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 Voltage applied over a load 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>

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′)e−j2πf′tdf′

Substituting this back into the original equation makes it

  • W=∫−∞∞(∫−∞∞X(f)ej2πftdf)(∫−∞∞X*(f′)e−j2πf′tdf′)dt
  • W=∫−∞∞X(f)∫−∞∞X*(f′)(∫−∞∞ej2π(f−f′)tdt)df′df

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

  • W=∫−∞∞X(f)∫−∞∞X*(f′)(δ(f−f′))df′df
  • 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