Integral of a function multiplied by a Heaviside step function

[fred2028]

This is me questioning a concept, so I did not follow the template.

I know that
\[\int {u(t-a)dt = tu(t-a)} \]
where u(t) is the unit step function. Does this mean that any arbitrary function multiplied by u(t) is just the integral of that function multiplied by u(t), and I don't have to do integration by parts? For example,
\[\int {u(t-a)\cos tdt = u(t-a)\int {\cos tdt} } \]
and not
\[\int {u(t-a)\cos tdt = u(t-a)\sin t - \int {tu(t-a)\cos tdt} } \]
?

And my 2nd question is how in the world do you solve
\[\int {u( - t + a)dt} \]
and how would you graph this?

Nextly, I assume that

\[f(t)u(t - a)u(t - b) = f(t),t \geqslant a\]

if a > b. Basically, the function only "turns on" when both unit steps are 1. Is this true? If so, would the following equal 0 since the 2 unit step functions do not overlap?

\[f(t)u(t - 1)u(t + 1)\]

 

[Pengwuino]

Yes. Think of the step function as just some number that turns off at a certain point. Let's say you have

\int^3 _{-2} \alpha t^2 dt

where
\alpha = 3 for 2 < t
\alpha = 0 otherwise.

When you integrate from -2 to 3, \alpha is 0 up until t = 2 and then it switches on to be 3. So your integral looks like

\int^2 _{-2} 0t^2 + \int ^3_2 3t^2dt.

The step function is basically the same idea except \alpha = 1 and you give it an argument u(f(t)) where the step function turns on when f(t) is a positive value. The simplest case being u(t) which means the step function is positive when t > 0. Do a quick google search for the step function and you'll see exactly how it works.

 

[fred2028]

Yes. Think of the step function as just some number that turns off at a certain point. Let's say you have

\int^3 _-2 \alpha t^2 dt

where
\alpha = 3 for 2 < t
\alpha = 0 otherwise.

When you integrate from -2 to 3, \alpha is 0 up until t = 2 and then it switches on to be 3. So your integral looks like

\int^2 _-2 0t^2 + \int ^3_2 3t^2dt.

The step function is basically the same idea except alpha = 1 and you give it an argument u(f(t)) where the step function turns on when f(t) is a positive value. The simplest case being u(t) which means the step function is positive when t > 0. Do a quick google search for the step function and you'll see exactly how it works.

Oh OK thanks, that makes my life a lot easier in many situations. Could you shed some light on my last 2 questions? Thanks!

 

[ZioX]

<br /> \[\int {u(t-a)\cos tdt = u(t-a)\int {\cos tdt} } \]<br />

This makes no sense though.

 

[fred2028]

<br /> \[\int {u(t-a)\cos tdt = u(t-a)\int {\cos tdt} } \]<br />

This makes no sense though.

This just means that when integrating a function times a Heaviside the Heaviside can be pulled out.

 

[ZioX]

Okay so what is
<br /> \[u(t-a)\int {\cos tdt} } \]<br />
when t=2?

 

[fred2028]

Okay so what is
<br /> \[u(t-a)\int {\cos tdt} } \]<br />
when t=2?

that depends on the value of a. This isn't the part I am confused about, it's been explained to me already. I need help with my last 2 questions.

 

[Pengwuino]

I had to edit my posts a few times since my latex is not perfect on my first go around.

For the second question, remember, the step function is basically a constant so you can pull it out of the integration if you're doing an indefinite integral. You integrate your function (which is just dt) and the result is multiplied by the step function. What it effectively does is it has your result "turn on" when the argument of the step function is greater than 0 or in other words, -t + 1 > 0.

Finally, you have your assumption of how the step function works right for the third question. However, for your final line, you got the idea but missed the mark. The first step function turns on when t > 1. The second turns on when t>-1. The range those two turn on certainly overlap! When t > 1, both conditions are satisfied. When t = 3, then t > 1 and t > -1. When t = 2453234, t is still greater than 1 and -1.

 

[ZioX]

You're missing the point.

You cannot pull a function that depends on the dummy variable outside of the integral.

 

[fred2028]

I had to edit my posts a few times since my latex is not perfect on my first go around.

For the second question, remember, the step function is basically a constant so you can pull it out of the integration if you're doing an indefinite integral. You integrate your function (which is just dt) and the result is multiplied by the step function. What it effectively does is it has your result "turn on" when the argument of the step function is greater than 0 or in other words, -t + 1 > 0.

Finally, you have your assumption of how the step function works right for the third question. However, for your final line, you got the idea but missed the mark. The first step function turns on when t > 1. The second turns on when t>-1. The range those two turn on certainly overlap! When t > 1, both conditions are satisfied. When t = 3, then t > 1 and t > -1. When t = 2453234, t is still greater than 1 and -1.

Oh thanks! My mistake in LaTeX for the 3rd question. I meant

\[f(t)u( - t + 1)u(t - 1)\]

so it only turns on when t < -1 and t > 1 at the same time, which is never. Then the graph would just be a constant 0 right?

 

[fred2028]

You're missing the point.

You cannot pull a function that depends on the dummy variable outside of the integral.

But the Heaviside function merely states that it equals a constant (1) when greater than t = a and another constant (0) when less, so can't you treat it like a constant?

 

[Pengwuino]

You're missing the point.

You cannot pull a function that depends on the dummy variable outside of the integral.

This is true but not really getting at the heart of the issue. As far as the integration is concerned, it's a constant and you simply use the dummy variable under the integration and it goes to the actual variable post-integration.

Oh thanks! My mistake in LaTeX for the 3rd question. I meant
f(t)u( - t + 1)u(t - 1)

so it only turns on when t < -1 and t > 1 at the same time, which is never. Then the graph would just be a constant 0 right?

Yes.