# Continuous-time signals  (Page 2/5)

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$x\left(t\right)=\frac{a\left(0\right)}{2}+\sum _{k=1}^{\infty }a\left(k\right)cos\left(\frac{2\pi }{T}kt\right)+b\left(k\right)sin\left(\frac{2\pi }{T}kt\right).$

where ${x}_{k}\left(t\right)=cos\left(2\pi kt/T\right)$ and ${y}_{k}\left(t\right)=sin\left(2\pi kt/T\right)$ are the basis functions for the expansion. The energy or power in an electrical,mechanical, etc. system is a function of the square of voltage, current, velocity, pressure, etc. For this reason, the natural setting for arepresentation of signals is the Hilbert space of ${L}^{2}\left[0,T\right]$ . This modern formulation of the problem is developed in [link] , [link] . The sinusoidal basis functions in the trigonometric expansion form a completeorthogonal set in ${L}^{2}\left[0,T\right]$ . The orthogonality is easily seen from inner products

$\left(cos\left(\frac{2\pi }{T}kt\right)\phantom{\rule{0.277778em}{0ex}},\phantom{\rule{0.277778em}{0ex}}cos\left(\frac{2\pi }{T}\ell t\right)\right)={\int }_{0}^{T}\left(cos\left(\frac{2\pi }{T}kt\right)\phantom{\rule{0.277778em}{0ex}}cos\left(\frac{2\pi }{T}\ell t\right)\right)\phantom{\rule{0.277778em}{0ex}}dt=\delta \left(k-\ell \right)$

and

$\left(cos\left(\frac{2\pi }{T}kt\right)\phantom{\rule{0.277778em}{0ex}},\phantom{\rule{0.277778em}{0ex}}sin\left(\frac{2\pi }{T}\ell t\right)\right)={\int }_{0}^{T}\left(cos\left(\frac{2\pi }{T}kt\right)\phantom{\rule{0.277778em}{0ex}}sin\left(\frac{2\pi }{T}\ell t\right)\right)\phantom{\rule{0.277778em}{0ex}}dt=0$

where $\delta \left(t\right)$ is the Kronecker delta function with $\delta \left(0\right)=1$ and $\delta \left(k\ne 0\right)=0$ . Because of this, the $k$ th coefficients in the series can be foundby taking the inner product of $x\left(t\right)$ with the $k$ th basis functions. This gives for the coefficients

$a\left(k\right)=\frac{2}{T}{\int }_{0}^{T}x\left(t\right)cos\left(\frac{2\pi }{T}kt\right)dt$

and

$b\left(k\right)=\frac{2}{T}{\int }_{0}^{T}x\left(t\right)sin\left(\frac{2\pi }{T}kt\right)dt$

where $T$ is the time interval of interest or the period of a periodic signal. Because of the orthogonality of the basis functions, afinite Fourier series formed by truncating the infinite series is an optimal least squared error approximation to $x\left(t\right)$ . If the finite series is defined by

$\stackrel{^}{x}\left(t\right)=\frac{a\left(0\right)}{2}+\sum _{k=1}^{N}a\left(k\right)cos\left(\frac{2\pi }{T}kt\right)+b\left(k\right)sin\left(\frac{2\pi }{T}kt\right),$

the squared error is

$\epsilon =\frac{1}{T}{\int }_{0}^{T}{|x\left(t\right)-\stackrel{^}{x}\left(t\right)|}^{2}dt$

which is minimized over all $a\left(k\right)$ and $b\left(k\right)$ by [link] and [link] . This is an extraordinarily important property.

It follows that if $x\left(t\right)\in {L}^{2}\left[0,T\right]$ , then the series converges to $x\left(t\right)$ in the sense that $\epsilon \to 0$ as $N\to \infty$ [link] , [link] . The question of point-wise convergence is more difficult. A sufficient condition that is adequate for mostapplication states: If $f\left(x\right)$ is bounded, is piece-wise continuous, and has no more than a finite number of maxima over an interval, the Fourierseries converges point-wise to $f\left(x\right)$ at all points of continuity and to the arithmetic mean at points of discontinuities. If $f\left(x\right)$ is continuous, the series converges uniformly at all points [link] , [link] , [link] .

A useful condition [link] , [link] states that if $x\left(t\right)$ and its derivatives through the $q$ th derivative are defined and have bounded variation, the Fourier coefficients $a\left(k\right)$ and $b\left(k\right)$ asymptotically drop off at least as fast as $\frac{1}{{k}^{q+1}}$ as $k\to \infty$ . This ties global rates of convergence of the coefficients to local smoothness conditions of the function.

The form of the Fourier series using both sines and cosines makes determination of the peak value or of the location of a particularfrequency term difficult. A different form that explicitly gives the peak value of the sinusoid of that frequency and the location or phase shift ofthat sinusoid is given by

$x\left(t\right)=\frac{d\left(0\right)}{2}+\sum _{k=1}^{\infty }d\left(k\right)cos\left(\frac{2\pi }{T}kt+\theta \left(k\right)\right)$

and, using Euler's relation and the usual electrical engineering notation of $j=\sqrt{-1}$ ,

${e}^{jx}=cos\left(x\right)+jsin\left(x\right),$

the complex exponential form is obtained as

$x\left(t\right)=\sum _{k=-\infty }^{\infty }c\left(k\right)\phantom{\rule{0.166667em}{0ex}}{e}^{j\frac{2\pi }{T}kt}$

where

$c\left(k\right)=a\left(k\right)+j\phantom{\rule{0.166667em}{0ex}}b\left(k\right).$

The coefficient equation is

$c\left(k\right)=\frac{1}{T}{\int }_{0}^{T}x\left(t\right)\phantom{\rule{0.166667em}{0ex}}{e}^{-j\frac{2\pi }{T}kt}dt$

The coefficients in these three forms are related by

${|d|}^{2}={|c|}^{2}={a}^{2}+{b}^{2}$

and

$\theta =arg\left\{c\right\}={tan}^{-1}\left(\frac{b}{a}\right)$

It is easier to evaluate a signal in terms of $c\left(k\right)$ or $d\left(k\right)$ and $\theta \left(k\right)$ than in terms of $a\left(k\right)$ and $b\left(k\right)$ . The first two are polar representation of a complex value and the last is rectangular. Theexponential form is easier to work with mathematically.

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Source:  OpenStax, Digital signal processing and digital filter design (draft). OpenStax CNX. Nov 17, 2012 Download for free at http://cnx.org/content/col10598/1.6
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