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

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Given Vs = 2e2t/to find the drop across the C R1 parallel combination

Find drop across CR1 pair given that Vs 2e2t/to.

V_s = 2e^{\frac{2t}{t_o}}
V_{RC} = ?

Outline of solution:

Transfer Function[edit]

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_1R_2} = \frac{\frac{R_1}{R_1 + R_2}}{1 + \frac{sCR_1R_2}{R_1 + R_2}}

-- or --

i_t = i_c + i_r = \frac{V_s - V_{RC}}{R_2} = C{d V_{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_1R_2} = \frac{\frac{R_1}{R_1 + R_2}}{1 + \frac{sCR_1R_2}{R_1 + R_2}}

Homogeneous Solution[edit]

First order so τ is:

\tau = -\frac{CR_1R_2}{R_1 + R_2}
V_{RC_h}(t) = A * e^{\frac{t}{\tau}} + C_1

Particular Solution[edit]

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

V_{RC_p}=\frac{V_sR_1}{R_1 + R_2} = \frac{R_1}{R_1 + R_2}

Evaluate Initial Conditions[edit]

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{d V_{RC} \over dt} = -\frac{CA}{\tau}e^{\frac{t}{\tau}}
i_C(0) = \frac{V_S}{R_2} = -\frac{CA}{\tau}

Since Vs = 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[edit]

The impulse solution is the derivative of the above:

V_{RC}(t)\delta(t) = \frac{1}{CR_2}e^{\frac{t}{\tau}}

Convolution Integral[edit]

V_{RC}(t) = \int_0^t \frac{e^{\frac{t-x}{\tau}}}{CR_2} 2 e^{\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[edit]

Know that VRC=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}})