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Homework Help: Laplace equivalent circuit

  1. Feb 20, 2014 #1
    The amplifier and filter of FIGURE 4 can be represented by the
    equivalent circuit of FIGURE 6. (Figs are in image 1 and 2 attached)

    (i) Draw the Laplace equivalent circuit and hence derive the equation
    for vout(s) in terms of R, L and C for a step input.

    (ii) Attempt, using the component values given in TABLE C, to
    determine the transient response of the amplifier and filter for a 10
    mV step input.

    Comment on the nature of the response and its suitability for this
    application. Assume the thermocouple gives out a voltage of 10 mV
    at 100ºC.

    I have managed i) but I am struggling with is ii)

    I get 100vi/LCs(s^2+s(1/RC)+1/LC

    Which gives

    Vo(s) = (2*10^5)/(s(s^2+10s+2000))

    I have tried to use partial fractions.
    Pulling 2*10^5 to the side gives 1//(s(s^2+10s+2000))

    1= A(s^2+10s+2000)+(B+C)(s)

    If s = 0 then A = 1/2000

    but then i can't solve for B and C

    Any help would be good

    Attached Files:

  2. jcsd
  3. Feb 20, 2014 #2
    sorry here is image 2

    Attached Files:

  4. Feb 20, 2014 #3


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  5. Feb 20, 2014 #4


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    Staff: Mentor

    Your last term should have an 's' associated with either B or C.

    Yes, fine.

    Plug your value for A back into eq. 1 above and see what simplifies. What do you get? (Remember to fix your (B + C) term).
  6. Feb 21, 2014 #5
    So 1= A(s^2+10s+2000)+(B+C)(s)

    Should be 1= A(s^2+10s+2000)+(Bs+C)s

    But when I plug the result for A in I get stuck.
  7. Feb 21, 2014 #6


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    Staff: Mentor

    Where do you get stuck? Can you show more of your work? If you show your steps we might be able to suggest something.
  8. Feb 23, 2014 #7

    so I have tried for s=1 but get stuck with solving Bs+c = -0.0055

    so I tried to simplify to ((s(s+10))/2000)+1+(Bs+C)s
  9. Feb 23, 2014 #8


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    Staff: Mentor

    ##1 = \frac{1}{2000}\left( s^2 + 10*s + 2000 \right) + (B s + C)s##

    Expanding the RHS:

    ##1 =\left( \frac{1}{2000} + B \right)s^2 + \left( \frac{1}{200} + C \right) s + 1##

    Cancel the 1's on either side and divide through by s:

    ##0 = \left( \frac{1}{2000} + B \right)s + C + \frac{1}{200}##

    Now suppose that s = 0 ....
  10. Feb 23, 2014 #9
    Thanks so



    Simplifying to


    Then we get to

    0=(1/2000+B)s^2+1/200+C by eliminating 1's and dividing by s.

    if s=0 then c=-1/200

    Then plugging in the value for (C) B must = -1/2000
    plugging these values into the original partial fraction with some simplifying gives


    Is this right so far, I am still working on the answer.
  11. Feb 23, 2014 #10


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    Staff: Mentor

    Not bad. Since it's time to write out the results of the partial fraction expansion, it's also time to bring back the multiplier that was ignored for a time; remember the 2 x 105 that lay outside the "interesting part" of the partial fraction? You can bring that back by multiplying A, B, and C by it, so that:
    $$V_o(s) = \frac{100}{s} - \frac{100 s}{s^2 + 10 s + 2000} - \frac{1000}{s^2 + 10 s + 2000} $$
  12. Feb 23, 2014 #11
    OK so I get this result now I need to get each component to fit the standard laplance table so I can invert back the (t)


    The first bit easily fits in the table but the other two look like damped sine and damped cosine but I cant get them to fit properly.
  13. Feb 23, 2014 #12


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    You should be able to make them fit with a bit of algebra. Concentrating on the last two terms, start by combining them into one:
    $$100\left(\frac{s + 10}{(s + 5)^2 + 44.44^2} \right)$$
    regroup the numerator appropriately and split apart into two terms again. Carry on adjusting things until they fit the patterns for decaying sine and cosine.
  14. Feb 24, 2014 #13
    Spiting it up was wher eI was going.


    So that's the damped cosine sorted.

    The second term I need to get th 44.44 as the numerator
    so puling the 5 to the side and multiply the top by 44.44 then multiplying the 5 by the inverse of 44.44
    Giving us a damped sine.

    so Vo(t) = 100[1-(e^-5t cos 44.44t + 0.112 e^-5t sin 44.44t)]

    so now
    Attempt, using the component values given in TABLE C, to
    determine the transient response of the amplifier and filter for a 10
    mV step input.

    Comment on the nature of the response and its suitability for this
    application. Assume the thermocouple gives out a voltage of 10 mV
    at 100ºC.

    I multiply the whole by 10mV?
  15. Feb 24, 2014 #14


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    Staff: Mentor

    Sure. Looks okay!

    That would be the idea, yes.
  16. Feb 24, 2014 #15
    How do I under stand the response and it suitability for its application?

