Hello,
how to I integrate something like this?: (I do understand how to integrate something just of , or some other variable)
Can someone give some guidlines? Maybe I will post more questions later, if I will not completly understand.
Thank you.
Hello,
how to I integrate something like this?: (I do understand how to integrate something just of , or some other variable)
Can someone give some guidlines? Maybe I will post more questions later, if I will not completly understand.
Thank you.
I replaced $\omega$ with $a$ and $\tau$ with $x$:
Wolfram|Alpha: Computational Knowledge Engine
Yes, i know what is the answer, I know how to use wolfram alpha.
I just want to learn this, to do with my hands.
Let's go back to the point:
If I integrate just the cosine function (without the unit step), the integral have to be larger then the integral of cosine function multiplied with unit step or delayed unit step. Why do I think that? Because we cut the left part, the x < 0 is zero. So the integral on the left part zero. Unless we consider the unit step as just one. So:
$\displaystyle 3\int_{-\infty}^\infty \cos (\omega \tau ) 1(\tau ) 1 (t-\tau ) d\tau = \begin{cases}\displaystyle 3\int_0^t \cos (\omega \tau ) 1(\tau ) 1(t-\tau ) d\tau & t\ge 0 \\ 0 & t < 0\end{cases}$
For $t\ge 0$, this is:
$\dfrac{3}{\omega} \left. \sin ( \omega \tau ) \right]_0^t = \dfrac{3}{\omega} \sin ( \omega t)$
But, the condition on $t$ is given by multiplying by $1(t)$.
Why do we integrate then between 0 and t?
Yes because of the condition.
On the start we have minus infinity to infinity.
How do we know that? How do we know that when we multiply with unit step function we must integrate between 0 and t?
If $t < 0$ then for all $\tau < 0$, we have $1(\tau ) = 0$ and for all $\tau \ge 0$ we have $1(t-\tau ) = 0$. For $t \ge 0$ we have:
For $\tau \in (-\infty, 0]$, $1(\tau ) = 0$. On $\tau \in (t,\infty)$, $1(t-\tau ) = 0$. So, $1(\tau )1(t-\tau ) = 1$ only on the interval $0 \le \tau \le t$. This gives us our bounds of integration.
For multiplying by the unit step function, we found that for $t < 0$, the integral is zero. For $t \ge 0$, the integral is times 1. That gives us the multiplication by $1(t)$.
But how? Let's say .
The graph is: https://www.wolframalpha.com/input/?i=H(5)*H(t-5).
So it's one from 5 to . So here we must integrate between 5 to .
Maybe I am missing something.
$t$ is a constant. $\tau$ is a variable. So, at $t=5$, it is 1 on $\tau \in (0,5)$.
https://www.wolframalpha.com/input/?i=H(x)*H(5-x)
Thanks Slipeternal, now I fully understand the unit step part.
But I have another question: If we had some integral like this:
How can I here expose the as a constant? It's inside the cosine function.
$u(\tau) = t-\tau$
$u(0) = t-0 = t$
$u(t) = t-t = 0$
$du = -d\tau$
$\displaystyle \int_0^t 3\cos(t-\tau) d\tau = \int_{u(0)}^{u(t)} 3\cos u (-du) = -3 \int_t^0 \cos u \, du = 3\int_0^t \cos u \, du = 3\sin t$
No, I did not miss a minus sign. When you swap the bounds of integration, it gives an extra negative sign.