1
\$\begingroup\$

I have the following circuit:

enter image description here

I have to find the transfer function by using mesh analysis. First of all I used the Laplace transformation.

enter image description here

Although I saw some similar exercises, I don't really see what I can get out of the mesh equations.

1st Mesh $$V(s) - I_{1}(s)R_1 - V_{L_1}(s) = 0$$

2nd Mesh $$V_{L_1}(s) - I_{2}(s)R_2 - I_{2}L_{2}s = 0$$

Also I assumed that $$V_{L_1}(s) = sL_{1}(I_{1}(s) - I_{2}(s))$$

\$\endgroup\$
  • \$\begingroup\$ I'd recommend checking by doing a triangle->star transformation (Delta-Y transformation) on your 2nd mesh. It greatly simplifies the problem. Remember: there's no current leaving the circuit at the right hand side. \$\endgroup\$ – Marcus Müller Aug 12 at 11:49
0
\$\begingroup\$

Well, I'll present a method that I like to use.

We have that:

$$\mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}\tag1$$

Now, we know that:

$$\text{V}_\text{i}\left(\text{s}\right)=\text{I}_\text{i}\left(\text{s}\right)\cdot\left(\text{R}+\frac{\text{sL}\cdot\left(\text{R}+\text{sL}\right)}{\text{sL}+\text{R}+\text{sL}}\right)\tag2$$

And:

$$\text{V}_\text{o}\left(\text{s}\right)=\text{I}_\text{o}\left(\text{s}\right)\cdot\text{sL}=\text{I}_\text{i}\left(\text{s}\right)\cdot\frac{\text{sL}}{\text{sL}+\text{R}+\text{sL}}\cdot\text{sL}\tag3$$

So:

$$\mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=\frac{\text{I}_\text{i}\left(\text{s}\right)\cdot\frac{\text{sL}}{\text{sL}+\text{R}+\text{sL}}\cdot\text{sL}}{\text{I}_\text{i}\left(\text{s}\right)\cdot\left(\text{R}+\frac{\text{sL}\cdot\left(\text{R}+\text{sL}\right)}{\text{sL}+\text{R}+\text{sL}}\right)}=\frac{\text{s}^2\text{L}^2}{\text{R}^2+\text{s}\left(\text{sL}^2+3\text{LR}\right)}\tag4$$

So, let's say that \$\text{R}=\text{L}=1\$, than we get:

$$\mathcal{H}\left(\text{s}\right)=\frac{\text{s}^2}{\text{s}^2+3\text{s}+1}\tag5$$

And so for the frequency response, we are looking at:

  • The magnitude: $$\left|\mathcal{H}\left(\text{j}\omega\right)\right|=\frac{\omega^2}{\sqrt{\omega^4+7\omega^2+1}}\tag6$$
  • The phase: $$\arg\left(\mathcal{H}\left(\text{j}\omega\right)\right)=\begin{cases} 0\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space\space\space\omega=0\\ \\ 0\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space\space\space\omega=1\\ \\ \arctan\left(\frac{3\omega}{\omega^2-1}\right)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space\space\space\omega>1\\ \\ \frac{\pi}{2}+\arctan\left(\frac{\left|\omega^2-1\right|}{3\omega}\right)\space\space\space\space\space\text{when}\space\space\space0<\omega<1 \end{cases}\tag7$$
\$\endgroup\$
  • \$\begingroup\$ could you further explain the "Now, we know that" part? Also, an edit happened and now the resistors and inductors are not equal in value. At last, the "So, let's say that R=L=1, than we get" and following part go way out of what is being asked. \$\endgroup\$ – jDAQ Aug 16 at 3:23
-1
\$\begingroup\$

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:

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

\$\endgroup\$
  • \$\begingroup\$ The final equation is not a TF - it has V1 on the right hand side. This question looks like homework, btw, so only guidance should be given. Finally, a TF means that initial conditions are zero. Apart from that, everything's ok! \$\endgroup\$ – Chu Aug 12 at 16:40
  • \$\begingroup\$ 1. There is not any V1 on the right hand side. 2. The question is not completely solved, so this is actually only a guidance. 3. It has been indicated that no IC are required. \$\endgroup\$ – Brethlosze Aug 12 at 21:55
  • \$\begingroup\$ Thank you so much for you help! I was trying to find I2 - I took the rearranged equation for I2 (4th equation in your post) and substituted the I1 with the 3d equation(rearranged I1). But I ended up getting I2 on the both sides, i don't understand how you eliminated I2 on the right side and got the 5th equation.. I would be very very grateful if you could point out my mistake. \$\endgroup\$ – vb_2004 Aug 13 at 5:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.