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Half Power frequencies for a parallel RLC circuit

  1. Aug 5, 2009 #1
    hello!
    I was in the lab doing the experiment of resonance frequency by plotting the graph of current vs frequency for a parallel RLC circuit. In this experiment i have also to find the Q factor and the bandwidth. The resonance frequency corresponds to the frequency with minimum intensity. Now question is when i am trying to find the half power frequencies using the relations I=0.707I_{max}, the parallel line crossing that intensity does not cross the curve so how do i have to find those half power frequencies?
     
  2. jcsd
  3. Aug 5, 2009 #2
    I do not completely understand your question. See thumbnail. Does this help?
     

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  4. Aug 7, 2009 #3
    Amazing! Here is what i did and what i got.
    Is really there any difference between my circuit and yours which could cause such divergences in our graphs?
     

    Attached Files:

  5. Aug 7, 2009 #4
    kthouz,

    You were correct in your first post that current is at minimum at parallel resonance. Your plot looks good to me, just that you need to continue it to higher frequencies. Also it would look better if you would use semi-log paper with the X axis being the log axis.
     
  6. Aug 7, 2009 #5
    ? One graph is for current and the other graph is for voltage.
     
  7. Aug 7, 2009 #6
    The difference is that he excited his circuit with a current source and plotted the voltage across the circuit.

    You excited yours with a voltage source and plotted the current through the circuit.
     
  8. Aug 8, 2009 #7
    Both of two graphs are currents but for different resistors. One is 47ohms and the other is 100ohms.
    Actually, my question is not yet answered: I am supposed to find the bandwidth and the Q factor using the relation Q=fr/df. where fr is the resonance frequency and df the bandwidth.
    When i am trying to find the half-power frequencies by taking the ones corresponding to I=0.707xI_max, i have to draw a horizontal line through I=0.707xI_max. The problem is that this line will never cross the current graph. So how do i have to proceed? is there any other alternative?
     
  9. Aug 8, 2009 #8

    f95toli

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    I haven't read the whole thread but looking at your graph I can say that you are definitly making a misstake somewhere since your graph is not showing a resonance; you are suppose to get a PEAK on resonance; either in voltage or current depending on if it is a parellell or series circuit you are working with.

    And yes; there are ways to extract the bandwith for the case when you actually DO have a resonance in the form of a dip and it is less than 3dB deep; but it is more complicated and generally requires numerical fitting (meaning you can assume this is not the method you are suppose to use).
     
  10. Aug 8, 2009 #9
    Suppose now i have a peak. Do you think that i will find a horizontal line that can cross the graph? me i don't think so and that's why i think that there should be another method.
    What if i use I=I_max + 0.707I_max? here i want to considere the parallel case as the reversed series case. because in parallel the resonance frequency is the minimum of the curve whereas in series is the maximum.
     
  11. Aug 8, 2009 #10

    uart

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    It's already been said but I'll re-iterate.

    1. You are plotting the circuits conductance whereas bob's plot was of impedance (hence your resonance dip and bob's resonance peak). You should get a conductance minimum at resonance exactly as you have shown, there is no real problem here. Yes your resonance dip is not very sharp but that just means your circuits Q factor is fairly low.

    2. You simply need to extend your measurement to higher frequencies and the conductance will continue to rise (and reach the 1.41 x minimum point you seek). You probably need to go out to about 2.0 to 2.5kHz that's all.
     
  12. Aug 8, 2009 #11

    uart

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    It's already been said but I'll re-iterate.

    1. You are plotting the circuits conductance whereas bob's plot was of impedance (hence your resonance dip and bob's resonance peak). You should get a conductance minimum at resonance exactly as you have shown, there is no real problem here. Yes your resonance dip is not very sharp but that just means your circuits Q factor is fairly low.

    2. You simply need to extend your measurement to higher frequencies and the conductance will continue to rise (and reach the 1.41 x minimum point you seek). You probably need to go out to about 2khz for the 100R circuit and about 4kHz for the 47R ciruit, that's all there is to it ok.
     
  13. Aug 8, 2009 #12
    thank you!
     
  14. Aug 8, 2009 #13

    f95toli

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    Yes, if you have a measurement that covers the whole resonance!
    A peak has -by definition- a maximum and assuming the values decrease before/after the resonance (as it should in a resonator ) the 0.707Imax line will cross the curve. The same is true for a dip. Look at Bob's example above
    Also, I assume this is a school project meaning it is extremely unlikely that they've given you a circuit where you need to do anything more sophisticated than draw a line through the 3dB point.

    There are -as I wrote above- other methods but they are all more complicated.
    I sometimes use microwave resonators which DO "dip" less than 3 dB (and sometimes look nearly as bad as your curves) and I have to use a weigthed non-linear least-square fitting routine written in Matlab to extract their parameters; numerical fitting is simply the only option if yu have "incomplete" resonances; there are as far as I know no analytical/graphical methods that can give you an accurate value.
     
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