Matthew's Asgn: Difference between revisions

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<math>\ \dfrac{s/L}{(s^2+w^2)(R/L+s)} = \dfrac{A(s+jw)(s+R/L)}{(s+R/L)(s^2+w^2)} + \dfrac{A*(s+jw)(s+R/L)}{(s^2+w^2)(s+R/L)} + \dfrac{B(s^2+w^2)}{(s+R/L)(s^2+w^2)} \,\!</math>
<math>\ \dfrac{s/L}{(s^2+w^2)(R/L+s)} = \dfrac{A(s+jw)(s+R/L)}{(s+R/L)(s^2+w^2)} + \dfrac{A*(s+jw)(s+R/L)}{(s^2+w^2)(s+R/L)} + \dfrac{B(s^2+w^2)}{(s+R/L)(s^2+w^2)} \,\!</math>



<math> \dfrac{S}{L} = A(s-jw)(s+R/L) + A*((s+jw)(s+R/L)) + B(s^2+w^2) \,\!</math>
<math> \dfrac{S}{L} = A(s-jw)(s+R/L) + A*((s+jw)(s+R/L)) + B(s^2+w^2) \,\!</math>


<math> \dfrac{S}{L} = A(s^2-jws+R/Ls-jwR/L) + A*(s^2+jws+R/L+jwr/L) + B(s^2+w^2) \,\!</math>

<math> \dfrac{S}{L} = s^2(A+A*+B) + s((A*-A)jws+(A*+A)R/L) + (A*-A)Rjw/L+Bw^2 \,\!</math>

After a lot of messy math and work:

<math> A = \dfrac{1/2}{R-jwL} \,\!</math>

<math> A* = \dfrac{1/2}{R+jwL} \,\!</math>

<math> B = \dfrac{-R}{R^2+w^2L^2} \,\!</math>

<math> I(s)= \dfrac{a+jb}{s+jw} + \dfrac{a-jb}{s-jw} + \dfrac{B}{s+R/L} \,\!</math>

<math> I(s)= \dfrac{(a+jb)(s-jw)+(a-jb)(s+jw)}{s^2+w^2}+\dfrac{B}{s^2+w^2} \,\!</math>

<math> I(s)= \dfrac{2as}{s^2+w^2} + \dfrac{2bw}{s^2+w^2} + \dfrac{B}{s+R/L} \,\!</math>

<math> a = \dfrac{1/2R}{R^2+w^2L^2} \,\!</math>

<math> b = \dfrac{1/2wL}{R^2+w^2L^2} \,\!</math>

<math> i(t) = \mathcal{L}^{-1} \{I(s)\} \,\!</math>

<math> i(t) = 2a \cos wt + 2b\sin wt + BE^{-Rt/L} -ie^{-Rt/L} \,\!</math>

<math> i(t) = 2 \dfrac{R}{R^2+w^2L^2} \cos wt + 2 \dfrac{wL}{R^2+w^2L^2} \sin wt + \dfrac{-R}{R^2+w^2L^2} e^{-Rt/L} -ie^{-Rt/L} \,\!</math>


This Solves the Series RL circuit for the current throughout the circuit, given initial conditions or no initial conditions
<math> = A(s^2-jws+R/Ls-jwR/L) + a*(s^2+jws+R/L+jwr/L) + B(s^2+w^2) \,\!</math>

Latest revision as of 18:40, 1 November 2010

I decided that I would attempt to perform a simple analysis of a series RL circuit, which could then be used to do a more complex analysis on a basic transformer. I have always had interest in electronics, and transformers are key to basic electronics.

I decided that i would do the analysis of a RL circuit with the variables instead of given values.


Given:

V(t)=

V(s)=

I(0)=i

RLcircuit.jpg

The Laplace transform for an inductor:

=

The Laplace transform for a resistor:

=

Therefore the Resulting Equation for the system after applying the Laplace Transform:

A series of algebraic manipulations follows to come up with I(s):



We can then use partial fraction manipulation to expand the expression:


After a lot of messy math and work:

This Solves the Series RL circuit for the current throughout the circuit, given initial conditions or no initial conditions