Need help with Impulse function and unit step function at singularity

• phoenixy
In summary, the conversation discusses the concept of singularity and how to prove that the unit function is not defined at 0. It also mentions the derivative of the impulse function and the code for integrating from negative infinity to positive infinity. Finally, it addresses confusion about the derivative of a given function and provides a helpful link for further understanding.
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

Of course.Click on this

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

and use {} wherever necessary.

Daniel.

thanks dex,

still need help with the signularity functions ...

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

phoenixy said:
thanks dex,

still need help with the signularity functions ...

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

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

tanujkush said:
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

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

1.

What is the impulse function and how is it different from the unit step function?

The impulse function, also known as the Dirac delta function, is a mathematical function that represents an instantaneous, infinitely large spike at a specific point. It has a value of zero everywhere except at the point of singularity, where it has a value of infinity. The unit step function, on the other hand, is a function that has a value of 1 for all positive inputs and 0 for all negative inputs. It represents a sudden change from 0 to 1 at the point of singularity.

2.

How are the impulse and unit step functions used in practical applications?

The impulse and unit step functions are commonly used in physics and engineering to model sudden impulses or changes in a system. For example, the impulse function can be used to model the impact of a hammer hitting a nail, while the unit step function can be used to model the sudden turn-on of a switch in an electrical circuit.

3.

Can the impulse function and unit step function be integrated or differentiated?

Yes, both functions can be integrated and differentiated. However, since they are both discontinuous at the point of singularity, special techniques such as the sifting property and Heaviside's step function are used to handle their integration and differentiation.

4.

Are there any real-life examples of the impulse and unit step functions?

Yes, there are many real-life examples of these functions. For instance, the sound of a single clap or gunshot can be modeled using the impulse function, while the turning on of a light switch can be modeled using the unit step function. In physics, the force of an object hitting another object can also be represented by the impulse function.

5.

Can the impulse and unit step functions be used to solve differential equations?

Yes, these functions are commonly used to solve differential equations in physics and engineering. They can be used to model sudden changes or impulses in a system, which can then be used to solve for the system's response over time.

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