Steady state solution using Final Value Theorem in s-domain.

Discussion in 'Homework Help' started by Suyash Shandilya, Sep 14, 2016.

  1. Suyash Shandilya

    Thread Starter New Member

    Sep 14, 2016
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    sdomaindoubt.PNG
    Q. The circuit is driven by Vi=VpCos(t/RC).The steady state output Vo will be :


    I can do this by normal method of taking omega as 1/RC, hence Xc = -jR ....
    But I thought of transforming the problem to s-domain and then using the final value theorem to find out the steady state solution.
    Let's say impedence of parallel RC is Zp and series is Zc.
    So output would be Vi(s)*Zp/(Zc+Zp)
    Where,

    Vi = Vp*s/(s^2+(1/RC)^2)
    Zc is R + 1/sC
    Zp = R/(1+sCR)

    But the overall polynomial (after multiplying with s, as FVT demands) has the degree of denominator greater than numerator. Which would mean infinite value at s->0 implying same at t ->infinite which we can see is wrong.
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello there,

    You are just going to have to show more work in order to see what you are doing wrong, if anything.

    For this example, the transfer function has the power of s as s^1 in the numerator and s^2 in the denominator, and the forcing function has the power of s^1 in the numerator and s^2 in the denominator, so the combination of the two produce s^2 in the numerator and s^4 in the denominator, and multiplication by s means s^3 in the top and s^4 in the bottom, which should produce 0 in the limit. However, that is not really the final value anyway because the output will be sinusoidal too, so to find out the steady state solution you've got to find the peak of the output and then it will be plus and minus that peak...a sinusoidal wave but of constant amplitude.
    Sinusoidal forcing functions usually produce sinusoidal responses, and step inputs produce either DC responses or oscillations in the limits. If the output was DC then it would have a constant final value.

    Show more work and we'll see what you have so far.
     
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  3. DGElder

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    Apr 3, 2016
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    Due to the fact this is multiple choice and because of the choices, this is a problem that is designed to be reasoned out, not calculated.
    If you understand RC circuits, not just how to plug in formulas, you should be able to pick the correct answer without any calculations, pencil or paper. If this is homework you will be expected to explain your reasoning.
     
  4. Suyash Shandilya

    Thread Starter New Member

    Sep 14, 2016
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    Well I certainly am quite less accustomed to RLC circuits as you are. Could you please tell how do you discern that the amplitude will be one-third and a 2pi phase shift?
    All I could decipher from this circuit was it's a high pass + low pass = band pass filter.
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi again,

    Yes it is like a bandpass but not a good one :)

    I thought you wanted to look into the final value theorem, not just find the answer. If you dont know the answer there are several approaches.

    Since you already have some information, first try to find out why the phase shift is zero from input to output.
     
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  6. DGElder

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    Apr 3, 2016
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    I don't see what the FVT has to do with this problem. There is no transient, it is a steady state problem. I think this problem was designed to get you to think qualitatively about what happens in a RC network.

    The problem comes down to determining two variables, the magnitude and phase of the output compared to the source.
    1. You can see that the magnitude of the output is the result of a voltage divider. Look at which, if either, impedance, |Zp| or |Zs|, is greater. By inspection you can see that one of the two output magnitude options is wrong so the other must be correct.

    2. We can see that this is a capacitive circuit so the current, I, must be leading the voltage source by something between 0 and 90 degrees (non inclusive). The output is equal to Zp*I. Since Zp is capacitive, the Vo phase will be lagging between 0 and 90 degrees behind the current. Given the possible resultants of the sum of the two phases, one of the two answers is clearly impossible so the other must be correct.

    I am assuming one of the answers is correct of course. You may want to solve the problem using phasors to check and satisfy yourself that you have the correct answer. That will also give you some additional insights into the relationship between a series and parallel arrangement of an R and C. Try it with a couple different frequencies as well - interesting relationship between the phases.
     
    Last edited: Sep 14, 2016
  7. Avyhe2956

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    Sep 14, 2016
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    Can you help me with this question (?) :
    http://forum.allaboutcircuits.com/threads/tskew-and-timings-between-flip-flops.127446/
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    He just wanted to see if the FVT could be used in some way to help figure out the circuit, that's about it. He did not really know whether it would work or not. I dont think the required conditions are met for this circuit though.
    For example, at least two poles are on the jw axis so it breaks one condition already.
     
