Second shifting theorem

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?

LCKurtz

Homework Helper
Gold Member
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?

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?

LCKurtz

Homework Helper
Gold Member
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?
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.

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