(b) y(t) is not periodic although sin t and 4 cos 2πt are independently periodic.

Size: px
Start display at page:

Download "(b) y(t) is not periodic although sin t and 4 cos 2πt are independently periodic."

Transcription

1 Chapter 7, Sluti. (a) his is peridic with ω which leads t /ω. (b) y(t) is t peridic althugh si t ad cs t are idepedetly peridic. (c) Sice si A cs B.5[si(A B) si(a B)], g(t) si t cs t.5[si 7t si( t)].5 si t.5 si7t which is harmic r peridic with the fudametal frequecy ω r /ω. (d) h(t) cs t.5( cs t). Sice the sum f a peridic fucti ad a cstat is als peridic, h(t) is peridic. ω r /ω. (e) (f) (g) he frequecy rati.6..5 makes z(t) peridic. ω. r /ω. p(t) is t peridic. g(t) is t peridic. Chapter 7, Sluti. (a) he frequecy rati is 6/5.. he highest cmm factr is. ω / r. (b) ω r /ω. (c) f (t) si 6 t (/)( cs t) ω r /ω /() /6. (d) f (t) e jt cs t jsi t. ω r /ω..

2 Chapter 7, Sluti., ω / / g(t) 5, < t <, < t <, < t < a (/) g(t)dt.5[ 5dt ] dt.75 a (/) g(t) cs(ω t) dt (/)[ 5 cs( t)dt cs( t)dt ].5[ 5 si t si t ] ( /())5 si(/) a (5/())( ) ()/, dd, eve b (/) g(t) si(ω t) dt (/)[ 5 si( t)dt si( t)dt ] x5.5[ cs t x cs t ] (5/())[ cs cs(/)] Chapter 7, Sluti. f(t) 5t, < t <,, ω / a (/) f (t)dt (/) ( 5t)dt.5[t (5t / )] 5 a (/) f (t) cs(ω t) dt (/) ( 5t) cs(t)dt ( ) cs(t)dt (5t) cs(t)dt 5 cs t 5t si t [ 5/( )](cs )

3 b (/) ( 5t) si(t)dt ( ) si(t)dt (5t) si(t)dt 5 si t 5t cs t [/()](cs ) /() Hece f(t) 5 si(t) Chapter 7, Sluti 5., ω / a z(t)dt [x x ].5 a z(t)cs ωdt cs tdt cs tdt si..t si t b hus, z(t).5 z(t)cs ωdt 6 dd si t si tdt si tdt cs t cs t 6, dd, eve

4 Chapter 7, Sluti 6., ω a y(t)dt (x x) 6 Sice thisis a dd fucti,a. b y(t) si(ω t)dt si(t)dt si(t)dt cs(t) cs(t) (cs() ) (cs() cs()) ( cs()) ( cs()) ( cs()),, eve dd y(t) dd si(t) Chapter 7, Sluti 7., ω /, a 6 a f (t)cs ωdt [ cs t / 6dt ( )cs t / 6dt] 6 si t / 6 si t / 6 b f (t)si ωdt [ si t / 6dt ( )si t / 6dt] 6 [ si / si / si 5 / ]

5 cs t / 6 cs t / 6 [ cs5 / cs / si / ] f (t) ( a cs t / 6 b si t / ) 6 where a ad b are defied abve. Chapter 7, Sluti 8. f (t) ( t), -< t <,, ω / a f (t)dt (t )dt t t t a f (t)cs dt (t )cs tdt cs t si t si t ω b t f (t)si ωdt (t )si tdt si t cst cst cs f (t) ( ) cs t Chapter 7, Sluti 9. f(t) is a eve fucti, b. 8, ω / / a f ( t) dt cs / ( )si / 8 t dt t.8

6 [ ]dt t t dt t t dt t f a / ) / ( cs ) / ( cs 5 ] / cs / cs [ 8 )cs ( ω Fr, / si 5 ] / [cs 5 t dt t dt t a Fr >, ) ( si ) ( ) ( si ) ( ) ( si ) ( ) ( si ) ( t a si si 6.66, / si si a a hus,, 6.6,,.8, b b b a a a a Chapter 7, Sluti. ω /, ω t j t j t j t j t j j e j e dt )e ( dt e dt h(t)e c [ ], eve, dd, 6j 6] [6cs j e e e j c j j j hus, t j dd e j6 f (t)

