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Vout(t) using laplace and differentiation

  1. Jun 19, 2013 #1
    1. The problem statement, all variables and given/known data
    2qi5kbm.png


    2. Relevant equations
    V(L)=L*di/dt
    laplace(u(t))=1/s

    3. The attempt at a solution
    was just wondering if i did this right. converted to the s domain, then wrote voltage equation around loop, in terms of current I(s):
    V(s)=R*I(s)+L([itex]\frac{dI(s)}{dt}[/itex]-iL(0-))
    V(s)=R*I(s)+L(s*I(s)-0)
    V(s)=I(s)[R+Ls]
    I(s)=[itex]\frac{V(s)}{R+Ls}[/itex]

    I(s)=[itex]\frac{\frac{20}{s}}{10+20s}[/itex]

    I(s)=[itex]\frac{\frac{20}{s}}{20(s+0.5)}[/itex]=[itex]\frac{1}{s}[/itex]*[itex]\frac{1}{s+0.5}[/itex]=e-0.5t u(t)
    V(L)=L*I(s)=20e-0.5t u(t), so C was my answer.
     
    Last edited: Jun 19, 2013
  2. jcsd
  3. Jun 19, 2013 #2
    Doesn't look correct to me. First find the Vout in s domain tgen convert it to time domain.
     
  4. Jun 19, 2013 #3
    is the equation for I(s) correct at least? is there another way to find Vout=V(L) other than doing it the way i did?
     
  5. Jun 19, 2013 #4
    Equation of I(s) is correct. All you havw to do is to find thw Voltage. You are very close.
     
  6. Jun 19, 2013 #5
    The other way is to solve the diffrential equation that is without going into the s-domain. You can try that if you want. But i recommend you first complete it with your first attempt.
     
  7. Jun 19, 2013 #6
    ok so looking at my book, i think my s-domain equation for I(s) is right but somehow my time-domain transformation is wrong. I think i should've split it into partial fractions.

    I(s)=[itex]\frac{1}{s}[/itex]*[itex]\frac{1}{s+0.5}[/itex]=[itex]\frac{A}{s}[/itex]+[itex]\frac{B}{s+0.5}[/itex]
    using the 'method of residues' in my book it becomes [itex]\frac{2}{s}-\frac{2}{s+0.5}[/itex] which i think then becomes I(s)=(2-2e-0.5t)u(t) in time domain.

    then rather than trying to take the derivative of this and multiply by the inductor, i should just multiply it by resistance:
    Vout=10*(2-2e-0.5t)u(t) =(20-20e-0.5t)u(t)

    is this right?
     
    Last edited: Jun 19, 2013
  8. Jun 20, 2013 #7
    You are still doing it wrong. See what i wrote in my forst post. Do not convert current in timw domain first. First find the Vout in s-domain, then convert Vout im time domain.

    Hint: Vout(s) = I(s)*Z(s).

    What is Z here?, then what is Z(s)?
     
  9. Jun 20, 2013 #8
    Find the transfer function or use voltage division to find the voltage across the output in terms of what you are given.


    V(o) = (V(s)*jωL)/(R+jωL)

    Now using ohm's law you know that the current flowing through this circuit is simply this output voltage divided (which you have all the values for) divided by the total impedance of the system.
     
  10. Jun 20, 2013 #9
    This is not correct. s jω, is only true for sinosidal transfer function. For any general input voltage one has the solve the Laplace and then the inverse laplace function
     
  11. Jun 20, 2013 #10
    i see...so just my equation for Vout was wrong. but why would i multiply the current by the entire impedence? i thought Vout was just the voltage across the inductor. but if I do Z(s)*I(s)=Vout like you said, then Z(s)=10+20s and [itex]\frac{10+20s}{(s)(s+0.5)}[/itex]=[itex]\frac{20}{s}[/itex] using partial fractions so 20u(t)?
     
  12. Jun 20, 2013 #11
    You have to find the Voltage across the inductor so multiply I(s) it by the impedance of the inductor only. You have multiplied the I(s) by the total impedance of the circuit, which gave you the voltage across the circuit that is input Voltage only!

    PS: Revise ohms law(V=IR), you are weak at it.
     
  13. Jun 20, 2013 #12
    i am not weak with ohms law, i thought it was indeed the impedence of inductor times the current, you kinda misled me when you said "Vout(s) = I(s)*Z(s)."

    I(s) times the impedence of the entire system, this is what i undrstood it to mean. Z(L) is what i understand to be the impedence of the inductor only. which is i believe L*s=20s in s-domain.

    if i do 20s/((s)(s+0.5)) and use partial fractions i get the answer i first got. 20e^-0.5t.
     
  14. Jun 20, 2013 #13
    You have have got the the correct answer in your first post, but it was accidentally correct due to mathematical error. You did a mistake in taking the inverse laplace transform of I(s), which you corrected in your post. And sorry if I misled you, we don't help you to get the correct answer what we try is to make you understand the concepts, so that you can do other problems it yourself.

    Now that you have got the correct answer by the correct method, i would say good work!
     
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