how I am supposed to change the following role in order to apply the laplace transform.

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$f(t) = t+2t$

I recognize that it needs to be prefer this

$Lf(t-t_0)u(t-t_0) = e^-st_0F(s), F(s) = Lf(t)$


You"re using the formula $$cal Lf(t-t_0),cal U(t-t_0) = e^-t_0sF(s).$$ where, $$cal U(t-t_0)=cases0,& $0 le t lt t_0$ cr 1,& $tge t_0$ .$$

As an example of utilizing the over formula,let"s take into consideration the transform of $ t , cal U(t-1)$. Keep in mind that this duty is not in a type where the formula is directly applicable. However, we can very first write$$t ,,cal U(t-1) = igl((t+1)-1igr),cal U(t-1).$$

Then, we can apply the formula v $f(t)=t+1$, $t_0=1$, and also $$F(s)=cal L t+1= cal L t +cal L 1 =1over s^2+1over s^vphantom2$$ come obtain$$cal Ligl t, cal U(t-1)igr=cal Ligl igl((t+1)-1igr), cal U(t-1) igr=e^-1sF(s)=e^- s Bigl( extstyle1over s^2+1over s^vphantom2 Bigr).$$

The above "trick" have the right to be generalised to create the formula: $$ cal Lf(t ),cal U(t-t_0) = e^-t_0scal Liglf(t+t_0) igr. $$

A couple of points to it is in made:

Note what the above formula says. The function $cal U (t-t_0)$essentially "switches on" the role $f$ at $t_0$, the is $f(t ),cal U(t-t_0)$ is $0$ for $0le tBy definition, we take Laplace transforms of functions that aredefined because that $tge 0$; therefore for any kind of such duty $f$, we have $$ cal L f(t), cal U(t)= cal L f(t) .

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Now, let"s take into consideration your problem, utilizing the formula in ~ the beginning of this short article (of course, you might use the shorcut formula obtained afterwards).

Your function, after expansion, becomes:$$eqalignt,cal U(t)-t,cal U(t-1)+2t,cal U(t-1) -2t,cal U(t-2)&=colormaroont,cal U(t) + colordarkgreent,cal U(t-1) -colordarkblue2t,cal U(t-2)cr $$The Laplace transform is linear; so, we can compute the Laplace transforms of each colored term over ind the invoke:$$ ag1cal Ligl\colormaroont,cal U(t) + colordarkgreent,cal U(t-1) -colordarkblue2t,cal U(t-2)igr=cal Ligl\colormaroont,cal U(t)igr + cal Ligl\colordarkgreent,cal U(t-1)igr -2cal Ligl\colordarkblue t,cal U(t-2)igr$$We have$$ ag2 colormarooncal Ligl t, cal U(t ) igr= cal L(t)= extstyle1over s^2 .$$

$$ ag3colordarkgreencal Ligl t, cal U(t-1)igr=cal Ligl igl((t+1)-1igr), cal U(t-1) igr=e^-1scal L +1=e^- s Bigl( extstyle1over s^2+1over s^vphantom2 Bigr),$$and$$ ag4colordarkbluecal Ligl t, cal U(t-2)igr= cal Ligl igl((t+2)-2igr), cal U(t-2) igr= e^-2s cal L +2= e^-2 s Bigl( extstyle1over s^2+2over s^vphantom2 Bigr),$$

So, substituting the outcomes of $(2)$, $(3)$, and $(4)$ right into $(1)$:$$eqaligncal L t,cal U(t) + t,cal U(t-1) -2t,cal U(t-2)&= extstyle1over s^2 +e^- s Bigl( extstyle1over s^2+1over s^vphantom2 Bigr)-2e^-2 s Bigl( extstyle1over s^2+2over s^vphantom2 Bigr)cr $$ A last remark:

It would certainly be simpler for this difficulty to simply use the definition of the Laplace change to find the transform of your function, as Unreasonable Sin does in his answer. Of course, the an approach used below proves to be the shorter one when dealing with more complicated functions.