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The solution in the Laplace Domain, with zero initial conditions, requires to solve the circuit simply replacing the inductors impedance effect \$v(t)=L\frac{di(t)}{dt}\$ with \$V(s)=sLI(s)\$

The circuits expressions are:

$$V_1=R_1I_1+sL_1(I_1-I_2)$$

$$sL_1(_I1-I_2)=R_2I_2+sL_2I_2$$$$sL_1(I_1-I_2)=R_2I_2+sL_2I_2$$

$$V_2=sL_2I_2$$

They can be simply rearranged as:

$$I_1=(V_1+sL_1I_2)/(R_1+sL_1)$$$$I_1={V_1+sL_1I_2 \over R_1+sL_1}$$

$$I_2=sL_1I_1/(R_2+sL_2+sL_1)$$$$I_2={sL_1I_1 \over R_2+sL_2+sL_1}$$

From here, I_1\$I_1\$ and I_2\$I_2\$ can be obtained:

$$I_2=V_1sL_1/((R_1+sL_1).(R_2+sL_2+sL_1)-(sL_1)^2)$$$$I_2={V_1sL_1 \over R_1+sL_1}{1 \over (R_2+sL_2+sL_1)-(sL_1)^2}$$

$$I_1={V_1\over (R_1+sL_1)}(1+(sL_1)^2/((R_1+sL_1).(R_2+sL_2+sL_1)-(sL_1)^2))$$$$I_1={V_1\over R_1+sL_1}{1+(sL_1)^2 \over (R_1+sL_1)((R_2+sL_2+sL_1)-(sL_1)^2)}$$

And finally, the transfer function is calculated as:

$$V_2/V_1=1-R_1I_1/V_1-R_2I_2/V_1$$

The solution in the Laplace Domain, with zero initial conditions, requires to solve the circuit simply replacing the inductors impedance effect \$v(t)=L\frac{di(t)}{dt}\$ with \$V(s)=sLI(s)\$

The circuits expressions are:

$$V_1=R_1I_1+sL_1(I_1-I_2)$$

$$sL_1(_I1-I_2)=R_2I_2+sL_2I_2$$

$$V_2=sL_2I_2$$

They can be simply rearranged as:

$$I_1=(V_1+sL_1I_2)/(R_1+sL_1)$$

$$I_2=sL_1I_1/(R_2+sL_2+sL_1)$$

From here, I_1 and I_2 can be obtained:

$$I_2=V_1sL_1/((R_1+sL_1).(R_2+sL_2+sL_1)-(sL_1)^2)$$

$$I_1={V_1\over (R_1+sL_1)}(1+(sL_1)^2/((R_1+sL_1).(R_2+sL_2+sL_1)-(sL_1)^2))$$

And finally, the transfer function is calculated as:

$$V_2/V_1=1-R_1I_1/V_1-R_2I_2/V_1$$

The solution in the Laplace Domain, with zero initial conditions, requires to solve the circuit simply replacing the inductors impedance effect \$v(t)=L\frac{di(t)}{dt}\$ with \$V(s)=sLI(s)\$

The circuits expressions are:

$$V_1=R_1I_1+sL_1(I_1-I_2)$$

$$sL_1(I_1-I_2)=R_2I_2+sL_2I_2$$

$$V_2=sL_2I_2$$

They can be simply rearranged as:

$$I_1={V_1+sL_1I_2 \over R_1+sL_1}$$

$$I_2={sL_1I_1 \over R_2+sL_2+sL_1}$$

From here, \$I_1\$ and \$I_2\$ can be obtained:

$$I_2={V_1sL_1 \over R_1+sL_1}{1 \over (R_2+sL_2+sL_1)-(sL_1)^2}$$

$$I_1={V_1\over R_1+sL_1}{1+(sL_1)^2 \over (R_1+sL_1)((R_2+sL_2+sL_1)-(sL_1)^2)}$$

And finally, the transfer function is calculated as:

$$V_2/V_1=1-R_1I_1/V_1-R_2I_2/V_1$$

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The solution in the Laplace Domain, with zero initial conditions, requires to solve the circuit simply replacing the inductors impedance effect \$v(t)=L\frac{di(t)}{dt}\$ with \$V(s)=sLI(s)\$

The circuits expressions are:

$$V1=R1I1+sL1(I1-I2)$$$$V_1=R_1I_1+sL_1(I_1-I_2)$$

$$sL1(I1-I2)=R2I2+sL2I2$$$$sL_1(_I1-I_2)=R_2I_2+sL_2I_2$$

$$V2=sL2I2$$$$V_2=sL_2I_2$$

They can be simply rearranged as:

$$I1=(V1+sL1I2)/(R1+sL1)$$$$I_1=(V_1+sL_1I_2)/(R_1+sL_1)$$

$$I2=sL1I1/(R2+sL2+sL1)$$$$I_2=sL_1I_1/(R_2+sL_2+sL_1)$$

From here, I1I_1 and I2I_2 can be obtained:

$$I2=V1sL1/((R1+sL1).(R2+sL2+sL1)-(sL1)2)$$$$I_2=V_1sL_1/((R_1+sL_1).(R_2+sL_2+sL_1)-(sL_1)^2)$$

$$I1=V1/(R1+sL1)(1+(sL1)2/((R1+sL1).(R2+sL2+sL1)-(sL1)2))$$$$I_1={V_1\over (R_1+sL_1)}(1+(sL_1)^2/((R_1+sL_1).(R_2+sL_2+sL_1)-(sL_1)^2))$$

