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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?
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...
