CHEE 319 Tutorial 3 Solutions. 1. Using partial fraction expansions, find the causal function f whose Laplace transform. F (s) F (s) = C 1 s + C 2

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1 CHEE 39 Tutorial 3 Solutions. Using partial fraction expansions, find the causal function f whose Laplace transform is given by: F (s) 0 f(t)e st dt (.) F (s) = s(s+) ; Solution: Note that the polynomial s(s + ) in the denominator of F (s) is of the form n i= (s p i), where the p i are all distinct. We can therefore use the so-called cover-up method. We want to write s + C s +. Let s find the constants C and C. We start with C. We have s + sc s +. It follows that C = (sf (s)), so that s C = s(s + ) = (s + ) = =. Similarly, we have (s + )F (s) = (s + )C s + C. It follows that C = ((s + )F (s)) s=, so that (s + ) C = s(s + ) = s= s =. s= Hence, F (s) = s s +. Using the Laplace table, as well as linearity of the inverse Laplace transform, we see that f is the causal function { e t, t 0 (.) F (s) = 0 s(s+)(s+0) ;

2 Solution: Since the polynomial s(s + )(s + 0) in the denominator of F (s) is of the form n i= (s p i), where the p i are all distinct, we can use the cover-up method as we did in (.). We want to write s + C s + + C 3 s + 0. Let s find the constants C, C, and C 3, starting with C. We have s + sc s + + sc 3 s + 0. It follows that C = (sf (s)), so that 0s C = s(s + )(s + 0) = 0 (s + )(s + 0) =. Similarly, for C we have (s + )F (s) = (s + )C s + C + (s + )C 3 s + 0, It follows that C = ((s + )F (s)) s=, so that 0(s + ) C = 0 s(s + )(s + 0) = s= s(s + 0) = 0 s= 9. Finally, for C 3 we have (s + 0)F (s) = (s + 0)C s + (s + 0)C s + + C 3. It follows that C 3 = ((s + 0)F (s)) s= 0, so that 0(s + 0) C 3 = 0 s(s + )(s + 0) = s= 0 s(s + ) = 0 s= 0 90 = 9. Hence, F (s) = s 0 9 s s + 0. As in (.), using the Laplace table, as well as linearity of the inverse Laplace transform, we see that f is the causal function { 0 9 e t + 9 e 0t, t 0 (.3) F (s) = 3s+ s +4s+0 ; Solution: Denoting + 4i by z, we see that the denominator of F (s) factors as s + 4s + 0 = (s + z)(s + z). (Here the overline denotes complex conjugation. That is, if

3 z = a + bi, then z a bi.) Since z, z are distinct, we can proceed as in (.) and (.). We want to write s + z + C s + z. Let s find C. We have C = ((s + z)f (s)), so that (s + z)(3s + ) C = (s + z)(s + z) = 3s + s + z = 3z + z + z or C = 3/ i/. Similarly, C = ((s + z)f (s)), so that (s + z)(3s + ) C = (s + z)(s + z) = 3s + s + z = 3z + z + z or C = 3/ + i/. Hence = 3( + 4i) + 4i + 4i = 3( 4i) + + 4i + + 4i 4 i =, 8i 4 + i =, 8i F (s) = 3/ i/ s + z + 3/ + i/. s + z As in (.) and (.), it follows that f is the function given by (3/ i/)e zt + (3/ + i/)e zt = (3/ i/)e ( 4i)t + (3/ + i/)e ( +4i)t = e t 3/ (e i4t + e i4t ) i/ (e i4t e i4t )] = e t 3 ei4t + e i4t i e i4t e i4t ] = e t 3 ei4t + e i4t + i ei4t e i4t ] = e t 3 ei4t + e i4t = 3e t cos(4t) e t sin(4t) ei4t e i4t i when t 0. (Here we have used Euler s identity for the sine and cosine functions in the last line.) Hence f is the causal function OR Solution : Noting the Laplace transform pairs { 3e t cos(4t) e t sin(4t), t 0 e at sin ωt ω (s + a) + ω e at cos ωt s + a (s + a) + ω ] 3

4 and completing the square on the denominator of F (s): we see that F (s) = 3s + F (s) = (s + ) + 4 = 3 ] s + = 3 (s + ) + 4 and the solution f(t) follows. 3s + (s + ) + 4 s + /3 (s + ) + 4 ] 4 (s + ) + 4 ] s + 4/3 = 3 (s + ) + 4 (.4) F (s) = 3s +9s+ (s+)(s +5s+) ; Solution: The denominator of F (s) factors as (s + )(s + 5s + ) = (s + )(s + z)(s + z), where z 5 + i 9. Since, z, z are distinct, we can proceed as in (.) (.3). We want to write s + + C s + z + C 3 s + z. Let s find C. We have C = ((s + )F (s)) s=, so that C = (s + )(3s + 9s + ) (s + )(s + 5s + ) = 3s + 9s + s= s + 5s + = 6 s= 5. For C, we have C = ((s + z)f (s)), so that C = (s + z)(3s + 9s + ) (s + )(s + z)(s + z) = 3s + 9s + (s + )(s + z) = i 9. Similarly, C 3 = ((s + z)f (s)), so that C 3 = (s + z)(3s + 9s + ) (s + )(s + z)(s + z) = 3s + 9s + (s + )(s + z) = i 9. It follows that f is the function given by 6 ( 9 5 e t ) ( 9 90 i 9 e zt = 6 ( 9 5 e t ) 90 i 9 e ( 5/ i 9/)t + ) 90 i 9 e zt ( i 9 ) e ( 5/+i 9/)t = e t + e 5t/ 0 (ei 9t/ + e i 9t/ ) i 9(e i 9t/ e )] i 9t/ = 6 5 e t + e 5t/ 9 5 ei 9t/ + e i 9t/ 5 ] 9 ei 9t/ e i 9t/ 95 i = 6 5 e t e 5t/ cos( 9t/) 5 9 e 5t/ sin( 9t/) 95 4

