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Basic Circuit Analysis

  1. Dec 29, 2011 #1
    Hi guys, I have a winter break homework assignment for my EE college course. I've been given what seems to be a simple circuit analysis problem, but feel that I am making it much harder on my self by using nodal analysis. I have to find the transfer function of the circuit. My work has been scanned and attached, any input as to whether I am on the right or wrong track would be greatly appreciated.

    1. The problem statement, all variables and given/known data

    Determine the transfer function, Av(s)=Vo(s)/Vi(s), in terms of s, R1, R2, R3, C1, and C2.


    2. Relevant equations

    Assumptions:
    ALF should be equal to (R3)/(R3+RZ) because the caps open at low frequencies.
    RZ is R1 and R2 in parallel.

    3. The attempt at a solution

    Attachment 1
     

    Attached Files:

  2. jcsd
  3. Dec 29, 2011 #2

    berkeman

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    The equations for the two nodes Vo and Vx look okay. Are you going to add the equation at Vi into the mix as well?
     
  4. Dec 29, 2011 #3
    I was thinking on doing nodal at that node, but if I were to do nodal I would be running into a dead end because I would have a vx term in my equation and I have no where to substitute that into. I would also have a vo and vi term in that equation, but I can't solve for those because I am trying to solve for those terms for my end result. Am I making any sense or am I just confusing myself?
     
  5. Dec 29, 2011 #4

    berkeman

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    No, I think that's reasonable. After I asked it, I realized that it might not really help, since you do not know the current through the Vi source, so writing an equation there would introduce an extra unknown Ii, not really helping to eliminate unknowns.
     
  6. Dec 29, 2011 #5
    Right, that's also another valid point as to why that nodal wouldn't be necessary. So I pretty much just have to grit my teeth and work this algebra out to find my transfer function?
     
  7. Dec 29, 2011 #6

    The Electrician

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    RZ is R1 and R2 in parallel.

    should be:

    RZ is R1 and R2 in series.

    which is what you have in the attachment; a small error.

    But, my question is, why haven't done some algebra on your last equation and whip into the form:

    Vo/Vi = an expression in s

    Isn't that what the problem asks for?

    Edit: I see that your last post occurred while I was working on mine. Apparently you have already decided that you need to do what I suggested.
     
  8. Dec 29, 2011 #7

    berkeman

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    I think so. The good news is that you already stated the low-frequency solution that you can check your final transfer function against, and you should be able to see what the high-frequency limit should be for the transfer function behavior as well, so you will be able to check that too.
     
  9. Dec 29, 2011 #8

    berkeman

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    Good catch! I looked at the Rz statment, and couldn't figure out what it meant.
     
  10. Dec 29, 2011 #9
    Not quite sure why I typed parallel when I wrote series, my fault :redface:

    Always a nice check I've been taught. When at high-frequency, s→∞, so this would then mean that the caps short I believe. But in part b we need to find ALF so the high-frequency gain doesn't need to be evaluated.

    ---------------------------------------------------------

    Glancing over my attachment without any scratch work, I am at a loss where to begin my algebra. I could either: 1. Separate the vo and vi terms on both the left and right, and then solve for vo's and vi's; or 2. Cross multiply the terms. This appears that it will be a disgusting problem...
     
  11. Dec 29, 2011 #10

    berkeman

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    Either way should work. I probably would take the cross-multiply approach, and try not to do much distribution of terms if not necessary. Then check your final Vo/Vi expression for the low- and high-frequency cases to get some confidence that there were no algebra errors along the way.
     
  12. Dec 29, 2011 #11
    I've tried the cross multiplication method, but may have botched up somewhere and made a silly mistake because this does not seem correct from what I've done in any previous circuit analysis' (although all circuits are different which I've learned the hard way). Any chance you can check my algebra on the work I've completed for any silly mistakes?
     

    Attached Files:

  13. Dec 30, 2011 #12

    The Electrician

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    If you're going to be doing a lot of this when you're back in class, you should learn to use a symbolic algebra program--your school may have one on the network. There are even some free ones on the web, such as Scilab or Macsyma. You'll avoid a lot of tedious algebra with its potential for error.

    Using a symbolic algebra program, I solved your equation and also solved the problem with an alternate method. Some of your writing is a little hard to read, but I think it matches up with my result. See the attachment.
     

