# 4.3: Classic Fourier Series

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- 1615

Learning Objectives

- Signals can be composed by a superposition of an infinite number of sine and cosine functions.
- The coefficients of the superposition depend on the signal being represented and are equivalent to knowing the function itself.

The classic Fourier series as derived originally expressed a periodic signal (period **T**) in terms of harmonically related sines and cosines.

\[s(t)=a_{0}+\sum_{k=1}^{\infty }a_{k}\cos \left ( \frac{2\pi kt}{T} \right )+\sum_{k=1}^{\infty }b_{k}\sin \left ( \frac{2\pi kt}{T}\right )\]

**The complex Fourier series and the sine-cosine series are identical**, each representing a signal's spectrum. The **Fourier coefficients**, **a _{k }**and

**b**, express the real and imaginary parts respectively of the spectrum while the coefficients

_{k}**c**of the complex Fourier series express the spectrum as a magnitude and phase. Equating the classic Fourier series to the complex Fourier series, an extra factor of two and complex conjugate become necessary to relate the Fourier coefficients in each.

_{k }\[c_{k}=\frac{1}{2}\left ( a_{k}-ib_{k} \right )\]

Exercise \(\PageIndex{1}\)

Derive this relationship between the coefficients of the two Fourier series.

**Solution**

Write the coefficients of the complex Fourier series in Cartesian form as:

\[c_{k}=A_{k}+iB_{k}\]

Now substitute into the expression for the complex Fourier series.

\[\sum_{k=-\infty }^{\infty }c_{k}e^{i\frac{2\pi kt}{T}}=\sum_{k=-\infty }^{\infty }\left ( A_{k}+iB_{k}\right ) e^{i\frac{2\pi kt}{T}}\]

Simplifying each term in the sum using Euler's formula,

\[\left ( A_{k}+iB_{k}\right ) e^{i\frac{2\pi kt}{T}}=\left ( A_{k}+iB_{k}\right )\left ( \cos \left ( \frac{2\pi kt}{T} \right )+i\sin \left ( \frac{2\pi kt}{T}\right )\right )\]

\[\left ( A_{k}+iB_{k}\right ) e^{i\frac{2\pi kt}{T}}=A_{k}\cos \left ( \frac{2\pi kt}{T}\right )-B_{k} \sin \left ( \frac{2\pi kt}{T}\right )+i\left ( A_{k}\sin \left ( \frac{2\pi kt}{T}\right )+B_{k} \cos \left ( \frac{2\pi kt}{T}\right )\right )\]

We now combine terms that have the same frequency index **in magnitude**. Because the signal is real-valued, the coefficients of the complex Fourier series have conjugate symmetry:

\[c_{-k}=\overline{c_{k}}\; or \; A_{-k}=\overline{A_{k}}\; and\; B_{-k}=\overline{B_{k}}\]

After we add the positive-indexed and negative-indexed terms, each term in the Fourier series becomes:

\[2A_{k}\cos \left ( \frac{2\pi kt}{T}\right )-2B_{k} \sin \left ( \frac{2\pi kt}{T}\right )\]

To obtain the classic Fourier series, we must have:

\[2A_{k}=a_{k}\; and\; 2B_{k}=-b_{k}\]

Just as with the complex Fourier series, we can find the Fourier coefficients using the **orthogonality** properties of sinusoids. Note that the cosine and sine of harmonically related frequencies, even the **same** frequency, are orthogonal.

\[\forall k,l,k\in \mathbb{Z}l\in \mathbb{Z}:\left ( \int_{0}^{T}\sin \left ( \frac{2\pi kt}{T} \right ) \cos \left ( \frac{2\pi lt}{T} \right )dt=0\right )\]

\[\int_{0}^{T}\sin \left ( \frac{2\pi kt}{T} \right ) \sin \left ( \frac{2\pi lt}{T} \right )dt=\begin{cases} \frac{T}{2} & \text{ if } (k=l)\wedge (k\neq 0)\wedge (l\neq 0) \\ 0 & \text{ if } (k\neq l)\vee (k=0=l) \end{cases}\]

\[\int_{0}^{T}\cos \left ( \frac{2\pi kt}{T} \right ) \cos \left ( \frac{2\pi lt}{T} \right )dt=\begin{cases} \frac{T}{2} & \text{ if } (k=l)\wedge (k\neq 0)\wedge (l\neq 0) \\ T & \text{ if } k=0=l \\ 0 & \text{ if } k\neq l \end{cases}\]

These orthogonality relations follow from the following important trigonometric identities.

\[\sin (\alpha )\sin (\beta )=\frac{1}{2}\left ( \cos (\alpha -\beta )-\cos (\alpha +\beta )\right )\]

\[\cos (\alpha )\cos (\beta )=\frac{1}{2}\left ( \cos (\alpha +\beta )+\cos (\alpha -\beta )\right )\]

\[\sin (\alpha )\cos (\beta )=\frac{1}{2}\left ( \sin (\alpha +\beta )+\sin (\alpha -\beta )\right )\]

These identities allow you to substitute a sum of sines and/or cosines for a product of them. Each term in the sum can be integrated by noticing one of two important properties of sinusoids.