5 when t 0. Hence f is the causal function { 6 5 e t e 5t/ cos( 9t/) e 5t/ sin( 9t/), t 0 (.5) F (s) =. s 6 Solution: We can simply look this one up in the table, which tells us that f is the causal function { t 5 5!, t 0. Find the causal function f whose Laplace transform is given by: (.) F (s) = s(s+) ; Solution: Note that the denominator s(s + ) of F (s) cannot be written in the form n i= (s p i), where the p i are all distinct. As we have seen in the lectures, we must make use of Heaviside s theorem to find the partial fraction expansion of F (s). We want to write s + C s + + C 3 (s + ). The constants C, C and C 3 are given by C = (sf (s)) = (s + ) = 4 ; C = d ds (s + ) F (s) = d s= ds s= s = 4 ; C 3 = ((s + ) F (s)) s= = s = s=, so that F (s) = /4 s /4 s + / (s + ) Consulting the table, we see that f is the causal function { 4 4 e t te t, t 0 (.) F (s) = (s+)(s+5) (s+)(s +4) ; Solution: As in (.), it is clear from looking at the denominator of F (s) that we must make use of Heaviside s theorem. The denominator of F (s) can be written as (s+)(s + 4) = (s + )(s + 4)(s + 4) = (s + )(s + i) (s i). We want to write s + + C s + i + C 3 (s + i) + C 4 s i + C 5 (s i). 5

6 The constants C, C, C 3, C 4 and C 5 are given by (s + )(s + 5) C = ((s + )F (s)) s= = (s + 4) = 3 s= 5 ; C = d ds (s + i) F (s) = 6 s= i i; C 3 = ((s + i) (s + )(s + 5) F (s)) s= i = (s + )(s i) = 83 s= i i; C 4 = d ds (s i) F (s) = 6 s=i i; C 5 = ((s i) (s + )(s + 5) F (s)) s=i = (s + )(s + i) = 83 s=i i, so that F (s) = 3/5 6/ i/ s + s + i 83/0 39i/0 + (s i). 83/0 + 39i/0 6/5 579i/00 (s + i) + s i After performing some straightforward manipulations as in (.4), and consulting the table, we see that f is the function given by 3 5 e t 3 ( 579 cos(t) sin(t) ) ( 0 i te it ) 0 i te it when t 0. fashion as Using Euler s identity again, we can write this in a more recognizable 3 5 e t cos(t) Hence f is the causal function { 3 5 e t sin(t) t cos(t) 0 0 t sin(t) cos(t) + 00 sin(t) 0t cos(t) 0 t sin(t), t 0 (.3) F (s) = s (s +). Solution: Proceeding as in (.), we write the denominator as (s + ) = (s + i) (s i). We want to write s + i + C (s + i) + C 3 s i + C 4 (s i). 6

7 The constants C, C, C 3 and C 4 are given by C = d ds (s + i) F (s) = 0; s= i C = ((s + i) F (s)) s= i = s (s i) = s= i ; C 3 = d ds (s i) F (s) = 0; s=i C 4 = ((s i) F (s)) s=i = s (s + i) = s=i, so that F (s) = / (s + i) + / (s i). Consulting the table, we see that f is the function given by te it + ( e it + e it ) teit = t = t cos(t) when t 0. Hence f is the causal function { t cos(t), t 0 3. Solve the following scalar initial-value problems using the Laplace transform: (3.) ÿ(t) + ẏ(t) + 4y(t) = 0, with initial conditions y(0) = and ẏ(0) = ; Solution: Taking the Laplace transform of both sides, and denoting the Laplace transform of y by Y, we have s Y (s) sy(0) ẏ(0) + sy (s) y(0) + 4Y (s) = 0. Plugging in the initial conditions, we obtain Isolating Y (s), we obtain where s Y (s) s + sy (s) + 4Y (s) = 0. Y (s) = s + 3 s + s + 4 = s + 3 (s + z)(s + z), z + 5 i. 7

8 Let s use a partial fraction expansion to find y, as we did in Question. Obviously we can use the cover-up method. We want to write Y (s) = C s + z + C s + z. As in Question, we can perform straightforward calculations to show that the constants C and C are given by and that C = ((s + z)y (s)) = 5 6 i; C = ((s + z)y (s)) = i, y(t) = e t/ cos( 5t/) 5 3 e t/ sin( 5t/). (3.) ÿ(t) + ẏ(t) = t, with initial conditions y(0) = and ẏ(0) =. Solution: Taking the Laplace transform of both sides as in (3.), we have s Y (s) sy(0) ẏ(0) + sy (s) y(0) = s. Plugging in the initial conditions, we obtain s Y (s) s + sy (s) = s. Isolating Y (s) and rearranging terms, we obtain Y (s) = s + s s + s = s3 + s 3 (s + ) = s + + s 3 (s + ). For the partial fraction expansion of the second term of Y (s), we want to write s 3 (s + ) = C s + C s + C 3 s 3 + C 4 s +. The constants C, C, C 3 and C 4 are given by C = d ds s + = (s + ) 3 = ; C = d ds s + = (s + ) = ; C 3 = s + = ; C 4 = ((s + ) s 3 (s + ) ) = s= s s= 3 =, 8

9 so that Consulting the table, we see that Y (s) = s + + s s + s 3 s + = s s + s 3. y(t) = t + t. 9