# 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.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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Sherica
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Sherica
right! what he said ⤴⤴⤴
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China
Cied
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
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silver nanoparticles could handle the job?
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not now but maybe in future only AgNP maybe any other nanomaterials
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