Facilitator Guide. Unit 10

Transcription

1 Faciliaor Guide Uni 0

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3 UNIT 0 Faciliaor Guide ACTIVITIES NOTE: A many poins in e aciviies for Maemaics Illuminaed, worksop paricipans will be asked o explain, eier verbally or in wrien form, e process ey use o answer e quesions posed in e aciviies. Tis serves wo purposes: for e paricipan as a suden, i elps o solidify any previously unfamiliar conceps a are addressed; for e paricipan as a eacer, i elps o develop e skill of eacing sudens wy, no jus ow, wen i comes o confroning maemaical callenges. NOTE: Insrucions, answers, and explanaions a are mean for e faciliaor only and no e paricipan are in grey boxes for easy idenificaion.

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5 UNIT 0 (45 minues) WAVE ARITHMETIC Maerials Wave Arimeic Worksee eac person needs four copies Graping calculaor (opional) Colored pencils Uni 0

6 UNIT 0 Wave Arimeic Worksee a s s s 4s 5s 6s 7s 8s -a a s s s 4s 5s 6s 7s 8s -a a a s s s 4s 5s 6s 7s 8s -a -a Uni 0

7 UNIT 0 Tis aciviy is o be done in groups of ree. In e ex for Uni 0, Harmonious Ma, we saw ow a periodic funcion could be modeled by a roaing weel. In is aciviy, you will use is model o make ime domain graps of periodic funcions. You will en use ese represenaions o perform simple arimeic, addiion and muliplicaion, using waves. Adding waves ogeer is e fundamenal idea a e ear of Fourier series and will serve as e basic skill required for all of e aciviies in is uni. A 50 Imagine a spinning bicycle weel of radius A wi one pained spoke (sown below). ω a Verical componen Hori onal componen. Te spoke complees eig revoluions every 64 seconds. Wa is is period in seconds? Answer: T = 8 seconds (one revoluion every 8 seconds). Te spoke sweeps ou an angle equal o π radians in every revoluion. Wa is is angular velociy in radians per second? (π rad) Answer: angular velociy = ω = (8 sec) = π 4 rad/sec. Wa size angle does e spoke sweep ou in seconds? Answer: sec x π 4 rad/sec = π 4 radians Uni 0

8 UNIT 0 4. Assume a e spoke sars a ime zero wi an angle of zero radians. Wa funcion gives e verical componen of e ip of e spoke afer ime? Hin : Draw a riangle. Answer: A sin ω 5. Complee e following car: Time (sec) Angle of spoke (rad) 0 π 4 5π 4 Verical componen of posiion of e ip of e spoke (as a porion of spoke leng, A) π ( 4 ) = ( ) a 0 A sin 0.7A Answer: Time (sec) Angle of spoke (rad) 0 π 4 π π 4 π 5π 4 π 7π 4 π Verical componen of posiion of e ip of e spoke (as a porion of spoke leng, A) A sin (0) = 0 A sin π ( 4 ) = ( ) a 0.7A A sin π = A x = A A sin π 4 = ( ) a 0.7A A sin (π) = 0 A sin (5π/4) = - ( ) a ~ -0.7A A sin π = A x - = -A A sin 7π 4 = ( ) - a ~ -0.7A A sin (π)= 0 Uni 0 4

9 UNIT 0 6. Use e grap you jus compleed, along wi e Wave Arimeic Worksee, (using a color oer an black) o make a grap of e verical componen of e posiion of e ip of e spoke in e ime domain (wic means a e orizonal axis of your grap sould be denominaed in seconds). Te domain of your grap 50 sould be [0,8], and e range sould be [-A, A]. Draw your grap on e worksee were i says grap. ω a Verical componen Hori onal componen B. Wrie a funcion a gives e orizonal componen of e posiion of e ip of e spoke a ime. Answer: A cos ω On your Wave Arimeic Worksee, were i says grap, use a differen color o grap e orizonal posiion of e ip of e spoke in e ime domain. Noe a is grap sould ave e same domain and range as e previous grap. If you need elp geing sared, make a car similar o e one you made for e previous grap. Answer: For answer, see e answer o A #6. Uni 0 5

10 UNIT 0 C On e ird se of axes on e Wave Arimeic Worksee, grap e sum of e wo waves a you graped in A and B: Asin ω + Acos ω. To do is, simply add ogeer e y-values a eac ime sep and grap is composie value. Connec e poins you grap using a smoo curve. For example, a ime zero, Asin ω = 0 and Acos ω = A, so: Asin ω + Acos ω = 0 + A = A Te firs poin graped sould erefore be (0,A).. Before you grap, wy is e range [-A,A] insead of [-A,A]? Answer: Since we are adding values, i is possible o ave range values greaer an A; if bo componens of e sum were a eir max, e sum would be A. Conversely, if bo funcions were a a minimum, e sum would be A. For e answer o is grap, see A #6.. Wa is e period of is combined funcion? Wa is is angular velociy? (π rad) Answer: Te period is 8 secs; e angular velociy is π rad/period = (8 sec) = π 4 rad/sec. How do e period and angular velociy of e combined funcion compare o e period and angular velociy of e componen funcions? Answer: Te period and angular velociy of a sum of periodic funcions aving e same period and angular velociy are equal o e periods and angular velociies of e componen funcions. Wen all groups are finised wi C, convene e large group and discuss e resuls. Noe a e period of e composie funcion is equal o e periods of e componen funcions. Uni 0 6