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  9. Suyash Shandilya

    Thread Starter New Member

    Sep 14, 2016
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    I know the
    I know how to get the answer using simple fundamental concepts. I do want to look into the FVT. I am gonna try it again today. I just want to know how do you look at it so intuitively :)

    Spot on. And thanks for stating that FVT condition. :) :)

    I need not re-attempt it using s-domain, now that you have shed the light upon the violation now. I will surely try to realise how the phase change is zero.
     
    Last edited by a moderator: Sep 15, 2016
  10. Suyash Shandilya

    Thread Starter New Member

    Sep 14, 2016
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    I am trying hard to understand the second point but can't exactly 'feel' how will the phase not change or change by 2pi after passing through the high pass filter. I am gonna pick pen and paper and will try to properly solve it mathematically, but I'd love to have you elaborate the 2nd point a little more.

    And I really disappointed myself when I realised how simple it was to guess the magnitude... :p Thank you for explaining it though..
     
  11. DGElder

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    Apr 3, 2016
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    There is no 2 pi phase shift ( though that would look like a 0 phase shift), I don't know where you got that idea. The answer options are 0 phase shift or a 90 degree phase shift by virtue of the 2 available output functions: cos() or sin() respectively.

    I am assuming at this point you are familiar with phase diagrams. Sketch out a phase diagram showing the input voltage at zero degrees, then rough in a sketch of the source current phasor (that will be somewhere between 0 and +90degrees, right?). You don't need to know the actual magnitude or phase angle as long as the phase is bounded in the realizable range. Then sketch in a voltage phasor that shows how Zp (= R||C) will produce a voltage across it lagging the current. Keep in mind an RC network can not produce a phase change equal to or greater than 90 degrees; this is what limits the range of possible phase for each subsequent phase change and the resultant output voltage phase. See how, regardless of your choice of realizable phase angles, you have a limit to the range for your answer.

    OK, since the cat is out of the bag, i.e. the Vo phase shift is zero. The above reasoning doesn't tell you that directly. It does tell you that a 90 degree phase shift of the output voltage is impossible and with the right combination of R and Xc the zero phase shift option is possible. Therefore, the zero phase shift option must be the correct answer.
     
    Last edited: Sep 15, 2016
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  12. DGElder

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    Apr 3, 2016
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    It is also possible, though not as intuitive, to know the phase shift is zero - explicity, but you do have to do just a little math. You need to calculate |Xc| and realize it is equal to |R|. From that you can reason out in your head that the phase shift of Vo must be zero.

    With some vector math in your head you know the phase of Zs must be -45 degrees since it is the addition of equal magnitude vectors , |Xc| and |R|, 90 degrees apart. Zp vector math is a lot more complicated, but uneccesary. If you know that in the case of Zs and Zp being composed of equal magnitudes R and Xc - differing only in topology, and Zs phase = x, then Zp phase = (-90-x). So in this circuit Zs and Zp both have a phase of -45 degrees. How nice.

    To get |Z| for the whole circuit you can just add |Zs| and |Zp| since they have the same phase angle and the resultant Z will also have a -45deg phase. Very convenient! So now you know the current will lead the source voltage by 45 degrees and Zp will have a voltage lagging the current by 45 degrees. Therefore the Vo output is +45-45=0 degrees.

    Yep, more convoluted, but maybe just a little bit edifying.
     
    Last edited: Sep 15, 2016
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  13. MrAl

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    Jun 17, 2014
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    Hello again,

    A couple interesting points you brought up here. Let me elaborate a little...

    First, i thought you had the answer to the phase shift already or else i would not have stated that. But in any case you should look into this.

