My circuit is

enter image description here

I have set of equations of the form

enter image description here

Now I just need a single expression for transfer function. i,e ratio of voltage across RL to voltage from AC power supply(in terms of RLC circuit elements).

[I1 is left loop mesh current, I2 is the right loop mesh current]

How can I achieve that? is Cramer's rule application is relevant here?

EDIT: The transfer function in S domain (for some value of circuit elements)

                        2.257e-05 s^3


  7.039e-07 s^4 + 5.091e-05 s^3 + 0.001942 s^2 + 0.05085 s + 1
  • \$\begingroup\$ What is a cramer? If it refers to a person's name it should be capitalised properly. Capitals matter. \$\endgroup\$
    – Transistor
    Commented Aug 24, 2016 at 13:11

2 Answers 2


Cramer's rule can be used to get equation (3) from equation (1). Since that is already done, we can use (3) to write

$$ U_{\text{out}}=I_2 R_L=\frac{i \omega M U_{\text{in}} R_L}{M^2 \omega ^2+Z_1 Z_2}$$

from which

$$ \frac{U_{\text{out}}}{U_{\text{in}}}=\frac{i \omega M R_L}{M^2 \omega ^2+Z_1 Z_2} $$

  • \$\begingroup\$ Sorry, that was an oversight. Fixed now. \$\endgroup\$ Commented Aug 25, 2016 at 16:25
  • \$\begingroup\$ I just exported the waveform data from LTspice (Which is Uin for this equation) and then calculated Uout, and find Uout is just plain similar to Uin, which is supposed to be behaving differently as per theory. So please don't mind "are you sure the equations hold good for Mutual Inductor circuit ? (I mean that Uout= I2*RL)", if you know then please let me also know. Thanks. \$\endgroup\$
    – cppiscute
    Commented Aug 25, 2016 at 16:42
  • \$\begingroup\$ I just found out that is the issue of real and imaginary. plot is considering only real part of the values. So in this case going for "S"domain is the only option ? or can we calculate the output value (real+imaginary) with the above method? \$\endgroup\$
    – cppiscute
    Commented Aug 25, 2016 at 16:53
  • \$\begingroup\$ The transfer function above will give the steady-state responses for various sinusoids. In general you want to use the s domain. \$\endgroup\$ Commented Aug 25, 2016 at 18:28
  • \$\begingroup\$ Thanks. I solved for transfer function in S domain and got the result and I have added it in my question now. The only question now is, how can I get a similar result like probing for voltage at output of circuit; in the case of transfer function. Please help if you know the answer. \$\endgroup\$
    – cppiscute
    Commented Aug 25, 2016 at 18:38

I don't know about Cramer's rule but I'd solve it like this.

  • The dot notation on the coupled inductors is of no consequence except to put a minus sign in front of the final TF. Don't forget to do that!
  • The coupled inductors can be made into three joined inductors: -

enter image description here

The fact that there is no galvanic isolation is of no consequence to calculating the TF.

Now, the circuit boils down to this: -


simulate this circuit – Schematic created using CircuitLab

Can you take it from here?

  • \$\begingroup\$ Thanks. You conveyed the same answer to me few days back, I got the TF and I did test it in matlab. Even though I did not get the satisfactory results, may be because of my input signal, I am testing on my input signal (Which is not a simple AC or DC signal but a modulated one). Meanwhile I found the above method more efficient in the long run (if the real time updating of circuit elements is implemented). Regards \$\endgroup\$
    – cppiscute
    Commented Aug 24, 2016 at 13:30
  • \$\begingroup\$ I meant like...the simulation results for the simplified approach, is same as that of the original approach, but my theoretical test was not satisfactory, I crosschecked equations and it was ok, but results are not matching with simulation output. so I am testing on the input signal part. \$\endgroup\$
    – cppiscute
    Commented Aug 24, 2016 at 13:53
  • \$\begingroup\$ I'm not sure what you mean by a theoretical test. Practical tests could be problematic if the output impedance of the signal generator were not zero. \$\endgroup\$
    – Andy aka
    Commented Aug 24, 2016 at 14:10
  • \$\begingroup\$ 1. I simulated the circuit in circuit simulator and got the result. 2. I built the TF theoretically and tested in matlab, and this one is not matching to (1) because of my input problem. Thant I am testing. So above method is another way I thought of testing the TF equations. Thanks. \$\endgroup\$
    – cppiscute
    Commented Aug 24, 2016 at 14:19

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