View source for Circuit Theory/TF Examples/Example34/Vc

==Starting Point==
Starting point of i<sub>L</sub> looks good. Can get the voltage across the inductor and the parallel components. Then a voltage divider is going to help get to the V<sub>c</sub> initial condition.
==Transfer Function==
:<math>H(s)= \frac{i_o}{i_s} = \frac{\frac{1}{R_2 + \frac{1}{sC}}}{\frac{1}{sL}+\frac{1}{R_1}+ \frac{1}{R_2 + \frac{1}{sC}}}</math>
<pre>
L :=1;
R1 := 1/2;
C := 1/2;
R2 := 1.5;
simplify((1/(R2 + 1/(s*C))/(1/(s*L) + 1/R1 + 1/(R2 + 1/(s*C))))
</pre>
:<math>\frac{i_o}{i_s} = \frac{2s^2}{8s^2 + 11s + 4}</math>

==Homogeneous Solution==
Set the denominator of the transfer function to 0 and solve for s:
<pre>
solve(8*s^2 + 11*s + 4)
</pre>
:<math>s_{1,2} = \frac{-11 \pm \sqrt{7}i}{16}</math>
So the solution is going to have the form:
:<math>i_{L_h} = e^{\frac{11t}{16}}(A\cos \frac{7}{16} + B\sin \frac{7}{16}) + C</math>
==Particular Solution==
After a very long time the inductor shorts, all the current flows through it so:
:<math>i_o = 0</math>

==Initial Conditions==
[[File:Example34mupad1.png|thumb|MuPad screen shot finding the first equation associated with the constants]]
Adding the particular and homogeneous solutions, get:
:<math>i_o(t) = e^{-\frac{11t}{16}}(A\cos \frac{7t}{16} + B\sin \frac{7t}{16}) + C</math>
Doing the final condition again, get:
:<math>i_o(\infty) = 0 = C \Rightarrow C = 0</math>
Let's try for V<sub>c</sub> first. From the terminal relation for a capacitor:
:<math>V_c(t) = \frac{1}{C}\int i_o dt</math>
<pre>
io := exp(-11*t/16)*(A*cos(7*t/16) + B*sin(7*t/16))
</pre>
<pre>
VC := int(io,t)
</pre>
We know that initially V<sub>c</sub> = 0.5 so at t=0 can find equation for A and B:
<pre>
t :=0;
solve(0.5 = VC)
</pre>
At this point mupad goes numeric and get this equation:
:<math>A = - 0.6363636364B - 0.4829545455</math>
[[File:Example34mupad2.png|thumb|MuPad code for finding the second equation associated with the constants]]
Need another equation. Can find V<sub>t</sub> by adding V<sub>r</sub> and V<sub>c</sub>. Then from V<sub>t</sub> can find expression for the current through the inductor and visit it's initial condition. 
<pre>
io := exp(-11*t/16)*(A*cos(7*t/16) + B*sin(7*t/16))
</pre>
<pre>
VC := int(io,t)
</pre>
The integration constant is going to be zero because after a long time V<sub>C</sub> is zero (the inductor shorts). 
<pre>
VT := VC + io*1.5
</pre>
<pre>
IL := int(VT)
</pre>
Mupad goes numeric. 

At this point have to figure out the integration constant. After a long time, the inductor's current is going to be 1 because it shorts the current source. Looking at IL in the mupad window can see that every term is multiplied by e<sup>-0.6875t</sup> which is going to zero as t goes to &infin;. This means the integration constant is 1. 

So add 1 to IL, then set t=0 and I<sub>L</sub> = 0.5 and again solve for A and B:
<pre>
t :=0;
solve (IL + 1 = 0.5)
</pre>
Get this equation:
:<math>A = 0.4107683001B + 0.5463551119</math>
Now need to solve the two equations and two unknowns:
<pre>
solve([A = - 0.6363636364*B - 0.4829545455,A = 0.4107683001*B + 0.5463551119],[A,B])
</pre>
Get:
:<math>A = 0.142578125, B = -0.9829799107</math>
So now have time domain expression for i<sub>o</sub> step response:
:<math>i_o(t)\mu(t) = e^{-\frac{11t}{16}}(0.143\cos \frac{7t}{16} - 0.983\sin \frac{7t}{16})</math>

==Impulse Response==
[[File:Example34mupad3.png|thumb|MuPad code for finding the impulse response and using the convolution integral]]
The impulse response is the derivative of the step response:
<pre>
i_u := exp(-11*t/16)*(0.142578125 * cos(7*t/16) - -0.9829799107*sin(7*t/16))
</pre>
<pre>
i_s := diff(i_u,t)
</pre>

==Convolution Integral==
The first step is to substitute into i_s for t:
<pre>
i_sub := subs(i_s, t = y-x):
</pre>
Now form the convolution integral:
<pre>
f := i_sub*(1 + 3*cos(2*x)):
io := int(f,x = 0..y)
</pre>
Replacing y with t:
<pre>
i_o :=subs(io, y=t)
</pre>
There is going to be an integration constant. This value can not be determined because the driving function oscillates. The initial conditions of the inductor and capacitor have already been visited. More information (like a specific value at a future time) is needed in order to compute an integration constant.

Thus the final answer is:
:<math> i_o =  0.344cos(2t) + 0.314sin(2t) - 0.201e^{-0.6875t}cos(0.438t) + 1.29e^{-0.6875t}sin(0.438t) - 0.143</math>
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