How I am supposed to transcreate the adhering to function in order to apply the laplace transdevelop.

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$f(t) = t+2t$

I know that it has to be prefer this

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


You"re utilizing 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 instance of using the over formula,let"s consider the transdevelop of $ t , cal U(t-1)$. Keep in mind that this feature is not in a type wbelow the formula is straight applicable. However, we can first write$$t ,,cal U(t-1) = igl((t+1)-1igr),cal U(t-1).$$

Then, we deserve to apply the formula via $f(t)=t+1$, $t_0=1$, and $$F(s)=cal L t+1= cal L t +cal L 1 =1over s^2+1over s^vphantom2$$ to 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" can be generalized 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 be made:

Keep in mind what the over formula states. The function $cal U (t-t_0)$fundamentally "switches on" the attribute $f$ at $t_0$, that is $f(t ),cal U(t-t_0)$ is $0$ for $0le tBy definition, we take Laarea transcreates of attributes that areidentified for $tge 0$; thus for any type of such attribute $f$, we have $$ cal L f(t), cal U(t)= cal L f(t) .

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Now, let"s consider your difficulty, making use of the formula at the start of this short article (of course, you can use the shorreduced formula obtained afterwards).

Your feature, after growth, 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 transcreate is linear; so, we can compute the Lalocation transdevelops 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 also $(4)$ 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 be easier for this trouble to just usage the interpretation of the Laarea transcreate to uncover the transform of your attribute, as Unreasonable Sin does in his answer. Of course, the approach supplied right here proves to be the shorter one as soon as taking care of even more complex attributes.