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Homework Help: Find V(s) and V(t)

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data
    everything's in the picture.



    2. Relevant equations

    V=I*R

    Ic=Iin*R/(R+C)

    3. The attempt at a solution
    Ok, I think I got the first part alright. I found the current in terms of Vin to be: Vin/(25+(22.5*225/s)/(22.5+225/s)) which simplifies to Vin*(s+10)/(225s+475). Now that I have current I can use the current divider rule to find Ic and then multiply Ic by C to get Vc. By doing that I get the equation [Vin*(s+10)/(225s+475)]*[22.5s/(225/s+22.5)]*225/s. After much simplification I get Vc=9*Vin/(s+19). I divide that by Vin and get 9/(s+19). Setting the bottom equal to zero I get a pole at s=-19. Setting the entire thing equal to zero I get a zero at s=+or- infinity.

    Once I get to the 2nd part I'm a bit confused... if Vin=45 at an angle of 0o, is my Vc(t) just simply equal to 45*9/(0+19) since no s is given?
     

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  2. jcsd
  3. Feb 23, 2010 #2
    O i accidentally forgot to multiply my 25 by s at the beginning. I actually end up w Vc/Vin=9/(s^2+10s+9) so that makes my poles s=-1,-9 and my zeroes still + or - infinity.
     
  4. Feb 24, 2010 #3

    vela

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    I'm unfamiliar with the notation used. What is "45 at 0 degrees" supposed to mean?
     
  5. Feb 24, 2010 #4
    well 45 at 0 degrees is still given in the s domain. 45 will be the magnitude and 0 will be the degree thats in the cos or sin if you have a jw anywhere, but since I dont have any imaginary numbers in my s then I shouldnt have a cos or sin anywhere. My problem is, I can solve for my poles and find s=-1, and -19. Does that mean that I should have a voltage for Vin that is = to 45*(e^-t + e^-19t), or is it simply 45?
     
  6. Feb 24, 2010 #5

    vela

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    Sorry, that still doesn't make any sense to me.

    I'm guessing it means [tex]V_{in}=45 e^{i0}=45[/tex]. In other words, [itex]V_{in}(t)=45\delta(t)[/itex].
     
  7. Feb 25, 2010 #6
    Also known as the Steinmetz notation:

    [tex]
    45 \angle 0^{\circ}\;\texttt{V}
    [/tex]

    which is basically a shorthand notation for complex numbers in polar form, whereas you in your post have used the exponential form. Both mean the same.
     
  8. Feb 25, 2010 #7

    vela

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    Oh, okay. From what I read about the notation, it's specifying [itex]V_{in}[/itex] in the phasor domain, so in the time domain, you have [itex]V_{in}(t)=45\sqrt{2} \sin(\omega t+0^\circ)[/itex].

    Edit: I read a bit more, and it seems some people use different conventions. In light of the rest of what the question asks, I think it means

    [tex]V_{in}(t)=45 \cos(\omega t+0^\circ)=Re(45 e^{i\omega t+0^\circ})[/tex].
     
    Last edited: Feb 25, 2010
  9. Feb 25, 2010 #8
    [tex]
    45 \cos(\omega t+0^\circ)=45 \angle 0^{\circ}
    [/tex]

    That's the thing about Steinmetz notation, it doesn't show the angular frequency. It only shows you the magnitude of the corresponding phasor and its phase shift w.r.t. to the reference.

    As far as I know, it's much more common amongst engineers than physicists (like the i and j conventions for complex numbers).
     
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