Calculating N-point DFT of a signal based on another signal's DFT

Question

Recently I have asked this answer.

Now I would like to know a little bit more about expressing N-point DFT's of signals in terms of one another.

Having N-point DFT X(k) of a certain signal x(n), how can I calculate N-point DFT of a signal $x_{s}=x(n)+(-1)^n \cdot{} x(n)$ . Assuming $N$ is even.

Having thought about it a little bit, I came to a conclusion that we cancel half of the samples out and multiply the value of the rest of samples by a factor of 2. The spectrum will now have greater amplitude of peak but I am not sure how it will look like in terms of the location of frequency peak.

Answer 1

You just need to apply the formula consistently to see what's happening. Noting that $(-1)^n=e^{-j\pi n}$ you get for the DFT of $x[n](-1)^n$

$$\begin{align}\sum_{n=0}^{N-1}x[n](-1)^ne^{-j2\pi nk/N}&=\sum_{n=0}^{N-1}x[n]e^{-j\pi n}e^{-j2\pi nk/N}\\&=\sum_{n=0}^{N-1}x[n]e^{-j2\pi n(k+N/2)/N}\\&=X[k+N/2]\tag{1}\end{align}$$

where $X[k]$ is the DFT of $x[n]$, and where I've assumed that $N$ is even. Consequently, the DFT of $x[n]+(-1)^nx[n]$ is given by

$$\begin{align}\text{DFT}\{x[n]+(-1)^nx[n]\}&=\text{DFT}\{x[n]\}+\text{DFT}\{(-1)^nx[n]\}\\&=X[k]+X[k+N/2]\tag{2}\end{align}$$

where the index $k+N/2$ has to be taken modulo $N$ if $X[k]$ is defined for $k\in[0,N-1]$.

Answer 2

Let $x[n]$ be a sequence of length $N$ even, whose DTFT is $X(e^{j \omega})$ and the DFT is $X[k]$ which is given by the frequency sampling relation $$X[k] = X(e^{j \frac{ 2 \pi }{N} k}) ,$$ for $k=0,1,2,...,N-1$.

It can be shown that the DTFT of the new seqeunce is: $$y[n] = (-1)^n x[n] = e^{j \pi n} ~x[n] \longleftrightarrow Y(e^{j \omega}) = X(e^{j (\omega - \pi)}) $$

and the DFT of $y[n]$ is: $$ Y[k] = Y(e^{j \frac{ 2 \pi }{N} k}) = X(e^{j (\frac{ 2 \pi }{N}k - \pi)})=X(e^{j \frac{ 2 \pi }{N}(k - N/2)}) = X[k-N/2] $$

Since you look for DFT of $z[n] = x[n] + y[n]$, you get it from its DTFT : $$ Z(e^{j \omega}) = X(e^{j \omega}) + Y(e^{j \omega}) $$

Then the DFT of $z[n]$ is

$$Z[k] = Z(e^{j \frac{ 2 \pi }{N} k}) = X[k] + X[k-N/2]$$, for $k=0,1,2,...,N-1.$

Note that you can treat DFT $Z[k]$ as a periodic sequence, or use a modulus $N$ in its argument to find those values $k-N/2$.

Answer 3

Do the problem for finding discrete Fourier transform for the sequence $X(n)={1,2,3,4}$