Zin/zout for high-pass filter help?

  • Thread starter xhuang23
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  • #1
xhuang23
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Homework Statement


An expression for Zin (impedance seen fromthe input) and Zout (impedance seen from the output) of the high-pass filter circuit with respect to frequency is: ?

The Attempt at a Solution


i have, bythe way, equation for transfer function already. Vout/Vin= R/(R+1/jwC). i am pretty sure Zin=Xc=1/jwC but not so sure about zout. would it be R+Xc or just R? i know at high frequency the imaginary part drops, so its just R.

or maybe i am doing this wrong. should i do the same thing for zout/zin as vout/vin?
 
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Answers and Replies

  • #2
The Electrician
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It would be much easier to help you if you would post a schematic of your filter circuit.
 
  • #3
xhuang23
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JqceA9e.png

it is typical high-pass filter circuit so i didnt think to post a pic. the Vin source was supposed to be 1vpk. the R value i calculated from the fc (cutoff freq) value =1.5kHz.
 
  • #4
The Electrician
Gold Member
1,344
188

Homework Statement


An expression for Zin (impedance seen fromthe input) and Zout (impedance seen from the output) of the high-pass filter circuit with respect to frequency is: ?

The Attempt at a Solution


i have, bythe way, equation for transfer function already. Vout/Vin= R/(R+1/jwC). i am pretty sure Zin=Xc=1/jwC but not so sure about zout. would it be R+Xc or just R? i know at high frequency the imaginary part drops, so its just R.

or maybe i am doing this wrong. should i do the same thing for zout/zin as vout/vin?

How do you get Zin=Xc=1/jwC? Zin is the impedance calculated at the Vin terminals. Isn't R1 part of that impedance?

Zout will depend on whether or not anything is connected to the Vin terminals. If the filter is driven by a voltage source (having zero ohms internal impedance), then Zout is the impedance seen at the Vout terminals with the Vin terminals shorted (by the impedance of the driving source).

Try calculating Zin and Zout with this information.
 

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