# Fourier series: review

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## Fourier series: review

A function or signal $x\left(t\right)$ is called periodic with period $T$ if $x\left(t+T\right)=x\left(t\right)$ . All “typical” periodic function $x\left(t\right)$ with period $T$ can be developed as follows

$x\left(t\right)=\frac{{a}_{0}}{2}+\sum _{k=1}^{\infty }\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}{a}_{k}\text{cos}\left(k\frac{2\pi }{T}t\right)+{b}_{k}\text{sin}\left(k\frac{2\pi }{T}t\right)$

where the coefficients are computed as follows:

${a}_{k}=\frac{2}{T}{\int }_{-T/2}^{T/2}x\left(t\right)\text{cos}\left(k\frac{2\pi }{T}t\right)dt\phantom{\rule{1.em}{0ex}}\left(k=0,1,2,...\right)$
${b}_{k}=\frac{2}{T}{\int }_{-T/2}^{T/2}x\left(t\right)\text{sin}\left(k\frac{2\pi }{T}t\right)dt\phantom{\rule{1.em}{0ex}}\left(k=1,2,3,...\right)$

The natural interpretation of [link] is as a decomposition of the signal $x\left(t\right)$ into individual oscillations where ${a}_{k}$ indicates the amplitude of the even oscillation $\text{cos}\left(k\frac{2\pi }{T}t\right)$ of frequency $k/T$ (meaning its period is $T/k$ ), and ${b}_{k}$ indicates the amplitude of the odd oscillation $\text{sin}\left(k\frac{2\pi }{T}t\right)$ of frequency $k/T$ . For an audio signal $x\left(t\right)$ , frequency corresponds to how high a sound is and amplitude to how loud it is. The oscillations appearing in the Fourier decomposition are oftenalso called harmonics (first, second, third harmonic etc).

Note: one can also integrate over $\left[0,T\right]$ or any other interval of length $T$ . Note also, that the average value of $x\left(t\right)$ over one period is equal to ${a}_{0}/2$ .

Complex representation of Fourier series

Often it is more practical to work with complex numbers in the area of Fourier analysis. Using the famous formula

${e}^{j\alpha }=\text{cos}\left(\alpha \right)+j\text{sin}\left(\alpha \right)$

it is possible to simplify several formulas at the price of working with complex numbers.Towards this end we write

$a\text{cos}\left(\alpha \right)+b\text{sin}\left(\alpha \right)=a\frac{1}{2}\left({e}^{j\alpha }+{e}^{-j\alpha }\right)+b\frac{1}{2j}\left({e}^{j\alpha }-{e}^{-j\alpha }\right)=\frac{1}{2}\left(a-bj\right){e}^{j\alpha }+\frac{1}{2}\left(a+bj\right){e}^{-j\alpha }$

From this we observe that we may replace the cos and sin harmonics by a pair of exponential harmonicswith opposite frequencies and with complex amplitudes which are conjugate complex to each other.In fact, we arrive at the more simple complex Fourier series :

$x\left(t\right)=\sum _{k=-\infty }^{\infty }{X}_{k}{e}^{j2\pi kt/T}\phantom{\rule{2.em}{0ex}}\text{with}\phantom{\rule{1.em}{0ex}}{X}_{k}=\frac{1}{T}{\int }_{0}^{T}x\left(t\right){e}^{-j2\pi kt/T}dt$

Note that $X$ is complex, but $x$ is real-valued (the imaginary parts of all the terms in $\sum {X}_{k}{e}^{j2\pi kt/T}$ add up to zero; in other words, they cancel each other out). The absolute value of ${X}_{k}$ gives the amplitude of the complex harmonic with frequency $k/T$ (meaning its period is $T/k$ ); the argument of ${X}_{k}$ provides the phase difference between the complex harmonics.If $x$ is even, ${X}_{k}$ is real for all $k$ and all harmonics are in phase.

To verify [link] note that by [link] and [link] we have for positive $k$

${X}_{k}=\frac{1}{T}{\int }_{0}^{T}x\left(t\right)\left[\text{cos},\left(,2,\pi ,k,t,/,T,\right),-,j,\text{sin},\left(,2,\pi ,k,t,/,T,\right)\right]dt=\frac{1}{2}\left({a}_{k}-{b}_{k}j\right).$

For negative $k$ we note that ${X}_{-k}={X}_{k}^{*}$ by [link] , where ${\left(\right)}^{*}$ denotes the conjugate complex. By [link] , the ${X}_{k}$ are exactly as they are supposed to be.

