Fourier series: review
A function or signal
$x\left(t\right)$ is called
periodic with period
$T$ if
$x(t+T)=x\left(t\right)$ .
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:
$${a}_{k}=\frac{2}{T}{\int}_{-T/2}^{T/2}x\left(t\right)\text{cos}\left(k\frac{2\pi}{T}t\right)dt\phantom{\rule{1.em}{0ex}}(k=0,1,2,...)$$
$${b}_{k}=\frac{2}{T}{\int}_{-T/2}^{T/2}x\left(t\right)\text{sin}\left(k\frac{2\pi}{T}t\right)dt\phantom{\rule{1.em}{0ex}}(k=1,2,3,...)$$
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
$[0,T]$ 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
$$a\text{cos}\left(\alpha \right)+b\text{sin}\left(\alpha \right)=a\frac{1}{2}({e}^{j\alpha}+{e}^{-j\alpha})+b\frac{1}{2j}({e}^{j\alpha}-{e}^{-j\alpha})=\frac{1}{2}(a-bj){e}^{j\alpha}+\frac{1}{2}(a+bj){e}^{-j\alpha}$$
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
$k$
$${X}_{k}=\frac{1}{T}{\int}_{0}^{T}x\left(t\right)\left[\text{cos},(,2,\pi ,k,t,/,T,),-,j,\text{sin},(,2,\pi ,k,t,/,T,)\right]dt=\frac{1}{2}({a}_{k}-{b}_{k}j).$$
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
$$\begin{array}{ccc}\hfill {Z}_{k}& =& \frac{1}{T/\lambda}{\int}_{0}^{T/\lambda}z\left(t\right){e}^{-2\pi jkt/(T/\lambda )}dt=\frac{\lambda}{T}{\int}_{0}^{T/\lambda}x\left(t\lambda \right){e}^{-2\pi jkt\lambda /T}dt\hfill \\ & =& \frac{1}{T}{\int}_{0}^{T}x\left(s\right){e}^{-2\pi jks/T}ds={X}_{k}\hfill \end{array}$$
- Shift: The Fourier coefficients of
$z\left(t\right)=x(t+d)$ 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
${Z}_{k}={X}_{k}{e}^{j\pi k}={(-1)}^{k}{X}_{k}$ .
- 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:
$${x}^{\text{'}}\left(t\right)=\sum _{k=-\infty}^{\infty}{X}_{k}\xb7k\frac{2\pi j}{T}\xb7{e}^{j2\pi kt/T}$$
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)
$$x\left(t\right)=\text{sin}(2\pi t/T)\phantom{\rule{2.em}{0ex}}{X}_{1}=-{X}_{-1}=\frac{j}{2}$$
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
$\text{sin}\left(\alpha \right)=({e}^{j\alpha}-{e}^{-j\alpha})/\left(2j\right)=(j/2)({e}^{-j\alpha}-{e}^{j\alpha})$ 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 $ :
$$x\left(t\right)=\left\{\begin{array}{cc}1\hfill & \text{for}\phantom{\rule{4.pt}{0ex}}0<t<\pi \hfill \\ -1\hfill & \text{for}\phantom{\rule{4.pt}{0ex}}-\pi <t<0\hfill \\ c\hfill & \text{for}\phantom{\rule{4.pt}{0ex}}t=0,\pi \hfill \end{array}\right)\phantom{\rule{2.em}{0ex}}{b}_{k}=2{B}_{k}=\frac{4}{\pi k}\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{odd}\phantom{\rule{4.pt}{0ex}}k\ge 1$$
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<t<\pi $
$$1=\frac{4}{\pi}(\text{sin}\left(t\right)+\frac{1}{3}\text{sin}\left(3t\right)+\frac{1}{5}\text{sin}\left(5t\right)+...)$$
Note that for
$t=0$ the value of the series on the
right is 0, which is equal to
$x(0-)+x(0+)$ , the middle of the jump
of
$x\left(t\right)$ at 0, no matter what
$c$ is. Similar for
$t=\pi $ .