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

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

${\displaystyle V_{RC}=?}$

Outline of solution:

Transfer FunctionEdit

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

-- or --

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

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

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

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

Homogeneous SolutionEdit

First order so τ is:

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

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

Particular SolutionEdit

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:

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

Evaluate Initial ConditionsEdit

Combining the homogeneous and particular:

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

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

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

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

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

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

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

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

In summary:

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

Find impulse solutionEdit

The impulse solution is the derivative of the above:

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

Convolution IntegralEdit

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

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

Evaluate Integration ConstantEdit

Know that V_RC=0 at t=0 so:

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

So:

${\displaystyle C_{1}=0}$

And finally:

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

This article uses material from the Wikipedia article  Metasyntactic variable, which is released under the  Creative Commons Attribution-ShareAlike 3.0 Unported License.