circuit analysis laplace

MrAl

Joined Jun 17, 2014
11,494
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.

Follow this...

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.
 
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t_n_k

Joined Mar 6, 2009
5,455
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.
fig10-4.jpeg
 
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WBahn

Joined Mar 31, 2012
30,076
hello everyone

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

MrAl

Joined Jun 17, 2014
11,494
``
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.
View attachment 76602
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.
 
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