Circuit Theory/Convolution Integral/Examples/Example42

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example 2nd order circuit with two inductors and two resistors .. solved using the convolution integral in wikibook circuit analysis

Given that Vs = (1 + 2t)μ(t), find io using the convolution integral.

Transfer function[edit]

H(s) = \frac{\mathbb{I}_o}{\mathbb{V}_s} = \frac{\mathbb{I}_T}{\mathbb{V}_s} * \frac{\mathbb{I}_o}{\mathbb{I}_T} = \frac{1}{s + \frac{1}{1 + \frac{1}{1+s}}} * \frac{\frac{1}{1+s}}{1 + \frac{1}{1+s}}
simplify((1/(1+s))/((s+1/(1+1/(1+s)))*(1 + 1/(1 + s))))
H(s) = \frac{1}{s^2 + 3s + 1}
solve(s^2 + 3*s + 1,s)

Homogeneous Solution[edit]

Two real, different roots:

s_{1,2} = \frac{-3 \pm \sqrt{5}}{2}

So Io has this form:

i_o(t) = Ae^{s_1t} + Be^{s_2t}

Particular Solution[edit]

After a very long time, inductors are shorts, the voltage across both 1 ohm resistors is 1 volt, so io is 1 amp:

i_{o_p}(t) = 1

Initial Conditions[edit]

Adding the particular and homogeneous solutions together, have this form:

i_o(t) = 1 + Ae^{s_1t} + Be^{s_2t} + C_1

After a long period of time (using the particular initial condition again), the current is going to be 1:

i_o(\infty) = 1 + C_1 = 1
C_1 = 0

Initially, the two inductors are going to be opens, thus io has to be 0:

i_o(0) = 0 = 1 + A + B

Initially, all the drop is going to be across the first inductor, leaving the voltage across the second zero. The current io flows through both the second inductor and it's serial resistor, so an equation for this voltage can be obtained by taking the derivative:

V_L(t) = L{d i_o(t) \over dt} = s_1Ae^{s_1t} + s_2Be^{s_2t}
V_L(0) = 0 = s_1A + s_2B
solve([1+A+B,s1*A + s2*B],[A,B])
A = \frac{s_2}{s_1-s_2}
B = -\frac{s_1}{s_1-s_2}

So:

i_0(t) = 1 + \frac{s_2e^{s_1t} - s_1e^{s_2t}}{s_1 - s_2}

Impulse Solution[edit]

Taking the derivative of the above get:

i_{o_\delta} (t) = \frac{s_1s_2}{s_1-s_2}(e^{s_1t} - e^{s_2t})

Convolution Integral[edit]

i_o(t) = \int_0^t \frac{s_1s_2}{s_1-s_2}(e^{s_1t} - e^{s_2t})*(1+2x) dx
syms t x
s1 = (-3 + sqrt(5))/2;
s2 = (-3 - sqrt(5))/2;
f = (s1*s2/(s1-s2)*(exp(s1*(t-x))-exp(s2*(t-x)))*(1+2*x));
S =int(f,0,t);
vpa(S, 3)
i_o(t) = 2t + 4.96e^{-0.382t} + 0.0403e^{-2.62t} - 5 + C_1

At t=0:

i_o(0) = 0 = 4.96 + .04 -5 + C_1

Which means that C_1 = 0 so finally:

i_o(t) = 2t + 4.96e^{-0.382t} + 0.0403e^{-2.62t} - 5