Circuit Theory/Convolution Integral/Examples/Example43



Given that is = 1 + cos(t), find io 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= &infin;:
 * $$i_o(\infty) = \frac{1}{2} = \frac{1}{2} + C_1$$
 * $$C_1 = 0$$

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:
 * $$V_L(t) = L*{d i_o(t) \over dt}$$

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