    Attached Files:

  14. Dec 30, 2011 #13
    Nice I wasn't even aware of those programs, I'll have to take a look at them and use them as checks, thank you!

    I've continued to work the problem out and when I try to get it in the form needed:
    Av(s) = ALF([itex]\frac{1+a_1 s+a_2 s^2}{1+b_1 s+b_2 s^2}[/itex])
    it does not seem like it will turn out correct. I tried to explain my thoughts on paper which may make it easier to understand. What I thought of is to divide both conductance terms in the numerator and denominator by their equivalent conductance, and do the same for the other terms throughout the numerator and denominator. The problem with this is that it does put it in the correct form, but there is no gain at low-frequency (ALF). Which leads me to my conclusion that what I am thinking is not a legal algebraic maneuver?

    Edit: Just posted my scan of my work to show visualization.
     

    Attached Files:

  15. Dec 30, 2011 #14

    The Electrician

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    I don't think there's any way to avoid just slogging through a lot of messy algebra. Using a symbolic algebra program, it's not too hard, but without it I don't think I could be sure I hadn't made a mistake.

    See the attachments for the method and result.

    The first image is a little tall, but if you view it with Firefox, just hover your mouse pointer over the image and you should see a magnifying glass symbol; click it.
     

    Attached Files:

  16. Dec 30, 2011 #15
    Apologies on the late response, I've been traveling for the holiday weekend today and just got a chance to sit down at my computer.

    Gotcha, I do understand how you arrived at that final answer, but my question when I was looking at my gain equation, which you also have, is that why doesn't my predicted ALF in my original post appear to be the same as my derived ALF? This could be because of a mistake when going from resistances to conductances.
     
  17. Dec 30, 2011 #16

    The Electrician

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    In the first post, you said "ALF should be equal to (R3)/(R3+RZ) because the caps open at low frequencies.
    RZ is R1 and R2 in parallel."

    In the attachment to post #13, I don't see a derived expression identified as ALF so I can't offer an opinion as to "why doesn't my predicted ALF in my original post appear to be the same as my derived ALF?"

    But the expression for ALF in the first image attached to post #14 does reduce to the same thing you had in post #1 when the substitutions relating conductances to resistances are made. See the attachment to this post.
     

    Attached Files:

  18. Dec 31, 2011 #17
    Here's my question in written form: Unless I've made an algebra mistake in doing what you've done using the computer, I come out with a similar ALF without a R3 in the denomenator :frown: although your solution matches my predicted low-frequency gain so I must have made a mistake somewhere.
     
    Last edited: Dec 31, 2011
  19. Dec 31, 2011 #18

    The Electrician

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    In the sixth line down, there are 3 terms in the denominator:

    [itex]\frac{R1 R2^2}{R1 R2 R3}+\frac{R1 R2 R3}{R1 R2 R3}+\frac{R1 R2}{R2 R3}[/itex]

    After some cancellation, the three terms appear to have become:

    [itex]\frac{R2}{R3}+0+\frac{R1}{R3}[/itex]

    But this is wrong; that middle term should have become 1 rather than zero:

    [itex]\frac{R2}{R3}+1+\frac{R1}{R3}[/itex]

    You will be kicking yourself for having made such a silly error, but we all do it. :uhh:

    If such a silly error can be made during a relatively simple reduction as this, what hope is there of getting the larger mass of algebra correct without using a symbolic algebra program?

    Years ago when I used to do this sort of thing by hand, I would get out my pad of engineering paper and work the problem once. I would then set that aside and solve it a second time without referring to the first worksheets. And then I would do it a third time, hoping for at least a 2 out of 3 agreement!
     
  20. Dec 31, 2011 #19
    Gahhh :grumpy: such a stupid mistake! My fault, I fixed my work and thankfully came out with the same low-frequency gain as I had predicted! Thank you very much for your help! It's very appreciated! One last question in regards to the two algebra simplifying programs you had suggested. Do you have a preferred program out of the two?
     

    Attached Files:

  21. Dec 31, 2011 #20

    The Electrician

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    I don't use either one of those myself, but there are others on the forum who do. Maybe someone will pipe up with a recommendation.
     
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