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

# FS coefficients; time scaling and conjugation

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2080 Asoj, 2+6 marks
- Video: [9. Continuous Time Fourier Series:  Properties (Conjugation and Conjugate Symmetry)](https://www.youtube.com/watch?v=7df6WYJJ-7E) — Momin Uppal
- 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-significance-scaling-conjugation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

What are the significance of Fourier series coefficients of a signal? State and prove time scaling and conjugation properties of continuous time Fourier series representation.

## Short answer

Fourier coefficients a_k give amplitude, phase, and power of each harmonic; a_0 is the DC value. Time scaling: x(αt) has same a_k with scaled fundamental frequency αω0. Conjugation: x*(t) yields a*_{-k}.

## Answer

#### Significance of Fourier series coefficients
The coefficients $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$ describe the signal in the frequency domain:
- $|a_k|$ gives the **amplitude** and $\angle a_k$ the **phase** of the $k$th harmonic (frequency $k\omega_0$); plotting them gives the magnitude and phase spectra.
- $a_0$ is the **dc (average) value** of the signal.
- $|a_k|^2$ is the **power** in the $k$th harmonic, and by Parseval $\sum|a_k|^2$ is the total average power.
- They show the **bandwidth** of the signal: how fast $|a_k|$ falls tells how many harmonics are needed (smooth signals need fewer).
- An LTI system acts on each harmonic separately: the output coefficients are $a_k H(jk\omega_0)$, which makes system analysis simple.

#### Time scaling property
**Statement:** if $x(t)$ has period $T$, $\omega_0 = 2\pi/T$, and $x(t) \overset{FS}{\longleftrightarrow} a_k$, then for $\alpha > 0$, $x(\alpha t)$ has period $T/\alpha$ and

$$
x(\alpha t) = \sum_{k=-\infty}^{\infty}a_k e^{jk(\alpha\omega_0)t}
$$

i.e. the coefficients are unchanged but the fundamental frequency becomes $\alpha\omega_0$.

**Proof:** let $y(t) = x(\alpha t)$, period $T' = T/\alpha$; put $\tau = \alpha t$:

$$
\begin{aligned}
b_k &= \frac{1}{T'}\int_0^{T'}x(\alpha t)e^{-jk\alpha\omega_0 t}dt \\
&= \frac{\alpha}{T}\int_0^{T}x(\tau)e^{-jk\omega_0\tau}\frac{d\tau}{\alpha} = a_k
\end{aligned}
$$

#### Conjugation property
**Statement:**

$$
x^*(t) \overset{FS}{\longleftrightarrow} a_{-k}^*
$$

**Proof:** take the conjugate of the synthesis equation:

$$
\begin{aligned}
x^*(t) &= \left[\sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}\right]^* = \sum_{k=-\infty}^{\infty}a_k^* e^{-jk\omega_0 t} \\
&= \sum_{m=-\infty}^{\infty}a_{-m}^*\,e^{jm\omega_0 t} \qquad (m = -k)
\end{aligned}
$$

So the coefficient of $e^{jm\omega_0 t}$ in $x^*(t)$ is $a_{-m}^*$.

**Consequence (conjugate symmetry):** if $x(t)$ is real, $x(t) = x^*(t)$, so $a_k = a_{-k}^*$. Hence $|a_k| = |a_{-k}|$ (even magnitude spectrum) and $\angle a_{-k} = -\angle a_k$ (odd phase spectrum). If $x(t)$ is also even, $a_k$ is real and even.

## Check yourself

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

1. What do |a_k| and angle a_k give?
   A. Amplitude and phase of the kth harmonic
   B. Period and duty cycle
   C. Power and energy of whole signal
   D. Peak and RMS
   Answer: A. They are the magnitude and phase spectra. (Key point: Significance of coefficients)

2. How is the output coefficient of an LTI system obtained?
   A. a_k + H(j k w0)
   B. a_k H(j k w0)
   C. a_k / H(j k w0)
   D. H(j w0)
   Answer: B. Each harmonic is scaled by the system frequency response at k w0. (Key point: LTI output)

3. What is the FS of x*(t)?
   A. a_k*
   B. a_(-k)*
   C. -a_k
   D. a_(-k)
   Answer: B. Conjugating the synthesis equation and substituting m = -k gives a_(-m)*. (Key point: Conjugation property)

4. What follows for real x(t)?
   A. Magnitude odd, phase even
   B. Magnitude even, phase odd (a_k = a_-k*)
   C. All a_k real
   D. a_0 = 0
   Answer: B. x = x* gives a_k = a_(-k)*. (Key point: Real signals)

## Related in Signals and Systems

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