# A switch can represent the unit step function

I have a some what basic question. I was just brushing up on some basics with a book on fundamentals of electric circuits. I came across this:

They say that Fig 7.40a with a switch can be replaced by figure 7.40b where u(t) is the unit step function. I don't understand that. For t < 0, that voltage source will be producing 0V -> hence a short between the resistor and the negative plate of the capacitor. Wouldn't that cause the initial voltage on the capacitor to flow through the resistor?

To me, it seems like it's only possible to jump from Fig 7.40a to 7.40b if the initial voltage on the capacitor is 0V. Am I missing something?

• “v” is the initial cap voltage, it does not change at t=0 but shortly after slowly due the current unlike the switch voltage , which is instant thus a “step” Commented May 25, 2021 at 23:22
• Be careful about accepting answers too quickly! Commented May 25, 2021 at 23:48

You're right, this specific substitution only works if the initial voltage of the capacitor is 0V.

You could of course replace the voltage source with "Vs u(t) + Vo", then it'd also work if there was an initial voltage Vo.

• they changed terms for Vo to v and that is already in the equation 7.40b. so you are missing that Commented May 26, 2021 at 2:05

In this case you can use the formulation in 7.40(b). The initial current through the capacitor doesn't affect the solution.

If C was an inductor the initial current would be a factor and you would see it as part of the solution.

To elaborate, if the initial current through the capacitor was significant you would see it as part of the solution. But the solution only depends on the initial voltage across the capacitor, so you can assume the initial capacitor current is anything you want -- 0 A or 3 A or ... -- and you'll still get the same answer.

The same holds for inductors and initial voltage. Only the initial current through an inductor will affect the time evolution of the system -- the initial voltage has no effect.

If you're still not convinced, just consider how you solve this problem... the circuit gives you a 1st-order differential equation for $$\v(t)\$$ -- the voltage across the capacitor. You solve this and obtain an equation like:

$$v(t) = ...\text{some function of t}... + C$$

To determine what $$\C\$$ is you substitute $$\t = 0\$$ and use the initial value $$\v(0)\$$. But nowhere did you have to use the initial current.

When you have an inductor in a circuit you obtain a differential equation for $$\i(t)\$$ -- the current through the inductor. Likewise, to completely solve it you need to know the initial current, but not the initial voltage.

So in this case you can assume any history for the capacitor's current -- it won't affect the evolution of the system after time 0.

• it's the initial voltage in question not current and that is v. Commented May 26, 2021 at 2:06
• It's the initial voltage of the voltage source that's in question. But that only sets the current which doesn't affect the solution. Commented May 26, 2021 at 2:31
• It’s a poorly written paragraph, but that is incorrect. Both Vsu(t)and v the initial cap voltage set the initial current and that was not the question. It was “ if the initial voltage on the capacitor is 0V. Am I missing something?” @ErikR Commented May 26, 2021 at 11:41