Circuit Theory/Convolution Integral/Examples/Example43

2R 1L 1C circuit to solve using the convolution integral .. for wikibook circuit analysis

Given that i_s = 1 + cos(t), find i_o using the convolution integral.

Outline:

Transfer Function

$H(s)={\frac {\mathbb {I} _{o}}{\mathbb {I} _{s}}}={\frac {\mathbb {V} _{T}}{\mathbb {I} _{s}}}*{\frac {\mathbb {I} _{o}}{\mathbb {V} _{T}}}={\frac {1}{{\frac {1}{1/s}}+{\frac {1}{1}}+{\frac {1}{s+1}}}}*{\frac {1}{s+1}}$

simplify(1/((s+1)*(s + 1 + 1/(s+1))))

$H(s)={\frac {1}{s^{2}+2s+2}}$

Homogeneous Solution

Setting the denominator to zero and find the values of s:

solve(s^2 + 2*s + 2)

$s=-1\pm i$

This means the solution has the form:

$i_{o_{h}}(t)=e^{-t}(A\cos t+B\sin t)+C_{1}$

Particular Solution

After a long time the cap opens and the inductor shorts. The current is split between the two resistors. Each will get 1/2 of the unit step function source which would be 1/2 amp:

$i_{o_{p}}(t)={\frac {1}{2}}$

Initial Conditions

$i_{o}={\frac {1}{2}}+e^{-t}(A\cos t+B\sin t)+C_{1}$

The particular solution still has to apply so at t= ∞:

$i_{o}(\infty )={\frac {1}{2}}={\frac {1}{2}}+C_{1}$

$C_{1}=0$

Initially the current has to be zero in this leg so:

$i_{o}(0)={\frac {1}{2}}+A=0$

$A=-{\frac {1}{2}}$

The initial voltage across the cap is zero, and across the leg is zero and across the inductor is zero. So:

$V_{L}(t)=L*{di_{o}(t) \over dt}$

f := 1/2 + exp(-t)*((-1/2)*cos(t) + B*sin(t));
g = diff(f,t)

$V_{L}(t)=e^{-t}(B(cos(t)-sin(t))+{\frac {1}{2}}(cos(t)+sin(t))$

$V_{L}(0)=B+{\frac {1}{2}}=0$

$B=-{\frac {1}{2}}$

So now:

$i_{o}={\frac {1}{2}}(1-e^{-t}(\cos t+\sin t))$

Impulse Response

Taking the derivative of the above

f := 1/2*(1-exp(-t)*(cos(t) + sin(t)));
g = diff(f,t)

get:

$i_{o_{\delta }}(t)=e^{-t}\sin t$

Convolution Integral

$i_{o}(t)=\int _{0}^{t}e^{-t+x}\sin(t-x)*(1+cos(x))dx$

f := exp(x-t)*sin(t-x)*(1 + cos(x));
S =int(f,x=0..t);

$i_{o}(t)=0.2\cos(t)+0.4\sin(t)-0.7\cos(t)e^{-t}-1.1\sin(t)e^{-1}+0.5$