11 UNIT 0 5 Answer: Wave Arimeic Worksee for Secions A roug C Grap a = a sin ω ω = π 8 = π 4 s s s 4s 5s 6s 7s 8s s s s 4s 5s 6s 7s 8s a sin π/4= ( )/a.7a a sin π/=a a sin π/4= ( )/a x.7a a sin π=o a sin 5π/4= -( )/a -.7a a sin π/= -a a sin 7π/4= -.7a a sin π=o -a Grap a = a cos ω s s s 4s 5s 6s 7s 8s ω = π 8 = π 4 s s s 4s 5s 6s 7s 8s a cos π/4= ( )/a.7a a cos π/=o a cos π/4= ( )/a -.7a a cos π= -a a cos 5π/4= -( )/a -.7a a cos π/= o a cos 7π/4= ( )/.7a a cos π=a -a a Grap Grap + a.4a 0+a=a = a sin ω 0+a=a.7a+.7a=.4a + a cos ω 0+a=a ω = π 4.7a+-.7a=0.7a-.7a=0 s s s 4s 5s 6s 7s 8s -a 0+-a=-a -a+0=-a -a -.7a+-.7a=-.4a Uni 0 7

12 UNIT 0 D Have pleny of copies of e Wave Arimeic Worksee available. In your group, use e Wave Arimeic Worksees o find e following sums and producs. Divide e work as you see fi (no one sould do all six eac person sould do wo, one addiion and one muliplicaion). Be ready o analyze and discuss your resuls as a group. Grap : Grap : Grap : Grap 4: = A sin ω = A cos ω = c (an arbirary consan) = A sin ω Sums: Problem : Grap + Grap Problem : Grap + Grap Problem : Grap + Grap 4 Noe: muliplicaion of graps works jus like addiion, excep a you muliply e y-values a eac ime-sep raer an add. Producs: Problem 4: Grap * Grap Wa is e area under is curve? Problem 5: Grap * Grap 4 Problem 6: Grap * Grap Compare e areas under e curves (one period) for producs of waves of e same frequency and producs of waves of differen frequencies. Hin : You can do is visually by couning squares. You can ink of squares above e x-axis as unis of posiive area and squares below e x-axis as unis of negaive area. Uni 0 8

13 UNIT 0 Anwser: Wave Arimeic Worksee for Problem : Grapic + Grap GraPH a c -a =c (consan) s s s 4s 5s 6s 7s 8s =c c c c c c c c c c GraPH a.7a =a sin ω s s s 4s 5s 6s 7s 8s a sin (0) = ax0 = 0 a sin (π/4) = a( )/ 0.7a a sin (π/)=ax=a a sin (π/4)=a( )/ 0.7a a sin (π)=a x 0 = 0 a sin (5π/4)= a(- )/ -0.7a a sin (π/)=ax-= -a a sin (7π/4)=a(- )/ -0.7a a sin (π)= a x 0 = 0 -.7a -a a GraPH GraPH +.4a a.7a c -c s s s 4s 5s 6s 7s 8s -.7a -a -.4a -a Uni 0 9

14 UNIT 0 Anwser: Wave Arimeic Worksee for Problem : Grapic + Grap GraPH a.7a -.7a =a sin ω s s s 4s 5s 6s 7s 8s a sin (0)=ax0=0 a sin (π/4)=a( )/ 0.7a a sin (π/)=ax=a a sin (π/4)=a( )/ 0.7a a sin (π)=a x 0 = 0 a sin (5π/4)= a(- )/ -0.7a a sin (π/)=ax-=-a a sin (7π/4)=a(- )/ -0.7a a sin (π)= a x 0 = 0 -a GraPH a.7a a= sin ω s s s 4s 5s 6s 7s 8s a sin (0)=ax0=0 a sin (π/4)=a( )/ 0.7a a sin (π/)=ax=a a sin (π/4)=a( )/ 0.7a a sin (π)=a x 0 = 0 a sin (5π/4)= a(- )/ -0.7a a sin (π/)=ax-=-a a sin (7π/4)=a(- )/ -0.7a a sin (π)= a x 0 = 0 -.7a -a a GraPH GraPH +.4a a a sin ω + a sin ω a sin ω s s s 4s 5s 6s 7 8s -a -.4a -a Uni 0 0

