--- title: "PSD–autocorrelation relation; stereo FM" subject: "Communication Systems" chapter: 3 key: "semester_6/communication_systems:psd-autocorrelation-stereo-fm" source: https://www.bctengineer.com/q/semester_6/communication_systems/psd-autocorrelation-stereo-fm site: BCT Engineering --- # PSD–autocorrelation relation; stereo FM Communication Systems · Chapter 3: Angle Modulation - Asked in: 2078 Chaitra, 3+7 marks - Video: [Relation between PSD and auto correlation || Wernier kincheff || Signals and Systems](https://www.youtube.com/watch?v=0j5voZJG79w) — Friend's Explanation - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/psd-autocorrelation-stereo-fm - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3 ## Question What is the relation between psdf and Autocorrelation function? Explain the Stereo FM encoder and decoder with spectral diagram. ## Short answer By the Wiener–Khinchin theorem, power spectral density and autocorrelation function are a Fourier-transform pair: S_X(f) = ∫ R_X(τ)e^{-j2πfτ}dτ. For stereo FM, an encoder combines L+R (0–15 kHz) and L−R DSB-SC centered at 38 kHz with a 19 kHz pilot. The decoder extracts the pilot to regenerate the 38 kHz carrier, coherently detects L−R, and a matrix recovers L and R. ## Answer #### Relation between PSD and autocorrelation By the **Wiener–Khinchin theorem**, the power spectral density (PSD) $S_X(f)$ of a wide-sense stationary process and its autocorrelation function $R_X(\tau)$ form a Fourier transform pair: $$ S_X(f) = \int_{-\infty}^{\infty}R_X(\tau)\,e^{-j2\pi f\tau}d\tau, \qquad R_X(\tau) = \int_{-\infty}^{\infty}S_X(f)\,e^{j2\pi f\tau}df $$ Consequences: - Total average power: $P = R_X(0) = \int S_X(f)\,df$. - $S_X(f)$ is real and even (because $R_X(\tau)$ is real and even), and it is never negative. - Example: white noise $R(\tau) = \frac{N_0}{2}\delta(\tau) \Leftrightarrow S(f) = \frac{N_0}{2}$. #### Stereo FM encoder Stereo FM sends left (L) and right (R) channels on one carrier while staying compatible with mono receivers. ```text L --+-->(+)--L+R-->[LPF]----------->( Σ )--> | ^ ^ ^ to FM R --|----+ | | mod. +-->(-)--L-R-->[LPF]-->[Bal.mod]--+ | ^ (R) ^ 38k | [19k osc]->[x2] | | | +--19 kHz pilot----+ ``` 1. Matrix forms $L+R$ and $L-R$; both are band-limited to 15 kHz and pre-emphasized. 2. $L-R$ DSB-SC modulates a 38 kHz sub-carrier in a balanced modulator. 3. The 38 kHz sub-carrier comes from a 19 kHz oscillator and a doubler; a small 19 kHz pilot is also added. 4. Composite signal: $$ m(t) = (L+R) + (L-R)\cos(2\pi\,38\text{k}\,t) + A_p\cos(2\pi\,19\text{k}\,t) $$ #### Spectrum ```text | L+R pilot L-R DSB-SC |______ | ____ ____ | | | | | | | +------+------+---+----+--+----+---> f (kHz) 0 15 19 23 38 53 ``` #### Stereo FM decoder ```text FM +->[LPF 0-15k]------------L+R--+->(+)--> 2L demod | | ^ out --+->[BPF 23-53k]-->( X )->[LPF]-+ | | ^ L-R | | | | 38k +--|-->(-)--> 2R +->[BPF 19k]-->[ x2 ] | (pilot) L-R ---+ ``` 1. FM discriminator gives the composite signal. 2. LPF (0–15 kHz) gives $L+R$ (all a mono set needs). 3. A narrow BPF picks the 19 kHz pilot; a doubler makes a phase-locked 38 kHz carrier. 4. BPF (23–53 kHz) selects the $L-R$ DSB-SC signal, which is coherently detected with the 38 kHz carrier and low-pass filtered. 5. Matrix: $(L+R)+(L-R) = 2L$ and $(L+R)-(L-R) = 2R$, then de-emphasis and audio amplifiers. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the relation between PSD and autocorrelation of a WSS process? A. They form a Fourier transform pair B. PSD is the derivative of the autocorrelation C. PSD is the square of the autocorrelation D. PSD is the inverse of the autocorrelation Answer: A. Wiener-Khinchin: S_X(f) is the Fourier transform of R_X(tau), and R_X(tau) is its inverse transform. (Key point: Wiener-Khinchin theorem) 2. How is the total average power obtained from these functions? A. P = R_X(infinity) B. P = S_X(0) C. P = R_X(0) = integral of S_X(f) df D. P = derivative of R_X at tau = 0 Answer: C. Setting tau = 0 in the inverse transform gives R_X(0), the area under the PSD. (Key point: Total power) 3. What PSD corresponds to R(tau) = (N0/2) delta(tau)? A. A delta function at f = 0 B. A constant N0/2 at all frequencies C. A rectangle limited to 15 kHz D. A PSD proportional to f^2 Answer: B. The Fourier transform of an impulse is a constant, so white noise has a flat PSD of N0/2. (Key point: White noise) 4. Which set of components forms the composite stereo FM baseband signal? A. L+R (0-15 kHz), 19 kHz pilot, and L-R DSB-SC around 38 kHz B. L+R at 38 kHz, L-R baseband, and a 19 kHz pilot C. L and R sent as two separate FM carriers D. L-R baseband, R at 38 kHz DSB-SC, and a 38 kHz pilot Answer: A. The sum L+R stays at baseband for mono compatibility, L-R is DSB-SC on 38 kHz (23-53 kHz), and a 19 kHz pilot is added. (Key point: Composite signal) 5. Why is a low-level 19 kHz pilot transmitted? A. To carry the L+R signal for mono sets B. To let the receiver regenerate a phase-locked 38 kHz carrier for detecting L-R C. To increase the FM deviation to 75 kHz D. To act as the pre-emphasis reference Answer: B. Doubling the recovered pilot gives a 38 kHz carrier in phase with the transmitter's, needed for coherent detection of L-R. (Key point: Purpose of 19 kHz pilot) 6. Why can a mono receiver still play a stereo broadcast properly? A. It demodulates the L-R DSB-SC signal and ignores L+R B. It subtracts the pilot from the composite signal C. Its 0-15 kHz low-pass filter picks out L+R, which is the full mono program D. It decodes the 38 kHz subcarrier with a fixed oscillator Answer: C. L+R occupies 0-15 kHz at baseband, so a simple LPF recovers the mono signal and the higher components are filtered out. (Key point: Mono compatibility) 7. In the decoder matrix, how are the left and right outputs obtained? A. (L+R)+(L-R) gives 2R and (L+R)-(L-R) gives 2L B. L+R and L-R are used directly as L and R C. (L-R) is divided by the pilot to obtain R D. (L+R)+(L-R) gives 2L and (L+R)-(L-R) gives 2R Answer: D. Adding the two signals cancels R and leaves 2L; subtracting cancels L and leaves 2R. (Key point: Decoder matrix) ## Related in Communication Systems - [Stereo FM encoder and decoder block diagram](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Astereo-fm-encoder-decoder-blocks) — Communication Systems, chapter 3 - [Stereo FM encoder and decoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Astereo-fm-encoder-decoder) — Communication Systems, chapter 3, asked 2× - [PSD expression and relation with autocorrelation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apsd-expression-autocorrelation) — Communication Systems, chapter 1 - [Parseval's theorem; PSD–autocorrelation relation for white noise](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aparseval-psd-autocorrelation-white-noise) — Communication Systems, chapter 9 - [Pre-emphasis in FM broadcasting; stereo encoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apre-emphasis-stereo-encoder) — Communication Systems, chapter 3, asked 2× - [White noise: relation between PSDF and AC function](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Awhite-noise-psdf-ac-relation) — Communication Systems, chapter 9 ## Related in other subjects - [Encoder vs decoder; 8-to-3 encoder](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aencoder-decoder-8to3) — Digital Logic, chapter 4 - [Encoder; 4×16 decoder from 2×4 decoders](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aencoder-4x16-decoder) — Digital Logic, chapter 4