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Practice Problem on Discrete-time Fourier transform computation

Compute the discrete-time Fourier transform of the following signal:

$ x[n]= u[n+2]-u[n-1] $

See these Signal Definitions if you do not know what is the step function "u[n]".

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Answer 1

$ x[n] = u[n+2]-u[n-1] $.

$ X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $

$ = \sum_{n=-2}^{0} x[n] e^{-j\omega n} $

$ = 1+ e^{j\omega} + e^{2j\omega} $

Answer 2

$ X_{2\pi}(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $

$ = \sum_{n=-2}^{0} x[n] e^{-j\omega n} $

$ = e^{2j\omega} + e^{j\omega} + 1 $

Answer 3

$ X(\omega) = \sum_{n=-\infty}^{+\infty} (u[n+2] -u[n-1]) e^{-j\omega n} $

$ = \sum_{n=-2}^{+\infty} u[n+2] e^{-j\omega n} - \sum_{n=1}^{+\infty} u[n-1] e^{-j\omega n} $

$ = \sum_{n=-2}^{0} (u[n+2] -u[n-1]) e^{-j\omega n} $

$ = e^{2j\omega} + e^{j\omega} + 1 $

Answer 4

$ x[n] = u[n+2]-u[n-1] $

$ x[n] = ((\delta[-2]) +(\delta[-1]) +(\delta[0])) $

so $ X[Z] = e^{2 j \omega} +e^{j \omega} +1 $

Answer 5

$ x[n] = u[n+2] - u[n-1] $

$ X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $

$ X_(\omega) = \sum_{n=-2}^{0} e^{-j\omega n} $

$ X_(\omega) = 1 + e^{j\omega}+ e^{2j\omega} $

Answer 6

from the equation we can get that

$ X(n) = u[n+2] - u[n-1] = \left\{ \begin{array}{l l} 1 & \quad when \quad n = -2,-1,0\\ 0 & \quad \text{else} \end{array} \right. $

Hence, substitute into the DTFT equation,

$ X_(\omega) = \sum_{n = -\infty}^{ \infty} x[n] e^{-j\omega n} $

change the limit to

$ X_(\omega) = \sum_{n = -2}^{ 0} e^{-j\omega n} $

Then, we expand to the normal expression.

$ X_(\omega) = 1 + e^{j \omega} + e^{j 2 \omega} $

Answer 7

$ X(\omega) = \sum_{n=-\infty}^\infty x[n]e^{-j2\omega n} $

$ X(\omega) = e^{-j2\omega n}(\delta (n+2)-\delta (n+1)+\delta (n)) $

$ X(\omega) = e^{-j2\omega} + e^{-j\omega} + 1 $

Answer 8

x[n] = u[n+2]-u[n-1]

$ X(\omega) = \sum_{n=-\infty}^\infty x[n]e^{-j\omega n} $

$ = \sum_{n=-2}^0 x[n]e^{-j\omega n} $

$ = e^{j\omega 2}+e^{j\omega}+1 $

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