Q5Digital Signal Processing

Question

Q.5. A signal x(t) = exp(-2πBt) u(t) is the input to an ideal low pass filter with bandwidth B Hz. The output is denoted by y(t). Evaluate:

\int_{- \infty}^{\infty} [y (t) - x (t)]^{2} d t

Answer 1

The input signal is x(t) = e^(-2πBt)u(t), a causal decaying exponential. An ideal low-pass filter with bandwidth B Hz passes all frequency content of x(t) within |f| ≤ B unchanged and completely removes (sets to zero) all frequency content with |f| > B, producing output y(t).

By Parseval's theorem, the mean-square error between input and output equals the energy of the removed (high-frequency, |f|>B) portion of the spectrum, since y(t) - x(t) in the frequency domain is exactly -X(f) for |f|>B and 0 for |f|≤B (the part of X(f) that the ideal LPF removed):

\int_{- \infty}^{\infty} [y (t) - x (t)]^{2} d t = \int_{∣ f ∣ > B} ∣ X (f) ∣^{2} df

The Fourier transform of x(t) = e^(-2πBt)u(t) is a standard pair:

X (f) = \frac{1}{2 π B + j 2 π f} \Rightarrow ∣ X (f) ∣^{2} = \frac{1}{4 π ^{2} B ^{2} + 4 π ^{2} f ^{2}} = \frac{1}{4 π ^{2} ( B ^{2} + f ^{2} )}

The removed energy (outside the passband) is:

E_{r e m o v e d} = 2 \int_{B}^{\infty} \frac{1}{4 π ^{2} ( B ^{2} + f ^{2} )} df = \frac{1}{2 π ^{2}} \int_{B}^{\infty} \frac{df}{B ^{2} + f ^{2}}

Using the standard integral ∫df/(B²+f²) = (1/B)tan⁻¹(f/B):

E_{r e m o v e d} = \frac{1}{2 π ^{2} B} [tan^{- 1} (\frac{f}{B})]_{B}^{\infty} = \frac{1}{2 π ^{2} B} (\frac{π}{2} - \frac{π}{4}) = \frac{1}{2 π ^{2} B} \cdot \frac{π}{4} = \frac{1}{8 π B}

Therefore:

\int_{- \infty}^{\infty} [y (t) - x (t)]^{2} d t = \frac{1}{8 π B} \approx \frac{0.0398}{B}

This result shows that the mean-square error introduced by ideally low-pass filtering this exponential signal at exactly its own natural bandwidth parameter B is inversely proportional to B, meaning that for larger bandwidth B (a faster-decaying, more broadband exponential), a proportionally smaller fraction of its total energy lies beyond the B Hz cutoff and is therefore lost, consistent with the intuition that the exponential's spectral energy becomes more concentrated within any fixed multiple of its own characteristic bandwidth B as B increases.

Q5Digital Signal Processing

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