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Magnitude and phase of DTFT of aⁿu[n]

Short answer

The DTFT is X(e^{jω}) = 1/(1 - a e^{-jω}). Magnitude: |X| = 1/√(1+a^2-2a cos ω), peaking at ω=0 with value 1/(1-a) for 0<a<1 (low‑pass). Phase: ∠X = -tan^{-1}(a sin ω/(1-a cos ω)), odd, zero at ω=0 and ±π.

Given discrete time signal is x[n] = aⁿu[n], |a| < 1. Given that Fourier transform of x[n] is continuous in nature, plot magnitude and phase response of the transformed signal.

DTFT:

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

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

Magnitude:

∣X(ejω)∣=11+a2−2acos⁡ω|X(e^{j\omega})| = \frac{1}{\sqrt{1 + a^2 - 2a\cos\omega}}

Phase:

∠X(ejω)=−tan⁡−1(asin⁡ω1−acos⁡ω)\angle X(e^{j\omega}) = -\tan^{-1}\left(\frac{a\sin\omega}{1 - a\cos\omega}\right)

Values for 0<a<10<a<1 (taking a=0.5a = 0.5 for the sketch):

ω\omega∣X∣\lvert X\rvert (general)∣X∣\lvert X\rvert, a=0.5a=0.5Phase, 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

The phase reaches its most negative value, −sin⁡−1a-\sin^{-1}a (−30∘-30^\circ for a=0.5a=0.5), where cos⁡ω=a\cos\omega = a (ω=π/3\omega = \pi/3 for a=0.5a = 0.5). By odd symmetry it reaches +30∘+30^\circ at ω=−π/3\omega = -\pi/3.

 |X(e^jw)|  (a = 0.5, low-pass shape)
 2.0 |              *
     |            *   *
     |          *       *
0.89 |        *           *
0.67 |  * * *               * * *
     +--+-----------+-----------+--> w
       -pi          0           pi

 phase (a = 0.5, odd function)
 +30 |        * * *
 +15 |    * *
   0 +--*-----------*-----------*--> w
 -15 |                    * *
 -30 |              * * *
       -pi          0           pi

 (both plots repeat every 2pi)

Observations:

  • The spectrum is continuous in ω\omega and periodic with period 2π2\pi.
  • The magnitude is even and the phase is odd (real signal).
  • For 0<a<10<a<1 the signal is low-pass (peak at ω=0\omega = 0). For −1<a<0-1<a<0 the peak moves to ω=π\omega=\pi (high-pass).

✓ Answer checked by Claude Opus 5.5 (high effort)

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