# 3.1 Complex fourier series

Definition of the complex Fourier series.

In an earlier module , we showed that a square wave could be expressed as a superposition of pulses. As useful asthis decomposition was in this example, it does not generalize well to other periodic signals:How can a superposition of pulses equal a smooth signal like a sinusoid?Because of the importance of sinusoids to linear systems, you might wonder whether they could be added together to represent alarge number of periodic signals. You would be right and in good company as well. Euler and Gauss in particular worried about this problem, and Jean Baptiste Fourier got the credit even though tough mathematical issues were notsettled until later. They worked on what is now known as the Fourier series : representing any periodic signal as a superposition of sinusoids.

But the Fourier series goes well beyond being another signal decomposition method.Rather, the Fourier series begins our journey to appreciate how a signal can be described in either the time-domain or the frequency-domain with no compromise. Let $s(t)$ be a periodic signal with period $T$ . We want to show that periodic signals, even those that haveconstant-valued segments like a square wave, can be expressed as sum of harmonically related sine waves: sinusoids having frequencies that are integer multiples of the fundamental frequency . Because the signal has period $T$ , the fundamental frequency is $\frac{1}{T}$ . The complex Fourier series expresses the signal as a superposition ofcomplex exponentials having frequencies $\frac{k}{T}$ , $k=\{\text{…}, -1, 0, 1, \text{…}\}$ .

$s(t)=\sum_{k=()}$ c k 2 k t T
with ${c}_{k}=\frac{1}{2}({a}_{k}-i{b}_{k})$ . The real and imaginary parts of the Fourier coefficients ${c}_{k}$ are written in this unusual way for convenience in defining the classic Fourier series.The zeroth coefficient equals the signal's average value and is real- valued for real-valued signals: ${c}_{0}={a}_{0}$ . The family of functions $\{e^{i\frac{2\pi kt}{T}}\}$ are called basis functions and form the foundation of the Fourier series. No matter what theperiodic signal might be, these functions are always present and form the representation's building blocks. They depend on thesignal period $T$ , and are indexed by $k$ .
Assuming we know the period, knowing the Fourier coefficientsis equivalent to knowing the signal. Thus, it makes no difference if we have a time-domain or a frequency-domain characterization of the signal.

What is the complex Fourier series for a sinusoid?

Because of Euler's relation,

$\sin (2\pi ft)=\frac{1}{2i}e^{i\times 2\pi ft}-\frac{1}{2i}e^{-(i\times 2\pi ft)}$
Thus, ${c}_{1}=\frac{1}{2i}$ , ${c}_{-1}=-\left(\frac{1}{2i}\right)$ , and the other coefficients are zero.

To find the Fourier coefficients, we note the orthogonality property

$\int_{0}^{T} e^{i\frac{2\pi kt}{T}}e^{-i\frac{2\pi lt}{T}}\,d t=\begin{cases}T & \text{if k=l}\\ 0 & \text{if k\neq l}\end{cases}$
Assuming for the moment that the complex Fourier series "works," we can find a signal's complex Fourier coefficients, its spectrum , by exploiting the orthogonality properties of harmonically related complexexponentials. Simply multiply each side of [link] by $e^{-(i\times 2\pi lt)}$ and integrate over the interval $\left[0,T\right]$ .
$\begin{array}{l}{c}_{k}=\frac{1}{T}\int_{0}^{T} s(t)e^{-(i\frac{2\pi kt}{T})}\,d t\\ {c}_{0}=\frac{1}{T}\int_{0}^{T} s(t)\,d t\end{array}$

Finding the Fourier series coefficients for the square wave ${\mathrm{sq}}_{T}(t)$ is very simple. Mathematically, this signal can be expressed as ${\mathrm{sq}}_{T}(t)=\begin{cases}1 & \text{if 0< t< \frac{T}{2}}\\ -1 & \text{if \frac{T}{2}< t< T}\end{cases}$ The expression for the Fourier coefficients has the form

${c}_{k}=\frac{1}{T}\int_{0}^{\frac{T}{2}} e^{-(i\frac{2\pi kt}{T})}\,d t-\frac{1}{T}\int_{\frac{T}{2}}^{T} e^{-(i\frac{2\pi kt}{T})}\,d t$
When integrating an expression containing $i$ , treat it just like any other constant.
The two integrals are very similar, one equaling the negative of theother. The final expression becomes
${c}_{k}=\frac{-2}{i\times 2\pi k}(-1^{k}-1)=\begin{cases}\frac{2}{i\pi k} & \text{if k\text{odd}}\\ 0 & \text{if k\text{even}}\end{cases}$
$\mathrm{sq}(t)=\sum_{k\in \{\dots , -3, -1, 1, 3, \dots \}} \frac{2}{i\pi k}e^{(i)\frac{2\pi kt}{T}}$
Consequently, the square wave equals a sum of complex exponentials, but only those having frequencies equal to odd multiples of thefundamental frequency $\frac{1}{T}$ . The coefficients decay slowly as the frequency index $k$ increases. This index corresponds to the $k$ -th harmonic of the signal's period.

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research.net
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Introduction about quantum dots in nanotechnology
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in general
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