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The function F (s) is the Laplace transform of f (t) and is a function of the complex frequency, s = + j , considered earlier in this section Note that the function f (t) is de ned only for t 0 This de nition of the Laplace transform applies to what is known as the one-sided or unilateral Laplace transform, since f (t) is evaluated only for positive t In order to conveniently express arbitrary functions only for positive time, we introduce a special function called the unit step function, u(t), de ned by the expression u(t) = 0 1 t <0 t >0 (652)
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EXAMPLE 68 Computing a Laplace Transform
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Find the Laplace transform of f (t) = e at u(t)
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Known Quantities: Function to be Laplace-transformed Find: F (s) = L{f (t)} Schematics, Diagrams, Circuits, and Given Data: f (t) = e at u(t) Assumptions: None Analysis: From equation 651:
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F (s) =
e at e st dt =
e (s+a)t dt =
1 (s+a)t e s+a
1 s+a
Comments: Table 61 contains a list of common Laplace transform pairs
EXAMPLE 69 Computing a Laplace Transform
Problem
Find the Laplace transform of f (t) = cos( t)u(t)
Solution
Known Quantities: Function to be Laplace-transformed Find: F (s) = L{f (t)} Schematics, Diagrams, Circuits, and Given Data: f (t) = cos( t)u(t) Assumptions: None Analysis: Using equation 651 and applying Euler s identity to cos( t) gives:
F (s) =
1 1 j t e + e j t e st dt = 2 2
e( s+j )t + e( s j )t dt
1 e (s+j )t s + j
1 e (s j )t s j
1 1 s + = 2 s + j s j s + 2
Comments: Table 61 contains a list of common Laplace transform pairs
Part I
Circuits
Check Your Understanding
619 Find the Laplace transform of the following functions:
a u(t) b sin( t)u(t) c tu(t)
Table 61 Laplace transform pairs f(t) (t) (unit impulse) u(t) (unit step) e at u(t) sin t u(t) 1 1 s 1 s+a s 2 + 2 s s 2 + 2 (s + a)2 + 2 s+a (s + a)2 + 2 1 s2 F(s)
620 Find the Laplace transform of the following functions:
a e sin t u(t) b e at cos t u(t)
From what has been said so far about the Laplace transform, it is obvious that we may compile a lengthy table of functions and their Laplace transforms by repeated application of equation 651 for various functions of time, f (t) Then, we could obtain a wide variety of inverse transforms by matching entries in the table Table 61 lists some of the more common Laplace transform pairs The computation of the inverse Laplace transform is in general rather complex if one wishes to consider arbitrary functions of s In many practical cases, however, it is possible to use combinations of known transform pairs to obtain the desired result
cos t u(t) e at sin t u(t) e at cos t u(t) tu(t)
EXAMPLE 610 Computing an Inverse Laplace Transform
Problem
Find the inverse Laplace transform of F (s) = 4 4 2 + + s + 3 s2 + 4 s
Solution
Known Quantities: Function to be inverse Laplace transformed Find: f (t) = L 1 {F (s)} Schematics, Diagrams, Circuits, and Given Data:
F (s) =
4 4 2 + + = F1 (s) + F2 (s) + F3 (s) s + 3 s2 + 4 s
Assumptions: None Analysis: Using Table 61, we can individually inverse-transform each of the elements of
F (s): f1 (t) = 2L 1 f2 (t) = 2L 1 f3 (t) = 4L 1 1 s+3 2 s 2 + 22 1 s = 2e 3t u(t) = 2 sin(2t)u(t)
= 4u(t)
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Thus f (t) = f1 (t) + f2 (t) + f3 (t) = 2e 3t + 2 sin(2t) + 4 u(t)
EXAMPLE 611 Computing an Inverse Laplace Transform
Problem
Find the inverse Laplace transform of F (s) = 2s + 5 s 2 + 5s + 6
Solution
Known Quantities: Function to be inverse Laplace transformed Find: f (t) = L 1 {F (s)} Assumptions: None Analysis: A direct entry for the function cannot be found in Table 61 In such cases, one must compute a partial fraction expansion of the function F (s), and then individually transform each term in the expansion A partial fraction expansion is the inverse operation of obtaining a common denominator, and is illustrated below
F (s) =
2s + 5 A B = + s 2 + 5s + 6 s+2 s+3
To obtain the constants A and B, we multiply the above expression by each of the denominator terms: (s + 2)B (s + 2)F (s) = A + s+3 (s + 3)F (s) = (s + 3)A +B s+2
From the above two expressions, we can compute A and B as follows: A = (s + 2)F (s)|s= 2 = 2s + 5 s+3
s= 2
B = (s + 3)F (s)|s= 3 = Finally, F (s) =
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