I want to calculate the closed-form analytical expression for DFT coefficients of the following problem

$$x[n]=\left\{ \begin{array}{ll} 1 & \mbox{if } 0\leq n \leq M-1 \\ 0 & \mbox{if }M \leq n \leq L-1 \end{array} \right.$$ Write out the closed-form analityical expression for its DFT coefficients $X[k]$.

I always get the wrong solution... My attempted solution is: $$\frac{1 - e^{-j\pi k \frac{M}{L}} }{ 1 - e^{-j\pi k \frac{1}{L}} }$$

What is the correct solution and why?


The DFT of a vector of size L is L samples of the Fourier transform of the signal:

$$ X[k] = X(e^{j\theta})|_{\theta = 2\pi k/L} $$

So now we need to calculate the FT of $x[n]$:

$$ X(e^{j\theta}) = \sum_{n=0}^{n=M-1} e^{-j n \theta} $$

This is a geometric series which results in:

$$ X(e^{j\theta}) = \frac{1 - e^{-jM\theta}}{1 - e^{-j\theta}} $$

Evaluate on the DFT samples to give:

$$ X[k] = \frac{1 - e^{-j M 2 \pi k/L}}{1 - e^{-j 2 \pi k / L}} $$

You can extract half phase from top and bottom to give:

$$ X[k] = e^{-j(M-1)\pi k/L} \frac{\sin(M \pi k/L)}{\sin(\pi k/L)} $$

  • $\begingroup$ E^(-j*(M-1)*pik/L)*sin(Mpik/L)/sin(pik/L) $\endgroup$ – Radha Kodali May 23 '20 at 17:41

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