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[SOLVED] Deriving transfer function for circuit

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RichardS

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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
Zload = Rload + (1/C2*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

View attachment 129484
 
Last edited by a moderator:

Fix
Zload = (Rload) / (1 + (Rload*C2*s))
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 ?

schemeit-project (1).png
 

Code: 
uout = i2*(Z2+Zload+Z3) - i1*Z3 ?

isn't right

It's actually
Code: 
i2*(Z2+Zload+Z3) - i1*Z3 = 0
uout = i2*Zload

- - - Updated - - -

Transfer function uout/uin as calculated by Sapwin (R1 + R2 combined to a single Resistor R12 for simplifaction)

Code: 
- (  C1 L3 Rload ) w^2
------------------------------------------------------------------------------
+ (  Rload + R3 )
+ (  C1 R12 Rload + C1 R12 R3 + C2 R3 Rload + L3 + L2 ) j w
- (  C2 C1 R12 R3 Rload + C1 L3 Rload + C1 L3 R3 + C1 L3 R12 + C1 L2 R12 + C1 L1 Rload + C1 L1 R3 + C2 L3 Rload + C2 L2 Rload ) w^2
- (  C2 C1 L3 R3 Rload + C2 C1 L3 R12 Rload + C2 C1 L2 R12 Rload + C2 C1 L1 R3 Rload + C1 L2 L3 + C1 L1 L3 + C1 L1 L2 ) j w^3
+ (  C2 C1 L2 L3 Rload + C2 C1 L1 L3 Rload + C2 C1 L1 L2 Rload ) w^4
 
FvM said:
Code: 
uout = i2*(Z2+Zload+Z3) - i1*Z3 ?

isn't right

It's actually
Code: 
i2*(Z2+Zload+Z3) - i1*Z3 = 0
uout = i2*Zload

- - - Updated - - -

Transfer function uout/uin as calculated by Sapwin (R1 + R2 combined to a single Resistor R12 for simplifaction)

Code: 
- (  C1 L3 Rload ) w^2
------------------------------------------------------------------------------
+ (  Rload + R3 )
+ (  C1 R12 Rload + C1 R12 R3 + C2 R3 Rload + L3 + L2 ) j w
- (  C2 C1 R12 R3 Rload + C1 L3 Rload + C1 L3 R3 + C1 L3 R12 + C1 L2 R12 + C1 L1 Rload + C1 L1 R3 + C2 L3 Rload + C2 L2 Rload ) w^2
- (  C2 C1 L3 R3 Rload + C2 C1 L3 R12 Rload + C2 C1 L2 R12 Rload + C2 C1 L1 R3 Rload + C1 L2 L3 + C1 L1 L3 + C1 L1 L2 ) j w^3
+ (  C2 C1 L2 L3 Rload + C2 C1 L1 L3 Rload + C2 C1 L1 L2 Rload ) w^4
Click to expand...

Thanks for help, I figured that mesh analysis mistake. I can't figure out what that calculation is but without expanding Zs it would be like uout/uin is
(i2*Zload)/(i1*(Z1+Z3)-i2*Z3)

Where
Z1=(1/Cin*s) + R1 + R2 + L1*s
Z2 = R3+L2*s
Z3 = L3*s
Zload = (Rload) / (1 + (Rload*C2*s))
 

it would be like uout/uin is
(i2*Zload)/(i1*(Z1+Z3)-i2*Z3)
Click to expand...

And further putting in
Code: 
i2*(Z2+Zload+Z3) - i1*Z3 = 0

then calculating the complex Zx terms should result in the same transfer function as given by Sapwin
 

For my opinion, the most simple method is as follows:

* Start with the second (simplified) diagram using Z1, Z2, Z3 and ZL.
It is easy to find the transfer function of this classical lattice network: Vout/Vin=R3RL/[RL(R1+R3)+R3(R1+R2)+R1R2]

* As a second step, introduce the corresponding expressions for Z1...ZL.
(Note that ZL is a parallel connection of R and C).
 
Transfer Function: Vout/Vin= 1/(1+((Z1.Z3+Z1.Z2+Z1.ZL)/(Z3.Z2+Z3.ZL)))
 

LvW said:
For my opinion, the most simple method is as follows:

* Start with the second (simplified) diagram using Z1, Z2, Z3 and ZL.
It is easy to find the transfer function of this classical lattice network: Vout/Vin=R3RL/[RL(R1+R3)+R3(R1+R2)+R1R2]
.
Click to expand...

Sorry for the typo: In the transfer function please read R1...RL as Z1...ZL.
 

Thanks for help I got

(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)))
 

I ended up with something a bit different

i1*(Z1+Z3) - i2*z3 - uin = 0
i2*(Z2+ZL+Z3) - i1*Z3 = 0

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
 

RichardS said:
I ended up with something a bit different
Click to expand...
Post #3 and #7 are correct. If you ended up with something different, then, it is incorrect.

The simple current divider rule takes you straight to the solution...
 
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    FvM

    FvM

    Points: 2
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My calculation from post #12 is way off. I realized too late the mistake I did. After fixing the mistake I again get same thing from post #7
so correct final thing is Uout/Uin=(Z3*ZL) / [ZL*(Z1+Z3)+Z3*(Z1+Z2)+Z1*Z2]

Thanks for help everyone.
 

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