Derivatives of singularity functions

In summary, the derivatives of the given functions are: 1) $$\delta(t-1)u(t+1)+u(t-1)\delta(t+1)$$2) $$r(t-6)\delta(t-2)+u(t-6)u(t-2)$$3) $$sin(4t)\delta(t-3)+cos(4t)u(t-3)$$Before attempting to differentiate, it is important to analyze and simplify the given functions. In this case, it is helpful to consider the graph of each function and determine when the unit step function will equal 0 or 1. This allows for rewriting the functions in a simpler form and avoiding incorrect operations on distributions.
  • #1
iRaid
559
8

Homework Statement


Hello, I'm having trouble understanding this, seemingly simple, concept. Any help or input is appreciated.

Evaluate the following derivatives:
$$\frac{d}{dt} u(t-1)u(t+1)$$
$$\frac{d}{dt} r(t-6)u(t-2)$$
$$\frac{d}{dt} sin(4t)u(t-3)$$

Homework Equations


The Attempt at a Solution


1) $$\delta(t-1)u(t+1)+u(t-1)\delta(t+1)$$
2) $$r(t-6)\delta(t-2)+u(t-6)u(t-2)$$
3) $$sin(4t)\delta(t-3)+cos(4t)u(t-3)$$

Now, I don't really understand what to do from here...
 
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  • #2
iRaid said:

Homework Statement


Hello, I'm having trouble understanding this, seemingly simple, concept. Any help or input is appreciated.

Evaluate the following derivatives:
$$\frac{d}{dt} u(t-1)u(t+1)$$
$$\frac{d}{dt} r(t-6)u(t-2)$$
$$\frac{d}{dt} sin(4t)u(t-3)$$


Homework Equations





The Attempt at a Solution


1) $$\delta(t-1)u(t+1)+u(t-1)\delta(t+1)$$
2) $$r(t-6)\delta(t-2)+u(t-6)u(t-2)$$
3) $$sin(4t)\delta(t-3)+cos(4t)u(t-3)$$

Now, I don't really understand what to do from here...

Before trying to differentiate the function ##f(t) = u(t-1)u(t+1)## you should first see if you can re-write ##f(t)## in a simpler form. For example, what does the graph of ##f(t)## look like?

Do the same type of preliminary analysis on all the other functions you are given.
 
  • #3
Ray Vickson said:
Before trying to differentiate the function ##f(t) = u(t-1)u(t+1)## you should first see if you can re-write ##f(t)## in a simpler form. For example, what does the graph of ##f(t)## look like?

Do the same type of preliminary analysis on all the other functions you are given.

Sorry, I'm honestly not sure what 2 singularity functions multiplied together would look like. I know that they're both step functions starting at t=-1 and stepping up at t=1.
 
  • #4
iRaid said:
Sorry, I'm honestly not sure what 2 singularity functions multiplied together would look like. I know that they're both step functions starting at t=-1 and stepping up at t=1.

The unit step function is defined on the whole real line, so if it "starts" anywhere, it starts at -∞. For what values of ##w## is ##u(w)## equal to zero? Equal to 1? So, for what values of ##t## is ##u(t-1)## equal to zero, and to 1? Same question for ##u(t+1)##. So, when is the product ##u(t-1)u(t+1)## equal to zero? When does it equal 1?
 
  • #5
One must be very careful multiplying distributions, and the unit-step function is in fact a distribution in this context. As Ray has already stated, you should first rewrite the product which has apparently a meaning when interpreted as usual functions rather than distributions in terms of a well-defined distribution. Note that there are no problem to use linear combinations of distributions. You only must not multiply two or more distributions!
 
  • #6
iRaid said:
Sorry, I'm honestly not sure what 2 singularity functions multiplied together would look like. I know that they're both step functions starting at t=-1 and stepping up at t=1.
Surely you can multiply 1s and 0s! If t< -1 then t-1< -2 and t+ 1< 0 so both u(t-1) and u(t+ 1) are 0. u(t-1)u(t+1)= 0. If -1< t< 1 then t-1< 0 but t+1> 0 so u(t-1)= 0 and u(t+ 1)= 1. u(t-1)u(t+1)= 0. if t> 1 then t-1> 0 and t+ 1> 2> 0 so both u(t-1) and u(t+ 1) are 0. u(t-1)u(t+1)= 1.

u(t-1)u(t+1)= 0 for t< 1, 1 for [itex]t\ge 1[/itex]. Equivalently, u(t-1)u(t+1)= u(t-1). With a little thought you should see that if a> b then u(t-a)u(t-b)= u(t- b).
 
  • #7
Again, that's a very subtle point! Of course, you can multiply 0's and 1's, but not unitstep distributions. You can multiply them as functions, but you cannot do operations of such products that only have meaning in the sense of distributions. This exercise is a very nice example. If you formally do so you get sometimes (not always) nonsensical results.

If in doubt, you have to check your distribution valued results with test functions!
 

What are singularity functions?

Singularity functions are mathematical functions that are defined differently at different points, often resulting in a discontinuity at a specific point. They are commonly used to model physical phenomena such as a sudden change in force or displacement.

What are derivatives of singularity functions?

Derivatives of singularity functions are the mathematical operations that determine the rate of change of a singularity function. They are useful in understanding the behavior of a function at a specific point and can represent physical quantities such as velocity or acceleration.

How are derivatives of singularity functions calculated?

The derivatives of singularity functions can be calculated using the standard rules of differentiation. However, since these functions are often discontinuous, special techniques such as the Heaviside step function or Dirac delta function may be required to accurately calculate the derivative at a specific point.

What are some real-world applications of derivatives of singularity functions?

Derivatives of singularity functions have various applications in physics and engineering, such as modeling the behavior of a mass-spring system or analyzing the stress and strain in a material undergoing sudden changes. They are also used in signal processing, control systems, and robotics.

Are there any limitations or challenges in using derivatives of singularity functions?

One limitation of using derivatives of singularity functions is that they can be mathematically challenging to work with due to their discontinuous nature. Additionally, in practical applications, it may be difficult to accurately measure or simulate the sudden changes represented by these functions. Care must also be taken when interpreting the results, as they may not always align with physical reality.

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