N, and I upsample it by a factor of 3. How would the DFT change or related?

For example:

x = [1 2 3 4 5 6 7];

y = [1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0 7];

What is the difference between $X[k]$ and $Y[k]$?

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Note that if you upsample a sequence $x[n]$ by a factor $M$ you have

$$y[Mn]=x[n]\tag{1}$$

The DFT of $x[n]$ is

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

The DFT length of $y[n]$ is $MN$, so the DFT of $y[n]$ is

$$Y[k]=\sum_{n=0}^{MN-1}y[n]e^{-j\frac{2\pi}{MN}nk}\tag{3}$$

Since only every $M^{th}$ sample of $y[n]$ is non-zero, $(3)$ can be written as

$$Y[k]=\sum_{n=0}^{N-1}y[Mn]e^{-j\frac{2\pi}{MN}Mnk}=\sum_{n=0}^{N-1}y[Mn]e^{-j\frac{2\pi}{N}nk}\tag{4}$$

With $(1)$ we get

$$Y[k]=X[k]\tag{5}$$

Note that from $(2)$ $X[k]$ is $N$-periodic, so the length $MN$ DFT of $y[n]$ is just an $M$-fold repetition of the DFT coefficients of $x[n]$.

A simple example:

x = [1,1];
X = fft(x)
ans =

   2   0

y = [1,0,1,0];
fft(y)
ans =

   2   0   2   0

y=[1,0,0,1,0,0];
fft(y)
ans =

   2   0   2   0   2   0
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  • $\begingroup$ Thank you! That was very explanatory! $\endgroup$
    – ahalol
    Oct 23 '19 at 18:31

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