Math 201 Lecture 17: Discontinuous and Periodic Functions

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1 Math 2 Lecture 7: Dicontinuou and Periodic Function Feb. 5, 22 Many example here are taken from the textbook. he firt number in () refer to the problem number in the UA Cutom edition, the econd number in () refer to the problem number in the 8th edition. Quiz: When olving uing Laplace tranform we get. Review t y + y t, y(), y () () Y C. (2) What i the olution y(t) to the original problem? I there any other way to olve it? Solution. Recall that we have two poible way to determine C. he firt one i to ue the fact that Y () ha to a. hi doen t work here a whatever value C take, Y + C 2 3 a. So we have to ue the econd method. We know that the olution y(t) ha to take the form t2 + C. (3) 4 Now check againt the initial condition: y(t) L C y() C ; y () C. (4) So C and the olution i y(t) t2 4. If we are not allowed to ue Laplace tranform, we can multiply the equation by t to get which i Cauchy-Euler (nonhomogeneou though) and can be olved. Function with jump. t 2 y + t y t 2 (5). Baic Information We have been uing definition to calculate Laplace tranform of function with jump, uch a <t < 3 f(t). t t >3 By definition (6) Lf () e t 3 f(t)dt e t dt + t e t dt ( )e 3 2. (7) We can do thi to all uch function: f (t) <t<t f(t) (8) f k (t) t k < t t Lf(t) f (t) e t dt + + f k (t)e t dt. (9) t k However thi i not a good idea becaue: It doe not take advantage of the poibility that Lf,, Lf k can be calculated efficiently or even found in the table.

2 2 Math 2 Lecture 7: Dicontinuou and Periodic Function More importantly, it look like whenever f ha dicontinuitie, the Lf ha factor like e a. Now if when olving an equation, we get then how do we figure out y? Unit Step Function. Y ()e a () Definition. (Unit tep function) he unit tep function u(t) i defined by t < u(t) 6 t >. () From the definition we ee that for any contant M, t < a M u(t a) M t > a. (2) Now uppoe we have a function f uch that f(a + ) f(a ) + M, it i eay to ee that f Mu(t a) i continuou at a. More generally, let u notice that the function f(t)+ g(t)u(t a) (3) take value f(t) for < t < a, and value f(t) + g(t) for t > a. Similar reult hold when there are three, four, or more term in the um. Repreentation of function with jump: In general, the repreentation of a function g (t) <t<t g(t) (4) g k (t) t k < t <t k i g(t) g (t) + [g 2 (t) g (t)] u(t t ) + [g 3 (t) g 2 (t)] u(t t 2 ) + + [g k (t) g k (t)] u(t t k ). One can check, uing the definition of the unit tep function u, that g(t) indeed take the correct va Example 2. (7.6 5; 7.6 5) Expre the given function uing unit tep function. <t < 2 <t < 2 g(t). (6) 2 <t < <t Solution. We have g (t), g 2 (t)2, g 3 (t), g 4 (t)3. hu g(t)2u(t ) u(t 2)+2u(t 3). Example 3. (7.6 6; 7.6 6) Expre (5) uing unit jump function. g(t) <t<2 t + 2<t (7) Solution. We have g(t)(t +)u(t 2). (8) lue in each interval [t i, t i+ ]. Laplace tranform of function with jump.

3 Feb. 5, 22 3 Remember that we introduce the unit jump function to compute Laplace tranform of dicontinuou function. From the above, we ee that we need to be able to compute Uing definition of Laplace tranform, we have Lg(t)u(t a) In particular, we have Alo, if we define f(t) uch that then which lead to he correponding invere relation i Lg(t)u(t a). (9) a e a e a Laplace and Invere Laplace tranform: Procedure. e t g(t)u(t a)dt e t g(t)dt a a e (t a) g(t)dt e v g(v + a)dv e a Lg(t + a)(). (2) Lu(t a) e a. (2) f(t a) g(t), (22) g(t +a) f(t) (23) Lf(t a)u(t a)e a F(). (24) L e a F() f(t a)u(t a). (25) Laplace tranform of function with jump:. Repreent the function uing unit jump function: g(t) g (t) +[g 2 (t) g (t)] u(t t )+ + [g k (t) g k (t)] u(t t k ). (26) 2. ranforming each g(t) u(t a) a follow: a. Obtain g(t + a). b. Obtain Lg(t + a). c. Lg(t)u(t a) e a Lg(t + a)(). Invere Laplace tranform of function involving e a : Example.. Write the function to be invere tranformed a e a F(). 2. Obtain f(t)l F(). 3. Obtain f(t a). 4. he invere tranform i f(t a)u(t a). Example 4. Compute the Laplace tranform of <t < 2 <t < 2 g(t). (27) 2 <t < <t

