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Thread: Measuring Impedance...is it correct?

  1. #1 Measuring Impedance...is it correct? 
    Forum Freshman
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    Feb 2011
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    Hi peeps~

    I have a question which I am struggling with need your professional help.

    I have a very simple series circuit, consisting a resistor and PZT material(I was told this can be viewed as a conductor? or capacitor I forgot..) connected using a function generator and an oscilloscope.

    Say, I input 1 volt using the function generator and goes through the circuit and shows 0.8 volt on my oscilloscope. Now I need to know how to find impedance. I searched around and was told that Total impedance Z, is R*Vin / Vout.

    Simple enough yes? but funny thing is that there is a phase difference between Vin and Vout, so I was told that there is a real part and an imaginary part where Vout is actually,

    Vout = 0.8Cos(phase value) + j*0.8Sin(phase value)

    where left one is real part and right part is imaginary part. I was wondering if this is correct? if it is correct, the impedance is usually a single value, so how can I add real and imaginary part to make into a single value...

    please help a poor student out.

    Cheers
    Sam


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  3. #2 Re: Measuring Impedance...is it correct? 
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    Quote Originally Posted by student Sam
    Hi peeps~

    I have a question which I am struggling with need your professional help.

    I have a very simple series circuit, consisting a resistor and PZT material(I was told this can be viewed as a conductor? or capacitor I forgot..) connected using a function generator and an oscilloscope.

    Say, I input 1 volt using the function generator and goes through the circuit and shows 0.8 volt on my oscilloscope. Now I need to know how to find impedance. I searched around and was told that Total impedance Z, is R*Vin / Vout.
    You seem to be using an equation without really understanding it. If Z is the impedance of the PZT material, and Vin is the voltage across the series combination of R and Z, and Vout is the voltage across Z, then no, that is not correct. Look up the voltage divider equation.
    Simple enough yes? but funny thing is that there is a phase difference between Vin and Vout, so I was told that there is a real part and an imaginary part where Vout is actually,

    Vout = 0.8Cos(phase value) + j*0.8Sin(phase value)

    where left one is real part and right part is imaginary part. I was wondering if this is correct? if it is correct, the impedance is usually a single value, so how can I add real and imaginary part to make into a single value...

    please help a poor student out.

    Cheers
    Sam
    Impedance is not "usually a single value." Impedance is a complex number. The magnitude of the impedance is found from the real and imaginary parts using the Pythagorean theorem.


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  4. #3  
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    Oh.. you are correct, Z is the impedance of the PZT. Would the equation for Vout still be wrong if the Vout is the voltage across the whole circuit and not the PZT only?
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  5. #4  
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    Yes, it would still be wrong.

    Take a look at this wikipedia article, which will tell you how to calculate the output voltage.

    http://en.wikipedia.org/wiki/Voltage_divider

    Now, if the power supply voltage is 1 volt, and you measure 0.8 volts on Z2, then

    Vout=Vin [Z2/(Z1+Z2)]
    Vin=1,
    Vout=0.8
    Z1=R
    You can solve this to get the magnitude of Z2. Then if you know the phase angle, you can use trigonometry to find the real and imaginary parts.
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  6. #5  
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    Thank you very much for your reply, it will definately help me to understand this problem
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