RC circuit with rectangular pulse

In summary: V_C(t) = 12\text{V} + 5.35e^{-100ms/\tau}In summary, The circuit above has a voltage response of 12V and 5.35e^{-100ms/\tau}.
  • #1
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Homework Statement


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.


Homework Equations





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
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
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
Maylis said:
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)?

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

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  • #6
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
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
Maylis said:
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?

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> :smile:
 
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Related to RC circuit with rectangular pulse

1. What is an RC circuit with rectangular pulse?

An RC circuit with rectangular pulse is an electrical circuit that consists of a resistor (R) and a capacitor (C) connected in series. This circuit is driven by a rectangular pulse input, which is a sudden change in voltage that rises and falls in a short period of time.

2. How does an RC circuit with rectangular pulse work?

When a rectangular pulse is applied to an RC circuit, the capacitor charges and discharges over time due to the flow of current through the resistor. The time constant, which is the product of the resistance and capacitance, determines the rate at which the capacitor charges and discharges.

3. What is the time constant of an RC circuit with rectangular pulse?

The time constant of an RC circuit with rectangular pulse is equal to the product of the resistance and capacitance (RC). It represents the time it takes for the capacitor to charge or discharge to approximately 63.2% of its maximum voltage or current.

4. How does the time constant affect the behavior of an RC circuit with rectangular pulse?

The time constant determines the rate at which the capacitor charges and discharges. A smaller time constant means the capacitor charges and discharges more quickly, while a larger time constant results in a slower charge and discharge. This affects the shape and duration of the rectangular pulse output.

5. What are the applications of an RC circuit with rectangular pulse?

An RC circuit with rectangular pulse is commonly used in electronic circuits for signal conditioning, timing, and filtering. It is also used in pulse-width modulation (PWM) circuits for controlling power to devices such as motors and LED lights. Additionally, it is used in pulse generators for producing precise and adjustable rectangular pulses.

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