I am trying to solve the above question. I am not sure how to proceed. I know the formula for 64 point DFT of $x[n]$.
$X[k]=\sum_{n=0}^{63} x[n] e^{-j2\pi nk/64}$
But how can I find $R[k]$ and $Y[k]$? Can anyone please tell the relation?
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$\begingroup$ What have you done so far? Are you at the absolute zero step? $\endgroup$ – Fat32 Nov 17 '18 at 15:57
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$\begingroup$ more specifically, it will be very helpful to investigate the behaviour of compressor and expander stages... And note that DFT exhibits time-frqeuency duality. $\endgroup$ – Fat32 Nov 17 '18 at 16:02
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$\begingroup$ I only know that the downsampler will take every second value of the $X[k]$ and $Y[k]$ will have an additional zero in every second sample of $R[k]$ $\endgroup$ – Christy Nov 17 '18 at 16:06
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$\begingroup$ ok. What about their effect on the other domain? (freq vs time) ? can you show that? Note that as I stated, DFT exhibits duality, you can takle advantage of this. $\endgroup$ – Fat32 Nov 17 '18 at 16:06
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$\begingroup$ I learnt DFT of downsampler from: ccrma.stanford.edu/~jos/ReviewFourier/…. So, by using duality, $r[n]=16(x[-n]+x[16-n])$ and $y[n]=r[2n]$? Please confirm. $\endgroup$ – Christy Nov 17 '18 at 17:09
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Given a finite length input sequence $x[n]$ of $N=64$ points, its 64-point DFT be $X[k]$. Then according to the block diagram you've provided, the signal $r[n]$ will be of $N/2$ points long and its $N/2$ - point DFT $R[k]$ will be related to $X[k]$ as :
$R[k] = X[2k]$, which will be manipulated as:
$$ \begin{align} X[k] &= \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} kn} ~~~,~~~ k=0,1,...,N-1 \\ R[k] & = X[2k] = \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} 2kn} ~~~,~~~ k=0,1,...,N/2-1\\ &=\sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N/2} kn} \\ &=\sum_{n=0}^{\frac{N}{2}-1} x[n] e^{-j \frac{2\pi}{N/2} kn} + \sum_{n=\frac{N}{2}}^{N-1} x[n] e^{-j \frac{2\pi}{N/2} kn} \\ &=\sum_{n=0}^{\frac{N}{2}-1} x[n] e^{-j \frac{2\pi}{N/2} kn} + \sum_{n=0}^{\frac{N}{2}-1} x[n+\frac{N}{2}] e^{-j \frac{2\pi}{N/2} kn} \\ &=\sum_{n=0}^{\frac{N}{2}-1} \left( x[n]+x[n+\frac{N}{2}] \right) e^{-j \frac{2\pi}{N/2} kn} \\ &= \text{N/2-point DFT} \{ r[n] \}\\ &= \sum_{n=0}^{\frac{N}{2}-1} r[n] e^{-j \frac{2\pi}{N/2} kn} \\ \end{align} $$
hence we see that $r[n]$ is the sequence whose length (or period) is reduced to $N/2$ and is generated by adding the second half of $x[n]$ onto the first half.
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$\begingroup$ thank you so much. Since $y[n]$ comes from 64 point IDFT, so the $y[n] = 16(x[-2n]+x[16-2n])$ ? Also I was not getting the correct relation for $r[n]$, since it comes from 32 point IDFT and not 64 point IDFT? $\endgroup$ – Christy Nov 18 '18 at 1:56
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$\begingroup$ First note that, as shown above, 32-point r[n] is given by $r[n] = x[n] + x[n+N/2] = x[n] + x[n+32]$ for $n=0,1,...,31$. Now what's the relation beetween r[n] and y[n] ? you can show by duality that y[n] is a subperiodic signal formed by repetition of r[n] (with a scale factor)... $\endgroup$ – Fat32 Nov 18 '18 at 10:42