15 UNIT 0 Wave Arimeic Worksee for Problem : Grap + Grap 4 GraPH a.7a -.7a =a sin ω s s s 4s 5s 6s 7s 8s a sin (0)=ax0=0 a sin (π/4)=a( )/ 0.7a a sin (π/)=ax=a a sin (π/4)=a( )/ 0.7a a sin (π)=a x 0 = 0 a sin (5π/4)= a(- )/ -0.7a a sin (π/)=ax-=-a a sin (7π/4)=a(- )/ -0.7a a sin (π)= a x 0 = 0 -a GraPH 4 a =a sin ω s s s 4s 5s 6s 7s 8s a sin (0)=0 a sin (π/4)=a sin(π/) = ax = a a sin (π/)=a sin(π) = 0 a sin (π/4)=a sin(π/) = ax- = -a a sin (π)=a sin(π) = 0 a sin (5π/4)= a sin(5π/) = ax = a a sin (π/) = sin (π) = 0 a sin (7π/4)= a sin (7π/) = ax- = -a a sin (π)= a sin (4π) = 0 -a a GraPH GraPH 4 +.7a.4a a =a sin ω + a sin ω.a -.a s s s 4s 5s 6s 7 8s -a -.4a -.7a -a Uni 0

16 UNIT 0 55 Wave Arimeic Worksee for Problem 4: Grap *Grap GraPH a c -a s s s 4s 5s 6s 7s 8s =c c c c c c c c c c =c GraPH a.7a s s s 4s 5s 6s 7s 8s a sin (0)=ax0=0 a sin (π/4)=a( )/ 0.7a a sin (π/)=ax=a a sin (π/4)=a( )/ 0.7a a sin (π)=a x 0 = 0 a sin (5π/4)= a(- )/ -0.7a a sin (π/)=ax-=-a a sin (7π/4)=a(- )/ -0.7a a sin (π)= a x 0 = 0 =a -.7a -a a GraPH GraPH x 7.4a a.7a c -c a -a -.4a -a Uni 0

17 UNIT 0 56 Wave Arimeic Workseefor Problem 5: Grap * Grap 4 GraPH a.7a =a sin ω s s s 4s 5s 6s 7s 8s a sin (0)=ax0=0 a sin (π/4)=a( )/ 0.7a a sin (π/)=ax=a a sin (π/4)=a( )/ 0.7a a sin (π)=a x 0 = 0 a sin (5π/4)= a(- )/ -0.7a a sin (π/)=ax-=-a a sin (7π/4)=a(- )/ -0.7a a sin (π)= a x 0 = 0 -.7a -a GraPH 4 a =a sin ω a sin (0)=0 a sin (π/4) = a sin(π/)=a x = a a sin (π/) = a sin(π)=0 a sin (π/4) = a sin(π/)=a x-=-a a sin (π)=a sin(π)=0 a sin (5π/4) = a sin(5π/)= a x =a a sin (π/) = sin(π) = 0 a sin (7π/4) = a sin(7π/) = a x - = -a a sin (π) = a sin (4π) = 0 s s s 4s 5s 6s 7s 8s -a a GraPH GraPH x 4.7a a a -a -a Uni 0

18 UNIT 0 57 Wave Arimeic Worksee for Problem 6: Grap * Grap GraPH.7a a =asin ω a sin (0)=ax0=0 a sin (π/4)=a( )/ 0.7a a sin (π/)=a x =a a sin (π/4)=a( )/ 0.7a a sin (π)=a x 0 = 0 a sin (5π/4)= a(- )/ -0.7a a sin (π/)=ax-=-a a sin (7π/4)=a(- )/ -0.7a a sin (π)= a x 0 = 0 -.7a a GraPH.7a a =acos ω a cos (0)=ax=a a cos (π/4)=a( )/ 0.7a a cos (π/)=0 a cos (π/4)=a(- )/ -0.7a a cos (π)=a x - = -a a cos (5π/4)=a(- )/ -0.7a a cos (π/)=0 a cos (7π/4)=a( )/ 0.7a a cos (π)= a x = a -.7a a GraPH.5a -.5a Uni 0 4

19 UNIT 0 (0 minues) MATERIALS Grap paper Colored pencils Worksee: Cards for Fourier componens for groups A roug E Worksee: Cards for Fourier componens for groups A roug E Be sure o cu ou e cards for eac group before e aciviy. Divide e paricipans ino five groups (A, B, C, D, and E) of approximaely four people eac, wi as lile overlap wi previous groups as possible. A (0 minues) Making Waves In e previous aciviy, you added waves ogeer grapically o ge new waves. In is aciviy, you will use e ecniques you used in e las aciviy o creae wo well-known periodic funcions. Hand ou cards A and A o group A, B and B o group B, C and C o group C, ec. Eac person in your group sould ave wo cards, one wi a number and one wi a number. Card Number : Grap e wave-funcion sown on your card (number ), in e suggesed domain and range, on a piece of grap paper. As a group, decide on nine specific, evenly spaced, domain values for wic eac person will find range values. For example, if all four people ave a suggesed domain of [-,] en e agreed upon values could be {-, -0.75, -0.5, -0.5, 0, 0.5, 0.5, 0.75, and }. Eac person would use ese values o plo eir individual graps. Feel free o use a calculaor. Uni 0 5

20 UNIT 0 Afer eac person as graped is or er wave, add e four waves ogeer using e ecnique from Aciviy o ge a composie wave. Be sure o increase e range of e composie wave. Noe: i is muc easier o find e new range if you firs simplify e expression on your card so a e coefficien is one number. Wen all groups ave creaed eir composie waves, make a large grap on eier e blackboard, wieboard, or overead. Lead e group in consrucing e sum of e five composie waves. Wi any luck, you sould ave a riangle wave! Uni 0 6

21 UNIT 0 Fourier componens for groups A roug E Cards o cu ou; page one of wo Group A n=0, (-5 < x < 5,. < y,.) y <.) A n=, (-5 < x < 5,. < y,.) y <.) A ( ) 0 sin (( 0+)x) ( 0+) ( ) sin (( +)x) ( +) n=, (-5 < x < 5,. < y,.) y <.) A n=, (-5 < x < 5,. < y,.) y <.) A ( ) sin (( +)x) ( +) ( ) sin (( +)x) ( +) Group B n=4,( < x <,.< y <.) B n=6,( < x <,.< y <.) B ( ) 4 sin (( 4 +)x) ( 4 +) ( ) 6 sin (( 6+)x) ( 6+) n=5,( < x <,.< y <.) B n=7,( < x <,.< y <.) B ( ) 5 sin (( 5+)x) ( 5+) ( ) 7 sin (( 7 +)x) ( 7 +) Uni 0 7

22 UNIT 0 Fourier componens for groups A roug E Cards o cu ou; page wo of wo Group C n=8, ( -.5 < x <.5, -.0< y <.0 ) C n=0, ( -.5 < x <.5, -.0< y <.0 ) C ( ) 8 sin (( 8 +)x) ( 8 +) ( ) 0 sin (( 0+)x) ( 0+) n=9, ( -.5 < x <.5, -.0< y <.0 ) C n=, ( -.5 < x <.5, -.0< y <.0 ) C ( ) 9 sin (( 9+)x) ( 9+) ( ) sin (( +)x) ( +) Group D n=, ( -. < x <., < y <.005 ) D n=4, ( -. < x <., < y <.005 ) D ( ) sin (( +)x) ( +) ( ) 4 sin (( 4 +)x) ( 4 +) n=, ( -. < x <., < y <.005 ) D n=5, ( -. < x <., < y <.005 ) D ( ) sin (( +)x) ( +) ( ) 5 sin (( 5+)x) ( 5+) Group E n=6, ( -. < x <., -.00< y <.00 ) E n=8, ( -. < x <., -.00< y <.00 ) E ( ) 6 8 ( 6+) sin (( 6+)x) ( ) sin (( 8 +)x) ( 8 +) n=7, ( -. < x <., -.00< y <.00 ) E n=9, ( -. < x <., -.00< y <.00 ) E ( ) 7 sin (( 7 +)x) ( 7 +) ( ) 9 sin (( 9+)x) ( 9+) Uni 0 8

23 UNIT 0 Answers: Fourier componens for groups A roug E riangle wave n=0, {-5<x<5, -.<y<.} n=0 (-) 0 (. sin((. 0+)x) 0+) n=, {-5<x<5, -.<y<.} n= (-) (. sin((. +)x) +) n=, {-5<x<5, -.<y<.} n= (-) (. sin((. +)x) +) n=, {-5<x<5, -.<y<.} n= (-) (. sin((. +)x) +) n=4, {-<x<, -.<y<.} n=4 (-) 4 (. sin((. 4+)x) 4+) Uni 0 9

24 UNIT 0 Answers: Fourier componens for groups A roug E riangle wave n=5, {-<x<, -.<y<.} n=5 (-) 0 (. sin((. 5+)x) 5+) n=6, {-<x<, -.<y<.} n=6 (-) 6 (. sin((. 6+)x) 6+) n=7, {-<x<, -.<y<.} n=7 (-) 7 (. sin((. 7+)x) 7+) n=8, {-.5<x<.5, -.0<y<.0} n=8 (-) 8 (. sin((. 8+)x) 8+) n=9, {-.5<x<.5, -.0<y<.0} n=9 (-) 9 (. sin((. 9+)x) 9+) Uni 0 0

25 UNIT 0 Answers: Fourier componens for groups A roug E riangle wave n=0, {-.5<x<.5, -.0<y<.0} n=0 (-) 0 (. sin((. 0+)x) 0+) n=, {-.5<x<.5, -.0<y<.0} n= (-) (. sin((. +)x) +) n=, {-.<x<., -.005<y<.005} n= (-) (. sin((. +)x) +) n=, {-.<x<., -.005<y<.005} n= (-) (. sin((. +)x) +) n=4, {-.<x<., -.005<y<.005} n=4 (-) 4 (. sin((. 4+)x) 4+) Uni 0

26 UNIT 0 Answers: Fourier componens for groups A roug E riangle wave n=5, {-.<x<., -.005<y<.005} n=5 (-) 5 (. sin((. 5+)x) 5+) n=6, {-.<x<., -.00<y<.00} n=6 (-) 6 (. sin((. 6+)x) 6+) n=7, {-.<x<., -.00<y<.00} n=7 (-) 7 (. sin((. 7+)x) 7+) n=8, {-.<x<., -.00<y<.00} n=8 (-) 8 (. sin((. 8+)x) 8+) n=9, {-.<x<., -.00<y<.00} n=9 (-) 9 (. sin((. 9+)x) 9+) Uni 0

