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RLC circuit.

  1. Oct 21, 2012 #1

    1f5

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    1. The problem statement, all variables and given/known data
    The problem says:
    For the next circuit obtain a particular solution for R3 when v1=v2=l1=l2=0 at t=0.
    vs= cos(wt) dc source= 10v *R3 is the 1k resistor.
    http://postimage.org/image/fxtu2blaf/ [Broken]



    2. Relevant equations


    Kirchoff current law



    3. The attempt at a solution


    With node analysis, there are two differential equations to solve, I solved the differential equation and got:

    probl5_1.jpg

    According to simulations the efect of frequency is not too strong, I think that the system of DE was wrong, and that's why solution isn't reallistic.


    PD. voltage v is noise and it equals to cos(wt). I Have done a simulation i'll add it.

    Simulations:
    W=1Hz
    W=1MHz


    http://postimage.org/image/s28jolkiv/ [Broken]

    http://postimage.org/image/elvpjifpx/ [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 24, 2012 #2
    That circuit is called a doubly terminated lossless ladder and it is a low pass filter. LC elements arranged in a ladder like that result from implementing filters from a given transfer function and such circuits can be almost automatically generated.

    You wouldn't know that at this point of course, but you can see the low pass character like this. Capacitors are open circuits at DC and inductors are short circuits at DC. This means a DC input signal will only see a resistor divider. At high frequencies, capacitors are short circuits (1/jwC goes to zero) and inductors are open circuits (jwL goes to infinity). These means there is no path from input to output for high frequency signals and the output will be zero. So at low frequencies you have output and high frequencies you have no output. That's a low pass filter.

    The differential equations you wrote give me a headache. Do you not transform the elements to complex impedances or into the s-domain to solve the circuit algebraically? The impedance of a capacitor is 1/sC (or 1/jwC) and the impedance of an inductor is sL (or jwL). This greatly simplifies analysis and avoids errors in coupled differential equations.
     
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