Properties

• Linearity: The Fourier coefficients of the signal $z\left(t\right)=cx\left(t\right)+y\left(t\right)$ are simply
${Z}_{k}=c{X}_{k}+{Y}_{k}$
• Change of frequency: The signal $z\left(t\right)=x\left(\lambda t\right)$ has the period $T/\lambda$ and has the same Fourier coefficients as $x\left(t\right)$ — but they correspond to different frequencies $f$ :
${Z}_{k}{|}_{f=\frac{k}{T/\lambda }}={X}_{k}{|}_{f=\frac{k}{T}}\phantom{\rule{2.em}{0ex}}\text{since}\phantom{\rule{2.em}{0ex}}z\left(t\right)=x\left(\lambda t\right)=\sum _{k=-\infty }^{\infty }{X}_{k}{e}^{j2\pi \lambda kt/T}=\sum _{k=-\infty }^{\infty }{Z}_{k}{e}^{j2\pi t\frac{k}{T/\lambda }}$

The equation on the right allows to read off the Fourier coefficients and to establish ${Z}_{k}={X}_{k}$ . (For an alternative computation see  Comment 1 )

Comment 1

$\begin{array}{ccc}\hfill {Z}_{k}& =& \frac{1}{T/\lambda }{\int }_{0}^{T/\lambda }z\left(t\right){e}^{-2\pi jkt/\left(T/\lambda \right)}dt=\frac{\lambda }{T}{\int }_{0}^{T/\lambda }x\left(t\lambda \right){e}^{-2\pi jkt\lambda /T}dt\hfill \\ & =& \frac{1}{T}{\int }_{0}^{T}x\left(s\right){e}^{-2\pi jks/T}ds={X}_{k}\hfill \end{array}$

• Shift: The Fourier coefficients of $z\left(t\right)=x\left(t+d\right)$ are simply
${Z}_{k}={X}_{k}{e}^{j2\pi kd/T}$

The modulation is much more simple in complex writing then it would be with real coefficients.For the special shift by half a period, i.e., $d=T/2$ we have ${Z}_{k}={X}_{k}{e}^{j\pi k}={\left(-1\right)}^{k}{X}_{k}$ .

• Derivative: The Fourier series of the derivative of $x\left(t\right)$ with development [link] can be obtained simply by taking the derivative of [link] term by term:
${x}^{\text{'}}\left(t\right)=\sum _{k=-\infty }^{\infty }{X}_{k}·k\frac{2\pi j}{T}·{e}^{j2\pi kt/T}$

Short: when taking the derivative of a signal, the complex Fourier coefficients get multiplied by $k\frac{2\pi j}{T}$ . Consequently, the coefficients of the derivative decay slower.

Examples

• The pure oscillation (containing only one real but two complex frequencies)
$x\left(t\right)=\text{sin}\left(2\pi t/T\right)\phantom{\rule{2.em}{0ex}}{X}_{1}=-{X}_{-1}=\frac{j}{2}$
or ${B}_{1}=1/2=-{B}_{-1}$ , or ${b}_{1}=1$ , and all other coefficients are zero.This formula can be obtained without computing integrals by noting that $\text{sin}\left(\alpha \right)=\left({e}^{j\alpha }-{e}^{-j\alpha }\right)/\left(2j\right)=\left(j/2\right)\left({e}^{-j\alpha }-{e}^{j\alpha }\right)$ and setting $\alpha =2\pi t/T$ .
• Functions which are time-limited , i.e., defined on a finite interval can be periodically extended. Example with $T=2\pi$ :
$x\left(t\right)=\left\{\begin{array}{cc}1\hfill & \text{for}\phantom{\rule{4.pt}{0ex}}0
and all other coefficients zero. Note that $c$ is any constant; the value of $c$ does not affect the coefficients ${b}_{k}$ . We have for $0
$1=\frac{4}{\pi }\left(\text{sin}\left(t\right)+\frac{1}{3}\text{sin}\left(3t\right)+\frac{1}{5}\text{sin}\left(5t\right)+...\right)$
Note that for $t=0$ the value of the series on the right is 0, which is equal to $x\left(0-\right)+x\left(0+\right)$ , the middle of the jump of $x\left(t\right)$ at 0, no matter what $c$ is. Similar for $t=\pi$ .

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