172.20.22.213: /* Determining the coefficient $\alpha_n \,$ */

2007-10-04T22:54:33Z

Determining the coefficient <math> \alpha_n \,</math>

← Older revision		Revision as of 15:54, 4 October 2007
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	<math> \int_{-T/2}^{T/2} x(t)\, dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi nt}/T} dt</math>		<math> \int_{-T/2}^{T/2} x(t)\, dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi nt}/T} dt</math>

	*Integrating both sides for one period. The range of integration is arbitrary, but using <math> \int_{-T/2}^{T/2} </math> scales nicely when extending the Fourier series to a non-periodic function		*Integrating both sides for one period. The range of integration is arbitrary, but using <math> \int_{-T/2}^{T/2} </math> instead of <math> \int_{0}^{T} </math>scales nicely when extending the Fourier series to a non-periodic function

	<math> \int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi nt}/T}e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi (n-m)t}/T} dt</math>		<math> \int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi nt}/T}e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi (n-m)t}/T} dt</math>
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	<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt </math>		<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt </math>



	==Linear Time Invariant Systems==		==Linear Time Invariant Systems==

24.10.77.69: /* Identities */

2006-11-21T07:04:32Z

Identities

← Older revision		Revision as of 00:04, 21 November 2006
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	<math>\cos x = \frac{e^{jx}+e^{-jx}}{2} \,</math>		<math>\cos x = \frac{e^{jx}+e^{-jx}}{2} \,</math>

	<math> \~~left \langle \ Bra~~ \~~mid Ket \ \right \rangle = Ket~~ \~~cdot Bra~~ </math>		<math> <u\|v> = \int_{-\infty}^\infty u^*(x) v(x) dx </math>

	<math> \alpha_{-m} = \alpha^* \,</math>		<math> \alpha_{-m} = \alpha^* \,</math>

	The dirac delta has an infinite height and an area of 1		The dirac delta has an infinite height and an area of 1

172.16.18.138 at 15:28, 4 November 2006

2006-11-04T15:28:53Z

New page

==Fourier series==
The Fourier series is used to analyze arbitrary periodic functions by showing them as a composite of sines and cosines.

A function is considered periodic if <math> x(t) = x(t+T)\, </math> for <math> T \neq 0 </math>.

The exponential form of the Fourier series is defined as <math> x(t) = \sum_{n=-\infty}^\infty \alpha_n e^{{j2\pi nt}/T} \, </math>

==Determining the coefficient <math> \alpha_n \,</math> ==

<math> x(t) = \sum_{n=-\infty}^\infty \alpha_n e^{{j2\pi nt}/T} \, </math>

*The definition of the Fourier series

<math> \int_{-T/2}^{T/2} x(t)\, dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi nt}/T} dt</math>

*Integrating both sides for one period. The range of integration is arbitrary, but using <math> \int_{-T/2}^{T/2} </math> scales nicely when extending the Fourier series to a non-periodic function

<math> \int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi nt}/T}e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi (n-m)t}/T} dt</math>

*Multiply by the complex conjugate

<math> \int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \frac{Te^{{j2\pi (n-m)t}/T}}{{j2\pi (n-m)}} \bigg|_{-T/2}^{T/2} = \sum_{n=-\infty}^\infty \alpha_n T\delta_{n,m} = T\alpha_m</math>

*<math> \frac{Te^{{j2\pi (n-m)t}/T}}{{j2\pi (n-m)}} \bigg|_{-T/2}^{T/2} = T\frac{e^{j\pi(n-m)}-e^{-j\pi(n-m)}}{j2\pi(n-m)} = T \frac{\sin\pi(n-m)}{\pi(n-m)} = \begin{Bmatrix} T, n=m \\ 0, n\ne m \end{Bmatrix} = T\delta_{n,m}</math>

** Using L'Hopitals to evaluate the <math>\frac{T\cdot 0}{0}</math> case. Note that n & m are integers

<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt </math>

==Linear Time Invariant Systems==

Must meet the following criteria

*Time independance
*Linearity
**Superposition (additivity)
**Scaling (homogeneity)

==The Dot Product, Complex Conjugates, and Orthogonality==
[[Image:300px-Scalarproduct.gif |thumb|300px|right]]

