[SOLVED] impulse input to an RC integrator??

[rsashwinkumar]

Consider an ideal impulse is given to a RC integrator. Now the output will be its impulse response which will exponentially decay to 0V (which can be worked out on paper)

Now my question is, since an impulse is nothing but a sudden rise in voltage just at the instant t=0 (ideally), and since a capacitor does not allow sudden changes in its voltage (assuming that initial capacitor voltage to be 0V), shouldnt the voltage across capacitor remain at 0V?
Also to charge to a given voltage a capacitor takes 'RC'sec. Since impulse exists at t=0 alone (ideal impulse), the capacitor should not charge to the applied voltage.

But the impulse response clearly shows that at t=0, the voltage is 1V (assuming unit impulse). Where have I gone wrong in the above argument?
please help me out...

 

[barry]

I think you're getting tangled up in math. An 'ideal impulse' has zero width and infinite amplitude, so you're dealing with limits and all that scary calculus stuff. A unit impulse is also defined by its integral from t=-infinty to t=+infinity, which is equal to 1.

That's all I know. Hope it helps.

 

[SherpaDoug]

It has been too many decades since college. Please define an "ideal impulse". Does it have amplitude of 1, or an area of 1? In either case the impulse voltage across the resistor will create a current impulse, which the capacitor will collect as charge.

 

[barry]

The 'ideal impulse' has an AREA of 1 and amplitude of infinity. I'm not sure why, but I can't get my signal generator to create this.

 

[Prototyp_V1.0]

The 'ideal impulse' has an AREA of 1 and amplitude of infinity. I'm not sure why, but I can't get my signal generator to create this.

And you'll never get a generator that can make an ideal pulse.

It's the same as to ask for an ideal current generator or an ideal voltage generator. Those thing just doesn't exists.

Or to ask for a plane with no mass to reduce fuel cost. Sounds good in theory but you won't get it in your hands - for a while at least :razz:

 

[rsashwinkumar]

lets talk theoretically, suppose i give an infinite voltage just at the instant t=0, will the capacitor charge instantaneously? or will it charge through the res., R with time constant T=RC ??

 

[zorro]

Yes, it would charge instantaneously.
Start thinking in a rectangular pulse with duration "T" and amplitude 1/T. Plot how the C charges (towards the asymptot 1/T) and discharges towards the asymptot 0) with exponential shapes.
Now make T more and more short. The amplitude will be more and more high. C will charge with more and more slope towards the asymptot 1/T.
In the limit, the charge would be instantaneous.
regards

Z

 

[rsashwinkumar]

@zorro : i get ur point, but wont the RC time constant determine the charging of the capacitor??

 

[barry]

And you'll never get a generator that can make an ideal pulse.

It's the same as to ask for an ideal current generator or an ideal voltage generator. Those thing just doesn't exists.

Or to ask for a plane with no mass to reduce fuel cost. Sounds good in theory but you won't get it in your hands - for a while at least :razz:

That was a joke, dude.

 

[SherpaDoug]

The RC time constant will shape the voltage rise from zero to infinity. After 1 RC period you will only be about 70% of the way to infinity...which is infinity. It is a distinction without a difference.

 

[zorro]

@zorro : i get ur point, but wont the RC time constant determine the charging of the capacitor??

Yes.
The charge during the pulse, when the pulse has amplitude A, has a shape v(t)=A(1-exp(-t/RC)).
For t<<RC (when the pulse is very narrow) holds the approximation v(t)~=At/RC [take the 2 first terms in the McLaurin expansion of the exp() ].
So the slope is A/RC. The voltage raises linearly for t<<RC.
At the end of the pulse, when t=T, v(t) reaches the value v(T)=AT/RC.
But as A=1/T (the pulse has unit area), then the voltage on C at the end of the pulse is 1/RC.
Now take T->0+. The "end" of the pulse is reached instantaneously.

Regards

Z

 

[rsashwinkumar]

Now i get it!! thanks a lot