Circuit Analysis: Equation Derivation Help

[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!

 

[Gokul43201]

Frankkk, we can't help you unless you show some effort first. Please read our posting guidelines for homework. Also read the sticky thread at the top of this subforum.

 

[piercas]

Sure, I can help you with this problem. First, let's break down the given equation Vc(t) = 5(1-e^(-t/T)) to understand what it represents. This equation is known as the transient response of a capacitor in an RC circuit. It shows how the voltage across the capacitor (Vc) changes over time (t) when a step input of 5V is applied to the circuit. The time constant (T) in this equation is equal to the product of the resistance (R) and capacitance (C) in the circuit, i.e. T = RC.

Now, to derive the given equations, we need to consider the behavior of the capacitor in two different cases: when T > 1/10f and when T < 1/10f. Let's start with the first case.

When T > 1/10f, it means that the time constant of the circuit is longer than the period (1/f) of the input square wave. In other words, the capacitor has enough time to fully charge and discharge during each cycle of the input square wave. In this case, the voltage across the capacitor can be simplified to Vc(t) = 5(1-e^(-t/T)) = 5 - 5e^(-t/T).

Next, we need to consider the 50% duty cycle of the input square wave, which means that the ON and OFF time of the square wave are equal. This implies that the capacitor is charged to 5V during the ON time and discharged to 0V during the OFF time. Therefore, the voltage across the capacitor can be written as:

Vc(t) = 5 - 5e^(-t/T) for t < T/2
Vc(t) = 0 for t > T/2

Combining these two equations, we get the final equation:

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

Now, let's consider the second case when T < 1/10f. In this case, the time constant of the circuit is shorter than the period of the input square wave. This means that the capacitor does not have enough time to fully charge and discharge during each cycle of the input square wave. As a result, the voltage across the capacitor will not reach 5V or