27 UNIT 0 Fourier riangle wave composie Wen all groups combine eir graps, is is one way i could look: Fourier componens Group resuls: Group A: n = 0 roug n = { -5<x<5, -.<y<. }; o accommodae: { -5<x<5, -.6<y<.6 } 6 Uni 0

28 UNIT 0 Group B: n = 4 roug n = 7 { -<x<, -.<y<. } 6 Group C: n = 8 roug n = { -.5<x<.5, -.0<y<.0 } Uni 0 4

29 UNIT 0 Group D: n = roug n = 5 { -.<x<., -.005<y<.005 }; Group E: n = 6 roug n = 9 { -.<x<., -.00<y<.00 }; Uni 0 5

30 UNIT 0 Wen all groups combine eir daa, e formula is is: y = 9 n =0 ( ) n ( n +) sin n + ( )x ( ) A grap could look like is: { -5<x<5, -<y< } Uni 0 6

31 UNIT 0 If we zoom ou, o e range and domain { -0<x<0, -<y< }, a grap could look like is: Uni 0 7

32 UNIT 0 Cards for Fourier componens for groups A roug E Cards o cu ou; page one of wo Group A n=6, ( -.5 < x <.5, -.05 < y <.05 ) A n=8, ( -.5 < x <.5, -.05 < y <.05 ) A 6+ sin (( 6+)x) 8 + sin (( 8 +)x) n=7, ( -.5 < x <.5, -.05 < y <.05 ) A n=9, ( -.5 < x <.5, -.05 < y <.05 ) A 7 + sin (( 7 +)x) 9+ sin (( 9+)x) Group B n=, ( -.5 < x <.5, -.08 < y <.08 ) B n=4, ( -.5 < x <.5, -.08 < y <.08 ) B + sin (( +)x) 4 + sin (( 4 +)x) n=, ( -.5 < x <.5, -.08 < y <.08 ) B n=5, ( -.5 < x <.5, -.08 < y <.08 ) B + sin (( +)x) sin (( 5+)x) 5+ Uni 0 8

33 UNIT 0 Cards o cu ou; page wo of wo Group C n=8, ( -.5 < x <.5, -.08 < y <.08 ) C n=0, ( -.5 < x <.5, -.08 < y <.08 ) C 8 + sin (( 8 +)x) 0+ sin (( 0+)x) n=9, ( -.5 < x <.5, -.08 < y <.08 ) C n=, ( -.5 < x <.5, -.08 < y <.08 ) C 9+ sin (( 9+)x) + sin (( +)x) Group D n=4, ( -.75 < x <.75, -.5 < y <.5 ) D n=6, ( -.75 < x <.75, -.5 < y <.5 ) D 4 + sin (( 4 +)x) 6+ sin (( 6+)x) n=5, ( -.75 < x <.75, -.5 < y <.5 ) D n=7, ( -.75 < x <.75, -.5 < y <.5 ) D 5+ sin (( 5+)x) 7 + sin (( 7 +)x) Group E n=0, ( -.4 < x <.4, -.6 < y <.6 ) E n=, ( -.4 < x <.4, -.6 < y <.6 ) E 0+ sin (( 0+)x) + sin (( +)x) n=, ( -.4 < x <.4, -.6 < y <.6 ) E n=, ( -.4 < x <.4, -.6 < y <.6 ) E + sin (( +)x) sin (( +)x) + Uni 0 9

34 UNIT 0 Answers: Fourier componens for groups A roug E square wave n=0, {-.4<x<.4, -.6<y<.6} n=0. sin((. 0+)x) 0+ n=, {-.4<x<.4, -.6<y<.6} n=. sin((. +)x) + n=, {-.4<x<.4, -.6<y<.6} n=. sin((. +)x) + n=, {-.4<x<.4, -.6<y<.6} n=. sin((. +)x) + n=4, {-.75<x<.75, -.5<y<.5} n=4. sin((. 4+)x) 4+ Uni 0 0

35 UNIT 0 Answers: Fourier componens for groups A roug E square wave n=5, {-.75<x<.75, -.5<y<.5} n=5. sin((. 5+)x) 5+ n=6, {-.75<x<.75, -.5<y<.5} n=6. sin((. 6+)x) 6+ n=7, {-.75<x<.75, -.5<y<.5} n=7. sin((. 7+)x) 7+ n=8, {-.5<x<.5, -.08<y<.08} n=8. sin((. 8+)x) 8+ n=9, {-.5<x<.5, -.08<y<.08} n=9. sin((. 9+)x) 9+ Uni 0

