RC circuit with rectangular pulse

1. Mar 19, 2014

Maylis

1. The problem statement, all variables and given/known data
Consider the circuit below. If the source υs represents a 12-V, 100-ms-long rectangular pulse that starts at t = 0 and the
element values are R1 = 6 kΩ, R2 = 2 kΩ, R3 = 4 kΩ, and C = 15 μF, determine the voltage response υ(t) for t ≥ 0.

2. Relevant equations

3. The attempt at a solution
I am trying to use this slide as a platform to start the problem, but I am not sure what the voltage response is vs. natural response, and where I should do the timeshift. Also, I wasn't sure if my thevenin resistance to find the time constant was correct.

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2. Mar 19, 2014

Staff: Mentor

Hi Maylis,

Your Thevenin resistance that the capacitor "sees" is not correct; the two 6k branches are not in series from the capacitor's point of view. This is throwing off your time constant value.

A suggestion: Consider first the case where the input is simply a unit step and write the expression for the capacitor voltage. Determine its value at time t = 100 ms. So, what are the then-current conditions of the circuit at t = 100+ ms?

3. Mar 19, 2014

Maylis

Is it okay to essentially treat it as if it was not a step function, solve it as such, and then at the end tweak the answer to include the time shift by subtracting the time (in this case 100ms)?

4. Mar 19, 2014

Staff: Mentor

I don't know how you could treat it as not being a step function initially. It clearly is. The applied voltage is said to be a rectangular pulse, so the initial transition is from 0V to 12V, the leading edge of a suitably scaled unit step.

An approach is to break the circuit activity into easily understood sections. First the input transitions from 0V to 12V. That invokes a certain behavior in terms of the capacitor voltage, at least for the ensuing 100 ms; it's charging from 0V towards a final value determined by the resistor network surrounding the capactitor. But at the end of 100 ms the input voltage drops from 12V to 0V, curtailing the charging process. So you have the capacitor with whatever voltage it had at that instant now discharging towards 0V.

So the capacitor will initially charge from an initial voltage of 0V towards some eventual peak. Before it gets there, at 100 ms the input changes to 0V so the capacitor now wants to discharge from whatever voltage it had attained at 100ms back down to 0V. So, sort of a sawtooth shape for the overall curve. You should be able to sketch the curve for both intervals, and calculate the voltage on the capacitor for t = 100 ms when the change between charging and discharging occurs.

5. Mar 20, 2014

Maylis

Okay, here is my 2nd attempt. I tried to follow your directions..hopefully I succeeded

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6. Mar 20, 2014

Staff: Mentor

Okay, that's a pretty good attempt! You've found the two "sections" of the response that are associated with the pulse rising and falling edges:

1) $6 (1 - e^{-t/\tau})$
2) $5.35e^{-(t - 100ms)/\tau}$

The only problem is that you need to add some mathematical "switches" to turn on and off their contributions to the overall function over the appropriate spans of t. For that the unit step function u(t) comes in handy... See what you can devise.

7. Mar 20, 2014

Maylis

Okay, a little caveat. He said that we will not be using the u(t) function. Is there a way to do it without including it?

8. Mar 20, 2014

Staff: Mentor

Well, you could write it as you would a mathematical function comprised of separate pieces:

$$V_C(t) = \begin{cases} \text{0 V}, & \text{for } t \leq 0 \\ \text{<first part>}, & \text{for } 0 < t \leq 100 ms \\ \text{<second part>}, & \text{for } t > 100 ms \end{cases}$$
Fill in the appropriate expressions for <first part> and <second part>

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