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Theta & Top Hat function question

mc_watson87

Could someone help me understand this question? :smile:

I've sketched the function, but not really sure how to do the rest?

Maths question.JPG14.1KB

Reply 1

DFranklin

We need to know what

\theta(x)

is.

Reply 2

mc_watson87

DFranklin

We need to know what

\theta(x)

is.

No more information is given in the question. I think its the Heaviside step function.

Reply 3

DFranklin

mc_watson87

No more information is given in the question. I think its the Heaviside step function.

Yes, so do I, to be honest, but I do hate having to guess.

Well, if you've drawn the sketch, it should be fairly obvious what to do next, but as a hint, what does

(\theta(x)-\theta(x-1))+2(\theta(x-1)-\theta(x-2))

look like?

Reply 4

mc_watson87

DFranklin

Yes, so do I, to be honest, but I do hate having to guess.

Well, if you've drawn the sketch, it should be fairly obvious what to do next, but as a hint, what does

(\theta(x)-\theta(x-1))+2(\theta(x-1)-\theta(x-2))

look like?


    _
  _| |
_|   |__

?

So f(x) would be:

(\theta(x)-\theta(x-1))+2(\theta(x-1)-\theta(x-2))+3(\theta(x-2)-\theta(x-3))

?

Is it easier to sketch the graph first and then represent using top hat functions? Or visa versa?

How do i go about doing the last bit of the question? :redface:

Thanx

Reply 5

DFranklin

mc_watson87

How do i go about doing the last bit of the question? :redface:

Well, an integral is the area under a curve...

Reply 6

mc_watson87

DFranklin

Well, an integral is the area under a curve...

I still don't really understand what its asking me to work out

Reply 7

AlphaNumeric

\int_{a}^{b} \theta(x-a) dx = b-a

for b<a.

Everything follows from that.

Reply 8

mc_watson87

i am still very confused about the last part. I am confused over what F(x) is, what f(y) and what its asking me to calculate.

Reply 9

AlphaNumeric

f(y) = \theta(y) + \theta(y-1) + \theta(y-2) - 3\theta(y-3)

F(x) = \int_{-\infty}^{x}\Big( \theta(y) + \theta(y-1) + \theta(y-2) - 3\theta(y-3) \Big)

F(x) = \int_{-\infty}^{x} \theta(y) + \int_{-\infty}^{x} \theta(y-1) + \int_{-\infty}^{x} \theta(y-2) - 3 \int_{-\infty}^{x} \theta(y-3)

You want to consider the following sections :

x in [-infty,0)
x in (0,1)
x in (1,2)
x in (2,3)
x in [3,infty)

Can you see why?

An important result is

\int_{-\infty}^{a} \theta(x-b) dx = 0

if a<b. This is because you're just integrating the zero function.

x<0 => F(x) = 0
x in (0,1) =>

F(x) = \int_{-\infty}^{x} \theta(y) + \int_{-\infty}^{x} \theta(y-1) + \int_{-\infty}^{x} \theta(y-2) - 3 \int_{-\infty}^{x} \theta(y-3) = \int_{-\infty}^{x} \theta(y) = x

x in (1,2) =>

F(x) = \int_{-\infty}^{x} \theta(y) + \int_{-\infty}^{x} \theta(y-1) = x + x-1 = 2x-1

x in (2,3) =>

F(x) = \int_{-\infty}^{x} \theta(y) + \int_{-\infty}^{x} \theta(y-1) + \int_{-\infty}^{x} \theta(y-2) - 3 \int_{-\infty}^{x} \theta(y-3) = \int_{-\infty}^{x} \theta(y) = x + x-1 + x-2 = 3x-2

x > 3 =>

F(x) = \int_{-\infty}^{x} \theta(y) + \int_{-\infty}^{x} \theta(y-1) + \int_{-\infty}^{x} \theta(y-2) - 3 \int_{-\infty}^{x} \theta(y-3) = \int_{-\infty}^{x} \theta(y) = x + x-1 + x-2 - 3(x-3) = 3x-3 - 3x+9 = 6

Can you see why the last answer is constant? And why it's equal to 6? If you can't, draw drawing the function and it'll become obvious.

/edit

Give or take a 'dy' in the integrals, before someone points it out.

Reply 10

mc_watson87

Thanx alot dude! Finally got it. There were a lot of gaps in my knowledge cus i was teaching myself it. :smile:

One slight thing tho, what does x refer to on the integral boundaries and how do u know its greater than 3.

Cheers DFranklin aswell! =)

Reply 11

DFranklin

mc_watson87

Thanx alot dude! Finally got it. There were a lot of gaps in my knowledge cus i was teaching myself it. :smile:

One slight thing tho, what does x refer to on the integral boundaries and how do u know its greater than 3.I'm not totally sure what you're trying to say, but I don't think you know it's greater than 3.

You want to find

F(x) = \int_{-\infty}{x} f(x)\,dx

, where f(x) is that staircase function we've been looking at.

You need to find F(x) for all x. But since f(x) looks quite different for the 5 regions x<=0, 0<x<=1, 1<x<=2, 2<x<=3, x>3, it's easiest to split our answer for F(x) into 5 regions as well.

As an example, suppose 2<x<3. f(t) = 0 for t<0, so

\int_{-\infty}^0 f(t) \,dt = 0

. f(t) = 1 for 0<t<1, so

\int_0^1 f(t) \,dt = \int_0^1 1 \,dt = 1

. f(t) = 2 for 1<t<2, so

\int_1^2 f(t) \,dt = \int_1^2 2 \,dt = 2

. Finally f(x) = 3 for 2<t<x, so

\int_2^x f(t)\,dt = \int_2^x 3 \,dt = 3(x-2)

.

Adding it all up, when 2<x<3, we have

F(x) = \int_{-\infty}^x f(t)\,dt = \int_{-\infty}^0 f(t)\,dt + \int_0^1 f(t)\,dt + \int_1^2 f(t)\,dt + \int_2^x f(t)\,dt

= 0 + 1 + 2 + 3(x-2) = 3+3(x-2) = 3+3x-6 = 3x-3

(there's an (obvious) mistake in Alphanumeric's answer of 3x-2).

Of course, when x>3, F'(x) = f(x) = 0, so for x>3 the answer is constant.

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