Topic: Discrete-time Fourier transform computation

Question

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]".

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

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

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

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

Instructor's comment: Short and sweet. I like that.

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

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

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

Instructor's comment: It does help to visualize the signal first.

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

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

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

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

TA's comment: When you write the summation from -2 to 0, you are implying that you've already used the information of the unit step function. So there's no point leaving that in the equation.

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

$x[n] = ((\delta[-2]) +(\delta[-1]) +(\delta[0]))$ ( Instructor's comment: The right-hand side does not depend on n. Unless your signal is constant, this means you made a mistake.)

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

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

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

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

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

Xiang Zhang

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.

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

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

$X(\omega) = e^{-j2\omega n}(\delta (n+2)-\delta (n+1)+\delta (n))$ (Instructor's comment: The right-hand-side depends on n, but the left-hand-side does not. This means you made a mistake.)

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

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

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

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

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

TA's comment: When you write the summation from -2 to 0, you are implying that you've already used the information of the inoput x[n]. So there's no point leaving that in the equation.

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva