... REFERENCE PAGE FOR EXAM 2. e -st f(t) dt. DEFINITION OF THE TRANSFORM F(s) = t=0. TRANSFORMS OF DERIVATIVES f'(t) Ø sf(s) - f(0),

Transcription

1 reference page for exam 2 REFERENCE PAGE FOR EXAM 2 DEFINITION OF THE TRANSFORM F() = t= e -t f(t) dt TRANSFORMS OF DERIVATIVES f'(t) Ø F() - f(), f"(t) Ø 2 F() - f() - f'() TRANSFORM OF A CONVOLUTION f(t) g(t) Ø F()G() TRANSFORM TABLE u(t) r(t) t n u(t) in at u(t) co at u(t) e at u(t) 2 n! n+ a 2 + a a 2 - a (t) T SHIFTING RULES f(t-a)u(t-a) Ø e -a F() e at f(t) Ø F(-a)... -e -T tranform of T INVERSE TRANSFORMS It' undertood that thee invere tranform are good for any value of a,b,c (including non-real value) a long a you don't end up dividing by. () (t) (2) -a e at u(t) (3) u(t) (4) 2 t u(t) (5) t n- n (n-)! u(t) (6) 2 +a 2 co at u(t)

2 reference page for exam 2 (7) 2 +a 2 in at u(t) a (8) ( 2 +a 2 ) 2 t in at u(t) 2a (9) ( 2 +a 2 ) 2 (in at - at co at) u(t) 2a3 () ( 2 +a 2 ( - co at) u(t) ) a2 () 2 ( 2 +a 2 (at - in at) u(t) ) a3 (2) 3 ( 2 +a 2 ( ) 2a 2 t2 + a 4 co at - a 4 )u(t) (3) 2 ( 2 +a 2 ) 2 2a (in at + at co at) u(t) (4) (5) (6) ( 2 +a 2 )( 2 +b 2 ) ( 2 +a 2 )( 2 +b 2 ) 2 (-a) b 2 -a 2 ( a in at - in bt )u(t) b b 2 (co at - co bt) u(t) -a2 ( a 2 eat - t a - a 2 )u(t) (7) (8) (-a)(-b) (-a)(-b) a-b (eat -e bt ) u(t) a-b (aeat -be bt ) u(t) (9) (2) (2) (22) (23) 2 -a 2 inh at u(t) (pecial cae of (7)) a 2 -a 2 coh at u(t) (pecial cae of (8) ) (-a) 2 (at+)e at u(t) e (-a)(-b)(-c) [ at (a-b)(a-c) ] + e bt (b-a)(b-c) + e ct u(t) (c-a)(c-b) ae (-a)(-b)(-c) [ at (a-b)(a-c) ] + be bt (b-a)(b-c) + ce ct u(t) (c-a)(c-b) (24) (25) 2 a 2eat (a-b)(a-c) + b 2 e bt (a-b)(c-b) + c 2 e ct (a-c)(b-c) (-a)(-b)(-c) (-a)( 2 +b 2 ) [ ] a 2 +b 2 e at - co bt - a in bt b u(t) u(t) (26) -e at (a-b) 2 + e bt (a-b) 2 + teat (-a) 2 (-b) a-b u(t)

3 page of Section 5. SECTION 5. INTRODUCTION the unit tep function u(t) The function u(t) i defined by CHAPTER 5 THE (ONE-SIDED) LAPLACE TRANSFORM u(t) = { if t < if t > Fig how the function u(t) and ome variation. u(t) 3 3u(t) FIG the function f(t)u(t) If f(t) i an arbitrary function then f(t) u(t) = if t < f(t) if t > -2-2u(t) In other word, multiplying a function f(t) by u(t) kill f(t) until time t= and thereafter leave it alone. Fig 2 how the function e t veru e t u(t) and Fig 3 how in t veru in t u(t). e t e t u(t) in t in t u(t) FIG 2 FIG 3 All function in thi chapter are intended to be until time t = ince they are pictured a input and output of a ytem which i initialized at t =. So the chapter i concerned with function uch a in t u(t) and e t u(t) rather than with plain in t and e t. the unit ramp r(t) The function tu(t) i called the unit ramp and denoted r(t). In other word, r(t) = if t t if t Fig 4 how the function r(t), 3r(t) and -2r(t). lope lope 3 lope -2 r(t) 3r(t) -2r(t) FIG 4

4 page 2 of Section 5. the function f(t) (t) and f(t) (t a) Multiplying (t-3) by f(t) leave the zero height on the (t-3) graph unchanged and multiplie the impule at t = 3 by f(3) o that the area become f(3) intead of. In fact, f(t) (t-3) implifie to f(3) (t-3). In general, f(t) (t-a) i the ame a f(a) (t-a), an impule of area f(a) occurring at time t = a (Fig 5). In particular, f(t) (t) i the ame a f() (t), an impule of area f() occurring at t= (Fig 6) For example, t 2 (t-4) i the ame a 6 (t-4), an impule of area 6 at t=4. For example, t (t) i the zero function (carrying zero area). It in't an impule anymore. f(t) δ(t-a) area f(a) f(t) δ(t) area f() a FIG 5 FIG 6 the ifting property of the delta function From the box above, the area under the graph of f(t) (t-a) i f(a) and it i all concentrated at t=a. So: () In particular above a a f(t) (t-a) dt = f(a) - f(t) (t) dt = f() f(t) (t) dt = f() - f(t) (t-a) dt = f(a) And if a i not in the interval of integration then interval f(t) (t-a) dt =. For example, π 2π π (t - 2 π) in t dt = in 2 π = (t - π) in t dt = 2 (π/2 i not in the interval of integration) t=- t t (t) dt = = 3 7

5 page 3 of Section 5. definition of the (one-ided) Laplace tranform and invere tranform Start with a function f(t) which i for t. Then t= (2) F() = t= e -t f(t) dt We're only conidering > becaue if, the integral probably doen't even converge. In fact, for ome f(t) the tranform will exit only for ay 5 or 2 or π. Don't worry about it. The live variable in (2) i, and the dummy variable of integration i t. F() i called the Laplace tranform of f(t) f(t) i called the invere tranform of F(). You can write f(t) Ø F() f(t) = F() - F() = f(t) ome baic tranform pair f(t) u(t) r(t) t n u(t) co at u(t) in at u(t) e at u(t) F() 2 n! n+ 2 + a 2 a 2 + a 2 - a for > a (t) proof for u(t) If f(t) = u(t) then F() = u(t) e -t dt = e -t dt = - e-t t= provided > o that lim t e -t = proof for r(t) If f(t) = r(t) = t u(t) then = + = F() = te -t dt = (- te-t - 2 e-t ) t= = - t e t t= - 2 e-t t= + te-t t= antideriv i on the ref page for exam + 2 e-t t= = 2 / 2

6 page 4 of Section 5. proof for in at u(t) If f(t) = in at u(t) then F() = e -t in at dt = t= e-t (- in at - a co at) 2 + a 2 t= = a 2 + a 2 proof for (t) (t) = e -t (t) dt = e - (ifting property) = linearity property of the tranform [f(t)+ g(t)] = f(t) + g(t) [ af(t) ] = a f(t) In other word, (3) f(t) + g(t) Ø F() + G() (4) a f(t) Ø a F() proof of (3) [ f(t) + g(t) ] = [ f(t) + g(t) ] e -t dt = f(t) e -t dt + g(t) e -t dt = f(t) + g(t) corollary of linearity (at 2 + bt + c)u(t) Ø a b 2 + c example If f(t) = (4 + 3t 2 )u(t) then F() = = example 2 6r(t) Ø 6 2-3u(t) Ø -3

7 page 5 of Section 5. warning Suppoe f(t) = in 3t u(t) 3 To indicate that the tranform i 2 either write in 3t u(t) Ø 2 OK + 9 or write 3 in 3t u(t) = 2 OK + 9 or write 3 F() = 2 OK + 9 but don't write in 3t u(t) = WRONG WRONG WRONG review coh t = et + e -t 2, inh t = et - e -t 2 PROBLEMS FOR SECTION 5.. Find the tranform (a) t 5 u(t) (b) t 3 u(t) (c) e 3t u(t) (d) e -4t u(t) (e) in 4t u(t) (f) co 5t u(t) 2. Derive the tranform formula for e -at u(t) 3. Find the tranform (a) coh t u(t) (b) in 2 4t u(t) (Ue the identity in 2 t = ( - co 2t) ) 2 (c) co(at + b)u(t) (Ue the identity co(x + y) = co x co y - in x in y) (d) (8t 2 + 2t - 3) u(t) (e) (2e 3t - in πt) u(t) (f) 5 (g) (h) lope /2 (i) lope 7-2 (j) e 3t+4 u(t) (ue a little algebra firt) (k) (apple) 6 (t) (4t 3-3t 2 + 5t + 2) u(t)

8 page 6 of Section Find (a) 2π (t - π 4 ) co t dt (b) 2π t 3 (t-7) dt 2π (c) t 3 (t-6) dt (d) - (t) co t dt (e) - 6 (t) dt (f) 6 (t) dt (g) - t 2 (t) dt (h) e t (t) dt (i) - t 2 (t-2) dt (j) e t (t-2) dt 5. Find by inpection (a) e -t t 4 dt (b) e -u (c) e -wt t 4 dt (d) t 4 e 32 t t 4 dt u 4 du

9 page 7 of Section 5. HONORS 6. (a) Ue integration by part to expre the tranform of t n u(t) in term of the tranform of t n- u(t). (b) I already derived the tranform of tu(t), i.e., of r(t). Ue the reult in (a) to find the tranform of t 2 u(t), t 3 u(t), t 4 u(t) and t n u(t).

10 page of Section 5.2 SECTION 5.2 FINDING TRANSFORMS the graph of f(t a)u(t a) The graph of f(t-a)u(t-a) i found by tranlating the graph of f(t)u(t) to the right by a unit (Fig -4), i.e., by delaying the ignal. 2 2 (t-3) u(t-3) t u(t) u(t-2) - 3 FIG FIG 2 2 4r(t-3) = 4(t-3)u(t-3) lope 4 3 in(t-4)u(t-4) - FIG 3 FIG π t-hifting rule Let a be a poitive number. Delaying f(t)u(t) until time a will multiply the tranform by e -a. In other word, if f(t)u(t) Ø F() then For example (Fig ), f(t-a)u(t-a) Ø e -a F() t 2 u(t) Ø 2 3 o (t-3) 2 u(t-3) Ø 2e-3 3 Similarly, for the delayed ignal in Fig 2-4, u(t-2) Ø e-2 4r(t-3) Ø 4e-3 2 in(t-4)u(t-4) Ø e proof of the t-hifting rule f(t-a)u(t-a) = f(t-a)u(t-a) e -t dt = a f(t-a) e -t dt. Now let v = t-a, dv = dt. When t = a, v =. When t =, v =. So

11 page 2 of Section 5.2 f(t-a)u(t-a) = = e -a f(v) e -(v+a) dv f(v) e -v dv = e -a F() Thi i the integral for F() (but with dummy variable v intead of t) adding and ubtracting ramp Firt note that if you add two line with lope m and m 2 you get a line with lope m + m 2 becaue m x + b + m 2 x + b 2 = (m + m 2 )x + b +b 2 Fig 5-7 how how to ue thi to combine ramp graphically. m=3 m=2 + 3 m= = m=2 ' 3 FIG 5 m= m=3 3-5 = m= m=-2 ' 3 5 FIG 6 m= + m= m=-4 = m=2 m= ' ' 3 5 m=-2 FIG 7 finding tranform by decompoing into ramp and tep Look at the function in Fig 8. Fig 9 how that 4- Fig 8 = 4r(t) - 5r(t-) + r(t-5). So the tranform of Fig 8 i 4 2-5e- 2 + e-5 2 ' 5 FIG 8

12 page 3 of Section 5.2 m=- m=4 m=4 ' 5 m= = FIG m=-5 5 m= warning In Fig 9, if you top with the firt two term in the decomp, i.e., if you top with 4r(t) - 5r(t-) then you have the function in Fig intead of the deired Fig 8. have enough term in your decompoition to get what you want. Make ure you m=4m=- 5 FIG example The function in Fig can be written a r(t) - r(t-3) of Fig i 2 - e-3 2 (Fig 2). So the tranform m= 3- m= m= ' 3 FIG FIG 2 ' 3 = + 3 m=- tranform of a jumpy f(t) I'll find the tranform of the function in Fig 3 which jump from 8 down to. 8-4 FIG 3 Fig 4 how the decompoition of Fig 3. m=2 4 m= m=2 = - m= o far the decomp goe up to ubtract to make it height 8 and level off there jump down to height FIG 4

13 page 4 of Section 5.2 So Fig 3 = 2r(t) - 2r(t-4) - 8u(t-4) and the tranform of Fig 3 i 2 2-2e-4 2-8e-4 decribing a pule Fig 5 how that u(t-4) - u(t-7) = { if 4 t 7 otherwie u(t-4) u(t-7) = FIG 5 u(t-4) - u(t-7) So f(t) [ u(t-4) - u(t-7) ] = { f(t) if 4 t 7 otherwie Similarly, f(t) [ u(t) - u(t-7) ] = { Fig 6 and 7 how ome ine pule. f(t) if t 7 otherwie π 2π π in t [ u(t-π) - u(t-2π) ] in t [ u(t) - u(t-π) ] FIG 6 FIG 7 uing algebraic maneuver and t-hifting to find the tranform of a pule Let f(t) = e 2t if t 6 otherwie 6 Then f(t) = e 2t [ u(t) - u(t-6) ] = e 2t u(t) - e 2t u(t-6) FIG 8 You can't ue the t hifting rule on e 2t u(t-6) becaue the exponent ha plain t in it, not t-6. But you can get it into a more ueful form a follow: f(t) = e 2t u(t) - e 2(t-6)+2 u(t-6) = e 2t u(t) - e 2 e 2(t-6) u(t-6) TRICK contant now can t-hift

14 page 5 of Section 5.2 So F() = -2 - e2 e-6-2 = -2 ( - e2-6 ) example 3 Find the tranform of the pule f(t) in Fig 9 We have y= FIG 9 y=t 2 y= f(t) = t 2 [ u(t) -u(t-4)] = t 2 u(t) - t 2 u(t-4) = t 2 u(t) - [ (t-4) + 4 ] TRICK 4 not ready for t-hifting 2 u(t-4) = t 2 u(t) - [(t-4) 2 + 8(t-4) + 6] u(t-4) So ready for t-hifting F() = e warning The t hifting rule doe not apply directly to t 2 u(t-4); the tranform i not 2e-4 3 becaue 2e-4 3 goe with (t-4) 2 u(t-4) and not with t 2 u(t-4). The t hifting rule applie in example 3 after you ue algebra to get [ (t-4) 2 + 8(t-4) + 6] u(t-4) example 4 Find the tranform of the ine pule in Fig 2. y=in t y= π FIG 2 y= method Fig 2 how the pule decompoed into in t u(t) + in(t-π) u(t-π) So the tranform i e -π 2 +

15 page 6 of Section 5.2 in t u(t) in(t- π) u(t- π) π = + FIG 2 π method 2 Fig 6 = in t [ u(t) - u(t-π) ] = in t u(t) - in t u(t-π) = in t u(t) - in [ (t-π) + π ] u(t-π) TRICK = in t u(t) + in(t-π) u(t-π) by the identity in(x+π) = - in x So the tranform of the function in Fig 2 i e -π 2 + example 5 Find the tranform of the pule in Fig FIG 22 The pule i 2u(t-3) - 2u(t-7) o it tranform i 2 [ e -3 - e -7 ]. review of geometric erie a + ar + ar 2 + ar = a -r provided that - < r < tranform of a periodic function Suppoe f(t)u(t) ha period T for t (Fig 23). f(t)u(t) f(t) [u(t) - u(t-t)]... T FIG 23 FIG 24 T To find it tranform, firt find the tranform of f(t) [ ] u(t) - u(t-t) (Fig 24), the ignal coniting of the firt period followed by zero. Then multiply by -e -T

16 page 7 of Section 5.2 proof Fig 24 Fig 24 hifted by T f(t)u(t) = T + T 2T Fig 24 hifted by 2T + 2T 3T +... If you let G() denote the tranform of Fig 24 then f(t)u(t) Ø G() + e -T G() + e -2T G() + e -3T G() +... = G() + e -T + (e -T ) 2 + (e -T ) The erie in the bracket i geometric with a =, r = e -T, Since T and are poitive, e -TS i between and o the erie converge to -e -T and f(t)u(t) Ø tranform of Fig 24 -e -T example 6 The function in Fig 25 ha period 2a for t. So tranform of Fig 25 = tranform of Fig 26 Fig 27 how the decompoition of Fig 26: -e -2a Fig 26 = 5u(t) - u(t-a) + 5u(t-2a) So tranform of Fig 26 = 5 - e-a + 5e-2a and Fig 25 Ø 5 - e -a + 5-2a ( - e -2a ) a 2a 3a 4a a 2a -5 - FIG 25 FIG 26

17 page 8 of Section a 2a = 5 - FIG 27 a + 5-2a warning Don't forget the lat term in the decompoition in Fig 27. If you leave it out you will end up with when you wanted to get Fig 26. -hifting rule e at f(t) Ø F(-a) for > a In other word, multiplying a function by e at hift the tranform to the right by a. proof of the -hifting rule ( ) e at f(t) = e at f(t) e -t dt = f(t) e -(-a)t dt The integral on the righthand ide i the ame a the integral for the Laplace tranform F() but with -a intead of. So the integral i F(-a). footnote The proof of the hifting rule doen't work unle > a o that -a > becaue the original definition of the Laplace tranform a f(t) e -t dt required >. And in the lat integral in ( ), -a i playing the role of the in the definition. So the hifting rule really hold only for > a. example 7 To find the tranform of e -3t in t u(t), find the tranform of in t u(t) and then hift: e -3t in t u(t) Ø (+3) 2 + example 8 Table lit the tranform pair coh at u(t) Ø 2 -a 2 So, by the hifting rule, e 3t coh at u(t) Ø -3 (-3) 2 - a 2

18 page 9 of Section 5.2 PROBLEMS FOR SECTION 5.2. Sketch the graph (a) 9-t 2 (b) (9-t 2 ) u(t) (c) (9-t 2 )u(t-3) (d) (9-t 2 ) u(t-5) 2. Draw a picture of the indicated um (a) m=4 3-5 m=4 (b) m=2 + 2 m=-5 (c) (d) (e) 7 m=-3 + m=3 7 (f) m=- 3 - m=2 + ' Find the tranform (a) (b) (c) 2π - 6- m= m=2-5 T m= π' (d) (e) (f) (g) ' 23 ' ' 2T 3 m= 4. Suppoe you want the tranform of the function f(t) in the lefthand diagram. What have you done wrong if you decompoe f(t) into 6r(t-2) - 6r(t-3) (ee the righthand diagram) and conclude that the tranform i 6e-2 2-6e m=6 f(t) 2 3 m= 6r(t-2) 6r(t-3) m=6 m=6-2 3 Problem 4 f(t) Problem 4 would be decompoition

19 page of Section Find the tranform (a) 3-2- (b) 2- ' 2 ' ' 2 3 (c) (d) 2-2- ' ' 3/ ' 2' ' Sketch the graph (a) 9-t 2 (b) (9 - t 2 ) [ u(t) - u(t-3) ] (c) (9 - t 2 ) [ u(t-2) - u(t-4) ] 7. Given the graph of f(t) in the diagram. Sketch the graph of (a) f(t) u(t) (b) f(t) u(t-2) (c) f(t-2) u(t-2) (d) f(t) [ u(t) - u(t-4) ] -2-2 Problem Find the tranform (a) e -t 2 (b) f(t) = t 3 if t 2 otherwie (c) ine 2 3 (d) t 2 mirror image 2 9. Find the tranform of tu(t-5).. Find the tranform of the following function which are periodic on [, ) (a) k a 2a 3a 4a... (b) - a 2a 3a... (c) rectified in t π 2π 3π... train of unit impule (d)... a 2a 3a

20 page of Section 5.2 (e) train of unit impule (f) a 3a 2a ' a 3a4a ' ' 5a 7a ' 8a.(a) Show that the tranform of the taircae in the diagram i (e - ) (b) Show that the tranform of the indicated function with envelope e -2t i e-( + 2) ' ' ' Problem (a) etc. envelope etc Problem (b) 2. Find the tranform (a) e 3t co 4t u(t) (b) t 2 e -4t u(t) 3. Find e -t (t-5) 7 dt by inpection (don't do any integration). t=5

21 page of Section 5.3 SECTION 5.3 FINDING INVERSE TRANFORMS uing linearity in revere If the invere of F() and G() are in the table of invere tranform, you can find the invere of 3F() and F() + G() uing the linearity rule in revere. For example the table lit and 2 + a 2 Ø a in at u(t) 2 Ø co at u(t) + a2 o = 2 5 in 5 t u(t) + 6 co 3 t u(t) uing the t-hifting rule in revere You ued the rule f(t-a)u(t-a) Ø e -a F() to find the tranform of the delayed ignal f(t-a)u(t-a). right to left to find invere tranform: It can alo be ued from To find the invere tranform of e -a F(), invert F() (remember to put in the u(t)) and then delay until time a. o For example, the table lit () 4 Ø t3 3! u(t) e -5 4 Ø (t-5) 3 3! u(t-5) (Fig ) 5 FIG 6 (t-5)3 u(t-5) warning If you t-hift t3 3! t-hift all of it: WRONG e -5 4 Ø t3 3! u(t) to find the invere tranform of e-5 4 u(t-5) WRONG make ure you WRONG WRONG e -5 4 Ø (t-5)3 u(t) 3! e -5 4 Ø (t-5)3 3! WRONG WRONG RIGHT e -5 4 Ø (t-5)3 3! u(t-5) RIGHT uing the -hifting rule in revere You can ue the rule e at f(t)u(t) Ø F(-a)

22 page 2 of Section 5.3 to find the tranform of a ignal multiplied by an exponential. It' more ueful read from right to left: hifting the tranform to the right by a correpond to multiplying the original function by e at For example, you know that = co 7 t u(t) o, by the -hifting rule, - -5 (-5) = e 5t co 7 t u(t) inverting imple fraction where the denominator contain linear factor Table lit 2-4 Ø 2e-4t So (2) = (rearrange) Ø 5 3 e-7t/3 u(t) Table lit t9 Ø 9! u(t) o, by -hifting, (3) (+4) Ø t 9 9! e -4t u(t) completing the quare to invert ome fraction with nonfactorable quadratic denominator (4) = = ( ) (complete the quare) Ø 2 7 in 2 7 t e-3t/2 (baic formula plu -hifting) (5) = 7 (-3) complete the quare But the fraction in't ready to be inverted yet becaue the denominator i -hifted but the numerator in't. So rearrange the numerator to match the hifted denom: (-3) + 2 = (-3) (-3) = (-3) (-3) Ø 7 co 5t e 3t u(t) in 5t e3t u(t)

