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# finding impedance between two point

#### akbarza

hi
Maybe this question is stupid and simple, but I'm stuck on it.
in the below pic, how can i calculate impedance between the two points a and b?
again, i think it is a stupid question, but i trapped in it for 1 hour.
thanks
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hi again
i wanted a formula that express equal impedance in c's and r's and how this formula is gained.
thanks

#### Attachments

• impedence_finding.png
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Hi,

Point A is driven by a device. This device also refers (impedance) to GND (and VCC) like R1 does.
So if you really want to know the complete impedance of A to B you need to know the source impedance (into A), too.

If you don´t want to involve the source, then you also probaly don´t want to involve R1, too (because both refer to GND).

In this case: impedance is : (XC1 + R2) || R3

Klaus

hi
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hi
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hi
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hi
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hi
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i tried several times to upload a file, but i am failed.
in the the pic , i want v(N)/v(T)

#### Attachments

• file2.png
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Transfer function according to Sapwin:

( R2 + R1) + ( Cf R1 R2 + Cf R R1) s
--------------------------------------------------------------------------
( R2 + R1) + ( Cf R1 R2 + Cf R R2 + Cf R R1 + Cn R R2 + Cn R R1) s + ( Cn Cf R R1 R2) s^2

i tried several times to upload a file, but i am failed.
What to do with this information? I don´t know what you did and what happened.

But uploding a picture shouldn´t be that difficult:
* press the [insert image] button

Klaus

If not specified IN- assumed to be infinite Z .....eg presents no load to network.

Do you want Vn(s) / Vt(s) or Vt(s) / Vn(s), as it appears to be a feedback network to IN-?

If former since Vt is a V source that means its Z is 0 therefore R & R2 junction is
grounded. There is no T(s) [transfer function] solution since Vt(s) = 0.

If latter Vt is driving point for transfer function then :

Since Vt(s) is the voltage source all you have to do is write node/loop equations to get at
Vn(s), assuming IN- is Zinfinite.

Regards, Dana..

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Hi,

In post#1 I guessed the signal flow is standard: from left to right

But now the signal flow is right to left.

Klaus

Hi,

In post#1 I guessed the signal flow is standard: from left to right

But now the signal flow is right to left.

Klaus