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Obtain DTFT of rectangular pulse, $ x(n)= \begin{cases}A, & 0 \leq n \leq L-1 \\ 0, & \text { otherwise }\end{cases} $
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written 23 months ago by |
Solution:
$ X\left(e^{j \omega}\right)=\sum_{n=0}^{L-1} A e^{-j \omega n}\\ $
$ =A\left[\frac{1-e^{-j \omega L}}{1-e^{-j \omega}}\right]\\ $
$ \sum_{n=0}^{N-1} x^n=\frac{1-x^N}{1-x}\\ $
$ X\left(e^{j \omega}\right)=A\left[\frac{\left(e^{\frac{j \omega L}{2}}-e^{\frac{-j \omega L}{2}}\right) e^{\frac{-j \omega L}{2}}}{\left(e^{\frac{j \omega}{2}}-e^{-\frac{j \omega}{2}}\right) e^{\frac{-j \omega}{2}}}\right]\\ $
$ =A\left[\frac{2 j \sin \frac{\omega L}{2}}{2 j \sin \frac{\omega}{2}}\right] e^{-\frac{j \omega(L-1)}{2}}\\ $
$ =A e^{-\frac{j \omega(L-1)}{2}}\left[\frac{\sin \frac{\omega L}{2}}{\sin \frac{\omega}{2}}\right]\\ $
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