23 page 3 of Section 5.3 warning To invert 7 (-3) don't forget that the numerator mut be rearranged to match the hifted denominator. If they don't match (the denom ha -3 but the numerator ha ) the fraction can't be inverted uing -hifting. (6) On the other hand, to invert e - (-3) don't try to turn e - into e -(-3). Treat the e - in (6) a the ignal to firt invert the fraction (-3) getting 5 e-3t in 5 t u(t) (-hfting rule) and then t-hift to get e - (-3) Ø 5 e-3(t-) in 5 (t-) u(t-) how to invert fraction where the denominator i cubic or wore Suppoe F() i of the form polynomial another poly of higher degree I put many of thee in the table o look there firt. If you get one that' not in the table then one way to find an invere tranform i to decompoe F() (ugh) into a um of impler partial fraction (you learned how to do that in calculu ee handout on decompoition) and then invert the piece. That' how many of the formula in the reference table were derived in the firt place. If you have acce to Mathematica you can take tranform and invere tranform of many function directly. example 2-5 Ø 5 inh 5 t u(t) (table (9)) ( + 4) 2-5 Ø 5 e-4t inh 5 t u(t) (-hifting rule) e Ø 5 inh 5(t-2) u(t-2) (t-hifting rule) e -2 (+4) 2-5 Ø e -4(t-2) inh 5(t-2) u(t-2) (-hifting and t-hifting)

24 page 4 of Section 5.3 review of factoring quadratic ax 2 + bx + c = a x - -b + b2-4ac 2a -b - x - b2-4ac 2a example example 3 Let F() = = Ø [ -(+ 3) ] [ -(- 3) ] [ ] 2 e(+ 3)t - e (- 3)t u(t) (table (7)) 3 ( ) The econd factor in the denominator i nonfactorable. The decompoition i ( ) = / The firt term invert to 5 u(t). The econd term i like (5) above: ( ) = (-) (complete quare) = - 5 (-) - (-) (rearrange numerator to match the hift in the denom) = (-) (-) So Ø - 5 et co 2t u(t) + 5 ( ) 2 et in 2t u(t) (-hifting) Ø [ 5-5 et co 2t + et in 2t ] u(t) example 4 Let F() = (+2)(-4) method for invere tranforming Split F() into three fraction and ue (8),(7) and (22) in the table. F() = = 2 (+2)(-4) + 2 (+2)(-4) - 2 (+2)(-4) (+2)(-4) + 2 (+2)(-4) - 2 (+2)(-4)

25 page 5 of Section 5.3 Ø - 6 (-2e-2t - 4e 4t )u(t) (e-2t - e 4t )u(t) - 2( e-2t + 24 e4t )u(t) = [ 4-6 e-2t + ] method 2 for invere tranforming F() decompoe into 2 e4t u(t) So /4 - / /2-4 f(t) = [ 4-6 e-2t + ] 2 e4t u(t) PROBLEMS FOR SECTION 5.3. Find the invere tranform (a) 5 2 (b) 5 4 (c) 5-3 (e) 4 (f) Find the invere tranform and draw it graph (d) (a) e-2 (b) e (c) e (d) e Find the invere tranform (a) (-3) 3 (b) (+2) 4 (c) (-5) 2 (d) + 6 (e) e-3 (+ 6) 8 4. (a) (d) (b) (e) 2 + (c) ( + 4) (f) e (g) e -2 (+4) (h) (i) (j) (a) Derive the invere tranform formula that you'll find in the table for thee fraction by decompoing into impler fraction. (i) 2 - a 2 (ii) (-a) 2 (b) The table are miing the invere tranform of (-a) 2 What i it? 6. Find the invere tranform of + (firt ue long diviion) 7. For thi problem jut find the form of the partial fraction decompoition and ue it to find the form of the invere tranform, all without actually computing the contant involved in the decompoition. (a) 3 (b) (-2) (-)(+2) 4

26 page 6 of Section Find the invere tranform. If a partial fraction decompoition i neceary, jut find the form of the anwer without actually computing the contant involved in the decompoition. (a) ( 2 (b) ) 2 ( 2 +) (c) (d) (e) (f) (g) (+4) 2 (h) (+4) 2 (i) (j) (k) - 4 (-2) 2 (apple) The anwer to #8(apple) wa coh 3 t. Do you remember what coh 3 t i in term of exponential function.. Let F() = (a) Find the invere tranform by completing the quare (b) Would it work to factor the denominator into non-real linear factor and ue (7) in the table.

27 page of Section 5.4 SECTION 5.4 SOLVING DIFFERENTIAL EQUATIONS USING TRANSFORMS tranform of derivative () f'(t) Ø F() - f() (2) f''(t) Ø 2 F() - f() - f'() proof f'(t) = t= f'(t) e -t dt Now ue integration by part with u = e -t, dv = f'(t), du = -e -t dt, v = f(t) to get f'(t) = e -t f(t) t= + t= f(t) e -t dt = F() - f() - f() F() ee footnote footnote I'm auming that f(t) ha a tranform in the firt place. So the improper integral e -t f(t) dt mut exit for ay t= >. It can't exit unle e -t f(t) a t (in fact it mut quickly). So plugging t= into e -t f(t) give. Thi prove (). To get (2), think of f'' a (f')'. Then f''(t) = (f' )' = f'(t) - f'() by () = [ F() - f() ] - f'() by () again = 2 F() - f() - f'() olving a DE with IC uing tranform Look at y'' - 3y' + 2y = 2e -t with IC y() = 2, y'() = -. The idea i to take tranform on both ide of the DE to get a new algebraic equation with unknown Y(). You can olve (eaily) for Y() and then (le eaily) take the invere tranform to get y(t), the olution to the DE. In the proce the IC will be ued automatically. Tranforming both ide of the DE give 2 Y - y() - y'() - 3[ Y - y() ] + 2Y = 2 +

