I have my DSP final soon and I have been reviewing some past exams. Here is a question from one of these exams:
Let x[n] be a real valued finite duration signal in n $\in [0,N]$. Another signal $x_1[n]$ is defined as: $$x_1[n] = \begin{cases} x[n], & \text{if } n=0 \; or \; n=N ,\\ c x[n], & \text{if } n \in [1, \: N-1],\\ c x[2N-n] & \text{if } n \in [N+1,\: 2N-1] \end{cases} $$ where $c$ is a real valued constant and $x_1[n]$ is a finite duration signal $\in [0, \: 2N-1]$
a) Show that $\mathrm{X}_1(k) = DFT_{2N}\{x_1[n]\}$ is real valued$.$
b) Write $\mathrm{X}_1(k) = DFT_{2N}\{x_1[n]\} = \sum_{n=0}^{N}x[n]\phi_k[n] \:, \; k = 0,...,2N-1$ where $\phi_k[n]$'s have real values. Find those $\phi_k [n]$'s$.$
c) Show that $\mathrm{X}_1(k) = \mathrm{X}_1(-k)_{\text{mod}\:2N}$
Now I know that this question is testing the conjugate/even symmetric property of the DFT. I can handle c) but I am not sure how to approach parts a) and b). Here is my attempt for a) and solution for c):
a) The DFT is defined as follows: $X[k] = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi nk/N}$
In our case let $2N=M$ and plug in $x_1[n]$ into the DFT equation:
$$X_1[k] = \sum\limits_{n=0}^{M-1} x_1[n] e^{-j 2 \pi\frac nMk}$$ $$X_1[k] = x[0] + c\cdot \sum\limits_{n=1}^{N-1} x[n] e^{-j 2 \pi\frac nMk} + (x[N]\cdot(-1)^k) + c\cdot\underbrace{\sum\limits_{n=N+1}^{2N-1} x[2N-n] e^{-j 2 \pi\frac nMk}}_{G}$$ Let's work on $G$: $$G = \sum\limits_{n=N+1}^{2N-1} x[2N-n] e^{-j 2 \pi\frac nMk} \quad\quad \text{Replace $n$ with $n+N$}$$ \begin{align} G &= \sum\limits_{n=1}^{N-1} x[N-n] e^{-j 2 \pi\frac {(n+N)}{M}k}\\ G &= (-1)^k\sum\limits_{n=1}^{N-1} x[N-n] e^{-j 2 \pi\frac nMk}\\ \end{align}
I don't know how to proceed from here. Perhaps it would be easier to prove that $x_1[n]$ is symmetric but I am not sure if that suffices.
Here is my solution for c) which I assume is wholly correct.
c) Just keep in mind that since $x[n]$ is real $x_1[n]$ is also real and also assuming that $X_1(k)$ is real as asked to prove in a). \begin{align} X_1(k) &= \sum_{n=0}^{M-1}x_1[n]e^{-j2\pi k\frac nM}\\ X_1^*(-k) &= \left(\sum_{n=0}^{M-1}x_1[n]e^{-j2\pi (-k)\frac nM}\right)^*\\ &= \sum_{n=0}^{M-1}\left(x_1[n]e^{-j2\pi (-k)\frac nM}\right)^*&\text{linearity}\\ &= \sum_{n=0}^{M-1}x_1[n]\left(e^{-j2\pi (-k)\frac nM}\right)^*&x_1[n]=x_1^*[n]\text{ since it's real}\\ &= \sum_{n=0}^{M-1}x_1[n]e^{-(-j2\pi (-k)\frac nM)}\\ &= \sum_{n=0}^{M-1}x_1[n]e^{-j2\pi k\frac nM}\\ &=X_1(k)_{\text{mod} \:2N} \end{align}
And help for a) and b) is much appreciated!