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Transfer Function of a circuit

  1. Oct 12, 2004 #1
    Ok, here is the question. I have a circuit that looks like a high pass filter except it has an inductor instead of a capacitor. There is a 100 mH inductor and then after that a 20 x Pi Ohm resistor. The question asks me to find the transfer function of the circuit, and then to sketch a Magnitude Bode and phase plot of the circuit. The only info it gives me is the inductor and resistor values and that is it. Can anyone help with this? I don't even know how to start it and my brain is fried.
     
    Last edited: Oct 12, 2004
  2. jcsd
  3. Oct 20, 2004 #2
    The soluiton is relatively easy, but understanding the solution may be a bit more challenging. So I'll take the longer route, assuming you'll need to know how to do this again in the future.

    Consider this circuit:

    Vin-----R1----------Vout
    ....................|
    ....................|
    ...................R2
    ....................|
    ....................|
    ...............GROUND

    Standard voltage division. Vout = Vin* [R2/(R1 + R2)]
    The volatge gain of the circuit (Av = Vout/Vin) is R2/(R1+R2). Pretty simple. An ideal resistor does not vary with frequency, so no matter what the frequency, the gain is always R2/(R1+R2).

    A circuit involving capacitors and/or inductors, however, will have a gain that changes with frequency. That is, you are after Av(f).

    A common way you bring frequency into play is to use the s-domain. This means replace all capacitors of value C by "1/sc" and all inductors of value L by "sL". So, your circuit becomes:

    Vin-----sL----------Vout
    ....................|
    ....................|
    ....................R
    ....................|
    ....................|
    ...............GROUND

    Using the same voltage division principle as above yeilds Av(s) = R/(sL + R). This is your transfer function in the s-domain.

    A full Bode approximation plot consists of two separate plots, one for gain and one for phase.

    The s-domain is set up so that, if you let s = j2*pi*f inthe transfer function, you will get the complex frequency response as a function of frequency.

    For reasons that may not beocme obvious until you've done this a few times, rearrange Av(s) into the form M/(s+N). That is, Av(s) = R/(sL + R) --> (R/L)/(s+R/L).

    I'm going to take a less rigorous approach here; there are some subtleties to the math that could be expanded upon if you desire (you will arrive at true gain(f) and phase(f) equations; what is shown below is called the Bode approximation).

    Let s = 2*pi*f. So now Av(s) becomes Av(f) = (R/L)/(2*pi*f+R/L). This circuit is said to have a pole where f causes the denominator to go to zero, i.e. at f = -R/(2*pi*L). The Bode gain approximation deals with magnitude only, so drop the negative sign (the more rigorous approach would yield a positive f). This frequency is called the corner frequency; fc = R/(2*pi*L). At f = fc for a single pole circuit, the gain is -3dB. Also, at this frequency, the output is 45 degrees out of phase form the input. This is also the point where the actual gain of the circuit will differ the most form the Bode approximation (by 3dB).

    The Bode approximation for gain of a single-pole circuit: for f<fc, the gain is 0dB; for f>=fc, the gain decays at -20dB per decade, starting at 0dB.

    Av
    ^
    |
    |...................(fc, 0dB)
    |____________
    |...................\
    |....................\.<..(-20 dB/decade)
    |.....................\
    --------------------------------> f

    The Bode approximation for the phase of a single-pole circuit: for f<fc/10, the phase is 0. For fc/10<=f<=10fc, the phase decays at -45 degrees per decade, starting at 0 degrees. For f>10fc, the phase is -90 degrees.

    Phase
    ^
    |
    |...................(fc/10, 0 degrees)
    |____________
    |...................\
    |....................\.<..(-20 dB/decade)
    |......................\
    |.......................\__________
    |
    |.......................(10fc, -90 degrees)
    |
    |
    --------------------------------> f

    This is a lot of words, but the Bode plots of phase and gain are very simple to draw. If what I have written here seems overly complex, perhaps somone you know could show you the solution by hand. Simply getting the answer is quite easy, but hard (for me at least) to put in just a few words.
     
    Last edited: Oct 20, 2004
  4. Oct 30, 2004 #3
    Hey thanks! Yeah I already had to turn in my work by the time you replied so my professor posted the answers, but you did expand on it quite a bit so thanks for the extra info. (the simplified answer was to just use the voltage divider principle and sketch how the plot would look). A lot of these things are starting to connect now (in my head that is), so I am starting to get the hang of looking at electronics qualitatively instead of just math on your paper. The problem with me is that when I took regular Physics all I had to do was do math on paper basically for the electronics section of the book. It wasn't as in depth as the electronics course itself. Thanks again!
     
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