36 UNIT 0 7 Answers: Fourier componens for groups A roug E square wave n=0, {-.5<x<.5, -.08<y<.08} n=0. sin((. 0+)x) 0+ n=, {-.5<x<.5, -.08<y<.08} n=. sin((. +)x) + n=, {-.5<x<.5, -.08<y<.08} n=. sin((. +)x) + n=, {-.5<x<.5, -.08<y<.08} n=. sin((. +)x) + n=4, {-.5<x<.5, -.08<y<.08} n=4. sin((. 4+)x) 4+ Uni 0

37 UNIT 0 7 Answers: Fourier componens for groups A roug E square wave n=5, {-.5<x<.5, -.08<y<.08} n=5. sin((. 5+)x) 5+ n=6, {-.5<x<.5, -.05<y<.05} n=6. sin((. 6+)x) 6+ n=7, {-.5<x<.5, -.05<y<.05} n=7. sin((. 7+)x) 7+ n=8, {-.5<x<.5, -.05<y<.05} n=8. sin((. 8+)x) 8+ n=9, {-.5<x<.5, -.05<y<.05} n=9. sin((. 9+)x) 9+ Uni 0

38 UNIT 0 Fourier componens Group resuls: Group A n = 6 roug n = 9, {-.5<x<.5, -.5<y<.5} Group B n = roug n = 5, {-.5<x<.5, -.<y<.} Uni 0 4

39 UNIT 0 Group C n = 8 roug n =, {-.5<x<.5, -.5<y<.5} Group D n = 4 roug n = 7, {-.75<x<.75, -.5<y<.5} Uni 0 5

40 UNIT 0 Group E n = 0 roug n =, { -.4<x<.4, -.6<y<.6 } Wen all groups combine eir daa, e formula is is: y = 9 n =0 n + A grap could look like is: { -.4<x<.4, -.6<y<.6 } sin (( n + )x ) Uni 0 6

41 UNIT 0 If we zoom ou, o e range and domain { -0<x<0, -7.5<y<7.5 }, a grap could look like is: Uni 0 7

42 UNIT 0 Faciliaor s noe: No all groups will ave seleced e same domain values o grap. Before e five composie waves can be added, you sould come o a consensus as a large group as o wic domain values o use. Te bes way o do is is o use e domain values from group C. Te oer groups will ave o do a few more calculaions so a every domain value of e sum of composie waves as a conribuion from every group. Tis demonsraion will ave e mos impac if you empasize e idea a you ave aken 0 curvy waves and made a more-or-less recilinear riangle wave. Tis demonsraes e power of Fourier analysis o represen ANY periodic funcion as a sum of sine waves. Card Number : Repea e process you used for card number. Group C s domain sould again provide e domain for e final group sum. Wen you are finised, you sould ave a square wave! Uni 0 8

43 UNIT 0 B (0 minues) Tinking Furer. Look a e funcions on e number cards. Tey sould all be very similar excep a one number is differen in eac. Tis number is e index value of e sum of waves a add ogeer o make a riangle wave. Use summaion noaion o wrie e general formula for e sum of waves a form e riangle wave. Answer: y = n max n =0 ( ) n sin (( n + )x ) ( n +). Do e same for e funcions on e number cards. Answer: n max y = n =0 n + ( ) sin ( n +)x Uni 0 9

44 UNIT 0 (40 minues) Faciliaor s noe: Be sure a bo Aciviy and Aciviy are complee before aemping is. Te power of Fourier analysis lies in is abiliy o ake a complicaed periodic signal and break i down ino e componen frequencies a make i up. I s a bi like a professional ase-eser wo can sample a complex sauce and ell e componen ingrediens a wen ino i ofen e or se can even ge a roug sense of e relaive amouns of eac componen used. In is aciviy, you will be given a relaively complicaed waveform and be asked o deduce is ingredien frequencies and eir relaive amouns, also known as Fourier coefficiens. In essence, you will be deducing a complicaed recipe by asing, and adding a das of reason. Take a look a is wave: Call i Wave. In e grap, is e ime, in seconds, and is e eig of e wave, in generic unis. Noice a is wave is neier a pure sine wave nor a pure cosine wave. I is acually a composie of ree ingredien waves. To find e equaions for e ingredien waves, we need o know ree ings: ) weer ey are sine or cosine waves; ) e frequencies of e ingredien waves, and ) e ampliudes of e ingredien waves. Uni 0 40

45 UNIT 0 To keep ings simple, we will assume a e ingredien waves are sine waves. A is poin, ave e paricipans work in groups. Par : Finding Frequencies. Recall from e ex a a general Fourier expansion looks like is: f() = a 0 + a cos ω + b sin ω + a cos ω + b sin ω + a cos ω + b sin ω + In is example we are concerned only wi sine waves, so we can simplify e expansion o: f() = a 0 + b sin ω + b sin ω + b sin ω + were ω is e fundamenal angular velociy in radians/second. To sar off, in your own words, explain e relaionsip beween a wave s period and is fundamenal angular velociy. Give an example. Answer: ω is e angular velociy or ow fas a wave oscillaes. Te period is e ime i akes a wave o make one full oscillaion, wic is e same as saying i s e ime i akes a spoke o sweep ou an angle of π radians. One can find ω by dividing π by e period. For example, if a wave as a period of 0 seconds, ω will be π = π rad/sec Look a e grap of Wave. Wa is e period of is wave? Answer: Te period is e ime i akes for e wave o complee one cycle. Te x-axis is denominaed in seconds, so one can see a i akes seconds for Wave o complee one cycle; erefore, e period is seconds.. Noe a ω is e fundamenal frequency e frequency of e firs sine erm in e Fourier expansion. Te frequencies of all e oer sine erms are wole number muliples of ω. Using e value you found for e period of Wave, find ω for Wave. Answer: ω = π/period = π secs = π rad/sec 4. Wrie e firs ree erms of e Fourier expansion for Wave using e value of ω you jus found. Leave e coefficiens as unknowns for e ime being. Answer: f() = a 0 + bsin π + bsin π Uni 0 4