And finally, the transfer function is calculated as:

$$V2/V1=1-R1I1/V1-R2I2/V1$$$$V_2/V_1=1-R_1I_1/V_1-R_2I_2/V_1$$

The solution in the Laplace Domain, with zero initial conditions, requires to solve the circuit simply replacing the inductors impedance effect \$v(t)=L\frac{di(t)}{dt}\$ with \$V(s)=sLI(s)\$

The circuits expressions are:

$$V1=R1I1+sL1(I1-I2)$$

$$sL1(I1-I2)=R2I2+sL2I2$$

$$V2=sL2I2$$

They can be simply rearranged as:

$$I1=(V1+sL1I2)/(R1+sL1)$$

$$I2=sL1I1/(R2+sL2+sL1)$$

From here, I1 and I2 can be obtained:

$$I2=V1sL1/((R1+sL1).(R2+sL2+sL1)-(sL1)2)$$

$$I1=V1/(R1+sL1)(1+(sL1)2/((R1+sL1).(R2+sL2+sL1)-(sL1)2))$$

And finally, the transfer function is calculated as:

$$V2/V1=1-R1I1/V1-R2I2/V1$$

The solution in the Laplace Domain, with zero initial conditions, requires to solve the circuit simply replacing the inductors impedance effect \$v(t)=L\frac{di(t)}{dt}\$ with \$V(s)=sLI(s)\$

The circuits expressions are:

$$V_1=R_1I_1+sL_1(I_1-I_2)$$

$$sL_1(_I1-I_2)=R_2I_2+sL_2I_2$$

$$V_2=sL_2I_2$$

They can be simply rearranged as:

$$I_1=(V_1+sL_1I_2)/(R_1+sL_1)$$

$$I_2=sL_1I_1/(R_2+sL_2+sL_1)$$

From here, I_1 and I_2 can be obtained:

$$I_2=V_1sL_1/((R_1+sL_1).(R_2+sL_2+sL_1)-(sL_1)^2)$$

$$I_1={V_1\over (R_1+sL_1)}(1+(sL_1)^2/((R_1+sL_1).(R_2+sL_2+sL_1)-(sL_1)^2))$$

And finally, the transfer function is calculated as:

$$V_2/V_1=1-R_1I_1/V_1-R_2I_2/V_1$$

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The solution in the Laplace Domain, with zero initial conditions, requires to solve the circuit simply replacing the inductors impedance effect \$v(t)=L\frac{di(t)}{dt}\$ with sL.\$V(s)=sLI(s)\$

The circuits expressions are:

$$V1=R1I1+sL1(I1-I2)$$

$$sL1(I1-I2)=R2I2+sL2I2$$

$$V2=sL2I2$$

They can be simply rearranged as:

$$I1=(V1+sL1I2)/(R1+sL1)$$

$$I2=sL1I1/(R2+sL2+sL1)$$

From here, I1 and I2 can be obtained:

$$I2=V1sL1/((R1+sL1).(R2+sL2+sL1)-(sL1)2)$$

$$I1=V1/(R1+sL1)(1+(sL1)2/((R1+sL1).(R2+sL2+sL1)-(sL1)2))$$

And finally, the transfer function is calculated as:

$$V2/V1=1-R1I1/V1-R2I2/V1$$

The solution in the Laplace Domain, with zero initial conditions, requires to solve the circuit simply replacing the inductors with sL.

The circuits expressions are:

$$V1=R1I1+sL1(I1-I2)$$

$$sL1(I1-I2)=R2I2+sL2I2$$

$$V2=sL2I2$$

They can be simply rearranged as:

$$I1=(V1+sL1I2)/(R1+sL1)$$

$$I2=sL1I1/(R2+sL2+sL1)$$

From here, I1 and I2 can be obtained:

$$I2=V1sL1/((R1+sL1).(R2+sL2+sL1)-(sL1)2)$$

$$I1=V1/(R1+sL1)(1+(sL1)2/((R1+sL1).(R2+sL2+sL1)-(sL1)2))$$

And finally, the transfer function is calculated as:

$$V2/V1=1-R1I1/V1-R2I2/V1$$

The solution in the Laplace Domain, with zero initial conditions, requires to solve the circuit simply replacing the inductors impedance effect \$v(t)=L\frac{di(t)}{dt}\$ with \$V(s)=sLI(s)\$

The circuits expressions are:

$$V1=R1I1+sL1(I1-I2)$$

$$sL1(I1-I2)=R2I2+sL2I2$$

$$V2=sL2I2$$

They can be simply rearranged as:

$$I1=(V1+sL1I2)/(R1+sL1)$$

$$I2=sL1I1/(R2+sL2+sL1)$$

From here, I1 and I2 can be obtained:

$$I2=V1sL1/((R1+sL1).(R2+sL2+sL1)-(sL1)2)$$

$$I1=V1/(R1+sL1)(1+(sL1)2/((R1+sL1).(R2+sL2+sL1)-(sL1)2))$$

And finally, the transfer function is calculated as:

$$V2/V1=1-R1I1/V1-R2I2/V1$$

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