636 LAPLACE TRANSFORMS- Use the Laplace transform of the second
derivative to derive the transforms:
(a)L{sinhat}=a
s^2 −a^2(b)L{coshat}=s
s^2 −a^265.4 The initial and final value
theorems
There are several Laplace transform theorems used
to simplify and interpret the solution of certain
problems. Two such theorems are the initial value
theorem and the final value theorem.
(a) The initial value theorem states:limit
t→ 0[f(t)]=limit
s→∞[sL{f(t)}]For example, iff(t)=3e^4 tthenL{3e^4 t}=3
s− 4from (iii) of Table 64.1, page 628.By the initial value theorem,limit
t→ 0[3e^4 t]=limit
s→∞[
s(
3
s− 4)]i.e. 3e^0 =∞(
3
∞− 4)i.e. 3 = 3 , which illustrates the theorem.Problem 7. Verify the initial value theorem
for the voltage function (5+2 cos 3t) volts, and
state its initial value.Let f(t)= 5 +2 cos 3tL{f(t)}=L{ 5 +2 cos 3t}=5
s+2 s
s^2 + 9from (ii) and (v) of Table 64.1, page 628.By the initial value theorem,limit
t→ 0[f(t)]=limit
s→∞[sL{f(t)}]i.e. limit
t→ 0[5+2 cos 3t]=limit
s→∞[
s(
5
s+2 s
s^2 + 9)]=limit
s→∞[
5 +2 s^2
s^2 + 9]i.e. 5 +2(1)= 5 +2 ∞^2
∞^2 + 9= 5 + 2i.e. 7 = 7 , which verifies the theorem in this case.The initial value of the voltage is thus7V.Problem 8. Verify the initial value theorem for
the function (2t−3)^2 and state its initial value.Let f(t)=(2t−3)^2 = 4 t^2 − 12 t+ 9Let L{f(t)}=L(4t^2 − 12 t+9)= 4(
2
s^3)
−12
s^2+9
s
from (vii), (vi) and (ii) of Table 64.1, page 628.By the initial value theorem,limit
t→ 0[(2t−3)^2 ]=limit
s→∞[
s(
8
s^3−12
s^2+9
s)]=limit
s→∞[
8
s^2−12
s+ 9]i.e. (0−3)^2 =8
∞^2−12
∞+ 9i.e. 9 = 9 , which verifies the theorem in this case.
The initial value of the given function is thus 9.(b) The final value theorem states:limit
t→∞[f(t)]=limit
s→ 0[sL{f(t)}]For example, iff(t)=3e−^4 tthen:limit
t→∞[3e−^4 t] =limit
s→ 0[
s(
3
s+ 4)]i.e. 3e−∞=(0)(
3
0 + 4)i.e. 0 = 0 , which illustrates the theorem.