Definition of the Laplace transform. 0 x(t)e st dt
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1 Definition of the Laplace transform Bilateral Laplace Transform: X(s) = x(t)e st dt Unilateral (or one-sided) Laplace Transform: X(s) = 0 x(t)e st dt
2 ECE352 1 Definition of the Laplace transform (cont.) We denote this relationship as x(t) L u X(s). The lower limit 0 is to include discontinuity and impulse at t = 0. Determine the unilateral LT of x(t) = u(t + 3). X(s) = x(t)e st dt 0 = u(t + 3)e st dt = 0 = 1/s 0 e st dt
3 ECE352 2 Unilateral LT examples Determine the unilateral LT of x(t) = u(t 3). X(s) = = 0 x(t)e st dt 0 u(t 3)e st dt = 3 e st dt = e 3s1 s
4 ECE352 3 Properties of the Unilateral Laplace Transform Linearity Let x(t) L u X(s) and y(t) L u Y (s). ax(t) + by(t) L u ax(s) + by (s).
5 ECE352 4 Properties of the Unilateral Laplace Transform (cont.) Time scaling x(at) L u 1 a X ( s a) for a > 0. 0 x(at)e st dt = x(l)e (s/a)l1 0 a dl = 1 a = 1 ( s ) a X, for a > 0 a 0 x(l)e (s/a)l dl
6 ECE352 5 Properties of the Unilateral Laplace Transform (cont.) Time shift x(t τ) L u e sτ X(s) for τ > 0. 0 x(t τ)e st dt = x(l)e sl e sτ dl 0 = e sτ X(s), for τ > 0
7 ECE352 6 Properties of the Unilateral Laplace Transform (cont.) s-domain shift e s0t x(t) L u X(s s 0 ). 0 e s 0t x(t)e st dt = x(t)e (s s0)t dt 0 = X(s s 0 ) Multiplication by a complex exponential in time introduces a shift in complex frequency s.
8 ECE352 7 Properties of the Unilateral Laplace Transform (cont.) Convolution x(t) y(t) L u X(s)Y (s) when x(t) = 0, y(t) = 0 for t < 0. Differentiation in the s-domain tx(t) L u d ds X(s).
9 ECE352 8 Examples Determine the unilateral LT of x(t) = ( e 3t u(t)) (tu(t)) u(t) L u 1 s. e 3t u(t) L u 1 s 3. tu(t) L u 1. s 2 Thus, ( e 3t u(t)) (tu(t)) L u 1 s 2 (s 3).
10 ECE352 9 Examples (cont.) Example 6.4, p 493. see figure 6.7 on the same page. Find the LT of the system output y(t) for the input x(t) = te 2t u(t). Applying x(t) = u(t) y(t) = (1 e t/(rc) )u(t). If x(t) = δ(t) = d dt u(t) y(t) = h(t) = d dt = 1 RC e t/(rc) u(t). [ ] (1 e t/(rc) )u(t)
11 ECE Examples (cont.) With RC = 0.2, h(t) = 5e 5t u(t) h(t) = 5e 5t u(t) L u 5 s+5 x(t) = te 2t u(t) L u 1. (s 2) 2 Thus, Y (s) = X(s)H(s) = 5 (s 2) 2 (s+5).
12 ECE Properties Differentiation in the time domain Suppose that LT of x(t) exists. Find the unilateral Laplace Transform of dx(t)/dt. d dt x(t) L u 0 ( ) d dt x(t) e st dt d dt x(t) L u x(t)e st 0 + s 0 x(t)e st dt If X(s) exists, then x(t)e st 0 as t. Thus, d dt x(t) L u sx(s) x(0 ).
13 ECE Properties Integration where t x(τ)dτ x ( 1) (0 ) = L u x( 1) (0 ) s 0 x(τ)dτ is the area under x(t) from t = to t = 0. + X(s s
14 ECE Example Problem 6.5, p 495 Determine the unilateral LT of tu(t). Method 1: u(t) L u 1 s tu(t) L u d ds ( ) 1 s tu(t) L u 1 s 2
15 ECE Example Problem 6.5, p 495 (cont.) Method 2: tu(t) = t t u(τ)dτ u(τ)dτ ( L u 1 s U(s) + 0 u(τ)dτ ) = 1 s 2
16 ECE Properties Initial- and final-value theorems Initial value: lim s sx(s) = x(0 + ) The initial-value theorem does not apply to rational functions X(s) in which the order of the numerator polynomial is greater than or equal to that of the denominator polynomial order. Final value: lim s 0 sx(s) = x( ) The final-value theorem applies only if all the poles of X(s) are in the left half of the s-plane, with at most a single pole at s = 0.
