Rayliegh's Theorem: Difference between revisions

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Rayliegh's Theorem is derived from the equation for Energy
Rayleigh's Theorem is derived from the equation for Energy
*<math> W = \int_{-\infty}^{\infty}p(t)\,dt </math>  
*<math> W = \int_{-\infty}^{\infty}p(t)\,dt </math>  
If we assume that the circuit is a Thevin equavalent then <math> p(t)=\frac{x^2(t)}{R_L}</math>
If we assume that the circuit is a Voltage applied over a load then <math> p(t)=\frac{x^2(t)}{R_L}</math>
for matters of simplicity we can assume <math>R_L = 1 \Omega</math>
for matters of simplicity we can assume <math>R_L = 1\, \Omega</math>
<br>
This leaves us with
*<math> W = \int_{-\infty}^{\infty}|x|^2(t)\,dt</math>
This is the same as the dot product so to satisfy the condition for complex numbers it becomes
*<math> W = \int_{-\infty}^{\infty}x(t)\,x^*(t)\,dt</math>
If we substitute <math> x(t) = \int_{-\infty}^{\infty}X(f)\,e^{j2\pi ft}\,df </math> and <math>x^*(t)= \int_{-\infty}^{\infty}X(f')\,e^{-j2\pi f't}\,df'</math>
<br>
<br>Substituting this back into the original equation makes it
*<math>W = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}X(f)\,e^{j2\pi ft}\,df\right) \,\left(\int_{-\infty}^{\infty}X^*(f')\,e^{-j2\pi f't}\,df'\right)\,dt</math>
*<math>W = \int_{-\infty}^{\infty}X(f)\,\int_{-\infty}^{\infty}X^*(f')\left(\int_{-\infty}^{\infty}e^{j2\pi (f-f')t}\,dt\right)\,df'\,df</math>
The time integral becomes <math> \delta (f-f') \,which \ is\ 0\ except\ for\ when\ f' = f</math>
This simplifies the above equation such that
*<math>W = \int_{-\infty}^{\infty}X(f)\,\int_{-\infty}^{\infty}X^*(f')\left(\delta (f-f') \right)\,df'\,df</math>
*<math>W = \int_{-\infty}^{\infty}X(f)\,X^*(f)\,df</math>
Proving that the energy in the time domain is the same as that in the frequency domain
*<math> W = \int_{-\infty}^{\infty}X(f)\,X^*(f)\,df = \int_{-\infty}^{\infty}x(t)\,x^*(t)\,dt</math>

Latest revision as of 02:35, 11 October 2006

Rayleigh's Theorem is derived from the equation for Energy

  • W=p(t)dt

If we assume that the circuit is a Voltage applied over a load then p(t)=x2(t)RL for matters of simplicity we can assume RL=1Ω
This leaves us with

  • W=|x|2(t)dt

This is the same as the dot product so to satisfy the condition for complex numbers it becomes

  • W=x(t)x*(t)dt

If we substitute x(t)=X(f)ej2πftdf and x*(t)=X(f)ej2πftdf

Substituting this back into the original equation makes it

  • W=(X(f)ej2πftdf)(X*(f)ej2πftdf)dt
  • W=X(f)X*(f)(ej2π(ff)tdt)dfdf

The time integral becomes δ(ff)whichis0exceptforwhenf=f This simplifies the above equation such that

  • W=X(f)X*(f)(δ(ff))dfdf
  • W=X(f)X*(f)df

Proving that the energy in the time domain is the same as that in the frequency domain

  • W=X(f)X*(f)df=x(t)x*(t)dt