1
\$\begingroup\$

I am watching this video where the guy is deriving the equations for the step function of a series RC circuit.

enter image description here

At some point he has the following equation:

\$ \frac{dt}{RC} = \frac{dv}{v_s - v} \$

and he suddenly multiplies both sides by -1 in order to get a minus sign on the left side and invert \$v \$ and \$v_s \$

\$ -\frac{dt}{RC} = \frac{dv}{v - v_s} \$

He justifies that by saying "lets do that because we want to write v minus v_s"

what?

Obviously he knows that that minus sign will be the exponential exponent, making the final result correct but this is not a mathematical justification.

Do you guys have any idea why?

NOTE: you can see the sign change on the right side, on the line the integral symbol appears for the first time.

\$\endgroup\$
  • 1
    \$\begingroup\$ I'm not sure what you're asking. Are you trying to say that this step is algebraically mysterious? There's nothing wrong with it. You don't need justification to do an algebraic manipulation \$\endgroup\$ – Scott Seidman Aug 9 '18 at 12:24
  • \$\begingroup\$ the problem is that the manipulation leads to the correct answer. Imagine you are the first scientist do arrive at that equation. You would not have this insight to suddenly multiply by -1. You would not know that. What I mean is that the minus sign should appear naturally, not because you know it will lead to the correct answer. \$\endgroup\$ – SpaceDog Aug 9 '18 at 12:27
  • 2
    \$\begingroup\$ Not doing the multiplication will also get you to a correct answer. \$\endgroup\$ – Scott Seidman Aug 9 '18 at 12:29
  • \$\begingroup\$ I don't see it. If you can post an answer showing that I appreciate. \$\endgroup\$ – SpaceDog Aug 9 '18 at 12:34
4
\$\begingroup\$

It's the same either way. You don't have to multiple by -1. Try it:

$$\begin{align} \frac{dt}{RC} &=\frac{dv}{v_s-v}\\ \\ \int_{v_0}^v \frac{dv}{v_s-v} &= \int_{0}^t \frac{dt}{RC}\\ \end{align}$$


let's forget about the right side, since it doesn't pertain to your question.

Using

$$\begin{align} \int \frac{1}{ax+b}dx = \frac{1}{a}ln|ax+b|\\ \end{align}$$

then \$a=-1\$ and \$b=v_s\$ so

$$\begin{align} \int_{v_0}^v \frac{dv}{v_s-v} \rightarrow(-1)ln|v_s-v|\\ \end{align}$$

now follow the rest of the tutorial and you'll eventually end up with

$$\begin{align} -v &= -v_s + (v_s-v_0)e^{\frac{-t}{RC}}\\ \end{align}$$

at which point, obviously, to solve for \$v\$, multiply both sides by -1.

\$\endgroup\$
  • \$\begingroup\$ BRILLIANT. I was forgetting that minus sign that comes with the ln. Now I see. THANKS!!!!!!!!!!!!!!! Anyway I still agree that the guy should not change the minus sign, without reason, at the time he did. Much better to let the sign come naturally. THANKS \$\endgroup\$ – SpaceDog Aug 9 '18 at 16:06
  • 1
    \$\begingroup\$ @SpaceDog No problem. And I agree with your points. Operations should be done only when there is a clear reason to do so. There is no obvious reason to change signs where he does, and it doesn’t save any steps - you either do it there, or at the end. But doing it for no obvious reason, in the middle of the formulation, makes it seem like some mathematical trickery is going on. \$\endgroup\$ – Blair Fonville Aug 9 '18 at 16:44
  • \$\begingroup\$ yes and makes you do not understand or believe on what he is explaining. Thanks again. \$\endgroup\$ – SpaceDog Aug 9 '18 at 17:07
1
\$\begingroup\$

Put simply, \$v - v_s\$ has an easier interpretation than \$v_s - v\$.

\$v - v_s\$ is only a translation.

\$v_s - v\$ is a translation and a mirror/sign inversion for \$v\$.

I personally don't see an issue in executing a step early if you know what you have to end with as an educator. It should not lead to a different answer, and it may make other steps easier to see. That minus will have appeared in there some way later on anyway.

$$\begin{align} \frac{dt}{RC} &= \frac{dv}{v_s-v}\\ &\Downarrow \\ \frac{1}{RC}\int_0^tdt &= \int_{v_0}^v \frac{1}{v_s - v}dv \\ &\Downarrow \\ \frac{t}{RC} &= -\int_{v_0}^v \frac{1}{v_s-v}d(-v) \\ &\Downarrow \\ \frac{t}{RC} &= -\left( \ln|v_s-v| - \ln|v_s-v_0| \right) \\ &\Downarrow \\ \frac{t}{RC} &= -\ln\left| \frac{v_s-v}{v_s-v_0} \right| \\ &\Downarrow \\ -\frac{t}{RC} &= \ln\left| \frac{v-v_s}{v_0-v_s} \right| \end{align}$$

No matter what you try, as long as you have equivalence you'll end with the same equations in one way or another. If you got a different result than the teacher, you've done something wrong.

\$\endgroup\$
  • \$\begingroup\$ Sorry but I an not seeing your point. Care to expand the answer? \$\endgroup\$ – SpaceDog Aug 9 '18 at 12:28
  • 1
    \$\begingroup\$ I'm not sure what kind of convincing you need. I added an example of how you can approach the equations trying to keep the minus sign for later, but you have to do it one way or the other. \$\endgroup\$ – Sven B Aug 9 '18 at 14:05
  • \$\begingroup\$ @SvenB Good answer, but get your math correct: $$\int\frac{1}{x}\space\text{d}x=\ln\left|x\right|+\text{c}$$ \$\endgroup\$ – Jan Aug 9 '18 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.