28 page 2 of Section 5.4 Ue the IC y() = 2, y'() = - to get 2 Y [ Y - 2 ] + 2Y = 2 + Then ( )Y = Y = (+)( ) = (+)(-)(-2) warning Don't leave out the bracket and write -3Y - 2 when it hould be -3[Y - 2] You can plit thi into 2 2 (+)(-)(-2) - 5 (+)(-)(-2) - 5 (+)(-)(-2) and ue (24),(23),(22) in the table or you can decompoe into partial fraction and then invert (+)(-)(-2) = / /3-2 Either way the anwer i y(t) = [ 3 e-t + 4e t e2t ] u(t) example Ue tranform to olve y'' + 4y' + 3y = with IC y() = 3, y'() = olution Take tranform on both ide of the DE to get 2 Y (Y - 3) + 3Y = (the tranform of i ) Then Y = = (+3)(+) = 3 (+3)(+) + 3 (+3)(+) Ue the table to get y(t) = (-2e -3t + 5e -t ) u(t) example 2 Ue tranform to olve y'' + 3y' + 2y = f(t) with IC y() =, y'() = where f(t) = if t 4 otherwie

29 page 3 of Section 5.4 olution f(t) can be written a u(t) - u(t-4). Take tranform in the DE: 2 Y + 3Y + 2Y = - e-4 (3) Y() = ( ) - e -4 ( ) = (+2)(+) - e -4 (+2)(+) Then (ue table and t-hifting) (4) y(t) = [ 2 - e-t + 2 e-2t ] u(t) -[ 2 ] - e-(t-4) + 2 e-2(t-4) u(t-4) writing without the u notation If y = $$$ u(t) + ### u(t-a) then y = if t $$$ if t a $$$ + ### if t a For example, if y = u(t) + (t-7)u(t-7) + t 5 u(t-8) then (5) y = if t 7 + t-7 = t-6 if 7 t 8 t-6 + t 5 if t 8 warning If y = u(t) + (t-7)u(t-7) + t 5 u(t-8) it i not correct to write WRONG WRONG y = if t 7 t-7 if 7 t 8 t 5 if t 8 WRONG WRONG The right verion i in (5).

30 page 4 of Section 5.4 example 2 continued The olution in (4) can be rewritten a for t (6) y(t) = 2 - e-t + 2 e-2t for t e-t + 2 e-2t - [ ] 2 - e-(t-4) + 2 e-2(t-4) for t 4 warning When you write (4), don't leave out the u(t) and epecially not the u(t-4). If you do then your "anwer" i WRONG y = e -t + 2 e-2t + e -(t-4) - 2 e-2(t-4) WRONG which i very different from the correct anwer in (6). review of Cramer' rule Conider the ytem of equation a x + b y + c z = d a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 The determinant of coefficient i a b c a 2 b 2 c 2 a 3 b 3 c 3 If thi determinant i non-zero then the olution i x = coeff determinant but with column replaced by coeff determinant d d 2 d 3 d y = z = coeff determinant but with column 2 replaced by coeff determinant coeff determinant but with column 3 replaced by coeff determinant d 2 d 3 d d 2 d 3 olving a ytem of DE with IC uing tranform Conider the ytem with IC x'' = -3x + 2y y'' = 6x - 7y x() =, y() =, x'() =, y'() = 2.

31 page 5 of Section 5.4 The unknown are the function x(t) and y(t). Take tranform and collect term: 2 X - = -3X + 2Y 2 Y = 6X - 7Y ( 2 + 3)X - 2Y = -6X + ( 2 + 7)Y = + 2 Ue Cramer' rule to olve for X and Y: X = = = ( 2 + 9)( 2 + ) Y = ( 2 + 9)( 2 + ) = ( 2 + 9)( 2 + ) To find invere tranform, decompoe (or better till, get a larger et of table or ue Mathematica): X = Y = Then x = [ - 4 co 3t - 2 in 3t + 4 co t ] y = [ 3 ] 4 co 3t + 4 in 3t + 4 co t + 5 in t u(t) 4 in t u(t) finding the impule repone uing tranform Look at the ytem where input f(t) and output y(t) are related by 2y'' - 4y' - 6y = f(t) I want to find the impule repone h(t) of the ytem. Thi mean olving (7) 2y'' - 4y' - 6y = (t) with IC y() =, y'() = Take tranform in (7) to get 2 2 Y - 4Y - 6Y = Y =

32 page 6 of Section 5.4 Thi i the tranform H() of the impule repone h(t). Factor and ue table (8) (or decompoe): H() = 2 (+)(-3) h(t) = ( - 8 e-t + 8 e3t )u(t) footnote For comparion, here' the method from Chapter 2 for finding the impule repone. Switch from (7) to 2y '' - 4y' - 6y = with IC y() =, y'() = 2 Then 2m 2-4m - 6 =, m = 3,-, y h = Ae 3t + Be -t. The IC make A = 8, B = - 8 o h(t) = 8 e3t - 8 e-t for t In general, the tranform H() of the impule repone h(t) i referred to a the ytem' tranfer function. If input f(t) and output y(t) are related by ay'' + by' + cy = f(t) then (8) H() = a 2 + b + c example 3 Solve 2y" + 3y' + y = co t with IC y() =, y'() =. method (a in example and 2) Take tranform on both ide of the DE: 2 Y + 3Y + Y = 2 + Y = ( )( 2 + ) Now take the invere tranform to get olution y(t). I'm not going to bother doing it. (Mathematica did it in a plit econd. It would take me 5 minute jut to type the invere tranform.) method 2 By (8) H() = Then by (9), Y() = H() Èco t = Same now a method.