- The integral of a sinusoid over an
**integer**number of periods equals zero. - The integral of the
**square**of a unit-amplitude sinusoid over a period**T**equals**T/2**.

To use these, let's, for example, multiply the Fourier series for a signal by the cosine of the **l ^{th}** harmonic:

\[\cos \left ( \frac{2\pi lt}{T} \right )\]

and integrate.

The idea is that, because integration is linear, the integration will sift out all but the term involving **a _{l}**.

\[\int_{0}^{T}s(t)\cos \left ( \frac{2\pi lt}{T} \right )dt=\int_{0}^{T}a_{0}\cos \left ( \frac{2\pi lt}{T} \right )dt+\sum_{k=1}^{\infty }a_{k}\int_{0}^{T}\cos \left ( \frac{2\pi kt}{T} \right )\cos \left ( \frac{2\pi lt}{T} \right )dt+\sum_{k=1}^{\infty }b_{k}\int_{0}^{T}\sin \left ( \frac{2\pi kt}{T} \right )\cos \left ( \frac{2\pi lt}{T} \right )dt\]

The first and third terms are zero; in the second, the only non-zero term in the sum results when the indices **k **and **l **are equal (but not zero), in which case we obtain:

\[\frac{a_{1}T}{2}\]

\[If\; \; k=0=l\; \; a_{0}T\; is\; obtained\]

Consequently,

\[\forall l,l\neq 0:\left ( a_{l}=\frac{2}{T} \int_{0}^{T}s(t)\cos \left ( \frac{2\pi lt}{T} \right )dt\right )\]

All of the Fourier coefficients can be found similarly.

\[a_{0}=\frac{2}{T} \int_{0}^{T}s(t)dt\]

\[\forall k,k\neq 0:\left ( a_{k}=\frac{2}{T} \int_{0}^{T}s(t)\cos \left ( \frac{2\pi kt}{T} \right )dt\right )\]

\[b_{k}=\frac{2}{T} \int_{0}^{T}s(t)\sin \left ( \frac{2\pi kt}{T} \right )dt\]

Exercise \(\PageIndex{1}\)

The expression for **a _{0}** is referred to as the

**average value**of

**s(t)**. Why?

**Solution**

The average of a set of numbers is the sum divided by the number of terms. Viewing signal integration as the limit of a Riemann sum, the integral corresponds to the average.

Exercise \(\PageIndex{1}\)

What is the Fourier series for a unit-amplitude square wave?

**Solution**

We found that the complex Fourier series coefficients are given by

\[c_{k}=\frac{2}{i\pi k}\]

The coefficients are pure imaginary, which means **a _{k}=0**. The coefficients of the sine terms are given by:

\[b_{k}=-\left ( 2\Im (c_{k}) \right )\]

\[b_{k}=\begin{cases} \frac{4}{\pi k} & \text{ if } k\; odd \\ 0 & \text{ if } k\; even \end{cases}\]

Thus, the Fourier series for the square wave is

\[sq(t)=\sum_{k\in \left \{ 1,3,... \right \}}\frac{4}{\pi k}\sin \left ( \frac{2\pi kt}{T} \right )\]

Example \(\PageIndex{1}\):

Let's find the Fourier series representation for the half-wave rectified sinusoid.

\[s(t)=\begin{cases} \sin \frac{2\pi t}{T} & \text{ if } 0\leq t\leq \frac{T}{2} \\ 0 & \text{ if } \frac{T}{2}\leq t< T \end{cases}\]

Begin with the sine terms in the series; to find **b _{k}** we must calculate the integral

\[b_{k}=\int_{0}^{\frac{T}{2}}\sin \left ( \frac{2\pi t}{T} \right ) \sin \left ( \frac{2\pi kt}{T} \right )dt\]

Using our trigonometric identities turns our integral of a product of sinusoids into a sum of integrals of individual sinusoids, which are much easier to evaluate.

\[\int_{0}^{\frac{T}{2}}\sin \left ( \frac{2\pi t}{T} \right ) \sin \left ( \frac{2\pi kt}{T} \right )dt=\frac{1}{2}\int_{0}^{\frac{T}{2}}\cos \left ( \frac{2\pi (k-1)t}{T} \right ) -\cos \left ( \frac{2\pi (k+1)t}{T} \right )dt\]

\[\int_{0}^{\frac{T}{2}}\sin \left ( \frac{2\pi t}{T} \right ) \sin \left ( \frac{2\pi kt}{T} \right )dt=\begin{cases} \frac{1}{2} & \text{ if } k=1 \\ 0 & \text{ if } otherwise \end{cases}\]

Thus,

\[b_{1}=\frac{1}{2}\]

\[b_{2}=b_{3}=...=0\]

On to the cosine terms. The average value, which corresponds to **a _{0}**, equals

**1/π**. The remainder of the cosine coefficients are easy to find, but yield the complicated result:

\[a_{k}=\begin{cases} -\left ( \frac{2}{\pi } \frac{1}{k^{2}-1}\right ) & \text{ if } k\in \left \{ 2,4,... \right \} \\ 0 & \text{ if } k\; odd \end{cases}\]

Thus, the Fourier series for the half-wave rectified sinusoid has non-zero terms for the average, the fundamental, and the even harmonics.

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