Derivation Help for Homework Problem | Solving Vc(t) Equations | 5V Square Wave

[frankkk]

I am trying to solve a homework problem. I need some help to get going in the right direction if possible. My equation I am starting with is:

Vc(t) = 5(1-e^(-t/T))

and I have to end up with these two equations:
If T > 1/10f then for a 5 V square wave with a 50% duty cycle, the capacitor voltage is given by

Vc(t) = 5 - (5 / 1+e^(-1/2*f * T)) * e^(-t/T)

and

Vc(t) = (5 / 1+e^(-1/2*f * T)) * e^(-t/T)

Thanks!

 

[berkeman]

Welcome to PF, Frank. Homework problems are supposed to be posted in the PF Homework forum area, and you need to show your initial work in order for us to help you. We're happy to help, we just need to be sure that you do the bulk of the work, and learn from the exercise.

This thread may get moved by the moderators, or you could start a new thread in one of the homework forums, or you could see if it is allowed to stay here and we just follow the homework forum guidelines here.

So to get you started, I don't quite understand the problem statement. How can Vc(t) be equal to three different things? Is maybe Vc(t) a forcing function, and you are asked to find some kind of circuit response from the forcing function through some source impedance?

Try cleaning up the equations some or describing what is going on, and then show us what your first few steps are in trying to solve the problem. We can offer hints at that point.

 

[ravenprp]

To solve this problem, we can start by breaking down the given equation for Vc(t) into smaller parts. Let's first focus on the term (1-e^(-t/T)). This term represents the voltage across the capacitor at any given time t. Now, let's look at the given conditions for the problem - a 5V square wave with a 50% duty cycle and a time period of T > 1/10f.

We know that a square wave with a 50% duty cycle has equal high and low voltage periods. In this case, the high voltage period is 5V and the low voltage period is 0V. We can represent this using the Heaviside function:

H(t) = 1 for t > 0 and H(t) = 0 for t < 0

Multiplying this function with our original equation, we can get the following expression for Vc(t):

Vc(t) = 5*H(t) - (5*H(t) / 1+e^(-t/T))

Next, we can use the fact that T > 1/10f to simplify the equation further. This condition implies that the time constant T is much larger than the period of the square wave, which means that the exponential term e^(-t/T) will be very close to 0 during the high voltage period. Therefore, we can simplify the equation to:

Vc(t) = 5*H(t) - (5*H(t) / 1+0)

Vc(t) = 5*H(t) - 5*H(t)

Vc(t) = 0 for t > 0

This equation satisfies the condition for the high voltage period, but we still need to consider the low voltage period. During this period, the exponential term e^(-t/T) will be close to 1, which means that the voltage across the capacitor will be close to 5V. We can represent this using another Heaviside function, but with a negative argument:

H(-t) = 1 for t < 0 and H(-t) = 0 for t > 0

Multiplying this function with our original equation, we can get the following expression for Vc(t):

Vc(t) = 5*H(-t) - (5*H(-t) / 1+e^(-t/T))

Again