Geometrically, the dot product is a scalar projection of a onto b

*<math> \vec a \cdot \vec b = \left | a \right \vert \left | b \right \vert \cos \theta</math>

Arthimetically, multiply like terms and add

*<math> (3,2,1)\cdot(5,6,7)=3\cdot5^*+2\cdot6^*+1\cdot7^*</math>

Lets imagine that we are only have one dimension

*<math> (a+jb)\hat i \cdot (a+jb)\hat i \ne a^2+b^2 </math>

In order to get the real parts and imaginary parts to multiply as like terms, we need to take the complex conjugate of one of the terms

*<math> (a+jb)\hat i \cdot (a-jb)\hat i = a^2+b^2 </math>

To test for orthogonality, take the complex conjugate of one of the vectors and multiply.

*<math> \int_{-\infty}^{\infty} \phi_n (t) \phi_m^* (t) dt = 0</math>

==Changing Basis Functions==

We'd like to change from <math> \sum_{n=-\infty}^{\infty} \alpha_n e^{j2\pi nt/T} </math> to <math> \sum_{m=0}^{\infty} c_m \cos \left (\frac{2\pi mt}{T}+\Theta_m \right) </math>

<math> x(t) = \sum_{n=-\infty}^{\infty} \alpha_n e^{j2\pi nt/T} = \underbrace{ \sum_{n=-\infty}^{-1} \alpha_n e^{j2\pi nt/T} }_{n'=-n} + \; \alpha_0 \; + \sum_{n=1}^{\infty} \alpha_n e^{j2\pi nt/T} = \underbrace{\sum_{n'=1}^{\infty} \alpha_n e^{j2\pi nt/T}}_{m=n'} + \; \alpha_0 \; + \underbrace{\sum_{n=1}^{\infty} \alpha_n e^{j2\pi nt/T}}_{m=n}</math>

<math> = \alpha_0 + \sum_{m=1}^{\infty} \left (\alpha_m e^{j2\pi mt/T} + \alpha_{-m} e^{-j2\pi mt/T}\right) </math>

If we assume <math> x(t) \in \Re \ \forall \ m</math>, then to make the imaginary parts cancel out

*<math> \alpha_{-m} = \alpha_m^* </math>
*<math> u + u^* = 2 \Re [u] </math>
*<math> \alpha_{m} = | \alpha_m |e^{j\phi m} \,</math>

<math> = \alpha_0 + \sum_{m=1}^{\infty} 2 \Re \left[\alpha_m e^{j2\pi mt/T}\right] = \alpha_0 + \sum_{m=1}^{\infty} 2 \Re \left[|\alpha_m| e^{j\phi m}e^{j2\pi mt/T}\right] = \alpha_0 + \sum_{m=1}^{\infty} |\alpha_m|2 \Re \left[e^{j(2\pi mt/T+\phi m)}\right]= \alpha_0 + \sum_{m=1}^{\infty} |\alpha_m|2\cos \left (\frac{2\pi mt}{T} + \phi_m \right ) </math>

Changing variables
*<math> c_0 = \alpha_0 \,</math>
*<math> c_m = 2 | \alpha_m| \,</math>
*<math> \Theta_m = \phi_m \,</math>

<math> = \sum_{m=0}^{\infty} c_m \cos \left (\frac{2\pi mt}{T}+\Theta_m \right) </math>

==Identities==
<math>e^{j \theta} = \cos \theta + j \sin \theta \, </math> Euler's identity linking rectangular and polar coordinates

<math>\sin x = \frac{e^{jx}-e^{-jx}}{2j} \,</math>

<math>\cos x = \frac{e^{jx}+e^{-jx}}{2} \,</math>

<math> \left \langle \ Bra \mid Ket \ \right \rangle = Ket \cdot Bra </math>

<math> \alpha_{-m} = \alpha^* \,</math>

The dirac delta has an infinite height and an area of 1

Fourier Series - Revision history

172.20.22.213: /* Determining the coefficient \alpha_n \, */

24.10.77.69: /* Identities */

172.16.18.138 at 15:28, 4 November 2006

172.20.22.213: /* Determining the coefficient $\alpha_n \,$ */