Impulse Response of a Circuit

Electricity and Magnetism Level 4

In the circuit above, at $t=0$ , the capacitor has no energy stored in it. The goal of this problem is to calculate the voltage across the capacitor as a function of time. One is expected to get a differential equation of the form:

$\dot{V}_c= -aV_c + b V_S$

The recommended route to solve the above is numerical integration, but without using a computer. Say a time step is chosen which is $\delta t = h$ . Perform the following steps:

Using the explicit Euler numerical technique convert the above differential equation to a difference equation. You will get something of the form:

$V_c(k+1) = AV_c(k) + BV_S(k)$

Where $k=0,1,2 \dots$ denotes the time index.

Use the following definition for the source voltage:

$V_S(0) = \frac{1}{h}$ $V_S(k) = 0 \ \forall \ k>0$

Derive the voltage across the capacitor for the above input. The result is of the form:

$\boxed{V_c(k+1) = \frac{\left(1- \frac{h}{m}\right)^k}{n}}$

Here $m$ and $n$ are positive integers. Enter your answer as $m+n$ .

Bonus: Can this circuit be used as a filter? If so, what kind? Also, plot the capacitor voltage.

Note: The source voltage is an approximation of the Dirac-delta function but the diagram indicates a sinusoidal input. This subtlety is to be ignored.

The answer is 525.

1 solution

Steven Chase
Oct 16, 2020

This is a good one. Let $R_1 = 10 \Omega$ and $R_2 = 30 \Omega$ . The differential equation is:

$\frac{V_S - V_C}{R_1} - \frac{V_C}{R_2} = C \dot{V}_C$

Doing a bit of manipulation on this results in:

$\dot{V}_C = - a V_C + b V_S \\ a = 1/225 \\ b = 1/300$

Now for the explicit Euler:

$V_C (k+1) = V_C(k) + \Big(- a V_C(k) + b V_S(k) \Big) h \\ V_C(k+1) = A V_C(k) + B V_S(k) \\ A = 1 - a h \\ B = b h$

Assume that $V_C(0) = 0$ and that $V_S(0) = 1/h$ . Write down a few terms and observe a pattern.

$V_C (1) = 0 + \frac{B}{h} \\ V_C(2) = \frac{A B}{h} + 0 \\ V_C(3) = \frac{A^2 B}{h} + 0 \\$

The pattern is:

$V_C(k+1) = A^k \frac{B}{h} = (1 - a h)^k b = \Big(1 - \frac{h}{225} \Big)^k \frac{1}{300}$

As for the bonus, I assume we're talking about filtering in the context of AC input signals (sinusoids). Assuming we take the output voltage from across the capacitor, the voltage gain is given below (ratio of output voltage magnitude to input voltage magnitude). The circuit acts as a low pass filter.

@Steven Chase Upvoted. How do you identify a filter whether it is low or high?

Talulah Riley - 7 months, 4 weeks ago

The filter has highest gain at low frequency, and the gain decreases as the frequency increases. The filter therefore lets through low frequency signals more readily than it lets through high-frequency signals.

Steven Chase - 7 months, 4 weeks ago

Thanks for the solution. I was recently watching a video on impulse responses that inspired me to frame this exercise. I do not have much practical experience with circuits, so am wondering if impulse voltage inputs are practical inputs for real circuits? I speculate that an instantaneous impulse may cause a damaging current surge in some circuits.

Karan Chatrath - 7 months, 4 weeks ago

@Karan Chatrath share that video with me.

Talulah Riley - 7 months, 4 weeks ago

I don't think sharing the video would be useful for you. Instead, I suggest that you read about the Dirac-delta function.

Karan Chatrath - 7 months, 4 weeks ago

@Karan Chatrath – @Karan Chatrath no, it would be useful. Please share it.

Talulah Riley - 7 months, 4 weeks ago

I am most familiar with impulse responses in the context of digital filters. There is a particular type of filter called a "finite impulse response" filter (FIR), where the output is a weighted sum of the inputs:

$y_k = \Sigma \alpha_n x_{k-n}$

If you put a single unit impulse in as the input, the output will successively read out the filter coefficients $\alpha_n$ .

Steven Chase - 7 months, 4 weeks ago

I looked up this type of filter. Interesting read. I will be posting a follow-up problem on impulse responses, in a few hours

Karan Chatrath - 7 months, 4 weeks ago

Impulse Response of a Circuit

The answer is 525.

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