Circuit Theory/Convolution Integral/Examples/Example42

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

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

Transfer function

$H(s)=\frac{{𝕀}_o}{{𝕍}_s}=\frac{{𝕀}_T}{{𝕍}_s}*\frac{{𝕀}_o}{{𝕀}_T}=\frac{1}{s+\frac{1}{1+\frac{1}{1+s}}}*\frac{\frac{1}{1+s}}{1+\frac{1}{1+s}}$

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

Two real, different roots:

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

So I_o has this form:

$i_o(t)=A{e}^{s_{1}t}+B{e}^{s_{2}t}$

Particular Solution

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

$i_{o_p}(t)=1$

Initial Conditions

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

$i_o(t)=1+A{e}^{s_{1}t}+B{e}^{s_{2}t}+C_1$

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

$i_o(\mathrm{\infty })=1+C_1=1$

$C_1=0$

Initially, the two inductors are going to be opens, thus i_o 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 i_o flows through both the second inductor and it's serial resistor, so an equation for this voltage can be obtained by taking the derivative:

$V_L(t)=L\frac{d{i}_o(t)}{dt}=s_{1}A{e}^{s_{1}t}+s_{2}B{e}^{s_{2}t}$

$V_L(0)=0=s_{1}A+s_{2}B$

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:

$i_0(t)=1+\frac{s_2{e}^{s_{1}t}-s_1{e}^{s_{2}t}}{s_1-s_2}$

Impulse Solution

Taking the derivative of the above get:

$i_{o_{\delta }}(t)=\frac{s_1{s}_2}{s_1-s_2}(e^{s_{1}t}-e^{s_{2}t})$

Convolution Integral

$i_o(t)=\underset{0}{\overset{t}{\int }}\frac{s_1{s}_2}{s_1-s_2}(e^{s_{1}t}-e^{s_{2}t})*(1+2x)dx$

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:

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