# Second shifting theorem

1. Aug 23, 2011

### Rubik

In my text book it says that if we have the function 5sin(t) F(s) = 5/(s2 + 1) Which I understand however, it then says that the shifted function is 5sin(t - 2)u(t - 2).. and I am failing to see how they do that I can not see where the 2 is coming from? This second shifting function is really confusing me so any help would be greatly appreciated?

2. Aug 24, 2011

### LCKurtz

Suppose you wish to take the LT of f(t-a)u(t-a). Since u(t-a) is 1 only when t > a you get:

$$\mathcal{L}(f(t-a)u(t-a)=\int_0^{\infty}e^{-st}f(t-a)u(t-a)\, dt =\int_a^{\infty}e^{-st}f(t-a)u(t-a)\, dt$$

Let x = t - a giving

$$\int_0^{\infty}e^{-s(x+a)}f(x)u(x)\, dx = e^{-as}\int_0^{\infty}e^{-sx}f(x)u(x)\, dx =e^{-as}\int_0^{\infty}e^{-st}f(t)u(t)\, dt=e^{-as}\mathcal{L}f(t)$$

(x is a dummy variable and can be changed to t for clarity).

This says that if F(s) is the transform of f(t) then e-asF(s) is the transform of f(t-a)u(t-a). Or to say it in another way, if you want to find the inverse transform of e-asF(s), you can find the inverse of F(s) by itself, call it f(t), and then f(t-a)u(t-a) will be the inverse of e-asF(s). Does that help?

3. Aug 24, 2011

### Rubik

Sort of helps.. however my problem is when I apply the step function to get the equation into the form f(t) I struggle with what to do next.. I understand the theorem (I think) it is just getting it into a form so I can apply the theorem

For example I am working on this f(t) = t2 - t2((u)(t - 4) + t(u)(t - 4)

Now someone told me that I need to think what g(t1) needs to be so that g(t - a) = t2 but I am not so sure I get it.. so g(t - 4) = t2 but I have no idea what that is telling me or what I am looking for?

4. Aug 24, 2011

### LCKurtz

You don't have to put your problem in the for f(t-a)u(t-a) to get its transform. More typically you want to find the transform of something like f(t)u(t-a). This result is sometimes given as a corollary to the shifting theorem:

$$\mathcal{L}(f(t)u(t-a)=\int_0^{\infty}e^{-st}f(t)u(t-a)\, dt =\int_a^{\infty}e^{-st}f(t)u(t-a)\, dt$$
Now let x = t-a as before:
$$\int_0^{\infty}e^{-s(x+a)}f(x+a)u(x)\, dx =e^{-as}\int_0^{\infty}e^{-st}f(t+a)\, dt =e^{-as}\mathcal L(f(t+a))$$

So to take the transform of f(t)u(t-a), instead take the transform of f(t+a) and multiply the result by e-as.