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Electric circuits methodology question

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-2-11_21-8-38.png

    2. Relevant equations
    upload_2017-2-11_21-8-54.png

    3. The attempt at a solution
    My problem lies in when I get equations like:
    upload_2017-2-11_21-10-2.png
    upload_2017-2-11_21-11-6.png

    and if I plug in Vi and its derivative, I get equations like:
    upload_2017-2-11_21-10-42.png

    How do I solve for t in these equations if I have a sine and a cosine term that contain t?
     

    Attached Files:

  2. jcsd
  3. Feb 12, 2017 #2

    NascentOxygen

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    Have you sketched what you think the waveform V3(t) will be looking like after a few seconds has elapsed?
     
  4. Feb 12, 2017 #3

    gneill

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    You're going to want to sketch the waveforms first as suggested by @NascentOxygen since you may want to adjust the input function (such as moving the start time and changing sin to cos) to make the math easier for examining a particular location on the curve. But that should become clear later.

    Regarding your equation issue, there's a trig identity that relates sin, cos, and tan....
     
  5. Feb 12, 2017 #4
    Yes something like this but not precise:
    upload_2017-2-12_9-32-31.png

    SOS. What's that darn identity. I tried setting tan( 2pi t) = -.1 2pi but I get a negative time?
     
  6. Feb 12, 2017 #5

    gneill

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    sin(θ)/cos(θ) = ?
     
  7. Feb 12, 2017 #6
    That's what I did: tan( 2pi t) = -.1 2pi from the equation I got up there( 0.1 (2 pi) cos(2 pi t) + sin( 2 pi t) = 0. But if I were to solve for it I get a negative time.
     
  8. Feb 12, 2017 #7

    gneill

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    How does your math distinguish what part of the curve it applies to? The problem with sine and cosine curves is that many angles satisfy a given function value (many to one functions). That's why I suggested that you might want to shift your input from a sine to a cosine. In that way the first part of the curve to satisfy the required condition is the downward sloping face of the cosine's first quarter cycle.
     
  9. Feb 12, 2017 #8
    If I were to shift the curve, would I be shifting both the d/dt (sine) and sine curves or just one curve? So the input would become
    sin((2 pi t) + pi/2)?

    Also my professor gave me this when I asked him:
    upload_2017-2-12_9-57-9.png
    Somehow he got that from the relationship I wrote down (the one with sine and cosine). Is this somehow derived from shifting?
     
  10. Feb 12, 2017 #9

    gneill

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    upload_2017-2-12_13-8-29.png

    Just set a new start time so that the function becomes a cosine. Now the curve starts out headed towards the point of interest rather than climbing the "backside" of the sine function towards its peak.
    I suppose it does. I see "0.5u" but "n = 0,1,2,3,..." I'm not sure the significance (if any) for having different variables (n and u).

    Rather than trying to add or subtract various multiples of ##pi/4## or ##\pi/2## to the arguments of the trig functions I generally find it easier to just change the function from sine to cosine or vice versa, which accomplishes the shift and leaves the argument intact.
     
  11. Feb 12, 2017 #10
    The n = 1,2,3 is supposed to signify that that the function oscillates and will repeat the behavior again.
    I guess he did just change the variable

    Also if I shift like you suggested, I get -0.1 2 pi sin (2 pi t) + cos ( 2 pi t) = 0 which gives me
    tan (2 pi t ) = 1/ (0.1 2 pi) which doesn't get the solution I need. Am I doing something wrong?

    Also, why would we subtract is by .5? Technically the cosine curve begins .25 seconds in compared to the original sine curve so wouldn't we add .25 to get the true time?

    EDIT: Did he get 0.5u - 0.8928 from the original equation where t would equal -0.8928 and he assumes it's periodic every .5 seconds? How does he assume this?
     

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    Last edited: Feb 12, 2017
  12. Feb 12, 2017 #11

    gneill

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    I believe that the idea is to find the amount of time between when the RC circuit becomes isolated from the source voltage (diode stops conducting), and the zero crossing of the source signal. It crosses zero at t = 0.5 s for the original sinewave.
     
  13. Feb 12, 2017 #12
    I'm not following. Also if I do translate the curve to cosine as you suggested, I still don't get the right answer.
     
  14. Feb 12, 2017 #13

    gneill

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    Here's a plot of the voltage waveforms from a simulation of the circuit.

    upload_2017-2-12_16-32-25.png

    I've annotated the screen capture to show some of the items of interest. The outputs of the two rectifying circuits are symmetrical (red vs green plots), just inverted and time shifted by a half cycle (0.5 seconds).

    You're looking to calculate the output voltage (voltage on a capacitor) when the input signal crosses zero. That first happens at t = 0.5 s on the plot, but it's Δt = 0.5 - 0.411 = 0.089 seconds from when the diode turned off and the RC circuit became isolated. That's how long the voltage on the capacitor has had to decay to that point.

    If you use a cosine function to find the separation instant, then the cosine's start time would begin when the sine is at its peak at 1/4 second. So a solution using the cosine would yield a time of 0.411 s - 0.25 s = 0.161 s.

    So let's say you solved the simultaneous equations using the cosine function and you found the time when the diode stopped conducting to be tx = 0.161 s. Then you can add 1/4 seconds to that to find the "actual" time for the sine function, t = tx + 0.25 = 0.411 s. You can go on from there to find the capacitor voltage at that instant and then how much it decays by time t = 0.5 s.
     
  15. Feb 12, 2017 #14
    Thank you so much for the graph. I understand what happens when we shift the graph using cosine instead and I get that answer as well (.411 s).

    In regards to if I used the original equations, I first obtain -0.089 seconds. Am I correct to assume that the regular cosine and sine curves periodically intercept each other at half a period, so if I get 0.089, the curves will intercept again at -0.089 + 0.5, and then once more at -0.089 + 1? At the first time, the first diode will turn off and exponentially decay until it reaches 0 where it will again become Vin. When Vin first intercepts 0 , diode 2 turns on because its conditions are satisfied and stays on until its respective equation is violated at -0.089 +0.5n, where in this case n = 2?

    Then when Vin again intercepts 0, the first diode turns back on since it exponentially decayed towards 0, and the cycle continues?

    upload_2017-2-12_14-43-19.png
    upload_2017-2-12_14-43-32.png
     
  16. Feb 12, 2017 #15

    gneill

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    It's not so much that the cosine and sine intercept as that they both have the same period and thanks to the symmetry of the circuit anything that happens in one half cycle occurs in mirror image form in the next half cycle. This is true whether you base the math on the cosine or the sine. I just use the switch of trig function to avoid the negative time result in the math and then having to remember how to shift it :smile:. I find it easier knowing that when the math spits out a value that it's for the intended location on the curves and I don't have to go through all the adding or subtracting fractions of a period to adjust things. In this case the cosine places the first solution at the minimum positive time occurrence. After that I know that the same feature repeats every cycle (every second) for the top circuit, while the mirror image takes place a half cycle displaced for the bottom circuit.

    So yes, the diodes will alternate turning off at times dictated by -0.089 + 0.5n. The top diode turns off when n is odd, the bottom one turns off when n is even.
     
  17. Feb 12, 2017 #16
    Alright thank you so much! This was according to my professor the hardest problem we'd see for this topic so thanks a lot for your guidance.
     
  18. Feb 12, 2017 #17

    gneill

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    Happy to help. Good luck in your studies!
     
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