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

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

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

$V_s=2e^{\frac{2t}{t_o}}$

$V_{RC}=?$

Outline of solution:

Transfer Function

$H(s)=\frac{V_{RC}}{V_s}=\frac{\frac{1}{\frac{1}{R_1}+Cs}}{R_2+\frac{1}{\frac{1}{R_1}+Cs}}=\frac{R_1}{R_1+R_2+sC{R}_1{R}_2}=\frac{\frac{R_1}{R_1+R_2}}{1+\frac{sC{R}_1{R}_2}{R_1+R_2}}$

-- or --

$i_t=i_c+i_r=\frac{V_s-V_{RC}}{R_2}=C\frac{d{V}_{RC}}{dt}+\frac{V_{RC}}{R_1}$

$\frac{{𝕍}_s-{𝕍}_{RC}}{R_2}=sC{𝕍}_{RC}+\frac{{𝕍}_{RC}}{R_1}$

${𝕍}_s={𝕍}_{RC}(sC{R}_2+\frac{R_2}{R_1}+1)$

$\frac{{𝕍}_{RC}}{{𝕍}_s}=\frac{1}{sC{R}_2+\frac{R_2}{R_1}+1}=\frac{R_1}{R_1+R_2+sC{R}_1{R}_2}=\frac{\frac{R_1}{R_1+R_2}}{1+\frac{sC{R}_1{R}_2}{R_1+R_2}}$

Homogeneous Solution

First order so τ is:

$\tau =-\frac{C{R}_1{R}_2}{R_1+R_2}$

$V_{R{C}_h}(t)=A*e^{\frac{t}{\tau }}+C_1$

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:

$V_{R{C}_p}=\frac{V_s{R}_1}{R_1+R_2}=\frac{R_1}{R_1+R_2}$

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:

$i_C(t)=C\frac{d{V}_{RC}}{dt}=-\frac{CA}{\tau }{e}^{\frac{t}{\tau }}$

$i_C(0)=\frac{V_S}{R_2}=-\frac{CA}{\tau }$

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

$A=-\frac{\tau }{C{R}_2}$ and $C_1=\frac{\tau }{C{R}_2}$

In summary:

$V_{RC}(t)\mu (t)=\frac{\tau }{C{R}_2}(1-e^{\frac{t}{\tau }})$

Find impulse solution

The impulse solution is the derivative of the above:

$V_{RC}(t)\delta (t)=\frac{1}{C{R}_2}{e}^{\frac{t}{\tau }}$

Convolution Integral

$V_{RC}(t)=\underset{0}{\overset{t}{\int }}\frac{e^{\frac{t-x}{\tau }}}{C{R}_2}2e^{\frac{2t}{t_o}}dx=\frac{2\tau t_o}{C{R}_2(2\tau -t_o)}(e^{\frac{2t}{t_o}}-e^{\frac{t}{\tau }})+C_1$

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:

$V_{RC}(t)=\frac{2\tau t_o}{C{R}_2(2\tau -t_o)}(e^{\frac{2t}{t_o}}-e^{\frac{t}{\tau }})$