46 UNIT 0 Now a you ave found e frequencies of e ingredien waves, i s ime o find e coefficiens, wic deermine e amoun a eac ingredien wave conribues o Wave. Convene e large group and discuss. Par : Finding Coefficiens Because ere are ree ingredien waves, we need o find ree coefficiens, a 0, b, and b. To find ese, we need filers a can deermine e effec a eac as on Wave. Imagine a Wave is a soup made of bro, large cunks of poaoes, and medium cunks of beef. To separae ou e poaoes, we could run e soup roug a srainer (a ype of filer) a as oles large enoug o le e beef cunks and bro pass roug. Ten o separae ou e beef, we could run e now poao-less soup roug anoer srainer, is one wi oles small enoug o deain e beef cunks and le only e bro pass roug. Once e poaoes, beef, and bro ave been separaed, i is en easy o figure ou e relaive amouns of eac. Te firs filer for Wave sould ell us wa a 0 is. To make is filer, le s ink abou e area under a curve.. On a piece of grap paper, grap eac of e ree ingredien waves separaely in e domain of [0,] seconds. Label e verical axes in erms of a 0, b, and b, respecively. Answer: (e ree ingredien waves sown on e following page) Uni 0 4

47 UNIT 0 Te value is a 0. Te Ampliude is b. Uni 0 4

48 UNIT 0 Te ampliude is b.. Find e areas undernea eac curve in e given domain. Coun area below e -axis as negaive. (Noe o ose wo mig be emped o use inegraion o find ese areas: you do no need o use calculus for is; i can be done jus by looking a e graps.) Answer: Te area under y = a 0 is a 0 (Tis can be found by looking a e recangle, wic is seconds in leng and a 0 in eig.) Te areas under bo y = b sin πx and y = b sin πx are zero, because ey ave equal areas above and below e x-axis, wic cancel ou eac oer.. To find e area of a complicaed sape, you can break i ino simple sapes, find e area of eac, and en add eir areas ogeer o find e area of e wole sape. Because Wave is made up of ree ingredien waves, only one of wic, a 0, as any area, i sands o reason a any area under Wave is equal o e area under y = a 0. Explain ow is could be used o find e value of a 0. Faciliaor s noe: Encourage paricipans o discuss is in eir groups before wriing an answer. Answer: If we could find e area under Wave for one period, is area mus be equal o e area under y = a 0 for one period, wic is in urn equal o a 0 x T, were T is e period. So, we can find e area under Wave in one period and divide is area by e period o find a 0. Uni 0 44

49 UNIT 0 Convene e large group and make sure everyone undersands e answer o e preceding quesion before coninuing. 4. Find a 0. Answer: Te area under Wave for one period is uni-seconds. Divide is by e period of seconds o ge a 0 = uni. 5. Wrie e Fourier expansion for Wave using e informaion you ave found so far. Answer: f() = + b sin π + b sin π 6. Now, o find b and b, we re going o need wo more filers. Unforunaely, we can use e same area filer a we used o find a 0 because bo bsin π and b sin π ave areas of zero. However, we can use a rick. Look a your conclusion from Aciviy regarding e areas under curves of producs of waves aving e same frequency and producs of waves aving differen frequencies. Summarize your findings. Answer: Te produc of wo waves aving e same frequency will yield a curve a as a non-zero area. Te produc of wo waves aving differen frequencies will yield a curve a as zero area. 7. Wrie ou f() x sin π. Answer: f() x sin π = ( + bsin π + b sin π)(sin π) = sin π + b (sin π) (sin π) + b (sin π)(sin π) 8. For eac erm of f() x sin π, discuss is conribuion o e area under e curve over one period. Tere is no need o find exac values ye jus say weer or no e erm conribues o e area. Answer: sin π as an area of zero; b (sin π)(sin π) as a non-zero area; and b (sin π)(sin π) also as an area of zero. Te only erm a conribues any area o f() x sin π is b (sin π)(sin π). 9. Look a e grap of y = (sin π)(sin π). How does e period of is wave compare o e period of y = sin π? How does i compare o e period of f()? Uni 0 45