7 Chapter 7, Sluti., ω / / c jω jt / jt / y(t)e t dt (t )e dt ()e dt c jt e / / ( jt / ) j jt e / j jt e j / j / j / e (j / ) e e j j j / j But e j / cs / jsi / jsi /, e j / cs / jsi / jsi / c [ j(j / )si / si / ] y(t) j / [ j(j / )si / si / ] e t Chapter 7, Sluti. A vltage surce has a peridic wavefrm defied ver its perid as v(t) t( - t) V, fr all < t < Fid the Furier series fr this vltage. v(t) t t, < t <,, ω / a (/) f (t)dt (t t ) dt ( t ( / ) t / )

8 a (t t )cs(t)dt t cs(t) si(t) [ t cs(t) si(t) t si(t ] ) ( ) cs() b (t t )si(t)dt (t t )si(t) dt (si(t) t cs(t)) (t si(t) cs(t) t cs(t)) Hece, f(t) cs(t) Chapter 7, Sluti., ω a (/) h(t)dt [ si t dt si(t ) dt] [ cs t cs(t ) ] a (/) h(t) cs(ωt) dt [/()] si t cs(t)dt si(t ) cs(t)dt Sice si A cs B.5[si(A B) si(a B)] si t cs t.5[si(( )t) si(( ))t] si(t ) si t cs cst si si t si(t )cs(t) si(t)cs(t)

9 a [si([ ]t) si([ ]t)]dt [si([ ]t) si([ ]t)]dt 5 cs([ ]t) cs([ ]t) cs([ ]t) cs([ ]t) a 5 cs([ ] ) cs([ ] ) But, [/()] [/(-)] /( ) cs([ ]) cs([]) cs cs si si cs a (5/)[(6/( )) (6 cs()/( ))] [/(( ))]( cs ) [ 6/(( ))], eve, dd b (/) h(t) si ω t dt [/()][ si t si t dt ( si t) si t dt But, si A si B.5[cs(A B) cs(ab)] si t si t.5[cs([ ]t) cs([]t)] b (5/){[(si([ ]t)/( )) (si([]t)/ ( )] [(si([-]t)/(-)) (si([]t)/( )] } 5 si([ ] ) si([ ] ) 6 hus, h(t) cs(kt) (k ) k

10 Chapter 7, Sluti. Sice cs(a B) cs A cs B si A si B. f(t) cs( / )cs(t) si( / )si(t ) Chapter 7, Sluti 5. (a) Dcs ωt Esi ωt A cs(ωt - θ) where A D E, θ ta - (E/D) 6 A 6 ( ), θ ta - (( )/( )) f(t) 6 ( ) 6 cs t ta (b) Dcs ωt Esi ωt A si(ωt θ) where A D E, θ ta - (D/E) f(t) 6 ( ) 6 si t ta Chapter 7, Sluti 6. If v (t) is shifted by alg the vertical axis, we btai v * (t) shw belw, i.e. v * (t) v (t). v * (t) t

11 Cmparig v * (t) with v (t) shws that v * (t) v ((t t )/) where (t t )/ at t - r t Hece v * (t) v ((t )/) But v * (t) v (t) v (t) v ((t)/) v (t) - v ((t)/) 8 t t t - cs cs cs v (t) 8 t cs 9 t cs 5 5t cs 5 v (t) 8 t si 9 t si 5 5t si Chapter 7, Sluti 7. We replace t by t i each case ad see if the fucti remais uchaged. (a) t, either dd r eve. (b) t, eve (c) cs (-t) si (-t) - cs t si t, dd (d) si (-t) (-si t) si t, eve (e) e t, either dd r eve.

12 Chapter 7, Sluti 8. (a) leads t ω / f (-t) -f (t), shwig that f (t) is dd ad half-wave symmetric. (b) leads t ω / f (t) f (-t), shwig that f (t) is eve. (c) leads t ω / f (t) is eve ad half-wave symmetric. Chapter 7, Sluti 9. his is a half-wave eve symmetric fucti. a b, ω / / / a t cs(ωt) dt [/() ]( cs ) 8/( ), dd, eve f (t) 8 dd t cs Chapter 7, Sluti. his is a eve fucti. b, 6, ω /6 / a / f (t)dt (t )dt 6 dt

13 (t t) ( ) / a f (t) cs(t / )dt (/6)[ ( t ) cs(t / )dt cs(t / )dt ] t cs t t si t si 6 t si 6 [/( )][cs(/) cs(/)] hus f(t) At t, t cs cs cs f() (/ )[(cs(/) cs(/))cs(/) (/)(cs(/) cs(/))cs(/) (/9)(cs() cs())cs() -----].( ) f().756 Chapter 7, Sluti. his is a eve fucti. b,, ω / /. f(t) t, < t <, < t < t a ( t)dt t.5 a / f (t) cs(ω t)dt ( t t) cs dt

