How to solve this even symmetry question?

Question

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!

Answer 1

What you have shown for a) is actually what you should be doing for b). For a) you simply need to show that your given $x_1[n]$ is conjugate symmetric or even in this case.

Now for b) if we takeover from your $G$ then we can proceed as follows:

\begin{align*} G &= \sum\limits_{n=N+1}^{2N-1} x[2N-n] e^{-j 2 \pi\frac{n}{2N}k} \quad\quad \text{c.o.v. $l=2N-n$}\\ G &= \sum\limits_{l=1}^{N-1} x[l] e^{-j 2 \pi\frac {(2N-l)}{2N}k}\\ G &= \sum\limits_{l=1}^{N-1} x[l] e^{j 2 \pi\frac{l}{2N}k} \underbrace{e^{-j2\pi k}}_{(1)^k} \end{align*}

Now there is nothing stopping me from using $l=n$, it's a dummy variable anyways.

\begin{align*} G &= \sum\limits_{n=1}^{N-1} x[n] e^{j 2 \pi\frac{n}{2N}k} \end{align*}

You already have a sum running from $n=1$ to $N-1$, so combine the two under one summation:

\begin{align*} X_1[k] = x[0] + c\cdot \underbrace{\sum\limits_{n=1}^{N-1} x[n] e^{-j 2 \pi\frac nMk}}_{(i)} + (x[N]\cdot(-1)^k) + c\cdot\underbrace{\sum\limits_{n=1}^{N-1} x[n] e^{j 2 \pi\frac{n}{2N}k}}_{(ii)} \end{align*}

Combine $(i)$ and $(ii)$ as such:

\begin{align*} (i) + (ii) &= \sum\limits_{n=1}^{N-1} x[n] (e^{j 2 \pi\frac{n}{2N}k} + e^{-j 2 \pi\frac{n}{2N}k})\\ (i) + (ii) &= 2\cdot \sum\limits_{n=1}^{N-1} x[n] \underbrace{\frac{(e^{j 2 \pi\frac{n}{2N}k} + e^{-j 2 \pi\frac{n}{2N}k})}{2}}_{\cos(\pi \frac nN k)} \end{align*}

So now we have shown that your $\phi_k[n]$'s are indeed real and we have also found all of them as such:

$$\phi_k[n] = \begin{cases} 1, & \text{if } n=0\\ 2c \cdot \cos(\pi \frac nN k), & \text{if } n \in [1, \: N-1]\\ (1)^k, & \text{if } n=N\\ \end{cases} $$

for $k \in \{0,1,...,2N-1\}$.

And your c) is correct!