I am having trouble finding the same answer as the solution manual for this sequence.

The problem asks to compute the DFT of $$ x[n] = \begin{cases} 1 & \text{for even } n \in \{0\ldots N-1\} \\ 0 & \text{for odd } n \in \{0\ldots N-1\} \end{cases} $$

Setting up the DFT equation I agree with the solution manual $$ X[k]=\sum_0^{N-1}x[n]W^{kn}_{N}=\sum_0^{\frac{N}{2}-1}W^{k2n}_{N}=\sum_0^{\frac{N}{2}-1}e^{\frac{-j2\pi k(2n)}{N}} $$

However when apply the closed form geometric series formula $\sum_0^Na^k=\frac{1-a^{N+1}}{1-a}$ I get $$ \frac{1-e^{-j2\pi k}}{1-e^{\frac{-j2\pi k}{N}}} $$

while the solution manual says its
$$\frac{1-e^{-j2\pi k}}{1-e^{\frac{-j\pi k}{N}}}$$

Note their lower exponent does not have a 2. Could someone explain where and how the 2 goes away? Or does someone agree that the provided answer is incorrect?

This is question 8.5c from Oppenheim's Discrete-Time Signal Processing 3e.


Assuming $N$ even :

$$ \begin{align} X[k] &= \sum_{n=0}^{N-1} x[n] ~ W_N^{kn} ~~~,~~~k=0,1,...,N-1\\ \\ &= \sum_{n=0}^{N/2-1} 1 ~ W_N^{k ~2n} \\ \\ &= \sum_{n=0}^{N/2-1} e^{-j 4\pi kn/N} \\ \\ & = \frac{ 1 - e^{-j \frac{4\pi k}{N}N/2 }}{ 1 - e^{-j 4\pi k/N}} \\ \\ & = \frac{ 1 - e^{-j 2\pi k}}{ 1 - e^{-j 4\pi k/N}}~~~,~~~k=0,1,...,N-1\\ \end{align} $$

The result equals: $$ X[k] = \begin{cases}{ ~~~ N/2 ~~~ ,~~~ k = 0, N/2 \\ ~~~~~~ 0 ~~~~~ ~ ,~ ~\text{otherwise} } \end{cases} $$

  • 1
    $\begingroup$ glad you posted this, I wrote an answer up and was hesitant to post it seeing that my answer didn't match either of OP's hah. $\endgroup$ – user67081 Jan 17 at 23:31
  • $\begingroup$ @Fat32 thanks for the explanation. I didn't realize either was not correct. The relationship between the big N and little n keeps confusing me. based on your steps 3-4 I can see the little n is based on the summation length while the big N is based on the overall sequence length.. am I correct with this assumption? Could you also explain how you got the result to equal $N/2$? Is there some formula or assumptions you are making? $\endgroup$ – scott Jan 18 at 0:35
  • $\begingroup$ @Fat32 I guess I may be misunderstanding something, for either one of them if k goes to $0$ or $N/2$ I see the exp going to either 1 (for $0$) or some other value (for $N/2$). $\endgroup$ – scott Jan 18 at 19:54
  • $\begingroup$ @Fat32 I am still unsure how you are getting the final result. How does final value result in $N/2$ when it is based on the exponent? Is there some property of the exponential I am missing? For example - case $k=0$ i see it resulting in $\frac{1-e^0}{1-e^0}=\frac{1-1}{1-1}=0$? Maybe I am misunderstanding some concept about $e$? $\endgroup$ – scott Jan 18 at 20:18
  • $\begingroup$ @Fat32 so after thinking a bit more, I am assuming it is connected to the fact $e$ is representing a sinusoidal signal. I will evaluate it using Euler’s equations and update. $\endgroup$ – scott Jan 18 at 20:25

Figured I'd post this since I wrote it anyway, just a confirmation of Fat32's answer.

Letting $N' = \frac{N}{2} - 1$ we have $\sum_{0}^{\frac{N}{2} - 1} e^{\frac{-j2\pi k (2n)}{N}} = \sum_{0}^{N'} e^{\frac{-j2\pi k n}{(N'+1)}}$

Then plugging in the geometric sum formula:

$=\frac{1 - e^{\frac{-j2\pi k(N'+1)}{(N'+1)}}}{1-e^{\frac{-j2\pi k}{(N'+1)}}}=\frac{1 - e^{\frac{-j2\pi k}{1}}}{1-e^{\frac{-j2\pi k}{(N'+1)}}}$

and finally reverting back to our original variable $N=2(N'+1)$ gives $=\frac{1 - e^{\frac{-j2\pi k}{1}}}{1-e^{\frac{-j4\pi k}{N}}}$


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