Questions: Parseval's Theorem and Energy/Power Spectral Density
5 questions to test your understanding
Score: 0 / 5
Question 1 Multiple Choice
A filter has |H(f)|² = 0.25 uniformly across 100–200 Hz and zero elsewhere. The input signal has energy spectral density |X(f)|² = 10 J/Hz uniformly across 100–200 Hz and zero elsewhere. What fraction of the input energy appears at the output?
A25% — because |H(f)|² = 0.25 means the filter passes 25% of the input energy in that band
B50% — because |H(f)| = 0.5 and the filter passes 50% of the signal amplitude
C100% — because the entire 100–200 Hz band passes through the filter
D6.25% — because we must square the 0.25 fraction again when computing energy
By Parseval's theorem, output energy = ∫|H(f)|²·|X(f)|² df. In the band 100–200 Hz, |H(f)|²·|X(f)|² = 0.25 × 10 = 2.5 J/Hz. Total output energy = 2.5 × 100 Hz = 250 J. Input energy = 10 × 100 = 1000 J. The fraction is 250/1000 = 25%. The key is that energy scales as |H(f)|² (power/magnitude-squared), not |H(f)|. Option B makes the classic error of using magnitude instead of magnitude-squared for energy calculations.
Question 2 Multiple Choice
A student computes |H(f)| = 0.7 at a frequency of interest and concludes the filter passes 70% of the signal energy at that frequency. Is this correct?
AYes — the magnitude of the frequency response directly gives the fraction of energy passed at each frequency
BNo — energy scales as |H(f)|² = 0.49, so the filter passes only 49% of the energy at that frequency
CNo — the student should use the phase of H(f), not the magnitude, for energy calculations
DYes, because Parseval's theorem states that a filter preserves total signal energy
Energy (and power) scale as the square of amplitude. If the magnitude response is |H(f)| = 0.7, the output amplitude at frequency f is 0.7 times the input amplitude, but the output energy density is |H(f)|² = 0.49 times the input energy density. This factor of two matters enormously in filter design: a filter specified as '–3 dB at cutoff' has |H(f)|² = 0.5 (half power), which corresponds to |H(f)| = 0.707, not 0.5. Option D is also wrong: Parseval's theorem says total energy is preserved by the Fourier transform, not by filtering.
Question 3 True / False
The total energy of a signal computed in the time domain equals the total energy computed in the frequency domain.
TTrue
FFalse
Answer: True
This is exactly Parseval's theorem: ∫|x(t)|² dt = ∫|X(f)|² df. The Fourier transform is a change of representation — it decomposes the signal into frequency components but does not alter the signal's total energy content. Just as rotating a coordinate system changes a vector's components but not its length, the Fourier transform changes how energy is distributed across the representation but preserves the total. This is why |X(f)|² is called energy spectral density — its integral equals the total energy.
Question 4 True / False
Applying the Fourier transform to a signal changes the signal's total energy.
TTrue
FFalse
Answer: False
The Fourier transform is a lossless change of representation — a bijection between time-domain and frequency-domain descriptions of the same signal. It does not add or remove energy; it re-expresses the same energy in a different basis. Parseval's theorem is the formal statement of this conservation: the L² norm is preserved under the Fourier transform. If the Fourier transform changed energy, it would not be a faithful representation of the original signal.
Question 5 Short Answer
How does Parseval's theorem make filter analysis more practical? What can you calculate without ever computing the filtered time-domain signal?
Think about your answer, then reveal below.
Model answer: Parseval's theorem lets you compute output energy (or power) entirely in the frequency domain, without ever performing an inverse Fourier transform. When a signal passes through a filter, the output spectrum is Y(f) = H(f)·X(f), so the output energy spectral density is |Y(f)|² = |H(f)|²·|X(f)|². The total output energy is ∫|H(f)|²·|X(f)|² df, and the energy in any specific frequency band is ∫_band |H(f)|²·|X(f)|² df. This means you can determine exactly how much energy a filter passes or rejects in any band — and verify that a filter design meets specifications — using only the frequency-domain representations, without computing the time-domain output. Filter specs are defined in terms of |H(f)|² (e.g., –3 dB cutoff, –40 dB stopband attenuation) for precisely this reason.
The practical value is computational and conceptual: frequency-domain energy analysis lets you reason about filtering effects directly from the spectra. You can ask 'how much noise (specified by its PSD) does this filter remove?' and get an exact answer without simulating the filtered output in time.