1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need help finding frequency response of a circuit

  1. Dec 5, 2013 #1
    I'm working on a problem which has two parts.

    1. Find the frequency response function H(f) for a circuit. A hint given is to find H(s) first and then plug in [itex]s=j2\pi f[/itex] to get H(f).

    2. Find [itex]v_{out}(t)[/itex], which is the response of the LTIC system to the periodic input signal [itex]v_{in}(t) = \sum_{n=1, n \ odd}^{\infty} \frac{8}{\pi^2n^2}\cos\left(\frac{\pi}{4}nt\right)[/itex] using H(f).

    I used KCL on the circuit to find the transfer function H(s):

    [itex]H(s) = \frac{1}{1*10^{-10}s^2+2*10^{-5}s+1}[/itex]

    Now, the problem hint said to evaluate the function at [itex]s=j2\pi f[/itex], so if I plug that in then the above equation becomes:

    [itex]H(f) = H(s)|_{s=j2\pi f} = \frac{1}{-4\pi^2*10^{-10}f^2+j4\pi*10^{-5}f+1}[/itex]

    So, although I don't really understand the above equation (for one thing, I don't get why there would be a "j" in an s-domain equation -- I've never seen that before), I guess I answered the first part of the problem.

    My question is, how should I go about solving the second part of the problem, finding [itex]v_{out}(t)[/itex] for the crazy input signal?

    Appreciate your advice.
     
  2. jcsd
  3. Dec 5, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    That is the standard way of evaluating a transfer function with sinusoidal inputs. You let s = jw. You will see a lot of this in the future.

    I strongly recommend you change f to w/2pi everywhere, right from the start. It's ridiculous to work with f instead of w. w = 2 pi f and is called the radian frequency. BTW I write w because I'm too lazy to write ω.

    As for your input: do one n at a time, starting with n = 1, 3 ,5 ... ; for each n, s = jnw1. Then you add all the n terms in a new series, then invert each new term one by one, add all those & you have your vout(t).

    Look at cos(pi/4 n t). Rewrite this as cos(nw1t) where w1 = pi/4 which you get by setting pi/4 = w1.

    BTW this voltage waveform isn't so screwy - it's probably a square wave or something similar.
     
  4. Dec 7, 2013 #3
    Thanks for your help. I'm stumped on this problem, so I really appreciate it.

    OK, so the equation I had was:

    [itex]H(f) = H(s)|_{s=j2\pi f} = \frac{1}{-4\pi^2*10^{-10}f^2+j4\pi*10^{-5}f+1}[/itex]

    Following your suggestion to let [itex]f = \frac{w}{2\pi}[/itex], the above equation becomes

    [itex]H(\omega) = H(s)|_{s=j\omega} = \frac{1}{-1*10^{-10}\omega^2+j2*10^{-5}\omega+1}[/itex]


    For some reason this just isn't clicking for me.. I'm still confused how I can use the rewritten equation, along with:

    [itex]v_{in}(t) = \sum_{n=1, n \ odd}^{\infty} \frac{8}{\pi^2n^2}\cos\left(\frac{\pi}{4}nt\right)[/itex]

    to find [itex]v_{out}(t)[/itex].

    You say to "do one n at a time," and "for each n, s = jnw1", but there isn't an [itex]s[/itex] in the [itex]v_{in}(t)[/itex] equation?
     
  5. Dec 7, 2013 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Take the 3rd harmonic (n = 3). What is v_in(t)?
    What is V_in, the phasor of v_in?
    Then, what is V_out, the phasor of the output voltage v_out?
    Then, what is v_out?

    Do this for all odd n, then v_out(t) = Ʃn v_out_n.
     
  6. Dec 7, 2013 #5
    Thanks again for your help.

    When [itex]n=3[/itex], the expression inside the Riemann sum is: [itex]\frac{8}{9\pi^2}\cos(\frac{3\pi}{4}t)[/itex].

    In phasor form, I believe that is: [itex]\frac{8}{9\pi^2}\angle0\deg[/itex]

    How can I find [itex]v_{out}(t)[/itex]? In the S-Domain, I know that:

    [itex]H(s)=\frac{V_{out}(s)}{V_{in}(s)}[/itex]

    But to use that I would have to find [itex]{V_{in}(s)}[/itex] in the S-Domain, which I don't know how to do since it's a Riemann sum.

    Also, what's the point of converting to the phasor domain?
     
  7. Dec 8, 2013 #6

    rude man

    User Avatar
    Homework Helper
    Gold Member

    right
    The phasor form is (8/9√(2)π2)ejπ/2 = j(8/9√2π2).
    The 1/√2 is to go from peak to rms voltages and the exp(jπ/2) reflects the fact that the input is cos, not sin. Remeber, cos x = sin (x + π/2).
    Riemann sum? You're dealing with one harmonic term at a time, then summing the output v(t) for all n.

    The phasor domain is what you have to use to get a v_out from a v_in. It's the s domain with s = jw. It's what you use to deal with steady-state sinusoidal voltages of one frequency.

    Your only other choices would be to deal with differential equations in the time domain, or use the Laplace transform, but the former is "forget it" and the latter is more complicated than necessary since that way you solve for the transient as well as the steady-state solution whereas here you're only intersted in the steady-state solution. So, back to phasors:

    If in general v(t) = v0 sin(w0t + θ), the corresponding phasor is V = (v0/√2)e.
    Note that w0 does not appear in the phasor, it is understood to be w0, and note also how the phase angle is expressed.

    Then the output voltage is v_out(t) = √2V |H(jw0)| sin(w0t + ψ).
    where ψ is the phase angle of H(jw0).

    You need to be comfortable in changing from polar to cartesian complex number description & back. For example, H(jw) is in cartesian form but to multiply with your input phasor you want to change H(jw) to |H(jw)|exp(jψ) where ψ = phase angle of H(jw). This makes complex number multiplication easy and also makes taking the inverse of the output phasor easy to get v_out(t).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted