How to Design a Circuit from its Transfer Function Graph

Sometimes all we know about a circuit is its transfer function graph.   The transfer function might look like the one in Figure 1.  How can we design a circuit so that its input-output behavior will match the graph?

Figure 1

The design starts with the mathematical form of the transfer function.  This is a linear function, with the general form of a first order polynomial function.

(1)

where (x1,y1) and (x2,y2) are two arbitrary points on the linear graph.

So, let’s write this function for the graph in Figure 1.  The pair (x1,y1) can be (-1V,+3.5V) and (x2,y2) can be (+1V, +1.5V), which are input/output voltage levels for our circuit.   Replacing x1, y1, x2, y2 in equation (1), and replacing y with Vout and x with Vin, the transfer function is as follows:

(2)

or

(3)

where Vout is the circuit output signal in volts and Vin is the input signal.

Now comes a little bit of reasoning.   The transfer function (3) represents the given graph in Figure 1.  How can we create a circuit to resemble this function?  Well, if you browse through this article Useful Operational Amplifier Formulas and Configurations, you can see that the differential amplifier has exactly this format.   The first term is positive and the second is negative as in the following equation:

(4)

The proof of this transfer function can be found here: How to Derive the Differential Amplifier Transfer Function.

Math tell us that two linear functions are identical when their coefficients are identical.  Therefore, let’s write the coefficients identities in the form of a system of equations:

(5)

In this system I considered that Vin is V2 in equation (4), in other words, we are going to connect the input signal to R3 (see Figure 2).

Figure 2

This system is easy to be solved.  It is a system of 2 equations with 2 unknowns: R4/R3 and R1/R2.  From the system second equation the ratio R4/R3 is already given as equal with one.  We can choose R3 = R4 = 10 kOhms.

For V1 we can choose some reference voltage we have available in the project.  Let’s say V1 = 2.5V.  By replacing R4/R3 in the system first equation, and after calculations, R1/R2 = 1 and R1 = R2 = 10 kOhms.

Here is the circuit we just designed starting with its transfer function:

Figure 3