LTI System freq. domain HELP Please

Discussion in 'Homework Help' started by ohwcomp, Oct 14, 2008.

  1. ohwcomp

    Thread Starter New Member

    Sep 16, 2008
    9
    0
    Howdy Everyone!

    Im having trouble with the following interesting question especially with part (c)

    [​IMG]

    (a) For this part I realised the steady state principles and applied open and short circuits etc and used the voltage divider rule to get 2V and 1 A, is this correct?

    (b) Redrawing this I got

    6V turned into 6/s
    4Ω stayed the same
    1H turned into s
    0.05F turned into 1/0.05s

    Also for the initial conditions i added a current source under each inductor i(0-) and capacitor vc(0-).

    (c) How do I find this Vout? I can see that its the voltage at the node connecting Ir,Il,Ic. So at this node Ir + Il + Ic = 0? From KVL/KCL?

    How do I take into the account the -ve sign of the current source from the inductor.

    Would it be Ir - Il + Ic = 0?

    Ir = (Vo - 6/s)*1/4 ?????

    Il = ?

    Ic = ?

    Then how would I get an expression for Vout? Hence the complete solution?

    Cheers for the help in advance!
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Maybe tomorrow or in a few days i will be able to answer you because we have just started to study Laplace transforms at university. :)
     
  3. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    I get 2V and 1A for part A.

    Im not sure exactly what you're doing with the voltage in part B.

    For part b, i'd get: capacitor --> 1/(.05s) and inductor --> s

    For part c, just use nodal analysis. I get the following equation:

    (V_0 - 6)/4 + V_0/(s+2)+V_0/(1/(sc)) = 0

    From there, you need to get it in the right form. Take the inverse laplace transform and find a solution that satisfies the equation...

    Hopefully that gets you started.
     
  4. ohwcomp

    Thread Starter New Member

    Sep 16, 2008
    9
    0
    Hi,

    (V_0 - 6)/4 + V_0/(s+2)+V_0/(1/(sc)) = 0

    took into account the 2Ω which is not there as it is short circuited,

    Ir + Il + Ic = 0

    V=IR
    I = V/R

    Il = V_0/s

    Ic = V_0*s0.05

    ok so rearranging I get some nasty

    V_0 = 2s+?/s^2+5s+20 which I cant simplify and get the inverse. Im doing this from memory,so please check my V_0.

    But Im sure the bottom bit s^2+5s+20 is correct, how do I simplify this so I can take the LT inverse?

    Cheers
     
  5. ohwcomp

    Thread Starter New Member

    Sep 16, 2008
    9
    0
    Correcting my mistakes,

    You Got,
    (V_0 - 6)/4 + V_0/(s+2)+V_0/(1/(sc)) = 0

    Shouldnt it be

    Ic = 0.05s ( V_0 - (2/s) )

    Il = V_0 - (-1) / s

    Ir = V_0 - (6/s) /4

    Leaving V_0 = 2s+10/s^2+5s+20?

    Is that correct? Then how would I go on from here?

    Thanks
     
  6. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    Oh good call heh.

    That denominator looks good. I'm not sure where the numerator comes from, but going back to the equation i had:

    (V_0 - 6)/4 + V_0/(s)+V_0/(1/(sc)) = 0

    => V_0(1/4+1/s+sc) = 6/4

    => V_0(s/4+1+s^2c) = 3s/2

    =>V_0(s^2+s(1/4c)+1/c) = 3s/2c

    The inverse laplace of that is fairly straightforward, i think. It has been quite a while since i have actually done them, so correct me if i'm wrong haha.

    hope that helps a little.
     
  7. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    Oops. I just saw your latest post.

    Here's what i get for the currents:

    Il = V_0/sL

    Ic = V_0*sC

    Ir = (6-V_0)/4

    I am assuming you're calling the bottom node ground...
     
  8. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    wowza, thats a lie...

    Ir = (V_0 - 6)/4
     
  9. ohwcomp

    Thread Starter New Member

    Sep 16, 2008
    9
    0
    Heya Mate,
    Yours:
    Il = V_0/sL
    Ic = V_0*sC
    Ir = (V_0 - 6)/4

    Mine:
    Il = V_0 - (-1) / s
    Ic = 0.05s ( V_0 - (2/s) )
    Ir = V_0 - (6/s) /4

    Yours is different to mine, L = 1, C = 0.05 hence yours turns into
    Il = V_0/s
    Ic = V_0*s0.05
    Ir = (V_0 - 6)/4

    I believe mine takes into account the initial conditions where as yours assumes initial conditions = 0. I wish I could do that, but unfortunately the complete soltution is taking account these conditions so =((((

    Your input is greatly appreciated! Any more ideas?
     
  10. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    I see... Hopefully i'm not causing more problems then i'm helping hah.

    Anyway, there very well may be other methods, but the way i was taught this method, the initial conditions are applied to the equations after they are transformed back into the time domain.

    Once you get your solution back into a nonhomogeneous equation, you can use elementry diff eqn methods to find the forced and natural solutions.

    Once again, thats how i was taught... you may have been taught other methods.
     
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