    Vo=[1-(e^-5t cos 44.44t + 0.112 e^-5t sin 44.44t)] at 100degC

    frequency is 44.44 rad/s
    Here is some more info if it helps:

    A thermocouple is a device commonly used to measure the temperature in, for
    example, boilers and engines. It has the advantage of directly giving out a
    voltage that is approximately proportional to the measured temperature. The
    thermo-voltage is, however, only a few millivolts.
    In the block diagram of FIGURE 4, the output of the thermocouple, vT(t), is
    fed into an amplifier to give a sufficiently large signal to drive a signal
    processing stage. The signal processing stage includes an analogue-to-digital
    converter (ADC) to give an output to drive a digital display. A low-pass filter
    (LPF) is placed between the amplifier and the signal processing stage to
    remove noise that may be picked up from adjacent ‘noisy’ equipment and from
    ‘mains’ interference.
  17. Feb 24, 2014 #16


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    Staff: Mentor

    That's a fairly open-ended sort of question, since no specifics are given as to the particular magnitudes of the noise or mains interference. But you might want to consider comparing the response magnitudes of the 'real' signal to those of 'noise' of similar size to the signal. That might mean doing some more analysis, since 'mains' noise is a sinusoid at line frequency, not a unit step, and machine noise often looks more like impulses from commutation (switching) and the like. Since temperatures typically don't change quickly and thermocouples can be pretty slow to respond, their signals are essentially DC compared to the other sorts of noise... So, maybe make a few assumptions about noise size and frequencies, do some circuit response analysis based upon those assumptions, then do a bit of arm waving comparison of the results?
  18. Feb 25, 2014 #17
    Thanks for all your help.
  19. Sep 8, 2014 #18
    I'm on the same question and I'm a bit stuck too.

    I've worked the initial equivalent circuit info as far as 100ViR/s[Ls(RCs+1)+R] but I'm having real trouble seeing how it gets to 100vi/LCs(s^2+s(1/RC)+1/LC

    Tried various things but I always seem to end up either with an s^3 term in there or two s^2 terms.....any help would be greatly appreciated.
  20. Sep 8, 2014 #19


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    Staff: Mentor

    It might be easier to first deal with the transfer function for vo(s)/vi(s) without inserting the step function input for vi(s). Get the transfer function into suitable shape before throwing in the 1/s input. At first you should end up with a form something like:

    $$v_o(s) = \frac{100 R}{s^2 L R C + s L + R}v_i(s)$$
    Then it's just a matter of some algebra; Start by dividing numerator and denominator by R ...
  21. Sep 8, 2014 #20
    I think I realised where I went wrong....I was on autopilot and instead of dividing num and den by 'R', I simply cancelled them out, so there no way it was going to work.

    Working past that I got to 100Vi/s(Ls/R + s^3LC + s)

    Working out 100vi/LCs(s^2+s(1/RC)+1/LC gives my above line (I hope) so looking ok.

    However, If I hadn't had that line to 'work back' from I think I would have struggled with factorising and finding a L.C.D for it. Guess I'll just have to practice more at these.
  22. Sep 12, 2014 #21
    I've worked further and managed to get to the point where

    [itex]V_o(s) = \frac{100}{s} - \frac{100 s}{s^2 + 10 s + 2000} - \frac{1000}{s^2 + 10 s + 2000}[/itex]

    The next line has me stumped a bit tho. I'm not sure if this is even the right way to go, but I can't seem to factorise it normally. It looks like it has to be resolved with the 'completing the square' method

    [itex]s^2 + 10s + 2000 [/itex]

    move the 'loose number' to other side

    [itex]s^2 + 10s = -2000 [/itex]

    dividing by the multiplier of the squared term leaves it the same.

    Take half of the s coefficient (5) and square it (25) and add to both sides

    [itex]s^2 + 10s + 25 = -2000 + 25[/itex]

    Taking the l.h.s. to squared form =

    [itex](s + 5)^2[/itex]

    and we get

    [itex](s + 5)^2 = -1975[/itex]

    But from here it goes all over. If we square-root both sides we get a plain old

    [itex](s + 5)[/itex] and we can't root the negative 1975. If it was positive 1975 though, we would get our [itex]44.44^2[/itex]

    So, have I gone off on a complete tangent from the start trying to get to the next step, or made a schoolboy error somewhere?

    Any help would be greatly appreciated.
  23. Sep 12, 2014 #22


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    Staff: Mentor

    Since the input is supposed to be a 10 mV step, you might want to use 1/(100s) for the input. This will change the numerators of your Vo expression (divide each by 100).

    But before you do that you might want to take a look through your inverse Laplace transform list to see if one of the forms matches your second term, the one with s/(quadratic) form. I think you'll find a closer match with the exponentially decaying cosine... which will mean splitting up your terms in a slightly different way.

    Completing the square is a good way to rewrite the denominator. The way I usually do it is to write out the two forms and compare terms:
    $$s^2 + 10s + 2000 = (s + \alpha)^2 + \omega^2$$
    $$s^2 + 10s + 2000 = s^2 + 2\alpha s+ \alpha^2 + \omega^2$$

    so ##2\alpha = 10## and ##\alpha^2 + \omega^2 = 2000## . Solve for the unknowns. I think you'll find that both turn out to be positive!
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