Transfer function of Instrumentation amplifier with DC suppression filter

In summary, the circuit tries to incorporate a high pass filter to DC-subtract from the input signal. However, this feedback causes a change in the output voltage which results in a highpass rather than lowpass filter.
  • #1
shaikss
33
0

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.

Please help me out to find the TF of the circuit.

Thanks!
 

Attachments

  • 1.JPG
    1.JPG
    13.7 KB · Views: 615
  • 2.JPG
    2.JPG
    7.5 KB · Views: 576
  • 3.JPG
    3.JPG
    14.3 KB · Views: 675
Physics news on Phys.org
  • #2
shaikss said:

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).
 
  • #3
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

  • opamp.png
    opamp.png
    912 bytes · Views: 622
  • #4
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:
  • #5
Hi,

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

Attachments

  • ecg_front_end_system_portable.pdf
    467.8 KB · Views: 431
  • #6
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.

[itex]T(s)=-\frac{R2}{R1}(1+\frac{2Rf}{RG})(\frac{1}{1+sC1R5})[/itex]

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.
 
Last edited:
  • #7
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.
 
  • #8
aralbrec said:
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.
 

1. What is a transfer function in the context of an instrumentation amplifier with DC suppression filter?

The transfer function of an instrumentation amplifier with DC suppression filter refers to the mathematical relationship between the input signal and the output signal of the amplifier. It describes how the amplifier processes and amplifies the input signal to produce the desired output signal.

2. Why is DC suppression necessary in an instrumentation amplifier?

DC suppression is necessary in an instrumentation amplifier to remove any undesired DC offset in the input signal. This is important because DC offset can affect the accuracy and reliability of the amplifier's output. It is especially important in applications where small signals need to be accurately amplified.

3. How does a DC suppression filter work in an instrumentation amplifier?

A DC suppression filter works by blocking or filtering out any low-frequency signals, including DC offset, from the input signal. This is typically achieved using capacitors and resistors in a low-pass filter configuration. The filtered output signal is then amplified by the instrumentation amplifier.

4. What are the advantages of using an instrumentation amplifier with DC suppression filter?

The main advantage of using an instrumentation amplifier with DC suppression filter is improved accuracy and precision in amplifying small signals. It also helps to reduce noise and interference from the input signal, resulting in a cleaner output signal. Additionally, the DC suppression filter can protect the amplifier from any potential damage caused by high DC offset in the input signal.

5. Are there any limitations to using an instrumentation amplifier with DC suppression filter?

One limitation of using an instrumentation amplifier with DC suppression filter is that it can introduce phase shift in the output signal, which can affect the accuracy of the amplified signal. Additionally, the filter may also introduce some attenuation in the signal, reducing the overall gain of the amplifier. Careful design and selection of components can help minimize these limitations.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
16
Views
908
  • Engineering and Comp Sci Homework Help
Replies
1
Views
236
  • Engineering and Comp Sci Homework Help
Replies
24
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
849
  • Engineering and Comp Sci Homework Help
Replies
8
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
16
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
802
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
18
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
993
Back
Top