Circuit Theory/Convolution Integral/Examples/Example43

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

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

Outline:

Transfer Function[edit]

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[edit]

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[edit]

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[edit]

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*{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[edit]

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[edit]

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