Circuit Theory/TF Examples/Example14

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Finding io using transfer functions

Find a partical solution to one of the re-occurring examples in this book. Given VS, find io. This has to be split into two transfer functions:

𝕀o𝕍s=𝕀o𝕀t𝕀t𝕍s

Which is of the form:

𝕀o𝕍s=current divider*admittance

The current divider is going to be:

𝕀o𝕀t=1R31R3+1sL2+1sC2=sL2+1sC2R3+sL2+1sC2

Substituting into the transfer function:

𝕀o𝕍s=sL2+1sC2R3+sL2+1sC2*1R1+1sC1+sL1+11R3+1sL2+1sC2

Substituting and solving using this mupad code:

R1 :=1000;C1 :=.0001;C2 :=.0002;R3 := 2000;L1 := .001;L2 := .001;
f := (s*L2 + 1/(s*C2))/(R3 + s*L2 + 1/(s*C2))*1/(R1 + 1/(s*C1) + s*L1 + 1/(1/R3 + 1/(s*L2 + 1/(s*C2)))):
factor(f);
𝕀o𝕍s=1000*(s2+5000000.0)*ss4+5000000.0*s3+2.000015*1012*s2+3.5*1013*s+5.0*1013

The differential equation for this would be:

1000(d2Vsdt2+5000000.0)dVsdt=d4Iodt4+5000000.0d3Iodt3+2.000015*1012d2Iodt2+3.5*1013dIodt+5.0*1013Io

Set the source Vs to zero and then find Io by adding together Second order techniques (Real roots, are underdamped. Equal roots are critically damped and imaginary roots are overdamped.

Solving for the zeros (setting denominator to zero):

solve(s^4 + 5000000.0*s^3 + 2.000015*10^12*s^2 + 3.5*10^13*s + 5.0*10^13 = 0, s)
zeros=4561551.036,438431.4637,15.93127461,1.569297305

So in this case the solution would be:

io=Ae4561551.036t+Be438431.4637t+Ce15.93127461t+De1.569297305+E

The next step would be to find four initial conditions of i0 (and final condition). Most texts don't attempt to solve this complex of a problem. The steady state particular solution was easily found long ago with phasors. Transfer functions help to find a general solution to any type of voltage source. And it helps find the homogeneous solution that describes a ramp up when the circuit is first turned on. What circuits do engineers design that is worth this effort? Filters. The fact that today most of this is done digitally just means that all this needs to be repeated in a digital context. (Simulation becomes reality!). So using math tools to solve these problems in preparation for understanding more advanced tools is reason for continuing this thread of content.

So on to filters (Bode Plots .. Fourier Analysis) where the constants disappear and we stay in the complex domain. We will need the poles for to analyze as a filter (setting numerator to zero):

solve(s*(1000.0*s^2 + 5000000000.0)=0,s)
poles=0,2236.067977i,2236.067977i

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