---
title: "FS coefficients; time shifting and conjugation"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:fs-information-shift-conjugation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/fs-information-shift-conjugation
site: BCT Engineering
---

# FS coefficients; time shifting and conjugation

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2070 Bhadra, 1+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/fs-information-shift-conjugation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

What is the information provided by Fourier series coefficients of a signal? State and prove time shifting and conjugation properties of continuous time Fourier series representation.

## Short answer

FS coefficients a_k give the amplitude |a_k| and phase of each harmonic; a0 is DC, and ∑|a_k|^2 is total power. Time shift: x(t-t0) ↔ a_k e^{-j k ω0 t0}. Conjugation: x*(t) ↔ a_{-k}^*, giving conjugate symmetry for real signals, even magnitude, odd phase.

## Answer

#### Information provided by Fourier series coefficients
The coefficient $a_k$ gives the amount (magnitude $|a_k|$) and the phase ($\angle a_k$) of the $k$th harmonic $e^{jk\omega_0 t}$ in the signal. This makes the coefficients the **frequency-domain picture** of the signal. $a_0$ is the DC (average) value, $|a_k|^2$ is the power at frequency $k\omega_0$ (by Parseval, $P = \sum|a_k|^2$), and the spread of the $a_k$ shows the bandwidth of the signal.

#### Time shifting property
**Statement:** if $x(t) \xleftrightarrow{FS} a_k$ with period $T$ and $\omega_0 = 2\pi/T$, then
$$
x(t - t_0) \xleftrightarrow{FS} e^{-jk\omega_0 t_0}a_k
$$
**Proof:** the coefficients of $y(t) = x(t - t_0)$ are
$$
b_k = \frac{1}{T}\int_T x(t-t_0)e^{-jk\omega_0 t}dt
$$
Put $\tau = t - t_0$ (so $dt = d\tau$). The range of $\tau$ is still one full period:
$$
\begin{aligned}
b_k &= \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0(\tau + t_0)}d\tau \\
&= e^{-jk\omega_0 t_0}\cdot\frac{1}{T}\int_T x(\tau)e^{-jk\omega_0 \tau}d\tau = e^{-jk\omega_0 t_0}a_k
\end{aligned}
$$
**Interpretation:** $|b_k| = |a_k|$, so the magnitude spectrum does not change. The phase changes by $-k\omega_0 t_0$, which is proportional to the harmonic number.

#### Conjugation property
**Statement:**
$$
x^*(t) \xleftrightarrow{FS} a_{-k}^*
$$
**Proof:** the coefficients of $x^*(t)$ are
$$
\begin{aligned}
b_k &= \frac{1}{T}\int_T x^*(t)e^{-jk\omega_0 t}dt = \left[\frac{1}{T}\int_T x(t)e^{+jk\omega_0 t}dt\right]^* \\
&= \left[\frac{1}{T}\int_T x(t)e^{-j(-k)\omega_0 t}dt\right]^* = a_{-k}^*
\end{aligned}
$$
**Consequence:** if $x(t)$ is real, then $x(t) = x^*(t)$, which gives $a_{-k} = a_k^*$ (conjugate symmetry). The magnitude spectrum is then even and the phase spectrum is odd.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What information does the coefficient a_k provide?
   A. Only the frequency of the kth harmonic
   B. The amount (magnitude) and phase of the kth harmonic in the signal
   C. Only the period of the signal
   D. The time at which the signal peaks
   Answer: B. |a_k| gives the amount and angle a_k the phase of the harmonic e^{jkω0 t}. (Key point: Meaning of a_k)

2. In proving the time-shift property, substituting τ = t - t0 gives the factor:
   A. e^{+jkω0 t0} outside the integral
   B. e^{-jkω0 t0} outside the integral
   C. 1/t0
   D. cos(kω0 t0)
   Answer: B. e^{-jkω0 t} = e^{-jkω0(τ + t0)}, which separates into e^{-jkω0 t0} times the integral for a_k. (Key point: Time-shift proof step)

3. After a time shift t0, the phase of the kth harmonic changes by:
   A. A constant independent of k
   B. -kω0 t0, proportional to the harmonic number
   C. +t0 for every harmonic
   D. Nothing; only the magnitude changes
   Answer: B. Magnitudes are unchanged; the phase change -kω0 t0 grows with k. (Key point: Interpretation of time shift)

4. If x(t) is real, the conjugation property implies:
   A. a_{-k} = a_k*, so the magnitude spectrum is even and the phase is odd
   B. a_{-k} = -a_k, so the magnitude is odd
   C. a_k = 0 for k < 0
   D. The phase spectrum is even
   Answer: A. x = x* gives a_{-k} = a_k* (conjugate symmetry). (Key point: Conjugation consequence)

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