# Transfer Function Conversion

I am working on a system which comprises an inverter with output LC filter feeding an induction motor

Its trivial to calculate the resonant frequency of an LC filter, when the filter is loaded with an induction motor the frequency response is altered (no surprise here!)

Looking at this paper

They treat the motor as purely reactive and calculate the Thevenin equivalent which works quite well

I am trying to calculate the effects more accurately, so I use the equivalent circuit for the motor to derive a transfer function which is then put in parallel with the filter components with the clearf goal of reducing it all to one R one L and one C which then allows me to calculate things like the -3db point

I have written a Matlab script which generates a transfer function Z3 but no matter what I try to do I can not reduce the transfer function to R + Xj

In fact I am confused by the fact that if I substitute s = jw I end up with the equivalent resistance depending upon the supply frequency

clc
clear all
close all

syms    C L R Lm Le1 Le2 Rs Rr
s=tf('s');

V = 72;
f = 1200;

Rs = 1.4;              % Stator resistance (ohms)
Rr = 0.918;            % Rotor resistance (ohms)
Rc = 23.87;            % Core loss resistance (ohms)
Le1 = 2.32e-3;          % Stator leakage reactance (H)
Le2 = 2.32e-3;          % Rotor leakage reactance (H)
Lm = 25.25e-3;         % Magnetising reactance (H)
Lr = Lm + Le2;
Ls = Lm + Le1;
P = 4;                  %Poles
Rn = 100000;

% Filter
R  = 0.05;               %Parasitic resistance
C  = 10e-6;              %Set C
L = 2.94e-3;             %Set L

Z1 = Rs + s*Le1 + s*Lm*(s*Le2 + Rr)/(s*Lm + s*Le2 + Rr);

Z2 = R + s*L;

Z3 = Z1*Z2/(Z1 + Z2);

Zc = 1/(s*C);

Hs = Zc/(Z3 +Zc);

Vos = minreal(Hs)

W=logspace(0,7,40000);
h=bodeplot(Vos,W); grid;
p = getoptions(h);           %Create a plot options handle p.
p.FreqUnits = 'Hz';          %Modify frequency units.
setoptions(h,p);             %Apply plot options to the Bode plot and
%render.


And if we type into Matlab

minreal(Z3)

0.00177 s^3 + 0.9529 s^2 + 34.25 s + 315.6
------------------------------------------
s^2 + 333.9 s + 6538


It gives us the transfer function for everything except the capacitor and my question is how do I convert this transfer function into its equvalent R and L. I tried replacing s with jw but I got no where

• Your expectations for accuracy and simplicity are to be applauded. However...... – Andy aka Feb 15 '18 at 17:53
• I dont expect perfect accuracy nor do I believe this system is a perfect representation of anything but itself. I do enjoy the philosophical ponderings, I could bore you to death but I came to get to an answer. Looking at it it looks doable to me but its beaten me to the point its bugging me – Jamie Lamb Feb 15 '18 at 19:25

I am trying to calculate the effects more accurately, so I use the equivalent circuit for the motor to derive a transfer function which is then put in parallel with the filter components with the clearf goal of reducing it all to one R one L and one C which then allows me to calculate things like the -3db point

Why don't you just draw the LC and equivalent load of the induction motor as a circuit for a simulator then ask the sim to do an AC analysis.

The formula you derive will be complex and you should forget about your: -

goal of reducing it all to one R one L and one C

Because it won't happen. Use more appropriate tools.

On another point you equate the induction motor TF to be: -

Z1 = Rs + sLe1 + sLm*(sLe2 + Rr)/(sLm + s*Le2 + Rr)

But I see nowhere in this equation the resistance that represents the mechanical output of motor and slip-factor.

• I have the system simulated in Matlab and a hardware implementation. The script plots the bode plot for output voltage its approximately 1200Hz. I am aware of the limitations with the equivalent circuit, none LTI systems etc. This is most accurate when the rotor speed is at steady state. My simulation uses the dynamic model and the Matlab script gives pretty accurate results. The rotor resistance is Rr and for brevity the slip division is included. Matlab calculates 1200Hz and its possible to back calculate the equivalent L as 1.77mH. By having the R and L values I can calculate the -3dB point – Jamie Lamb Feb 15 '18 at 19:33
• When I said more accurately I meant more accurate than the paper attached, they simply forget the resistance and do do the Thevenin equivalent which does work with pretty good accuracy. To me I dont see why we cant reduce it to RLC for a operating fixed point which will give us the bode plot sure if we change things it could alter it drastically but thats another discussion getting to the answer to this is bugging me! – Jamie Lamb Feb 15 '18 at 19:38
• I’m not sure Rr represents the motor mechanical load impedance. – Andy aka Feb 15 '18 at 19:57
• The Rr here is actually Rr/slip but it looks hideous in the equations its not a good idea to denote slip with s. en.wikipedia.org/wiki/Induction_motor, see Steinmetz equivalent circuit in the link – Jamie Lamb Feb 15 '18 at 20:09