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How do I find the Discrete Fourier Transform of the sequence below?

$$ x = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$$

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1 Answer

Apply Discrete Time Fourier Transformation on the vector

$$ x = \begin{bmatrix} x[0] \\ x[1] \\ x[2] \\ x[3] \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} $$

to find the frequency domain vector

$$ \mathcal{F} \left \{ x \right \} = X = \begin{bmatrix} X[0] \\ X[1] \\ X[2] \\ X[3] \end{bmatrix}. $$

Use the formula

$$ X[k] = \sum_{n=0}^{N-1} x[n] \cdot e^{-\dfrac{j 2 \pi k n}{N}}, $$

where $N$ is the number of elements in the time domain vector (i.e.; $\;N\!=\!4\;$ in our case). If we rewrite the sum in the open form:

$$ \begin{array}{llll} X[k] & = 1e^{-j\dfrac{0\pi}{4}k} & + 2e^{-j\dfrac{2\pi}{4}k} & + 3e^{-j\dfrac{4\pi}{4}k} & + 4e^{-j\dfrac{6\pi}{4}k} \\ & = 1 & + 2e^{-j\dfrac{\pi}{2}k} & + 3e^{-j\pi k} & + 4e^{-j\dfrac{3\pi}{2}k} \\ \end{array} $$

Next, find the elements of $X$ one by one:

$$ \begin{array}{lllll} X[0] & = 1 & + 2 & + 3 & + 4 & = \mathbf{+10} \\ \\ X[1] & = 1 & + 2e^{-j\dfrac{\pi}{2}} & + 3e^{-j\pi} & + 4e^{-j\dfrac{3\pi}{2}} \\ & = 1 & - j2 & - 3 & + j4e & = \mathbf{-2 + j2} \\ \\ X[2] & = 1 & + 2e^{-j\pi} & + 3e^{-j2\pi} & + 4e^{-j3\pi} \\ & = 1 & - 2 & + 3 & - 4 & = \mathbf{-2} \\ \\ X[3] & = 1 & + 2e^{-j\dfrac{3\pi}{2}} & + 3e^{-j3\pi} & + 4e^{-j\dfrac{9\pi}{2}k} \\ & = 1 & + j2 & - 3 & - j4 & = \mathbf{-2 - j2} \end{array} $$

Hence, the Discrete Fourier Transform of $x$ is

$$ X = \mathcal{F} \left \{ x \right \} = \mathcal{F} \left \{ \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \right \} = \begin{bmatrix} 10 \\ −2+i2 \\ −2 \\ −2−i2 \end{bmatrix}. $$

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This was supposed to be a question. However, half the way I was writing my steps I found out my mistake. Then I decided to share it as a question/answer pair. I hope this example helps out other people. –  hkBattousai May 3 '13 at 13:50
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