don't reply if u r not above 2ed year college EE level

Discussion in 'Homework Help' started by wasimaxchao, Apr 15, 2009.

  1. wasimaxchao

    Thread Starter New Member

    Apr 15, 2009
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    all the laplace transform questions deal with merely asking u to find out the final "transformed t domain" equation.

    Well, a question in my homework asked about the same for the first part,
    but I was asked to give the specific value of V(t) when at the time of 1 of the "S domain" .......

    *the correct t domain equation is V(t)=1.414e^(-t) *cos(t-45)u(t) V

    45 is in degree.

    don't be naive by plugging "1" of s domain into the equation.....t and s r not the same domain......and don't say anything about changing one from radian to degree or changing 45 from degree to radian. I already did that......and it's no use since it's not the same domain 1.......


    *here is the circuit and how the V(t) equation is obatained....

    http://ee.cramster.com/basic-engineering-circuit-analysis-8th-problem-9-257317.aspx

    but on my homework, it asks what is V(t) at t=1 (dun just plug in 1 into the given solution equation....like i said...s and t r not the same domain......)

    original question:

    http://www.wretch.cc/album/show.php?i=momodeedee&b=21&f=1665764173&p=16
     
    Last edited: Apr 15, 2009
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    V(t) looks like the response to a step input. Say the voltage across R in a series RLC circuit.

    I think what you want is the response to a sinusoid of ω=1rad/sec. Perhaps if you gave us the full details of the question we might be more helpful.

    I guess I fit the > 2nd year EE level. Do you need my CV?
     
  3. wasimaxchao

    Thread Starter New Member

    Apr 15, 2009
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    tried that already.....making 45 degree into .785 radian

    and V(t)=1.414e^(-t) *cos(1-.785)u(t) Volts

    which is wrong.......this question is nothing to do with degree or radian convertion.......

    u need to fully understand wat happened at S and T domain while doing the laplace transform to the circuit.......
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Perhaps you've missed my point.

    Your circuit is currently excited by a source of a particular type - let's say it's a pulse or step function. (You still haven't divulged the question details.)

    You solve for the required unknown in the Laplace domain and look up the time domain equivalent from the Laplace tables.

    Suppose your source is no longer a pulse or step - rather a sinusoidal function such as ...

    e(t) = 1.0 x cos(t) [i.e. 1 volt peak cosine function at 1 rad/sec]

    For e(t), E(s)=s/(s^2+1)

    Now solve for the unknown variable in the Laplace domain - you'll get a different time domain result ....

    As I suggested - share the entire problem with us. Otherwise it's difficult to offer any advice.
     
  5. DrNick

    Active Member

    Dec 13, 2006
    110
    2
    If you look at the problem it is looking for a voltage at time = 1. If you plug one into your time domain solution you do not get "0.51" exactly. Did you try 0.508 for the answer? Perhaps the reason your solution is "wrong" is due to rounding problems...
     
  6. wasimaxchao

    Thread Starter New Member

    Apr 15, 2009
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    lol.... .51-0.508/.508 = .004 = .4% that will be within tolerence....
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Just so ....

    That's why posting the problem always helps.

    By the way "wasimaxchao" - re your original post statement

    "I was asked to give the specific value of V(t) when at the time of 1 of the "S domain" ......."

    It's worth noting for future reference that time has no relevance in the 'S' domain.
     
  8. Vishg14

    New Member

    Apr 16, 2009
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    V(s)=(1.414/(s+1))(se^(-45s)/(s^2 +1))
    This is the answer in laplace tarnsform, but i think the value of 45 should be phi/4 and in S-domain it does not contain the term t ,so im unable obtain the value of V(s) at t=1;.
     
  9. DrNick

    Active Member

    Dec 13, 2006
    110
    2
    What is the tolerance of the online grading system? I have found some to require accuracy up to 3 significant digits...I would suggest you try that. If that doesn't work your time domain solution is incorrect.
     
  10. wasimaxchao

    Thread Starter New Member

    Apr 15, 2009
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    the tolerence is 2%......if i typed .51, the system reads as .510.....so it's will be fine.....by the way, the solution was 1.02 and can anyone tell me why 1.02 is the correct answer??
     
  11. DrNick

    Active Member

    Dec 13, 2006
    110
    2
    your solution for the time domain voltage is incorrect. you should not be getting an exponentially decaying sinusoid. It should be the sum of two exponentials with real time constants.

    The solution that I got is pretty close to 1.02...about 10% off of that...

    When you solve the current loop you get:

    Vocurrent = 2*s^2/(s^2+2*s+1/2)

    the voltage across the output due to the voltage step is:

    Vovoltage = 4*s/(s^2+2*s+1/2)

    To get the full solution you sum the contribution due to the current source and the voltage source:

    Vo = 2*s(s+2)/(s^2+2*s+1/2)

    Next get the roots of the characteristic equation (ie find the poles).

    s^2+2*s+1/2 = (s+1-sqrt(2))*(s+1+sqrt(2)).

    Next you perform a PFE:

    A = 2*s*(s+2)/(s+1-sqrt(2)) evaluated at -1-sqrt(2) = 1/sqrt(2)
    B = 2*s*(s+2)/(s+1+sqrt(2)) evaluated at -1+sqrt(2) = 1/sqrt(2)

    Therefore the time domain signal should be:

    Vo(t) = 1/sqrt(2)*exp((-1+sqrt(2))*t)+1/sqrt(2)*exp((-1-sqrt(2))*t).

    When you evaluate this at t = 1, the solution is 1.13.

    this look right?
     
    Last edited: Apr 16, 2009
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