Circuit Theory/Convolution Integral/Examples/2R1LExample/2 Resistor, 1 inductor example

Find drop across CR₁ pair given that V_s 2e^2t/to.

Given V_s = 2e^2t/to find the drop across the C R₁ parallel combination

$V_{s}=2e^{\frac {2t}{t_{o}}}$

$V_{RC}=?$

Outline of solution:

Transfer Function

$H(s)={\frac {V_{RC}}{V_{s}}}={\frac {\frac {1}{{\frac {1}{R_{1}}}+Cs}}{R_{2}+{\frac {1}{{\frac {1}{R_{1}}}+Cs}}}}={\frac {R_{1}}{R_{1}+R_{2}+sCR_{1}R_{2}}}={\frac {\frac {R_{1}}{R_{1}+R_{2}}}{1+{\frac {sCR_{1}R_{2}}{R_{1}+R_{2}}}}}$

-- or --

$i_{t}=i_{c}+i_{r}={\frac {V_{s}-V_{RC}}{R_{2}}}=C{dV_{RC} \over dt}+{\frac {V_{RC}}{R_{1}}}$

${\frac {\mathbb {V} _{s}-\mathbb {V} _{RC}}{R_{2}}}=sC\mathbb {V} _{RC}+{\frac {\mathbb {V} _{RC}}{R_{1}}}$

$\mathbb {V} _{s}=\mathbb {V} _{RC}(sCR_{2}+{\frac {R_{2}}{R_{1}}}+1)$

${\frac {\mathbb {V} _{RC}}{\mathbb {V} _{s}}}={\frac {1}{sCR_{2}+{\frac {R_{2}}{R_{1}}}+1}}={\frac {R_{1}}{R_{1}+R_{2}+sCR_{1}R_{2}}}={\frac {\frac {R_{1}}{R_{1}+R_{2}}}{1+{\frac {sCR_{1}R_{2}}{R_{1}+R_{2}}}}}$

Homogeneous Solution

First order so τ is:

$\tau =-{\frac {CR_{1}R_{2}}{R_{1}+R_{2}}}$

$V_{RC_{h}}(t)=A*e^{\frac {t}{\tau }}+C_{1}$

Particular Solution

After a long time, due to V_s = 1, the capacitor opens. So V_RC is part of a voltage divider consisting of just two resistors:

$V_{RC_{p}}={\frac {V_{s}R_{1}}{R_{1}+R_{2}}}={\frac {R_{1}}{R_{1}+R_{2}}}$

Evaluate Initial Conditions

Combining the homogeneous and particular:

$V_{RC}(t)={\frac {R_{1}}{R_{1}+R_{2}}}+A*e^{\frac {t}{\tau }}+C_{1}$

At t=0, the voltage across the capacitor is zero so:

$V_{RC}(t)={\frac {R_{1}}{R_{1}+R_{2}}}+A+C_{1}=0$

Initially the cap is a short, so the current through the cap is limited by R₂ so:

$i_{C}(t)=C{dV_{RC} \over dt}=-{\frac {CA}{\tau }}e^{\frac {t}{\tau }}$

$i_{C}(0)={\frac {V_{S}}{R_{2}}}=-{\frac {CA}{\tau }}$

Since V_s = 1 (doing this for the unit step function because using convolution integral):

$A=-{\frac {\tau }{CR_{2}}}$ and $C_{1}={\frac {\tau }{CR_{2}}}$

In summary:

$V_{RC}(t)\mu (t)={\frac {\tau }{CR_{2}}}(1-e^{\frac {t}{\tau }})$

Find impulse solution

The impulse solution is the derivative of the above:

$V_{RC}(t)\delta (t)={\frac {1}{CR_{2}}}e^{\frac {t}{\tau }}$

Convolution Integral

$V_{RC}(t)=\int _{0}^{t}{\frac {e^{\frac {t-x}{\tau }}}{CR_{2}}}2e^{\frac {2t}{t_{o}}}dx={\frac {2\tau t_{o}}{CR_{2}(2\tau -t_{o})}}(e^{\frac {2t}{t_{o}}}-e^{\frac {t}{\tau }})+C_{1}$

f := (exp((t-y)/x)/(C*R2))*2*exp(2*y/z);
S :=int(f,y=0..t)

Evaluate Integration Constant

Know that V_RC=0 at t=0 so:

$0=X(1-1)+C_{1}$

So:

$C_{1}=0$

And finally:

$V_{RC}(t)={\frac {2\tau t_{o}}{CR_{2}(2\tau -t_{o})}}(e^{\frac {2t}{t_{o}}}-e^{\frac {t}{\tau }})$