Laplace Transforms with Maple

>	with(inttrans); # Loads a package of commands.

>	u3:= x-> Heaviside(x-3);

>	plot(u3(x),x=0..7,y=-1..3,discont=true,thickness=3);

>	laplace(u3(t),t,s); # FInd the laplace transform. t is the orgonial variable, s is the new new.

>	invlaplace(exp(-3*s)/s,s,t); # FInd the inverse Laplace fransform!

Next example.

>	f:= t -> sin(t)*Heaviside(t) - sin(t)*Heaviside(t-Pi)+ Heaviside(t-4)-Heaviside(t-6);

>	plot(f(t),t=0..7,y=-1..3,discont=true,thickness=3);

>	laplace(f(t),t,s);

Now we consider an LRC circuit. Recall the diffeq for the charge on the capacitor is LQ''+RQ'+(1/C)Q = E(t). Will ask what will Q(t) be if we use f(t) above as the external applied voltage. We assume Q'(0)=Q(0)=0. Here are the values for L, R and C.

>	L:=1;C:=1;R:=10;

Then we solve Lap(LQ''+RQ'+(1/C)Q + Lap(f(t)) for Lap(Q). We get Lap(Q) = Lap(f)/(Ls^2+Rs+1/C). Then Q(t) = Inverse Lap (Lap Q).

>	Q:= t->invlaplace(((1+exp(-s*Pi))/(s^2+1)+(exp(-4*s)-exp(-6*s))/s)/(L*s^2+R*s+1/C),s,t);

>	plot(Q(t),t=0..20,y=0..0.50,color=blue,thickness=3);

>	evalf(Q(10));

Let's see what the formula for Q(t) is.

>	invlaplace(((1+exp(-s*Pi))/(s^2+1)+(exp(-4*s)-exp(-6*s))/s)/(L*s^2+R*s+1/C),s,t);

Yuck!

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