# IDFT of upsampling in frequency domain

Suppose I have a discrete time signal $$x[n]$$ of length 32 and I take it's 32 point DFT $$X[k]$$. Then I upsample this by a factor of 2 to get $$Y[k]=X[k/2]$$ and then take 64 point IDFT to get $$y[n]$$.

What will be the the relation between $$y[n]$$ and $$x[n]$$?

By upsampling in frequency domain, I know that $$x[n/2]->X[2k]$$. So by duality, $$X[k/2]->Nx[-n2]$$? What will $$N$$ be here, $$32$$ or $$64$$? I am confused on how to apply duality and whether it is required or not?

• Hi! You have asked a very similar (or the same) question last week. There I have talked about duality for a quick answer. However you do not necessarily need it. Though it would be a simplification. By the way $Y[k] = X[k/2]$ is not upsampling but expansion, there are zero samples between $X[k/2]$ s. Nov 24, 2018 at 21:33

This answer does not require any elaborate mathematics. Rather it requires a careful interperation of periodicity of DFT sequences considering their lengths.

Lets display the signal chain below:

$$x[n] \longrightarrow \boxed{ 32DFT} \longrightarrow X[k] \longrightarrow \boxed{ \uparrow 2 } \longrightarrow Y[k] \longrightarrow \boxed{ 64-IDFT } \longrightarrow y[n]$$

Now the signals are:

Time sequence $$x[n]$$ has a length of $$32$$ samples defined for $$n=0,1,...,31$$. For any other values of $$n$$, we interpret $$x[n]$$ as the periodic extension by the help of modulus operator as $$x[n] = x[(n)_{32}]$$. For example $$x[35] = x[ (35)_{32} ] = x[3]$$.

The frequency sequence $$X[k]$$ has a length of $$32$$ samples defined for $$k=0,1,...,31$$. For any other values of $$k$$, we interpret $$X[k]$$ as the $$32$$ point periodic extension of $$X[k]$$ similarly.

After the expansion block we have:

DFT frequency sequence $$Y[k]$$ has a length of $$64$$ samples defined for $$k=0,1,...,63$$. For any other values of $$k$$, we interpret $$Y[k]$$ as the periodic extension by $$64$$ samples similarly.

The IDFT signal $$y[n]$$ has a length of $$64$$ samples defined for $$n=0,1,...,63$$. For any other values of $$n$$, we interpret $$y[n]$$ as the periodic extension by $$64$$ samples similarly.

Now, simply write the $$64$$ point inverse DFT equation to define $$y[n]$$ :

\begin{align} y_{64}[n] &= \frac{1}{64} \sum_{k=0}^{63} Y[k] e^{-j \frac{2\pi}{64} k n } ~~~&,~~~ \text{ for } k,n = 0,1,...,63 \\ &= \frac{1}{64} \sum_{k=0}^{63} X[k/2] e^{-j \frac{2\pi}{64} k n } ~~~&,~~~ \text{ for } k,n = 0,1,...,63 \\ &= \frac{1}{64} \sum_{m=0, k=2m}^{31} X[m] e^{-j \frac{2\pi}{64} 2m n } ~&,~ \text{ for } m=0,1,..,31 ~~ n = 0,1,...,63 \\ &= \frac{1}{2} \left( \frac{1}{32} \sum_{m=0}^{31} X[m] e^{-j \frac{2\pi}{32} m n } \right) ~~&,~\text{ for } m=0,1,..,31 ~~~ n = 0,1,...,63 \\ y_{64}[n] &= \frac{1}{2} x[(n)_{32}] ~~~&,~~~ \text{ for } n = 0,1,...,63 \\ \end{align}

The expression in the parenthesis above is simply the $$32$$ point inverse DFT of the sequence $$x[n]$$. Note that this sequence has a length of $$32$$ points and we can express it with the modulus notation $$x[n] = x[(n)_{32}]$$ as in the last line.

And the last line describes the $$64$$ point signal $$y_{64}[n]$$ as a two times repetition of $$32$$ point $$x[n]$$, divided by $$2$$. In other words, $$y[0] = x[0], y[15]=x[15], y[40] = x[8], y[60]= x[28]$$ etc...