# Need help with Impulse function and unit step function at singularity

1. May 14, 2005

### phoenixy

I have some trouble wrapping my head around singularity

One of assignment question is to show that the unit function is not defined at 0. To do that, I need to show

$$\lim_{\Delta\to0}[u_{\Delta}(t)\delta(t)]=0$$
$$\lim_{\Delta\to0}[u_{\Delta}(t)\delta_{\Delta}(t)]=\frac{1}{2}\delta(t)$$

Also, I need to show that the following is identical to u(t)
$$g(t)=\int u(t)\delta(t-\tau)d\tau$$
integrating from negative infinity to positive infinity

One more question, what's the derivative of the impulse function?

PS: what's the tex code for integration from infinity to infinity? I tried \int_-\infty^+\infty, but the tex output is messed up

2. May 14, 2005

### dextercioby

Of course.Click on this

$$\int_{-\infty}^{+\infty}$$

and use {} wherever necessary.

Daniel.

3. May 16, 2005

### phoenixy

thanks dex,

still need help with the signularity functions ...

4. Sep 27, 2009

### circuits83

Hello very body
i got confused understanding how we can get the derivative of V(t) equal to Hv(t) :

V= (1/R) e^(-t/RC) U(t) its derivative V' = Hv(t) = -(1/R²C)e^(-T/RC) U(t) +(1/R)&(t)
where : U(t) is the unite step fonction and &(t) is the delta fonction

5. Sep 29, 2009

### tanujkush

Second part:

The Kronecker delta is non zero only when the parameter in its brackets is equal to zero. So for the integral, its only when tao is equal to t that delta(t) is non zero. Set tau=t, and you have the result.

The derivative of the step function is the Kronecker delta function. check this out
http://mathworld.wolfram.com/HeavisideStepFunction.html

6. Sep 29, 2009

### trambolin

It is Dirac Delta function that is mentioned here. Kronecker Delta is different.

7. Sep 30, 2009

### tanujkush

Oh sorry, I meant the Dirac delta function, not the Kronecker, my bad.