# circuit analysis laplace

Discussion in 'General Electronics Chat' started by full, Dec 4, 2014.

1. ### full Thread Starter Member

May 3, 2014
225
2
hello everyone

what is mean the source 4u(-t)? why t in negative?

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Last edited: Dec 4, 2014
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Could it perhaps be the impulse notation?

3. ### full Thread Starter Member

May 3, 2014
225
2
this is solution ,why don't use source 4u(-t) ?

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4. ### MrAl Distinguished Member

Jun 17, 2014
2,555
515
Hello,

Because the 4 u(-t) "goes away" at t=0+.

I interpret u(-t) to be a 'backwards' step that starts at t=negative infinity and ends at t=0.

when a<0, u(a)=0
when a=0, u(a)=1
when a>0, u(a)=1

and after a simple change of variable:
when -a<0, u(-a)=0
when -a=0, u(-a)=1
when -a>0, u(-a)=1

so:
when t>0, u(-t)=0
when t=0, u(-t)=1
when t<0, u(-t)=1

This last one is easy to see because if u(t)=1 for t>0 then u(-t)=1 when t<0 because for example for t=-5 we have u(-(-5))=u(5)=1. So it is a reflection of the step function about the y axis.

That's the way they seem to be interpreting it too, so it just "goes away" just after t=0 and so the response depends on the other things in the circuit alone, and Vo is taken from across the initial condition generator and the capacitor as usual.

Last edited: Dec 4, 2014
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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
This attachment might be of interest. Supports MrAl's interpretation.
The 24/s voltage source can be interpreted as the consequence of the 4u (-t) current source pre-charging the capacitor in conjunction with the two resistors which have an initial total voltage drop of 24V @ t <0.

Last edited: Dec 4, 2014
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6. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
All you have to do is understand what the unit step function is.

u(x) is defined to be 0 for x<0 and 1 for x>0. (In general it is undefined at x=0 but that almost never matters. You can generally pick one or the other or split the difference and call it 0.5 and it will have no effect at all unless it is multiplied by an impulse that fires at the same time.)

So now just apply this definition to 4u(-t)

u(-t) is 0 when -t<0 and 1 when -t>0.

This immediately becomes

u(-t) is 0 when t>0 and 1 when t<0.

Just add t to both sides or multiply both sides by -1 remember to reverse the inequality.

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7. ### MrAl Distinguished Member

Jun 17, 2014
2,555
515
``
Hi,

Thanks for posting that. I was sort of thinking about something like this also, where the source was used to set the initial conditions for the circuit and that would occur before t=0 (so from t=minus infinity all the way to t=0). I didnt want to press this though because it is also common to see initial conditions simply given without regard for where they came from (sort of funny when you think about it like that). For example, "The initial capacitor voltage is 15 volts", without stating how it got that way.
But on the other hand, if we calculate the circuit nodes with the current that started at t=minus infinity and ended at t=0 and we got the required initial conditions in the circuit as stated with the problem (24v it think it was) then i think we should assume that the author was using that to set the initial conditions. Interesting in any case though.