50 UNIT 0 Grap: = (sin π)(sin π) Answer: Te period is alf a of sin π. I is also alf e period of f() 0. Wa is e area under = (sin π)(sin π) from = 0 o = seconds (one period of f()? Hin : Tink of a jigsaw puzzle no calculus is required! Try o make a recangle. Answer: Te area is uni-second. To find is, leave e ump beween 0 and second. Take e ump beween and seconds and spli i in alf along is verical axis of symmery. Fi ese wo pieces (you ll need o urn em upside down firs) ino e inerval from 0 o second o make a x square. Uni 0 46

51 UNIT Wa is e area under = (sin π)(sin π)? How abou e area under = (sin π)(sin π)? Grap: = (sin π)(sin π) Uni 0 47

52 UNIT 0 Grap: = (sin π)(sin π) Answer: uni-seconds for e firs and uni-seconds for e second, by e meod used in e previous problem.. Wa is e relaionsip beween n and e area undernea = n(sin π)(sin π)? Answer: Te wo quaniies are e same.. So, e area undernea f() sin π is enirely due o e conribuion from b (sin π)(sin π). Also, as you jus discovered, e area under b (sin π)(sin π) is equal o b. Use ese wo pieces of informaion, in conjuncion wi e given grap of f() sin π, o find b. Hin : To find e area under f() x sin π, jus coun squares! Uni 0 48

53 UNIT 0 Grap: = (sin π)( + b sin π + b sin π) Answer: Te area under = (sin π)( + b sin π + b sin π) is equal o unisecond. Tis means a b equals. 4. Now a you ave b, wrie ou e Fourier expansion of Wave so far: Answer: f() = + ()sin π + b sin π Uni 0 49

54 UNIT 0 5. So, if e filer o find b was o muliply f() by sin π and find e area, by wa sould we muliply f() o find b? Wy? Answer: We sould muliply f() by sin π, because i will make all e erms excep for b (sin π)(sin π) give an area of zero. 6. Before you find b, ake a look a e grap of = (sin π)(sin π). Grap of = (sin π)(sin π). 5 5 f() Uni 0 50

55 UNIT 0 7. Wa is e area under is curve for e period of f() (0-> secs)? Use e jigsaw meod. Answer: Te area is of a uni-second Use e grap of f() sin π o find b. Grap of = (sin π)( + b sin π + b sin π) Uni 0 5

56 UNIT 0 Answer: By couning squares, e paricipans sould find a e area under = (sin π)( + b sin π + b sin π) is ½ of a uni-second. Tis makes b = ½. 9. A las, you finally ave all of e informaion necessary o wrie e complee Fourier expansion of f(). Do i! Answer: f() = + ()sin π + ½ sin π You now know e frequencies and amouns of all e ingredien waves in oer words, you ave e complee recipe o make Wave! IF TIME ALLOWS: Use e following graps o find e recipe for Wave. Sar wi four erms of e general Fourier expansion (using only sine waves again). f() = a 0 + b sin ω + b sin ω + b sin ω Be careful wen finding b, b, and b ; e fundamenal frequency is differen in is example, wic as implicaions for e area filers. Faciliaor s noe: o find e b n, e paricipans will ave o find e areas of e various produc graps over one period and en divide e resul by ½ of a period. Te period is second. Te fundamenal frequency is π. Te answer is: f() = + sin π + (0)sin 4π + sin 6π Uni 0 5

57 UNIT 0 Le s call i Wave. Answer: Ingredien Waves Grap: = 5 Te value is a 0 Grap: = 5 sin π Ampliude is b Uni 0 5

58 UNIT 0 Grap: = sin 6π 5 Te ampliude is b. Paricipan, Produc Graps: Grap: f() x sin π Uni 0 54

59 UNIT 0 Grap: f() x sin 4π Grap: f() x sin 6π Uni 0 55

60 UNIT 0 CONCLUSION (0 minues) DISCUSSION HOW TO RELATE TOPICS IN THIS UNIT TO STATE OR NATIONAL STANDARDS Faciliaor s noe: Have copies of naional, sae, or disric maemaics sandards available. Maemaics Illuminaed gives an overview of wa sudens can expec wen ey leave e sudy of secondary maemaics and coninue on ino college. Wile e specific opics may no be applicable o sae or naional sandards as a wole, ere are many connecions a can be made o e ideas a your sudens wresle wi in bo middle scool and ig scool ma. For example, in Uni, In Sync, e relaionsip beween slope and calculus is discussed. Please ake some ime wi your group o brainsorm ow ideas from Uni 0, Harmonious Ma could be relaed and broug ino your classroom. Quesions o consider: Wic pars of is uni seem accessible o my sudens wi no fronloading? Wic pars would be ineresing, bu mig require some amoun of preparaion? Wic pars seem as if ey would be overwelming or inimidaing o sudens? How does e maerial in is uni compare o sae or naional sandards? Are ere any overlaps? How mig cerain ideas from is uni be modified o be relevan o your curriculum? WATCH VIDEO FOR NEXT CLASS (0 minues) Please use e las 0 minues of class o wac e video for e nex uni: Connecing wi Neworks. Worksop paricipans are expeced o read e accompanying ex for Uni, Connecing wi Neworks before e nex session. Uni 0 56

61 UNIT 0 NOTES Uni 0 57