# Circuit Theory/TF Examples/Example14

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:

${\displaystyle {\frac {\mathbb {I} _{o}}{\mathbb {V} _{s}}}={\frac {\mathbb {I} _{o}}{\mathbb {I} _{t}}}{\frac {\mathbb {I} _{t}}{\mathbb {V} _{s}}}}$

Which is of the form:

${\displaystyle {\frac {\mathbb {I} _{o}}{\mathbb {V} _{s}}}={\text{current divider}}*{\text{admittance}}}$

The current divider is going to be:

${\displaystyle {\frac {\mathbb {I} _{o}}{\mathbb {I} _{t}}}={\frac {\frac {1}{R_{3}}}{{\frac {1}{R_{3}}}+{\frac {1}{sL_{2}+{\frac {1}{sC_{2}}}}}}}={\frac {sL_{2}+{\frac {1}{sC_{2}}}}{R_{3}+sL_{2}+{\frac {1}{sC_{2}}}}}}$

Substituting into the transfer function:

${\displaystyle {\frac {\mathbb {I} _{o}}{\mathbb {V} _{s}}}={\frac {sL_{2}+{\frac {1}{sC_{2}}}}{R_{3}+sL_{2}+{\frac {1}{sC_{2}}}}}*{\frac {1}{R_{1}+{\frac {1}{sC_{1}}}+sL_{1}+{\frac {1}{{\frac {1}{R_{3}}}+{\frac {1}{sL_{2}+{\frac {1}{sC_{2}}}}}}}}}}$

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);

${\displaystyle {\frac {\mathbb {I} _{o}}{\mathbb {V} _{s}}}={\frac {1000*(s^{2}+5000000.0)*s}{s^{4}+5000000.0*s^{3}+2.000015*10^{12}*s^{2}+3.5*10^{13}*s+5.0*10^{13}}}}$

The differential equation for this would be:

${\displaystyle 1000({d^{2}V_{s} \over dt^{2}}+5000000.0){dV_{s} \over dt}={d^{4}I_{o} \over dt^{4}}+5000000.0{d^{3}I_{o} \over dt^{3}}+2.000015*10^{12}{d^{2}I_{o} \over dt^{2}}+3.5*10^{13}{dI_{o} \over dt}+5.0*10^{13}I_{o}}$

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)
${\displaystyle {\text{zeros}}=-4561551.036,-438431.4637,-15.93127461,-1.569297305}$

So in this case the solution would be:

${\displaystyle i_{o}=Ae^{-4561551.036t}+Be^{-438431.4637t}+Ce^{-15.93127461t}+De^{-1.569297305}+E}$

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 prepration 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)
${\displaystyle {\text{poles}}=0,-2236.067977i,2236.067977i}$