# finding transfer function of this circuit

Discussion in 'Homework Help' started by ckboii89, Aug 5, 2012.

1. ### ckboii89 Thread Starter New Member

Aug 1, 2012
5
0
i know the transfer functions is H(s) = Vo/Vi

The main problem im having is how to analyze this circuit...

This is not how the actual circuit looks like( due to limitations of circuit lab),

The capacitor and Resistor wires make a "X" to their respective nodes..

hints and help would be appreciated!

thank you!

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
It's not too hard.

Consider Vi(-) terminal as a reference potential.

Vo(+) is found from the voltage divider formed by Vi across R1 & R3 with Vo(+) being across R3.
Vo(-) is found from the voltage divider formed by Vi across C1 & R2 with Vo(-) being across R2.

Note: In the 's' domain C1 impedance would be represented by 1/(C1s)

And then finally the output Vo is found from.

Vo=Vo(+)-Vo(-)

If you redraw the schematic in a bridge arrangement the outcome is somewhat more obvious.

3. ### ckboii89 Thread Starter New Member

Aug 1, 2012
5
0
Just to clear up a few things:
Reference potential is ground right?
Could you also explain why R3 and R2 are the voltages for v+ and v-?

4. ### mlog Member

Feb 11, 2012
276
36
I agree with t_n_k. I solved it using the same approach as he (or she).

The reference is the low side of the input Vi.

If Vi is the input, then what is the value at Vo+ with respect to the low side of Vi? R1 and R3 form a voltage divider.

If Vi is the input, then what is the value at Vo- with respect to the low side of Vi? C1 and R2 form a voltage divider.

5. ### WBahn Moderator

Mar 31, 2012
18,086
4,917
You can pick the reference (commonly called 'ground', but just the node that we assign the value 0V to) to be anywhere you want. Putting it at the negative side of the input is a very common (and commonsense) place to put it.

It makes no sense to say that a resistance is a voltage (and neither t_n_k nor mlog made such a claim).

The circuit appears to have been drawn so as to be deliberately misleading. As suggested by t_n_k, redraw the circuit. Try this:

Put a horizontal bus along the top and label this Vin (this is Vi+).
Put a horizontal bus along the bottom and label it GND (this is Vi-).
Now hang two vertical branches from Vin, one starting with R1 and one starting with C1.
Now finish connecting these two branches to GND by hooking up R2 and R3 appropriately.
Now mark where Vo+ and Vo- are on this diagram.

At this point you should see how you can calculate Vo+ and Vo- independently of each other and subtract them to get Vo (in terms of Vi).