---
title: "Harmonic exponentials; CTFS conjugate symmetry"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:harmonic-exponentials-conjugation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/harmonic-exponentials-conjugation
site: BCT Engineering
---

# Harmonic exponentials; CTFS conjugate symmetry

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2070 Magh, 4+4 marks
- Video: [Complex Exponential Fourier Series](https://www.youtube.com/watch?v=oXrbcRyXj84) — 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/harmonic-exponentials-conjugation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

How could you represent a signal x(t) with harmonically related exponentials? State and prove conjugation and conjugate symmetry property of CTFS.

## Short answer

A periodic signal x(t) is represented by harmonically related exponentials: x(t)=∑ a_k e^{j k ω0 t}, with a_k=(1/T)∫_T x(t)e^{-j k ω0 t}dt. Conjugation: x*(t) ↔ a_{-k}^*. For real x(t), a_{-k}=a_k^*, giving even magnitude and odd phase.

## Answer

#### Representation with harmonically related exponentials
Take a periodic signal $x(t)$ with fundamental period $T$ and $\omega_0 = 2\pi/T$. The set of **harmonically related complex exponentials**
$$
\phi_k(t) = e^{jk\omega_0 t}, \quad k = 0, \pm1, \pm2, \dots
$$
are all periodic with period $T$, since each frequency is an integer multiple of $\omega_0$. The term for $k = 0$ is the DC part, $k = \pm1$ are the fundamental components, and $k = \pm N$ are the $N$th harmonics. A linear combination of them is periodic with period $T$:
$$
x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t} \quad \text{(synthesis equation)}
$$
The functions are orthogonal over one period: $\int_T e^{jk\omega_0 t}e^{-jn\omega_0 t}dt$ equals $T$ when $k = n$ and 0 otherwise. Multiplying the synthesis equation by $e^{-jn\omega_0 t}$ and integrating over $T$ therefore leaves only one term. This gives
$$
a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt \quad \text{(analysis equation)}
$$
The series converges to $x(t)$ when $x(t)$ meets the Dirichlet conditions. **Example:** $\cos\omega_0 t = \frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omega_0 t}$, so $a_{\pm1} = 1/2$.

#### Conjugation property
**Statement:** if $x(t) \xleftrightarrow{FS} a_k$, then
$$
x^*(t) \xleftrightarrow{FS} a_{-k}^*
$$
**Proof:** let $b_k$ be the coefficients of $x^*(t)$:
$$
b_k = \frac{1}{T}\int_T x^*(t)e^{-jk\omega_0 t}dt = \left[\frac{1}{T}\int_T x(t)e^{jk\omega_0 t}dt\right]^* = \left[\frac{1}{T}\int_T x(t)e^{-j(-k)\omega_0 t}dt\right]^* = a_{-k}^*
$$

#### Conjugate symmetry property
**Statement:** if $x(t)$ is real, its coefficients are conjugate symmetric:
$$
a_{-k} = a_k^*
$$
**Proof:** for a real signal $x(t) = x^*(t)$, so the two signals have the same coefficients: $a_k = b_k = a_{-k}^*$. Taking the conjugate of both sides gives $a_k^* = a_{-k}$.

**Consequences for real $x(t)$:**
- $|a_{-k}| = |a_k|$, so the magnitude spectrum is even.
- $\angle a_{-k} = -\angle a_k$, so the phase spectrum is odd.
- $\mathrm{Re}\{a_k\}$ is even and $\mathrm{Im}\{a_k\}$ is odd.
- Real and even $x(t)$ gives real, even $a_k$. Real and odd $x(t)$ gives purely imaginary, odd $a_k$.

For example, $\sin\omega_0 t$ has $a_1 = 1/(2j)$ and $a_{-1} = -1/(2j) = a_1^*$.

## Check yourself

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

1. How is the analysis equation for a_k obtained from the synthesis equation?
   A. Differentiate both sides with respect to t
   B. Multiply by e^{-jnω0 t} and integrate over one period; orthogonality leaves only one term
   C. Set t = 0
   D. Take the conjugate of both sides
   Answer: B. The integral of e^{j(k-n)ω0 t} over T is T for k = n and 0 otherwise, so only a_n survives. (Key point: Orthogonality used to find a_k)

2. The coefficients of x*(t) are found as b_k = [ (1/T) ∫ x(t) e^{jkω0 t} dt ]*. This equals:
   A. a_k*
   B. a_{-k}*
   C. a_{-k}
   D. -a_k
   Answer: B. e^{jkω0 t} = e^{-j(-k)ω0 t}, so the bracket is a_{-k}, and its conjugate is a_{-k}*. (Key point: Conjugation property)

3. For real x(t), why does a_{-k} = a_k* follow?
   A. Because x(t) = x*(t), so a_k = b_k = a_{-k}*; conjugating both sides gives a_k* = a_{-k}
   B. Because real signals have only cosine terms
   C. Because a_0 is zero for real signals
   D. Because the integral over T vanishes
   Answer: A. A real signal equals its conjugate, so their coefficients are equal: a_k = a_{-k}*. (Key point: Conjugate symmetry proof)

4. A real and odd periodic signal x(t) has Fourier coefficients that are:
   A. Real and even
   B. Purely imaginary and odd
   C. Purely imaginary and even
   D. Real and odd
   Answer: B. Real and odd x(t) gives purely imaginary, odd a_k, e.g. sin(ω0 t) has a_1 = 1/(2j) and a_-1 = -1/(2j). (Key point: Real and odd signal)

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