Circuit Theory/Convolution Integral/Examples/Example43

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

Outline:

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

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

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

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

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

${\displaystyle s=-1\pm i}$

This means the solution has the form:

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

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:

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

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

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

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

${\displaystyle C_{1}=0}$

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

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

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

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

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

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

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

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

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

So now:

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

Taking the derivative of the above

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

get:

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

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

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

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