# Transfer function of Instrumentation amplifier with DC suppression filter

## Homework Statement

I want to find the overall transfer function of Instrumentation amplifier with a DC-Suppression filter. I know how to calculate the TF of a normal Instrumentation amplifier. But when feedback is given from the Filter to the IA, I couldn't find it.

## Homework Equations

I used Superposition theorem and found the voltage V01 and V02 for the circuit given in attachment 1.
I have posted the equations for voltage V01 and V02 in attachment 2.
The final equation which I need to find is in attachment 3.

## The Attempt at a Solution

I got the equations for voltages V01 and V02 as given in attachment 2.

Thanks!!

rude man
Homework Helper
Gold Member

## Homework Statement

I have posted the equations for voltage V01 and V02 in attachment 2.

Problem: V01 and V02 have to be the same!
(I assume all amplifiers are operational amplifiers).

Find the voltage at the outputs of each opamp. The opamps are ideal so the output voltage at these points is unaffected by whatever they feed.

I've drawn a diagram showing the middle op amp isolated from the rest. Once you've found the output of this middle opamp, it will be a simple matter to find the output of the next stage.

#### Attachments

rude man
Homework Helper
Gold Member
First off, picture 1 has a lot more resistor designations than it should have. You can see that from your picture 3 which gives the gain with only about half the R's in picture 1.

So, this circuit should first be simplified as follows:

RG1 + RG2 = RG
RF1 = RF2 = RF
R1 = R3
R2 = R4

Whoever gave you this problem should have specified this up front, making the analysis a lot easier.

Now, set Vin2 = 0, apply Vin1 and compute the output of each op amp using KVL or whatever you prefer. Do not use the nodes labeled Vo1 and Vo2 in your KVL, they are the same voltage and not meaningful in a transfer-function sense.

EDIT: It finally dawned on me that V01 and V02 are mislabeled on picture 1. They should be the outputs of the front two op amps respectively.

Your transfer function is then Vo/Vin1. You should finish off by similarly computing Vo/Vin2 (set Vin1 = 0 etc.) and show that this = -Vo/Vin1. Thus the overall transfer function is Vo/(Vin1 - Vin2).

Last edited:
Hi,

I am sending you the paper, where I got this circuit.
Can you please check it out and let me know.

#### Attachments

• ecg_front_end_system_portable.pdf
467.8 KB · Views: 273
I don't think this circuit does what they say it does. I see no way it incorporates any sort of high pass dc suppression. I understand what they tried to do -- grab any dc component with the low pass filter in the last op amp and then feed it back to subtract from the incoming signal -- but that's not what happens. This feedback is only a complicated way of adding a feedback resistor across the capacitor. The feedback pushes the dc bias (Vdd/2) back to the second op amp as well.

You can see their own magnitude plot shows no sign of a highpass filtering (see figure 7) and they were careful not to mention any actual C or R5 values.

The transfer function I get is similar but with a change in sign and lowpass rather than highpass.

$T(s)=-\frac{R2}{R1}(1+\frac{2Rf}{RG})(\frac{1}{1+sC1R5})$

To get it, you need to write equations for the output of each opamp. The Vo1 and Vo2 are mislabelled in the paper and should be at the outputs of the two first op-amps. Equation 3 in the paper is actually the output of the second op-amp stage in terms of Vo1 and Vo2 and is missing a Vo term. They state one assumption that R4/R3=R2/R1 but do not mention another assumption they make that R2/R1 >> 1. At each step, arrange the equation for the output in terms of the differential (Vo2-Vo1).

Keep in mind this is a homework forum so we cannot hand you answers. You have to attempt them first so that people can help with any mistakes or oversights you may make.

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Ah, they are taking the output at the second stage op-amp.

If you subtract a lowpass transfer function from 1, a highpass transfer function results. This is because whenever the lowpass transfer function gain is low, 1 minus that will be high and vice versa.

In this circuit the desired output is at the second stage op-amp. This is passed through a lowpass filter to get Vo. Vo is fed back to be subtracted and a high pass results.

The output at the second op amp stage is as claimed.

rude man
Homework Helper
Gold Member
Ah, they are taking the output at the second stage op-amp.

.

That looks right, as indicated by their paper's eq. 3, which cavalierly ignores the dynamic behavior of the integrator (which effect is added in eq. 4).

All voltages are to be referred to Vdd/2, apparently because they only had one power supply (Vdd) for the circuit.