# Help with transfer function for my circuit

Discussion in 'Homework Help' started by RichardS11, Jun 1, 2016.

1. ### RichardS11 Thread Starter New Member

Jun 1, 2016
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Hi,

I need help with calculating my circuit, I've done something but not sure if that is correct. Any help or direction where to look at is welcome. In the end I need transfer function of the circuit.

The first circuit is the one I start with, the second picture is "simplified" circuit with impedance for easier calculation.
Where
Z1=(1/Cin*s) + R1 + R2 + L1*s
Z2 = R3+L2*s
Z3 = L3*s

What I need is
uout/uin =
i1/uin =
i2/uin =
i2/i1 =

where uout is voltage on load and uin is entry voltage u1

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2. ### RichardS11 Thread Starter New Member

Jun 1, 2016
8
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Fix
and wrong source symbol - fixed the picture

I'm a bit stuck at the point I came so far. There are now 2 branches, first one with i1 and second one with i2 and when writing the equation for first one the i2*z3 has to be subtracted from first one?

uin = i1*(Z1+Z3) - i2*z3
and
uout = i2*(Z2+Zload+Z3) - i1*Z3 ?

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3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Uin = I1*Z1 + I3*Z3

Where I1 is a Z1 currant and I3 is a Z3 current.

4. ### RichardS11 Thread Starter New Member

Jun 1, 2016
8
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Wouldn't the second branch actually be
i2 * ( Z2 + Z3 + Zload ) - i1 * Z3 = 0
and

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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What is I2 and I1 then ??
Are Z3 is in series with Z2 and Zload ?

6. ### RichardS11 Thread Starter New Member

Jun 1, 2016
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It's the current in certain branch. If we observe what happens in each branch then they are in series?

Is that not correct? Please correct me if I'm doing something wrong

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7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Ah , so you are doing the "mesh analysis". You should have told me that at the begining. In this case you are correct.

8. ### RichardS11 Thread Starter New Member

Jun 1, 2016
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I've been told for TF I don't need to do mesh analysis and the function comes as

Uout/Uin=(Z3*ZL) / [ZL*(Z1+Z3)+Z3*(Z1+Z2)+Z1*Z2] ?

9. ### RichardS11 Thread Starter New Member

Jun 1, 2016
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=(L3*RL*s)/((C2*RL*s + 1)*((R3 + L2*s)*(R1 + R2 + R3 + L1*s + L2*s + 1/(C1*s)) + L3*s*(R3 + L2*s) + (RL*(R1 + R2 + L1*s + L3*s + 1/(C1*s)))/(C2*RL*s + 1)))

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
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For such a circuit we can use any circuit analysis techniques we have in our toolbox. So where is the problem ?

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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335
If you will complete a mesh analysis, you will have expressions for i1 and i2 and from those you can get everything you need.

In post #1 you have a couple of equations which should apply to the image in post #6. Because your circuit consists of only two exterior meshes, the mesh currents I and II are the same as the branch currents i1 and i2. The equations used in the mesh analysis are derived by doing a "KVL walk" around each mesh. The voltages across the impedances encountered in the walk, and across any voltage sources encountered add up to zero.

The equation for the left hand mesh then should be: i1*(Z1+Z3) - i2*z3 - uin = 0, which is what you have.

The equation for the right hand mesh should be: i2*(Z2+Zload+Z3) - i1*Z3 = 0, which is a little different from what you have. You shouldn't include a term "uout" in the second equation because there is no "uout" source in that mesh. uout will be determined from i2 as the product of i2 and Zload; that is, uout = i2*Zload.

You can solve these two simultaneous equations for i1 and i2. You will obtain expressions like this:

i1 = uin * expression1
i2 = uin * expression2

A simple algebraic manipulation will then give you:

i1/uin = expression1
i2/uin = expression2

which are two of the things your problem asks you to find. The fourth of the required items is easily given as:

i2/i1 = expression2/expression1

Finally, since uout = i2*Zload, starting from i2/uin = expression2, we can multiply that by Zload and get this:

So go ahead and solve your two equations and see what you get.

12. ### MrAl Distinguished Member

Jun 17, 2014
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515
Hi,

If you can use any analysis then you can just use Thevenin and Norton. It makes it so much easier.

To start, turn U1 into a current source:
I1=U1/Z1

and now we have a current source in parallel with three impedances:
1. Z1
2. Z3
3. Z2+ZL

that gives us the voltage Vn at the center node, from which we get the transfer function by just using the voltage divider formula on Z2 and ZL:
VL=Vn*ZL/(ZL+Z2)

See how easy that is?

13. ### RichardS11 Thread Starter New Member

Jun 1, 2016
8
0
Thanks for detailed explanation, I got following results and I hope they are correct
i1 = -(uin*(z3 - (z1 + z3)*(z2 + z3 + zL)))/((z1 + z3)^2*(z2 + z3 + zL))
i1/uin=-(z3 - (z1 + z3)*(z2 + z3 + zL))/((z1 + z3)^2*(z2 + z3 + zL))

i2 = (uin*z3*(- z3^2 + z3 + z2 + zL))/(z2 + z3 + zL)^2
i2/uin = (z3*(- z3^2 + z3 + z2 + zL))/(z2 + z3 + zL)^2

i2/i1 = -(z3*(z1 + z3)^2*(- z3^2 + z3 + z2 + zL))/((z3 - (z1 + z3)*(z2 + z3 + zL))*(z2 + z3 + zL))

uout/uin = (z3*zL*(- z3^2 + z3 + z2 + zL))/(z2 + z3 + zL)^2

14. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
These results are not correct. You had the correct transfer function in post #8. Something has gone wrong in your solution of the two equations.

15. ### RichardS11 Thread Starter New Member

Jun 1, 2016
8
0
You are correct, this calculation is way off. I realized too late the mistake I did. After fixing the mistake I again get same thing from post #8
so correct final thing is Uout/Uin=(Z3*ZL) / [ZL*(Z1+Z3)+Z3*(Z1+Z2)+Z1*Z2]

Thanks for help everyone.