Voltage Transfer Function (Part 2)

Electricity and Magnetism Level pending

A sinusoidal $AC$ voltage source $V_S (t)$ excites an $RLC$ electrical network as shown. The source voltage is defined as follows:

$V_S(t) = 10 \sin(t) \,\,\,\,\, \text{for} \,\, 0 \leq t < 10 \pi \\ V_S(t) = \sin(t) \,\,\,\,\, \text{for} \,\, t \geq 10 \pi$

Define an expected "steady state" resistor voltage signal, which is related to the source voltage by parameters $\alpha$ and $\theta$ , which can be derived from the circuit transfer function (see Part 1 ).

$V_{RSS}(t) = 10 \alpha \sin(t + \theta) \,\,\,\,\, \text{for} \,\, 0 \leq t < 10 \pi \\ V_{RSS}(t) = \alpha \sin(t + \theta) \,\,\,\,\, \text{for} \,\, t \geq 10 \pi$

Let $V_R(t)$ be the actual resistor voltage, assuming that the capacitors and inductor are de-energized at time $t = 0$ . Determine the following integral:

$\int_0^{20 \pi} \Big(V_{RSS}(t) - V_R (t) \Big) \, dt$

Bonus: Plot $V_{RSS}(t)$ and $V_R (t)$ on the same graph and comment on their relationship

Note: $V_{RSS}(t)$ is discontinuous in time, but the effect of this discontinuity on the integral is inconsequential

Details and Assumptions:
1) $R = L = C_1 = 1$
2) $C_2 = 2$

The answer is 0.2308.

1 solution

Karan Chatrath
Oct 24, 2020

Let the current through the inductive branch be $I_1$ and through $C_2$ be $I_2$ . The source current is denoted as $I$ . Let the charge on the capacitor $C_1$ be $Q$ and that through $C_2$ be $Q_2$ . Applying Kirchoff's laws:

$-V_S + Q + \frac{Q_2}{2}=0$ $\dot{I}_1 + I_1 = \frac{Q_2}{2}$ $\dot{Q}_2 = I_2$ $\dot{Q} = I$ $I = I_1 + I_2$

Bearing in mind that all initial conditions are zero, each equation can be Laplace transformed into the $s$ domain as follows:

$-V_S(s) + Q(s) + \frac{Q_2(s)}{2}=0$ $sI_1(s) + I_1(s) = \frac{Q_2(s)}{2}$ $sQ_2(s) = I_2(s)$ $sQ(s)= I(s)$ $I(s) = I_1(s) + I_2(s)$

Solving for $RI_1(s) = V_R(s)$ gives:

$\frac{V_R(s)}{V_S(s)} = \frac{s}{3s^2+3s+1}$

The above is the required transfer function. By replacing $s = j\omega$ in the above transfer function and calculating the magnitude and phase of the resulting complex number gives values of $\alpha$ and $\theta$ . Having found $\alpha$ and $\theta$ , the above transfer function can be written as such:

$(3s^2+3s+1)V_R(s) = sV_S(s)$

Switching back to the time domain leads to the ordinary differential equation:

$3\ddot{V}_R + 3\dot{V}_R + V_R = \dot{V}_S$

It can also be shown from the first set of equations that $V_R(0)=\dot{V}_R(0) = 0$ . This equation can be solved in a number of ways. I have chosen a numerical route. In general the solution of this differential equation comprises of two parts: the null solution ( $V_N$ ) and the steady state solution. By computing $\alpha$ and $\theta$ , the steady state solution is already computed. $V_R = V_{RSS} + V_N$

$\implies V_{RSS} -V_R = -V_N$

The required integral is therefore:

$-\int_{0}^{20 \pi}V_N \ dt \approx 0.2308$

All these steps are performed numerically. These results can also be found analytically, although the solution has to be computed piecewise, which is a bit more tedious. The plot of $V_R$ and $V_{RSS}$ makes the relationship between them apparent. Their difference is simply the transient solution which inevitably dies out as time progresses. The transient solution again becomes non-zero because of the change in source voltage. This again dies out to zero with time.

Greetings! Off late, my attention has been directed towards numerically solving partial differential equations. I have run into a few difficulties. If you are comfortable or interested in this topic, I could post a note on it. I could use some help in this area, and I look forward to your inputs. Please let me know.

Karan Chatrath - 7 months, 2 weeks ago

I attempted your recent problem and got it wrong. I feel silly for not being able to solve it. Thanks for the follow up, anyway. It was a good one!

Karan Chatrath - 7 months, 1 week ago

Voltage Transfer Function (Part 2)

The answer is 0.2308.

1 solution

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