- 2077 Chaitra · 4 marks
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:
Write .
Magnitude:
Phase:
Values for (taking for the sketch):
| (general) | , | Phase, | |
|---|---|---|---|
| 2 | |||
| 0.894 | |||
| 0.667 |
The phase reaches its most negative value, ( for ), where ( for ). By odd symmetry it reaches at .
|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 and periodic with period .
- The magnitude is even and the phase is odd (real signal).
- For the signal is low-pass (peak at ). For the peak moves to (high-pass).
✓ Answer checked by Claude Opus 5.5 (high effort)
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