33 page 7 of Section 5.4 PROBLEMS FOR SECTION 5.4. Tranform the DE, olve for Y and then top (o that omeone who had a large et of invere tranform table and/or a computer to decompoe could eaily finih the problem) 2y'' + 3y' + 4y = e -8t in 3t with IC y() = -5, y'() = 6 2.Ue tranform to olve (a) y'' + y = in 3t with IC y() =, y'() = (b) y'' + y = 2 co t with IC y() = 2, y'() = (c) i'(t) + 5i(t) = 25 in 5t with IC i() = In particular, find the teady tate olution (d) y'' + 3y' + 2y = e -t with IC y() =, y'() = 3. Ue tranform to olve (a) y '' + 2y = f(t) with IC y() =, y'() = where f(t) = if t otherwie (b) y'' + 4y = in t if t π otherwie with IC y() =, y'() = 4. Solve the ytem of DE uing tranform (a) x' = 7x + 6y y' = 2x + 6y with IC x() = 2, y() = (b) x' = 2x - 2y, y' = x with IC x() = 2, y() = 2 (c) y ' = y 2-2y +, y ' 2 = y - 2y 2 with IC y () =, y 2 () = 5. If the input f(t) and repone y(t) are related by the given DE, ue tranform to find (and then ketch) the impule repone. (a) 2y'' + 3y = f(t) (b) y'' + 5y' + 6y = f(t) (c) y' + y = f(t) 6. Let f(t) be the triangular wave in the diagram. It derivative i the quare wave g(t). Suppoe you know G(). How would you find F(). f(t) Problem 6 g(t)

34 page 8 of Section 5.4 HONORS 7. A function f(t) ha many antiderivative. The partiular antiderivative whoe value i when t= i t f(t) dt. Analogou to the tranform rule for the derivative f'(t) and f''(t) it can be hown that there i a tranform rule for the antiderivative t f(t) dt, namely t ( ) f(t) dt Ø F() (a) Ue it to olve the following integral equation where 2y + t y(t) dt = f(t) f(t) = 4 for a t b otherwie (b) The rule in ( ) read from from right to left ay that to find the invere of omething, invert the omething and then take t. Ue thi to how how () in the invere tranform table can be derived from ().

35 page of Section 5.5 SECTION 5.5 CONVOLUTION tranform of a convolution Remember that f(t) g(t) = u=- f(t-u)g(u) du Now, in addition, let f(t) and g(t) be for t < (a ha been the cae throughout thi chapter). Then f(t) g(t) Ø F()G() In other word, convolving function (that tart at t = ) in the t world correpond to multiplication in the tranform world. So to find the convolution f(t) g(t) tep find the tranform F() and G() tep 2 multiply the tranform tep 3 take the invere tranform proof (lippery) f(t) g(t) = t= f(u)g(t-u) du u= e -t dt It' OK to ue u= a the lower limit in the convolution integral intead of - ince f(u) = for u. Now rewrite e -t a e -u e -(t-u) and rearrange to get f(t) g(t) = u= g(t-u)e -(t-u) dt t= f(u) e -u du Subtitute w = t-u, dw = dt in the inner integral to get f(t) g(t) = u= g(w) e -w dw w= -u f(u) e -u du Since g(w) = for w we can change the lower limit on the inner integral from w = -u to w =. So f(t) g(t) = u= g(w) e -w dw w= f(u) e -u du G() = G() u= f(u) e -u du = G()F() QED

36 example Let f(t) = t u(t) and g(t) = in t u(t). Find f(t) g(t) uing tranform Firt take the tranform of f(t) and g(t): page 2 of Section 5.5 in t u(t) tu(t) Ø Ø Multiply the tranform to get F()G() = 2 ( 2 + ) From the tranform table, 2 ( 2 + ) Ø (t - in t) u(t) So f(t) g(t) = (t - in t) u(t) = if t t - in t if t example 2 Ue tranform to find the convolution of the following function and write the final anwer without uing u(t) notation. 8 f(t) 4 g(t) 6 4 Firt find the tranform and multiply them. f(t) = 8u(t) - 8u(t-6) and g(t) = 4u(t) - r(t) + r(t-4) F() = 8-8 e-6 and G() = e-4 F()G() = e e e e- Now invert. f g = - 8t2 + 32t 2! u(t) + 8(t-4)2 2! u(t-4) + 8(t-6) 2-32(t-6) 2! u(t-6) So - 8(t-)2 2! u(t-)

37 page 3 of Section 5.5 if t 4 then f g = - 8t2 + 32t = -4t 2! t if 4 t 6 then f g = - 4t t + 8(t-4)2 = 64 2! if 6 t then f g = (t-6)2-32(t-6) = 4t 2! 2-8t + 4 if t then f g = 4t 2-8t + 4-8(t-)2 =, 2! i.e., if t -4t t if t 4 f g = 64 if 4 t 6 4t 2-8t + 4 if 6 t if t PROBLEMS FOR SECTION 5.5. Let g(t) = e -t u(t) and f(t) = 2-t for t 2 (f i otherwie) Find f(t) g(t) uing tranform. Give the anwer with the u notation and then again without the u notation. 2. Ue tranform to find f(t) g(t) f(t) 4-5 g(t) Ue tranform to convolve f(t) = e -t u(t) and g(t) = tu(t) 4. Let f(t) = e - t u(t) (where i jut a fixed contant.) Find f(t) f(t) f(t) f(t) 5. Take tranform to how that the olution to i y'' + a 2 y = f(t) with y() = K, y'() = K 2 y = a in at f(t) + K co at + K 2 a in at for t

38 page 4 of Section 5.5 HONORS 6. Let f(t) be an arbitrary function (tarting at t=). (a) Find the convolution (t) f(t) directly (uing the definition of convolution) and then again with tranform. Interpret the reult phyically by thinking of f(t) a an input into an initially at ret ytem which ha impule repone (t). (b) Find (t-a) f(t) directly and then again with tranform. Interpret the reult phyically.

