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DTFT of aⁿu[n] with spectrum plot

Short answer

For 0 < a < 1, X(e^(jω)) = 1/(1 - ae^(-jω)), with magnitude 1/sqrt(1 - 2a cosω + a²) and phase -arctan(a sinω/(1 - a cosω)). The magnitude is even and low-pass, phase is odd, and both repeat every 2π.

Find Fourier transform of signal x[n] = aⁿu[n] (where 0 < a < 1) and plot magnitude and phase spectrum.

DTFT

X(ejω)=∑n=−∞∞anu[n] e−jωn=∑n=0∞(ae−jω)n=11−ae−jω,since ∣ae−jω∣=a<1\begin{aligned} X(e^{j\omega}) &= \sum_{n=-\infty}^{\infty} a^n u[n]\,e^{-j\omega n} = \sum_{n=0}^{\infty}\left(a e^{-j\omega}\right)^n \\ &= \frac{1}{1 - a e^{-j\omega}}, \quad \text{since } |ae^{-j\omega}| = a < 1 \end{aligned}

Writing 1−ae−jω=(1−acos⁡ω)+j asin⁡ω1 - ae^{-j\omega} = (1 - a\cos\omega) + j\,a\sin\omega:

∣X(ejω)∣=11−2acos⁡ω+a2,∠X(ejω)=−tan⁡−1asin⁡ω1−acos⁡ω|X(e^{j\omega})| = \frac{1}{\sqrt{1 - 2a\cos\omega + a^2}}, \qquad \angle X(e^{j\omega}) = -\tan^{-1}\frac{a\sin\omega}{1 - a\cos\omega}

Spectrum plots

ω\omega∣X∣\lvert X\rvert∣X∣\lvert X\rvert for a=0.5a=0.5Phase for a=0.5a=0.5
0011−a\frac{1}{1-a}20∘0^\circ
π/2\pi/211+a2\frac{1}{\sqrt{1+a^2}}0.894−26.57∘-26.57^\circ
π\pi11+a\frac{1}{1+a}0.6670∘0^\circ
 |X|
 1/(1-a) |                        *
         |                     *     *
         |
         |                  *           *
         |               *                 *
         |         *  *                       *  *
 1/(1+a) |*  *  *                                   *  *  *
         +-------------------------------------------------> w
         -pi                      0                      pi

 phase  (odd; max |phase| = asin(a), reached where cos w = a)
+asin(a) |            *  *  *
         |         *           *
         |      *
         |   *
       0 |*                       *                       *
         |                                             *
         |                                          *
         |                           *           *
-asin(a) |                              *  *  *
         +-------------------------------------------------> w
         -pi                      0                      pi

(Plots drawn for a=0.5a = 0.5; one period −π-\pi to π\pi is shown.)

The magnitude is even and low-pass (peak at ω=0\omega = 0), the phase is odd, and both are periodic with period 2π2\pi.

✓ Answer checked by Claude Opus 5.5 (high effort)

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