# How to calculate the DFT for this sum of cosine's in the form $\sum A_i \cos(\omega_i n + \phi_i)$ for fixed $N$

I am stuck trying to calculate the DFT for a given $$N$$

Given the signal $$x[n] = \cos(\frac{\pi}{2}n - \frac{\pi}{2})+2 \cos(\pi n + \frac{\pi}{2})$$ and $$N = 4$$ I tried to calculate the DFT $$X[k] = \sum\limits_{n=0}^{N-1}x[n]e^{\frac{-2 \pi ink}{N}}$$

Now since

• $$X[k] = \sum\limits_{n=0}^{3}x[n]e^{\frac{-2 \pi ink}{4}} = \sum\limits_{n=0}^{3}e^{\frac{-2 \pi ink}{4}} \cos(\frac{\pi}{2}n - \frac{\pi}{2}) + \sum\limits_{n=0}^{3}e^{\frac{-2 \pi ink}{4}} 2 \cos(\pi n + \frac{\pi}{2})$$

I can have a look at both signals seperatly and it would come down on how to solve the dft for the cosine. For this matter I looked at different sources like here and here

I tried to solve it using the geoemetric series for $$\cos(\frac{\pi}{2}n - \frac{\pi}{2})$$ which led to

• $$\sum\limits_{n=0}^{3}e^{\frac{-2 \pi ink}{4}} \cos(\frac{\pi}{2}n - \frac{\pi}{2}) = \dots = \frac{1}{2} e^{-\frac{\pi}{2}} \frac{1 - e^{\frac{\pi}{2}3i(1-k)}}{1 - e^{\frac{\pi}{2}i(1-k)}} + \frac{1}{2} e^{\frac{\pi}{2}} \frac{1 - e^{\frac{\pi}{2}3i(1+k)}}{1 - e^{\frac{\pi}{2}i(1+k)}}$$

Now something seems wrong because now I get $$X[0] \neq 0$$ which is false given my plots using this code

import numpy as np
import matplotlib.pyplot as plt

time   = np.arange(0.,5,1)
pnts   = len(time)
signal = np.cos(np.pi/2*time - np.pi/2) + 2*np.cos(np.pi*time + np.pi/2)

fourTime = np.array(range(pnts))/pnts
fCoefs   = np.zeros((len(signal)),dtype=complex)

for freq in range(pnts):
csw = np.exp( -1j*2*np.pi*freq*fourTime )
fCoefs[freq] = np.sum(np.multiply(signal,csw) )

ampls = 2*np.abs(fCoefs)

plt.plot(time, signal)
plt.xlabel('Time (s)'), plt.ylabel('Amplitude')
plt.xlim(0,10)
plt.show()

plt.stem(ampls)
plt.xlabel('Frequency (Hz)'), plt.ylabel('Amplitude (a.u.)')
plt.xlim(0,10)
plt.show()


Anyways this way of calculating actually seems a little bit overcomplicated as well. I looked different fourier tables and their derivation as well and found contradicting results. I am confused and it seems as if I am still missing something crucial.

### updated using the hint from the first answer

$$\sum\limits_{n=0}^3 x[n] \cdot e^{\frac{-2 \pi i k}{4}} = x[0] e^{\frac{- \pi i 0 k}{2}} + x[1] e^{\frac{- \pi i 1 k}{2}} + x[2] e^{\frac{- \pi i 2 k}{N}} + x[3] e^{\frac{- \pi i 3 k}{2}}$$ $$0 \cdot e^{\frac{- \pi i 0 k}{2}} + e^{\frac{- \pi i 1 k}{2}} + 0 \cdot e^{\frac{- \pi i 2 k}{N}} - e^{\frac{- \pi i 3 k}{2}} = e^{\frac{- \pi i k}{2}} - e^{\frac{- \pi i 3 k}{2}}$$

So now when calculate $$X[k]$$ for $$k = 0, 1, 2, 3$$

• $$X[0] = 0$$
• $$X[1] = e^{\frac{- \pi i \cdot 1}{2}} - e^{\frac{- \pi i 3 \cdot 1}{2}} = - i - i = - 2 i$$
• $$X[2] = e^{\frac{- \pi i \cdot 2}{2}} - e^{\frac{- \pi i 3 \cdot 2}{2}} = -1 - (-1) = 0$$
• $$X[3] = e^{\frac{- \pi i \cdot 3}{2}} - e^{\frac{- \pi i 3 \cdot 3}{2}} = i - (-i) = 2i$$

So it is a little bit confusing for me that $$X[2] = 0$$ I expected it to be bigger than $$0$$ because I thought it would be corresponding to frequency of $$1/2$$.

The sequence $$x[n]$$ is written in an unnecessarily complicated way. Figure out what the values of the sequence are for $$n=0,1,2,3$$, and then it should be extremely straightforward to compute its DFT.
• So how do you now that this is the Nyquist? I am not yet able to see it, also because I expected two non zero values for two different frequencies (though the other cosine is always $0$) Okay maybe something with my code is wrong as well which produced the shown plots, I have added it as well. Jan 24 at 21:12
• @OuttaSpaceTime: For even $N$, $N/2$ is the Nyquist bin. Also note that $x[n]$ can be written as $x[n]=\sin(n\pi/2)$, so it's no surprise that there's only a single frequency component. Jan 24 at 21:17