Help with Heaviside unit functions

[phillyj]

Hi, I'm not really understanding how to convert the piecewise heaviside into a single function. This was on our quiz and the professor didn't really explain how he got the answer.

I can't get the tex to work so:
f(t)=
{ 0, 0 <= t < 1
{ (t-1) 1 <= 2
{ 1, t >= 2

The answer is (t-1)u₁(t)-(t-2)u₂(t)

I understand the first part of the statement but I have trouble with the second part.

I tried to work it out and i got (t-1)u₁(t) + u₂(t) - (1-t-1)u₂(t) but its not right

 

[yus310]

If t<0, H=0, t>0, H=1... that is all about you need to know, keep in mind the limits of integration in respect to t, and you can easily apply this rule to solving laplace or Fourier related integrals. They are used to model discontinous behavior...

best

 

[LCKurtz]

Hi, I'm not really understanding how to convert the piecewise heaviside into a single function. This was on our quiz and the professor didn't really explain how he got the answer.

I can't get the tex to work so:
f(t)=
{ 0, 0 <= t < 1
{ (t-1) 1 <= 2
{ 1, t >= 2

The answer is (t-1)u₁(t)-(t-2)u₂(t)

I understand the first part of the statement but I have trouble with the second part.

I tried to work it out and i got (t-1)u₁(t) + u₂(t) - (1-t-1)u₂(t) but its not right

An easy way to do these types of problems is like this. Say you want f(t) on (0,a), g(t) on (a,b), h(t) on (b,c) and 0 elsewhere.

Start with u(t)f(t) giving f(t) for t > 0
At t = a you want to get rid of f and get g so add this: u(t-a)(g(t)-f(t))
At t = b you want to lose the g and get h so add u(t-b)(h(t)-g(t))
At t = c you want to lose the h so add -u(t-c)h(t).

The result:

u(t)f(t) + u(t-a)(g(t)-f(t)) + u(t-b)(h(t)-g(t)) -u(t-c)h(t).

Depending on what you plan to do next you might want it like that or group it on the functions:

f(t)(u(t)-u(t-a)) + g(t)(u(t-a)-u(t-b)) + h(t)(u(t-b)-u(t-c))

Expressions like u(t-a)-u(t-b) are sometimes called filter functions because they pick up your function on (a,b) and zero elsewhere. You can use that to write the function in this last form directly from your piecewise definition.