# Transfer Function for Complex Frequencies

Discussion in 'Homework Help' started by tquiva, Feb 1, 2011.

1. ### tquiva Thread Starter Member

Oct 19, 2010
176
1
Could someone please assist me with this problem:

Here's my work so far towards all three solutions:

Will someone please give me feedback on my solution?
I did not finish with part (b) as I got stuck on the circuit on the bottom right corner. What this circuit simplification correct? And was my answer for (a) correct? Any feedback is greatly appreciated.

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
The endpoints of the resistor R2 are short circuited, thus it can be removed without any effect on the circuit. Continue your calculations with the remaining 3 resistors, R1, R3 and R4.

The schematic on the bottom right isn't the same as the one on the bottom left. Just remove R2 and try again.

3. ### tquiva Thread Starter Member

Oct 19, 2010
176
1
Why is R2 short circuited?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I don't think it would be.

R2 is a bit of a curiosity, in that it is a positive feedback element.

At DC, R2 feedback would place a value of Vo at the op-amp's positive input terminal. But the negative input terminal is biased via the voltage divider to a value of ...

Vo*R4/(R3+R4)

The op-amp would then presumably be driven into saturation at DC input, given V+ exceeds V- by a significant difference.

I haven't thought too much about the problem otherwise.

5. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
In the high-frequency analysis, where the capacitors are replaced by shorts, the resistance R2 has its two pins short circuit. That means that it has 0 voltage applied on it and 0 current flowing trough it. In other words it is a "ghost" part for this circuit and can be removed without any effect on the rest of the circuit. I would be 100% sure about that, unless t_n_k didn't question me.

6. ### tquiva Thread Starter Member

Oct 19, 2010
176
1
alright, I will take your word for it.
Thank you very much.

However, was my answer for part a correct?

7. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I will say yes. Keep in mind however that in real-world, the offset voltage of the OpAmp will throw its output on either Vcc or Ground. This will give make the TF either 0 or Vcc.