4 4 Math 2 Lecture 7: Dicontinuou and Periodic Function Solution. We have already found out that hu g(t)2u(t ) u(t 2)+2u(t 3). (28) Lg() 2Lu(t ) Lu(t 2)+2Lu(t 3) 2 e e 2 + 2e 3. (29) Example 5. Compute Laplace tranform of <t<2 g(t) t + 2<t Solution. We have already olved Let g (t)t+. We have (3) g(t)(t +)u(t 2). (3) Lg (t)u(t 2)e 2 Lg (t + 2) e 2 Lt +3e 2 [ ]. (32) Example 6. (7.6 ; 7.6 ) Determine the invere Laplace tranform of Solution. We have e 2. (33) L e a F() f(t a)u(t a). (34) Comparing with the problem, we have a 2, and F(). Inverting F() we have f(t) L e t. (35) hu So finally f(t 2)e t 2. (36) e L 2 e t 2 u(t 2). (37) Example 7. (7.6 5; 7.6 5) Compute the invere Laplace tranform of Solution. Comparing with we have a 3, F() e (38) L e a F() f(t a)u(t a). (39). We compute f(t) L L + 2 ( + 2) ( + 2) 2 + e 2t [co t 2 in t]. (4) hu e L f(t 3)u(t 3) e 2(t 3) [co (t 3) 2 in (t 3)] u(t 3). (4)

5 Feb. 5, 22 5 Periodic function. Definition 8. (Periodic function) A function f(t) i aid to be periodic of period if for all t in the domain of f. f(t + ) f(t) (42) uch function i determined by the value of f in any one period [a, a + ). In particular, all information of f i contained in the value of f(t) for t <. Naturally, we would like to compute it Laplace tranform uing thee value only. Introduce the windowed verion of f(t): f(t) <t< f (t) 6 otherwie. (43) hen we can compute the Laplace tranform of f : F () 6 e t f (t)dt Now we would like to know the relation between F and F Lf. If f ha period and i piecewie continuou on [, ], then e t f(t)dt. (44) Lf () F () e. (45) Example 9. Determine Lf where f ha period 2 and i given by Solution. We have f(t)t, on < t <2. (46) Lf () F () e 2 t e t dt e 2 ( e 2 ) 2 t de t [ ] ( e 2 t e t N t t2 2 e t dt ) [ ( e 2 2e 2 + ] ) e t t2 N t [ ( e 2 2e 2 + ) e 2 ] e 2 2e 2 2 ( e 2. (47) ) 2. Example Function with jump. A quick ummary. he common ituation are and Lg(t) u(t a)e a Lg(t + a)(), (48) L e a F() f(t a)u(t a). (49) Example. (7.6 29; ) Solve and ketch the olution. y + y u(t 3); y(), y (). (5)

6 6 Math 2 Lecture 7: Dicontinuou and Periodic Function Solution. We follow the ame old three tep.. ranform the equation. Compute hu the equation i tranformed into 2. Solve the tranformed equation. 3. ranform back. Compute Ly 2 Y y() y () 2 Y. (5) ( 2 +)Y e 3 Y y L Y L +. (52) e 3 ( 2 + ) (53) e 3 ( 2 + ) L ( 2 +) + L 2 + (t 3)u(t 3) + in t. (54) he invere tranform of can be computed via method of partial fraction, ( 2 +) L ( 2 L + ) 2 co t. (55) + hu y [ co (t 3)] u(t 3)+in t. (56) Proof of 3. Note and Comment heorem. If f ha period and i piecewie continuou on [, ], then Proof. We compute Lf () Lf () F () e. (57) n e t f(t) dt e t f(t)dt + 2 e t f(t)dt + I n (58) where (n+) I n e t f(t)dt. (59) Note that I F. n o evaluate I n, we introduce a new variable t n 6 t n. hen I n e (tn+n) f(t n + n ) dt n e n e tn f(t n ) dt n e n I n e n F. (6) herefore Lf () n I n F ( n e n ) F e. (6) hu end the proof.