# Finding the discrete time Fourier series for signal

I think I did everything correctly here, but I must be missing something still.

We have the following signal: My approach: We are told that the signal has period $$N = 4$$ We know $$Y[k] = \frac{1}{N}\sum_{n = 0}^{3}y[n]e^{-j\frac{\pi}{2}nk}$$ The signal only has two non-zero terms at $$n \mod 4 = 0$$ and $$n \mod 4 = 2$$

So, $$Y[k] = \frac{1}{N}\left(e^{-j\frac{\pi}{2}k} -\frac{1}{2} e^{-j \pi k}\right)$$

$$Y[k] = \frac{1}{4}\left( (-j)^{k} - \frac{1}{2} (-1)^{k}\right)$$

$$Y[k] = \begin{cases} \frac{1}{8} &\text{if}\, k = 0\\ \frac{1}{4}(.5 - j)&\text{if}\, k = 1\\ \frac{-3}{8} &\text{if}\, k = 2\\ \frac{1}{4}(j + .5) &\text{otherwise} \end{cases}$$

However the reported answer is actually

$$Y[k] = \begin{cases} \frac{1}{8} &\text{if}\, k = 0\\ \frac{3}{8} &\text{if}\, k = 1\\ \frac{1}{8} &\text{if}\, k = 2\\ \frac{3}{8} &\, k = 3 \end{cases}$$

Where am I going wrong? Any help is very much appreciated :)

Your first term in the second equation is wrong for $$n=0$$. You put in the term for $$n=1$$ . It should be
$$Y[k] = \frac{1}{N}\left(1 -\frac{1}{2} e^{-j \pi k}\right)$$