    Second, intuition should be considered only second best. The best is pure analysis from first principles. You can develop intuition from that, but should not rely on it too much because it is very often deceiving. In fact, it should be regarded as a heuristic rather than any true form of analysis because it is always an attempt to jump to the correct answer without an exhaustive analysis.
    The best comparison i've seen for comparing these two comes from the game of chess, where the possible number of moves in one game can be astronomical. The number of possible moves in the first move of the game is roughly 1000. That means the possible number of moves for the second move is about 1,000,000. Third move, 1,000,000,000 (1 billion), roughly. So you see how quickly the number of combinations increases and so the total number possible is a very huge number. Now if we use a CPU to calculate every exact combination to find out what the best or one of the best moves is for each move, it would take decades to figure out every single move, if not longer, and it would probably never finish because it just takes too long. This is why chess programs only do the analysis to a certain depth and then from there rely on heuristics that are based solely on the position, not on any more successive moves. But heuristics are imperfect, so in certain positions some programs do better than others. The ones that have better heuristics for that particular position will win the game or at least draw, while the other can only hope for a draw.
    Intuition is like a heuristic for humans. It works sometimes and sometimes it doesnt. The worst part though is when we get confident and then start to rely on these heuristics and that's when the mistake start to roll in.
    When time is a factor we sometimes have to rely on heuristics, just like in the chess program, but otherwise it's not a good idea.

    So intuition should come from analytics, not the other way around, given that the time frame allows.

    There are certain things we can almost be sure of, but they come from experience. This will also fail us however when we get into some novel situations. For an example, most RC networks never produce an actual zero or infinity because their outputs are usually damped. One exception i happen to know of though is the twin T notch filter, which can produce an actual zero that produces an actual zero output, and it is a network made from only resistors and capacitors, although it does assume zero source impedances but that's true for many analyses.

    Thirdly, FVT is stated as it is because that is probably the most general view that will handle every case, which includes the conditionals. However, i have found that sometimes we can have jw axis poles and be able to consider that a limit because the output is bounded. Just like with a DC level, the output only assumes one value but that comes from an AC perspective. From the standard viewpoint the output may vary from -1 to +1 but that variation is continuous and never changes or gets larger. FVT would tell us that the FVT does not work for this case, and to be exact to the way it is written, it doesnt work. But, we have to also realize that there is a final value nonetheless, and that value is 1 when viewed from the AC perspective, which means plus and minus 1, and never goes under that lower value and never goes over that upper value, as t goes toward infinity.
    Can we modify the FVT to handle these cases? Perhaps we can, but if we cant, then the FVT does not work, and we know it does not work exactly as stated so far.
    I have ran into problems where the limit does not exist mathematically, but if we accept a bounded form of the limit, then it can be said to have a bounded limit. If there was no limit, then we could not say that for example:
    x*limit=0
    or:
    x*limit=1

    or make any other claim like that, but if the limit was bounded then we could say that:
    x*limit=0

    if x was zero, just for example, when normally we could not say this if the limit did not exist.
    So in some problems, it would help to realize that the limit was bounded even though it did not settle on a single solitary value. That is because it would simplify other problems and provide a single result that is definitely valid and can be proved to be valid using other methods.

    BACK TO FVT
    Another view is that with s=0 we are basically looking at the DC response, and with the cap in series, how can it be anything other than zero. The cap attains a charge and then stops conducting, which allows the output to 'ramp' down to zero. With an AC input this does not happen, therefore if we hope to find a FV with some FVT then it has to include the AC part of the response, namely, s=a+jw where jw>0 even if a=0. With jw>0 then that means that w>0 and so we are back to the regular AC response, which will not be zero for w>0 because it is a linear network. The output will be bounded however for w not infinity.
    This of course means that for a network 1/(s+1) the output would be bounded as:
    1/(jw+1)
    which is simply the AC response.

    SUMMARY
    In summary, if i had to say just one thing here, i would say dont rely on intuition unless you have to. Rather, gain intuition after the direct analysis, and be very careful in the future as you try to apply that. Also, try to use intuition to gain access to more complicated results. That is, when you come to a more complex problem try to use your intuition to figure out the best way to analyze the new problem and then check it to see if it is correct.
    If you have a second way to check your results you will seldom make a mistake.
     
    Last edited: Sep 16, 2016
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  14. Suyash Shandilya

    Thread Starter New Member

    Sep 14, 2016
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    A very eye-opening advise sir. I have been given this before and I often mischievously wander towards the shortcuts.
    I will keep this in mind. Thank you. :)
     
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