--- title: "CTFS shifting, scaling and conjugation" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:ctfs-shift-scale-conjugation" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-shift-scale-conjugation site: BCT Engineering --- # CTFS shifting, scaling and conjugation Signals and Systems · Chapter 2: Fourier Series - Asked in: 2071 Bhadra, 6 marks - Video: [Properties of Fourier Series (Part 1)](https://www.youtube.com/watch?v=ED--r1lByl4) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-shift-scale-conjugation - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Discuss the following properties of continuous time Fourier series. (a) Time shifting (b) Time scaling (c) Conjugation ## Short answer Time shifting: x(t-t0) ↔ a_k e^{-j k ω0 t0}, magnitude unchanged. Time scaling: x(αt) has same coefficients a_k but period T/α and harmonic spacing αω0. Conjugation: x*(t) ↔ a_{-k}^*; for real signals, conjugate symmetry yields even magnitude and odd phase. ## Answer Let $x(t)$ be periodic with period $T$ ($\omega_0 = 2\pi/T$) and $x(t) \xleftrightarrow{FS} a_k$, where $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$. #### (a) Time shifting $$ x(t - t_0) \xleftrightarrow{FS} a_k e^{-jk\omega_0 t_0} $$ **Proof:** put $\tau = t - t_0$: $$ b_k = \frac{1}{T}\int_T x(t-t_0)e^{-jk\omega_0 t}dt = e^{-jk\omega_0 t_0}\frac{1}{T}\int_T x(\tau)e^{-jk\omega_0 \tau}d\tau = e^{-jk\omega_0 t_0}a_k $$ The magnitudes $|b_k| = |a_k|$ stay the same. Only a linear phase $-k\omega_0 t_0$ is added. #### (b) Time scaling If $\alpha > 0$, then $x(\alpha t)$ is periodic with period $T/\alpha$ and fundamental frequency $\alpha\omega_0$. Its Fourier series is $$ x(\alpha t) = \sum_{k=-\infty}^{\infty} a_k e^{jk(\alpha\omega_0)t} $$ **Proof:** with period $T' = T/\alpha$, put $\tau = \alpha t$: $$ b_k = \frac{\alpha}{T}\int_{T/\alpha} x(\alpha t)e^{-jk(\alpha\omega_0)t}dt = \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0\tau}d\tau = a_k $$ The coefficients do not change. What changes is the spacing of the harmonics, which becomes $\alpha\omega_0$ instead of $\omega_0$. Compressing a signal in time spreads its spectral lines further apart. #### (c) Conjugation $$ x^*(t) \xleftrightarrow{FS} a_{-k}^* $$ **Proof:** take the conjugate of the synthesis equation: $$ x^*(t) = \sum_{k} a_k^* e^{-jk\omega_0 t} = \sum_{k} a_{-k}^* e^{jk\omega_0 t} \quad (k \to -k) $$ So the coefficients of $x^*(t)$ are $a_{-k}^*$. **Consequence (conjugate symmetry):** if $x(t)$ is real, then $x(t) = x^*(t)$, so $a_k = a_{-k}^*$. This gives $|a_k| = |a_{-k}|$ (even magnitude) and $\angle a_{-k} = -\angle a_k$ (odd phase). If $x(t)$ is also even, the $a_k$ are real and even. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. If x(t) has Fourier series coefficients a_k, what are the coefficients of x(t - t0)? A. a_k e^{-jkω0 t0} B. a_k e^{+jkω0 t0} C. a_{k-t0} D. a_k / t0 Answer: A. Substituting τ = t - t0 pulls out the factor e^{-jkω0 t0}, so b_k = a_k e^{-jkω0 t0}. (Key point: Time-shift property) 2. For α > 0, what happens to the Fourier series coefficients of x(αt)? A. They are multiplied by α B. They are divided by α C. They stay the same a_k, but the harmonic spacing becomes αω0 D. They stay the same and so does the fundamental frequency Answer: C. x(αt) = Σ a_k e^{jk(αω0)t}: same coefficients, with fundamental frequency αω0 and period T/α. (Key point: Time scaling property) 3. If a periodic signal is compressed in time (α > 1), how does its line spectrum change? A. Lines move closer together B. Lines spread further apart C. Lines keep the same spacing but get taller D. Lines disappear and the spectrum becomes continuous Answer: B. The spacing between harmonics becomes αω0, which is larger than ω0. (Key point: Compression spreads spectrum) 4. For a real x(t), which statement about a_k is correct? A. a_k = a_{-k}* so the magnitude is even and the phase is odd B. a_k = -a_{-k}* so the magnitude is odd C. a_k is always real D. a_k = a_{-k} so the phase is even Answer: A. x = x* gives a_k = a_{-k}*, hence |a_k| = |a_{-k}| and angle a_{-k} = -angle a_k. 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