# Realization of basic transfer function using op-amps

Discussion in 'Homework Help' started by Trip1, Jan 28, 2012.

1. ### Trip1 Thread Starter New Member

Jan 28, 2012
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Hello, I have a question relating to a particular set of problems that i'm having trouble with.

I need to implement the following transfer function using a single op-amp, capacitors and resistors.

T(S) = -4/(s+3)

-a very basic transfer function. But I am asked to do it using the following circuit, and i'm not sure where to start. (see attached)

any help is appreciated.

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2. ### Georacer Moderator

Nov 25, 2009
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You may replace any resistance with a capacitor, resulting in a complex impedance of value Z=1/(sC), where C is the value of the capacitor.

$\frac{-4}{s+3}=-\frac{Z_1 Z_3}{Z_2}$

Some of your impedances will be resistors and some will be capacitors.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Another option would be to add a capacitor in parallel with one of the resistors. For instance, consider adding a capacitor in parallel with R3.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Keep in mind the circuit converts an input current to a related output voltage. If you want the transfer to be from voltage to voltage you would have to add a scaling resistance in series with the input.

Georacer likes this.
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The other thing I'm unsure of is whether the input current-to-output voltage transfer function given on the diagram is correct.

It would be worth double checking ...

Perhaps it should be

$V_o=-\frac{(R_1+R_2)R_3}{R_2}i_{in}$

6. ### hgmjr Moderator

Jan 28, 2005
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My calculation of the transfer function for the opamp circuit in trip1's initial post agrees with the expression that t_n_k has posted.

In any event, georacer's hint toward a solution was a good one.

hgmjr

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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My suggestion is that merely replacing any of the R values with a 1/Cs term wont give the required transfer function

T(s)=Vo/Iin=-4/(s+3).

That's why I thought a parallel addition of a capacitor was more likely to work.

8. ### Georacer Moderator

Nov 25, 2009
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Yes, I might have jumped to a conclusion too fast. But at any rate, taking one or two elements as resistors, the third one would end up as a complex impedance anyway.

9. ### panic mode Senior Member

Oct 10, 2011
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as t_n_k suggested, i derived transfer function from scratch and got same result as he posted. transfer function posted with circuit is incorrect, derived one is correct:

T(s)=Vo/Iin= -(R2+R1)*R3/R2

next i tried to see if replacing any of resistors with capacitor would give desired form. it didn't work for any individual substitution, but when i added capacitor in parallel with R3, it turned out ok

replace R3 with (1/sC)*R3/(1/sC + R3)
which becomes (R3/sC)/[(1+sC*R3)/(sC)]
cancel sC in double fraction and result is
R3/(1+sC*R3)
now factor R3 in denominator and cancel it as well
we get
1/(sC + 1/R3)

plug that into above transfer function in place of R3 and you get

T(s)=Vo/Iin= -(R2+R1)*R3/R2

T(s)= -(1+R1/R2)/(sC+1/R3)

all it is left is pick proper values to get

-4/(s+3)

10. ### hgmjr Moderator

Jan 28, 2005
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Trip1,

Typically when the goal is to derive the T(s) of a circuit, I generally understand that to mean T(s) = Vout/Vin. I don't normally think of T(s) = Vout/Iin. Can you please confirm that you are being asked to provide T(s) = Vout/Iin and not T(s) = Vout/Vin.

hgmjr

11. ### panic mode Senior Member

Oct 10, 2011
1,320
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transfer function is simply
output / input

where input and output can be anything.
in case of amplifier circuits you can have V/V, V/I, I/I and I/V
(although most of the times we are interested in V/V)

12. ### panic mode Senior Member

Oct 10, 2011
1,320
304
btw. while
T(s)= -(1+R1/R2)/(sC+1/R3)
does satisfy form
-4/(s+3)

it is not practical to use 1F capacitor.
but you can factor C in denominator and move (1/C) to numerator to get

T(s) = -[(1+R1/R2)/C] / [s + 1/(C*R3)]

then pick values using:
[(1+R1/R2)/C] = 4
[1/(C*R3)] = 3

13. ### hgmjr Moderator

Jan 28, 2005
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This is true.

I just wanted to have trip1 confirm which one he actually wanted since he did not state it outright.

hgmjr

14. ### steveb Senior Member

Jul 3, 2008
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Why do we go so far in helping people solve homework problems?

The OP has not even responded whether the hints and suggestions are useful or not, and we keep giving more and more and more information.

Why is this? It seems to me to be against the spirit of the homework forum rules.

I can understand giving hints, and then giving clarifications to hints if we think we might have not been clear and might cause more confusion, but at some point we need to wait for a response to know if the hints are sufficient. Also, at some point the OP needs to show some work.

I see many examples of this kind of thing. I generally ignore it, but occasionally I feel a need mention something.

15. ### hgmjr Moderator

Jan 28, 2005
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214
Valid point steveb. Our newer members are not that familiar with our preference here at AAC to provide guidance and hints to the homework questions posed rather than providing outright answers.

At least that is the general rule.

hgmjr

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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It certainly is interesting how we go to exceptional lengths to help people here. Judging by the comments one gets about the behavior of some members of other forums the lazy / impatient (etc.) posters would simply be abused or ignored.

I agree we probably do too much - particularly in the homework forum. What irks me particularly are folk who are too lazy to say "thanks" - even if it is just a matter of the single click thank-you.

Fortunately people are generally appreciative.

These days if the the OP doesn't come back to the discussion I lose interest very quickly unless there's a broader issue to resolve.

I guess also there is some value in providing input for those people other than the OP who might view the thread & find something that helps their understanding. There are a lot of observers ..... if the 'views' information is anything to go by.

timekeeper likes this.