39 page of review problem for Chapter 5 REVIEW PROBLEMS FOR CHAPTER 5. Rewrite the following without the tep notation and ketch the graph. (a) f(t) = e -5t [u(t-) - u(t-3)] (b) f(t) = e -5t u(t) + e -5(t-2) u(t-2) 2. Solve uing tranform. (a) y'' + y = co t with IC y() =, y'() = (b) y'' + 4y' + 5y = 5 with IC y() =, y'() = 2 3. Find the tranform. (a) 3 - ' 2 (b) 6 - ' y = 4-(t-2) 2 (c) 4 4. (a) Let a. Show that È f(at) = a F( ) (called the caling rule). a (b) Suppoe È in t inh t = Ue part (a) to find È in at inh at. 5. Ue tranform to find the impule repone of the ytem whoe input f(t) and repone y(t) are related by y'' + y' + 7y = f(t). 6. Find the tranform of the following function which i periodic for t. The non zero piece are ine. 6-2π 4π 6π 8π 7. Find the invere tranform (a) 4 (b) (+2) 3 (c) 5 (-4) 2 (d) (e) e-4 3 (f) (+)( 2 + ) (g) (h) 2 - (i) 2 (-2) 8. Find thee integral by inpection. (a) e -3t t 4 dt (b) e -t e -3t t 4 dt

40 page 2 of review problem for Chapter 5 9. Solve for x(t) and y(t) if x'' = -5x + 4y, y'' = 4x - 5y with IC x() =, y() = -, x'() =, y'() =. Ue tranform to find h(t) f(t). 4 - h(t) - f(t) 4 5

41 page of appendix to Chapter 5 APPENDIX FINDING TRANSFORMS AND INVERSE TRANSFORMS WITH MATHEMATICA Load the tranform package and the package containing the unit tep function. <<:Calculu:LaplaceTranform.m <<:Calculu:DiracDelta.m For convenience, introduce horthand notation for the unit tep and unit ramp function u[t_]:= UnitStep[t]; r[t_]:= t u[t]; Here' the graph and the tranform of a function built out of ramp. Plot[ 2r[t] - 3r[t-] + r[t-5],{t,-,7},tick->{{,5},none}, PlotStyle->{{GrayLevel[.5], Thickne[.2]}}]; 5 LaplaceTranform[2r[t] - 3r[t-] + r[t-5],t,] E E Here' the graph of an exponential pule and it tranform Plot[E^(2t) (u[t] - u[t-2]),{t,-2,}, Tick->{{2},None}, PlotStyle->{{GrayLevel[.5], Thickne[.2]}}]; 2 LaplaceTranform[E^(2t) (u[t] - u[t-2]),t,] (-2 + ) E (-2 + ) But Mathematica couldn't take the tranform of a ine pule. It gave an incomplete anwer LaplaceTranform[Sin[t] (u[t] - u[t-2]),t,] LaplaceTranform[Sin[t] UnitStep[-2 + t], t, ] 2 +

42 page 2 of appendix to Chapter 5 Here' the invere tranform of e-5 (t-5) 3. Mathematica give the anwer 4 3! but all cubed out. u(t-5) InvereLaplaceTranform[E^(-5)/^4,,t] t 5 t t (-(---) ) UnitStep[-5 + t] Here' the invere tranform of multiple out the factorial. (-a). Anwer i e at t 9 9! but Mathematica InvereLaplaceTranform[/(-a)^,,t] a t 9 E t Here' one where Mathematica doen't get the implet poible anwer until you do a little algebra yourelf and then make it ue ome trig. InvereLaplaceTranform[Pi E^(-)/(^2 + Pi^2),,t] Sin[Pi (- + t)] UnitStep[- + t] Sin[-Pi + Pi t] u[- + t]//trigreduce -(Sin[Pi t] UnitStep[- + t])

43 page of appendix 2 to Chapter 5 APPENDIX 2 PARTIAL FRACTION DECOMPOSITION decompoition with non-repeated linear factor Let F() = (+2)(-4). There i a decompoition of the form (3) (+2)(-4) = A + B +2 + C -4 Since the factor are not repeated (i.e., there are no factor in the denominator like (+2) 2 or 3 ) it' eay to get A,B,C. You can ue any method you may remember from calculu but here' how I do it. To find A, delete the factor from the left ide of (3) and et = To find B, delete the factor +2 from the left ide of (3) and et = -2 To find C, delete the factor -4 from the left ide of (3) and et = 4. So A = (+2)(-4) = = -2-8 = 4 warning Thi method only work for non-repeated linear factor. So and B = (-4) = -2 C = (+2) =4 F() = /4 = -2 2 = - 6 = 2 - / /2-4 f(t) = [ 4-6 e-2t + ] 2 e4t u(t) decompoition with non-repeated non-factorable quadratic factor Let F() = ( ) The factor doen't factor (becaue b 2-4ac < ). decompoition of the form There i a (4) Then ( ) = A + B + C A = ( ) = = 5 To find B and C, multiply by ( ) in (4) to get = A( ) + (B + C) Equate coeff of 2 : = A + B, B = - A = - 5 Equate coeff of : = -2A + C, C = 2A = 2 5 So

44 page 2 of appendix 2 to Chapter 5 F() = / The firt term invert to 5 u(t) For the econd term, complete the quare in the denom ( ) = (-) and then rearrange the numerator a follow to ''match'' : (-) = - 5 (-) - (-) = (-) (-) Ue -hifting to get the invere tranform - 5 et co 2t u(t) + 5 Then the final invere tranform i 2 et in 2t u(t) [ 5-5 et co 2t + et in 2t u(t) ] decompoition with repeated linear factor (-5) 4 = A -5 + B (-5) 2 + C (-5) 3 + D (-5) 4 = A(-5) 3 + B(-5) 2 + C(-5) + D et = 5: equate 3 coeff: 5 = D = A equate 2 coeff: (who care) A + B =, B = et = 6: 6 = A + B + C + D, C = So (-5) 4 = (-5) (-5) 4 Ø t 2 + 5t3 2 3! e 5t u(t) decompoition with repeated non-factorable quadratic factor Too ugly.