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Finding the Transfer function of an RL series circuit

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Attached image has full question and information.
    Determine the transfer function of the circuit. I've done this part, what I am struggling with is converting it into the form requested.
    R = 10 KΩ
    L = 10 H

    2. Relevant equations

    Voltage divider: Vo = R2/(R1 + R2) Vin
    Impedance of an Inductor = JωL

    3. The attempt at a solution

    T(ω) = jωL/(R+jωL)
    = 1/(1+R/jωL)
    = 1/(1+(1000/jω)
    Methods I have attempted at to turn this into the required form
    • Multiply by complex conjugate, just got even messier
    • Graphics calculator (i think this worked but will not be allowed in my test)
     

    Attached Files:

  2. jcsd
  3. Apr 21, 2015 #2

    gneill

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    Staff: Mentor

    Hi Stephengilbert1, Welcome to Physics Forums!

    The form they're looking for is essentially a phasor: magnitude and phase angle. So you need to take your transfer function and derive from it expressions for magnitude and phase angle, ##A(\omega)## and ##\phi(\omega)##.
     
  4. Apr 22, 2015 #3

    psparky

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    Gold Member

    Few things, I would take this (= 1/(1+(1000/jω) and put it in proper form.

    = jw/(jw+1000)

    In other words, you had it in proper form then "simplified" into a improper form. Remember that "S" and "JW" can be interchanged at will on these transfer functions.
    So you can see that we can transfer this to S/(S+1000). Looks a lot like a high pass filter. Thats a form you likely recognize and can now possibly transform into the form they want.

    Also, plug in values for 0 and infinity into your w and solve. This will always give you a general shape of your gain or bode plots. Also this will give you your phase angle at any given frequency. Same thing if you want frequencies in between the extremes, you can just plug and chug.
     
  5. Apr 22, 2015 #4
    Ok I understand some of this:
    • It is a high pass filter because at high frequencies the inductor acts as an open circuit
    • At ω = 0, T(ω) =0
    • At ω = ∞, T(ω) = 1
    • I can also figure out the break/cutoff frequency, which is asked in the following question, ωb = R/L = 1000 rad/s
    The part I am still stuck on is manipulating S/(S+1000) (from rectangular?) to the polar form the question asks for.
    Is it a matter of algebraic manipulation or am I not recognising something obvious?
     
  6. Apr 22, 2015 #5

    gneill

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    Staff: Mentor

    You're looking for expressions for the magnitude and phase angle. Plug in jω for S. Then:
    $$\left| \frac{jω}{jω + 1000} \right| = ?$$

    It'll be some expression involving ω.

    How might you find the phase angle? Hint: rearrange your transfer function into separate real and imaginary parts (the complex conjugate method you mentioned earlier will be helpful; don't worry if it looks a bit messy due to the denominators of the two parts, they should disappear when you find the angle).
     
  7. Apr 22, 2015 #6
    Using complex conjugate I get:
    (ω^2)/(ω^2+1000000) + (1000 j ω)/(ω^2+1000000)
    This is at least separate in real and imaginary parts.

    For the answer to this question my attempt is
    A(ω) = √ (ω^2/(ω^2+1000000))^2 + ((1000 ω)/(ω^2+1000000))^2
    Φ(ω) = arctan((1000 ω)/(ω^2+1000000) / (ω^2/(ω^2+1000000))

    Is this the full answer to the question? (Obviously sub into the Ae^j form)
     
  8. Apr 22, 2015 #7

    gneill

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    Staff: Mentor

    Okay.
    No, you've overcomplicated things for the amplitude, and failed to simplify the phase expression. The idea of splitting the transfer function into separate real and imaginary parts was in order to get at the phase angle, not the magnitude.

    Starting with the phase, within the arctan function you have a ratio of two fractions. The fractions have identical denominators. Shouldn't you be able to cancel them? What's left? Any more cancellation possible?

    For the amplitude, just work from the original transfer function. What's the magnitude of its numerator, jω? What's the magnitude of its denominator? What does that leave you with?
     
  9. Apr 22, 2015 #8

    Okay i understand the phase simplification,
    Φ = arctan(1000/ω)

    For amplitude the numerator magnitude is ω?
    But I'm not quite sure how to get the magnitude of the denominator, can i go:
    √(ω^2 + 1000^2)

    Which would make the total amplitude ω/√(ω^2 + 1000^2)
     
  10. Apr 22, 2015 #9

    gneill

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    Staff: Mentor

    Looks good.
    Yup. It's a simple complex number, so that's its amplitude.
    Yup. That looks good too!
     
  11. Apr 22, 2015 #10
    Thanks for all the help.
     
  12. Apr 23, 2015 #11

    psparky

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    Gold Member

    Thanks guys, that was the part that I did not know. I sure danced around it though!
     
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