--- title: "Parseval's theorem; PSD–autocorrelation relation for white noise" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:parseval-psd-autocorrelation-white-noise" source: https://www.bctengineer.com/q/semester_6/communication_systems/parseval-psd-autocorrelation-white-noise site: BCT Engineering --- # Parseval's theorem; PSD–autocorrelation relation for white noise Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2080 Chaitra (CS I), 2+6 marks - Video: [WIENER-KHINCHIN RELATION [Relationship Between Power Spectrum and Auto Correlation Function]](https://www.youtube.com/watch?v=Yt7ZhjeNQHQ) — PRAKASAM TUTORIALS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/parseval-psd-autocorrelation-white-noise - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question State Parseval's Theorem. Explain the relation between power spectral density function and autocorrelation function with the example of white noise. ## Short answer Parseval's theorem states that signal power or energy is equal whether computed in time or frequency domain. The Wiener–Khinchin relation shows that power spectral density and autocorrelation are Fourier transform pairs. For white noise, the constant PSD N0/2 yields an impulse autocorrelation (N0/2)δ(τ), so samples at different times are uncorrelated, and R(0) equals total power. ## Answer #### Parseval's theorem **Parseval's theorem** states that the power (or energy) of a signal is the same whether computed in the time domain or in the frequency domain. - Periodic power signal with Fourier coefficients $c_n$: $P=\frac{1}{T_0}\int_{T_0}\lvert g(t)\rvert^2dt=\sum_{n=-\infty}^{\infty}\lvert c_n\rvert^2$ - Energy signal (Rayleigh form): $E=\int\lvert g(t)\rvert^2dt=\int\lvert G(f)\rvert^2df$ #### Relation between PSD and autocorrelation **Wiener–Khinchin relation (PSD $\leftrightarrow$ AC):** for a power signal, $R_g(\tau)=\lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}g_T(t)g_T(t+\tau)dt$. The time-domain integral is $g_T(\tau)*g_T(-\tau)$ scaled by $1/T$, and $g_T(-t)\leftrightarrow G_T^*(f)$, so $$ \begin{aligned} \mathcal{F}\{R_g(\tau)\} &= \lim_{T\to\infty}\frac{1}{T}G_T(f)G_T^*(f)=\lim_{T\to\infty}\frac{\lvert G_T(f)\rvert^2}{T}=S_g(f) \end{aligned} $$ Hence $$ S_g(f)=\int_{-\infty}^{\infty}R_g(\tau)e^{-j2\pi f\tau}d\tau,\qquad R_g(\tau)=\int_{-\infty}^{\infty}S_g(f)e^{j2\pi f\tau}df $$ and at $\tau=0$: $R_g(0)=\int S_g(f)df=P$. **Example: white noise.** The PSD is flat, $S_W(f)=N_0/2$. Then by the relation above: $$ R_W(\tau)=\int_{-\infty}^{\infty}\frac{N_0}{2}e^{j2\pi f\tau}df=\frac{N_0}{2}\delta(\tau) $$ ```text S_W(f) R_W(tau) N0/2+----------------- ^ (N0/2) delta(tau) | | ----+-----------------> f ------+------> tau ``` A flat (infinitely wide) spectrum corresponds to an impulse autocorrelation: white noise is uncorrelated with itself at any non-zero shift. Conversely, if white noise is band-limited to $B$, $R(\tau)=N_0B\,\text{sinc}(2B\tau)$ spreads out (with $\text{sinc}(x)=\sin\pi x/\pi x$); the narrower the spectrum, the wider the autocorrelation. **Link with Parseval:** putting $\tau=0$ in $R(\tau)=\int S(f)e^{j2\pi f\tau}df$ gives $R(0)=\int S(f)df=P$, which is exactly Parseval's theorem for power: the time-average power equals the area under the PSD. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does Parseval's theorem state? A. Power is always zero in the frequency domain B. Power or energy is the same in the time and frequency domains C. Energy equals the peak amplitude squared D. PSD equals the signal squared Answer: B. Integral |g|^2 dt equals integral |G|^2 df. (Key point: Parseval's theorem) 2. How are PSD and autocorrelation related? A. They are a Fourier transform pair B. PSD is the derivative of R(tau) C. PSD equals R(tau) squared D. They are unrelated Answer: A. S(f) is the Fourier transform of R(tau) and vice versa. (Key point: Wiener-Khinchin) 3. A flat PSD corresponds to what autocorrelation? A. A constant for all tau B. A sinc function C. An impulse at tau = 0 D. An exponential decay Answer: C. The inverse transform of a constant is a delta function. (Key point: White noise example) 4. What happens to R(tau) as the noise band shrinks? A. It becomes narrower B. It stays an impulse C. It becomes zero D. It spreads wider Answer: D. A narrower spectrum gives a wider autocorrelation; R = N0 B sinc(2B tau). 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