17 ECE Example example 6.6, p 495 Determine x(0 + ) and x( ), if the unilateral Laplace Transform of x(t) is given as 7s + 10 X(s) = s(s + 2) x(0 + ) = lim s sx(s) = lim s 7s + 10 s + 2 = 7 x( ) = lim s 0 sx(s) = lim s 0 7s + 10 s + 2 = 5
18 ECE Example Problem 6.6(a), p 496 Determine the initial and final values of x(t) corresponding to the unilateral Laplace Transform ( ) 2 X(s) = e 5s s(s + 2) x(0 + ) = lim s sx(s) = lim s x( ) = lim sx(s) = lim s 0 s 0 2 (s + 2)e 5s = 0 2 = 1 (s + 2)e5s
19 ECE Inversion of the unilateral Laplace Transform Through definition: x(t)e σt = 1 2π x(t) = 1 2π X(σ + jω)e jωt dω X(σ + jω)e (σ+jω)t dω Substituting s = σ + jω and dω = ds/j, we get the inverse Laplace Transform: x(t) = 1 2πj σ+j σ j X(s)e st ds
20 ECE Inversion of the unilateral Laplace Transform (cont.) It requires contour integration. Difficult to do in general. Focus on rational functions Rational function: a ratio of polynomials in s, e.g. X(s) = B(s) A(s) = 3s + 4 (s + 1)(s + 2). Proper function: order of B(s) order of A(s).
21 ECE Inversion of the unilateral Laplace Transform (cont.) Many commonly encountered LTI systems can be described by integro-differential equations. Their transfer function can be expressed as the ratio of two polynomials in s.
22 ECE Inversion of the unilateral Laplace Transform (cont.) Focus: partial fraction expansion applied on rational Laplace Transform. Requires knowledge of several basic transform pairs, e.g., δ(t) L u 1 c k δ k d k (t) = c k dt kδ(t) L u c k s k A k e dkt u(t) L u At n 1 (n 1)! ed kt u(t) L u A k s d k A (s d k ) n
23 ECE Inversion of the unilateral Laplace Transform (cont.) Partial fraction expansion: X(s) = B(s) A(s) = b Ms M + b M 1 s M b 1 s + b 0 s N + a N 1 s N a 1 s + a 0 If M N: X(s) = X(s) = M N k=0 B(s) Ã(s) c k s k + X(s)
24 ECE Inversion of the unilateral Laplace Transform (cont.) A(s) = N (s d k ). Poles: s = d k k=1 If all poles are distinct: X(s) = N k=1 A k s d k If a pole d i is repeated r times, then there are r terms in the partial-fraction expansion associated with that pole: A i1 s d i, A i2 (s d i ) 2,, A ir (s d i ) r
25 ECE Examples Example 6.7, p 497 : find x(t) for: X(s) = 3s + 4 (s + 1)(s + 2) 2 X(s) is a proper rational function. Poles at s = 1 and s = 2 (double pole). Partial fraction expansion of X(s): X(s) = A 1 s A 2 s A 3 (s + 2) 2.
26 ECE Examples (cont.) A 1 = X(s)(s + 1) s= 1 = 1 A 2 = d [ ] X(s)(s + 2) 2 ds s= 2 = 1 A 3 = X(s)(s + 2) 2 s= 2 = 2 Thus, X(s) = 1 s s (s + 2) 2.
27 ECE Examples (cont.) Pole at -1: Pole at -2: Double pole at -2: e t u(t) L u 1 s + 1 e 2t u(t) L u 1 s + 2 2te 2t u(t) L u 2 (s + 2) 2 Thus, x(t) = ( e t e 2t + 2te 2t) u(t).
28 ECE Examples (cont.) Read Example 6.8, p 498 Problem 6.7(c), p 498 : find x(t) given X(s) as follows. X(s) = s2 + s 3 s 2 + 3s + 2 = 1 + 2s 5 s 2 + 3s + 2 = 1 + 2s 5 (s + 1)(s + 2) = 1 + A 1 s A 2 s + 2 A 1 = X(s)(s + 1) s= 1 = 3 A 2 = X(s)(s + 2) s= 2 = 1 x(t) = δ(t) 3e t u(t) + e 2t u(t)
29 ECE Inversion of the unilateral LT: complex pole pair Suppose that α + jω 0 and α jω 0 make up a pair of complex-conjugate poles. The partial-fraction expansion is written as A 1 s α jω 0 + A 2 s α + jω 0 L u A 1 e αt e jω 0t u(t) + A 2 e αt e jω 0t u(t) = e αt u(t) ( A 1 e jω 0t + A 2 e jω 0t ) = e αt u(t) ((A 1 + A 2 )cos(ω 0 t) + j(a 1 A 2 )sin(ω 0 t)) = C 1 e αt cos(ω 0 t)u(t) + C 2 e αt sin(ω 0 t)
30 ECE Inversion of the unilateral LT: complex pole pair (cont.) where C 1 = A 1 + A 2 C 2 = j(a 1 A 2 ) Read Example 6.9, p 500. Problem 6.8(a), p 500 : Find x(t) of X(s) = 3s + 2 s 2 + 4s + 5
31 ECE Example Poles at: s 1 = 2 + j1, s 2 = 2 j1 (α = 2, ω 0 = 1). X(s) expressed as A 1 X(s) = s + 2 j + A 2 s j 3s + 2 A 1 = s j = 3 s= 2+j 2 + j2 A 2 = 3s + 2 s + 2 j = 3 s= 2 j 2 j2 C 1 = 3(= A 1 + A 2 ) C 2 = 4(= j(a 1 A 2 ) x(t) = C 1 e αt cos(ω 0 t)u(t) + C 2 e αt sin(ω 0 t)u(t) = 3e 2t cos(t)u(t) 4e 2t sin(t)u(t)
32 ECE Solving differential equations with initial conditions Primary application of unilateral Laplace transform in systems analysis: solving differential equations with initial conditions. Initial conditions are incorporated into the solutions as the values of the signal and its derivatives that occur at time zero in the differentiation property.
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