Plotting input impedance vs frequency

Discussion in 'Homework Help' started by jegues, Oct 31, 2010.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    See figure attached for problem statement.

    We've been studying RC, LC and RLC circuits in class, but never have we discussed "resonance", or plotting things such as input impedance versus frequency, hence my confusion.

    I don't have matlab or anything so I've be trying to plot the following on wolframalpha,

    "plot y = log( |100+1/(2200*10^{-12}x)+(10*10^{-3})x| )"

    Can someone confirm that this is indeed what I want to plot?

    I simply plugged in the values for R, L and C given into our equation for Zin.

    It says to use the logarthimic unit for impedance but I'm not entirely sure how to do that.

    If that equation is indeed what I am being asked to plot could anyone be so kind as to post a matlab graph? I don't know why they ask for it using matlab, the students were not provided with it nor do we have any access to it.

    Is there any other tools out there that I can use to graph this?

    Thanks again!
     
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  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    The instructions say you can plot the graph manually,

    The Log statement involves the graph type, e.g. 1, 10, 100, 1000 isntead of 1,2,3,4, 5

    Plug in the numbers given, and calculate the impedance for 20 Hz, 60 Hz, 400Hz, 1khz, 2kHz, 4kHz, 8kHz, etc.

    A programmable calculator (HP or TI) will be able to run that equation for different inputs quite readily.
     
  3. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    43

    Okay, I've been plotting

    y =| 100+1/(2200*10^{-12}x)+(10*10^{-3})x |

    In my TI-83 Plus graphing calculator but I can't get the y axis to jump by powers of 10.

    Is there a way I can do that on my calculator?

    EDIT: See figure attached for my graph. Does this look correct? I still can't get it to plot the y axis high enough to see the resonant frequency
     
    Last edited: Oct 31, 2010
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Maxima is free and can plot mathematical functions:

    http://maxima.sourceforge.net/
     
  5. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Graph doesn't look right, capacitor and inductor in series make a bandpass filter, so it should be a "U" shaped graph. The graph above appears to only calculate the capacitance portion of the impedance. Equation using f for frequency (remember multiply by 2pi) -> R+j(L*f-1/C*f)

    MiscEl (freeware) gave me this info on the LCRs series circuit with 100 Ohm output Z

    Fo: 33.9319 kHz
    Zin min: 200.001 ohm
    Zout min: 50.0002 ohm
    Zout max: 100 ohm

    Bandwidth:
    Passband gain: -6.02 dB (500m) at Fo
    -18 dB low freq.: 28.35 kHz
    -18 dB high freq.: 40.61 kHz
    -18 dB bandwidth: 12.27 kHz

    Small signal analysis

    At: 10 Hz

    Gain: -97.2 dB (13.8µ)
    Phase: 90 °
    Zin: 7.23 Mohm
    Zout: 100 ohm

    At: 100 Hz
    Gain: -77.2 dB (138µ)
    Phase: 90 °
    Zin: 723 kohm
    Zout: 100 ohm

    At: 1 kHz
    Gain: -57.2 dB (1.38m)
    Phase: 89.8 °
    Zin: 72.3 kohm
    Zout: 100 ohm

    At: 100 kHz
    Gain: -34.9 dB (18m)
    Phase: -87.9 °
    Zin: 5.56 kohm
    Zout: 100 ohm

    At: 1 MHz
    Gain: -56 dB (1.59m)
    Phase: -89.8 °
    Zin: 62.8 kohm
    Zout: 100 ohm
     
    Last edited: Oct 31, 2010
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Don't ignore the advice in your problem that says "Note that the values are complex..."

    You should get a plot like the one shown in the attachment.
     
  7. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    So is the dip in the middle of the graph going to be my resonance frequency?
     
  8. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Yes, the point where inductive impedance and capacitive impedance match.
     
  9. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I've calculated the resonance frequency to be 33931.95 Hz, is this correct? It seems to agree with the graph as well.

    Also, how do I calculate half power bandwidth?

    The only formula I'm given is,

    \beta = \omega_{H} - \omega_{L}

    Where \omega_{H},\omega_{L} are the frequencies that correspond to \frac{1}{sqrt{2}} times the maximum voltage value.

    But we don't have any voltage values, so how do we find it?

    Thanks again!
     
  10. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Fo is correct.

    Here is a link to a page describing the 3dB down point (half power) with formulas and a graph to get the idea of what the bandwidth is between the two -3dB points.
     
  11. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I've been trying to make the graphs on maxima, but I can't figure it out. Can you tell me what steps you did to get that graph?

    I want it to only have a log scale on the Y axis and not such a large range on the x axis.

    Help?
     
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