14 [8/( )][ cs(/)] f(t) 8 t cs cs Chapter 7, Sluti. Calculate the Furier cefficiets fr the fucti i Fig f(t) t Figure 7.6 Fr Prb. 7. his is a eve fucti, therefre b. I additi, ad ω /. a f (t)dt tdt t a f (t) cs( ω t)dt t cs(t / ) dt t cs(t / ) si(t / ) 6 8 a (cs( / ) ) si( / ) Chapter 7, Sluti. f(t) is a dd fucti. f(t) t, < t < a a,, ω /

15 b / f (t) si(ω t)dt t si(t) dt [ si(t) t cs(t ] ) [/()]cs() () /() f(t) ( ) si(t) Chapter 7, Sluti. (a) his is a dd fucti. a a,, ω / / b f (t) si( ωt) dt f(t) t/, < t < b ( t / ) si(t) dt t cs(t) si(t) cs(t ) [/()][ cs()] [/()][ () ] a, b [/()][ ()] /.8 (b) ω ω r a, b [/()][ cs()] /(5) hus the magitude is A a b /(5).666 ad the phase is φ ta (b /a ) 9

16 (c) f(t) [ cs()] si(t) f(/) [ cs()] si( / ) Fr, f (/)( ) 6/ Fr, f Fr, f [/()][ cs()]si(/) 6/() Fr, f Fr 5, f 5 6/(5), ---- hus, f(/) 6/ 6/() 6/(5) 6/(7) (6/)[ / /5 / ] f(/).8 which is withi 8% f the exact value f.5. (d) Frm part (c) f(/).5 (6/)[ / /5 / ] (/)(/6) [ / /5 / ] r / / /5 / Chapter 7, Sluti 5. his is a dd fucti sice f(t) f(t). a a,, ω /. b / f (t) si(ω t)dt t si(t / ) dt

17 9 t si t t cs 9 si cs f(t) si t cs si Chapter 7, Sluti 6., ω / / a f (t)dt dt dt dt a f (t) cs( t) dt ω a cs(t / )dt cs(t / )dt cs(t / )dt t si si si t si t si b f (t) si(ωt) dt t si dt t si dt t si dt t cs t cs t cs

18 [ cs() ] Hece f(t) [(si( / ) si( / )) cs(t / ) (cs() ) si(t / ) ] Chapter 7, Sluti 7. (a) dd symmetry. (b) a a,, ω / / f(t) t, < t <, < t < b t t t t t si dt si cs si cs () ()/ /( ), dd () / /(), eve a, b ()/(9 ).5 (c) b /, b /, b /(9 ), b /(), b 5 (5 ) F rms a ( a b ) F rms.5σb [/( )][(6/ ) (6/(8 )) (/) (6/(65 ))] (/9.79)( ) F rms Cmpare this with the exact value f F rms t dt / 6.8

19 Chapter 7, Sluti 8. his is half-wave symmetric sice f(t /) f(t). a,, ω / a / f (t) cs(ω t)dt ( t) cs(t) dt si(t) t cs(t) si(t) [/( )][ cs()] 8/( ), dd, eve b ( t) si(t)dt cs(t) t si(t) cs(t) /(), dd f(t) 8 k cs(t) si(t ), k Chapter 7, Sluti 9. his fucti is half-wave symmetric., ω /, f(t) t, < t < Fr dd, a ( t) cs(t)dt cs(t) t si(t) [ ] /( ) b ( t) si(t)dt [ si(t) t cs(t) ] /

20 hus, f(t) cs(t) si(t), k k Chapter 7, Sluti. / jω / / t c f (t)e dt f (t) cs ω ω / tdt j f (t)si / tdt () / (a) he secd term the right had side vaishes if f(t) is eve. Hece c / f (t)cs ωtdt (b) he first term the right had side f () vaishes if f(t) is dd. Hece, c / j f (t)si ωtdt Chapter 7, Sluti. If h(t) f ( αt), ' / α ω ' ' / α αω a ' ' ' h(t) cs ω ' tdt ' ' f ( αt) cs ω ' tdt Let α t λ,, d t dλ / α, α' α a ' f ( λ)cs ωλdλ / α a Similarly, b ' b

21 Chapter 7, Sluti. Whe i s (DC cmpet) i /( ) / Fr, ω, I s / I [/( jω )]I s I s /( j6) hus, ta() ta (6 / ) i(t) cs( ta ()) Chapter 7, Sluti. Fr the DC case, the iductr acts like a shrt, V. Fr the AC case, we btai the fllwig: V V s V j jv 5 j.5 V V s V Vs j.5 5 A Θ j.5 5 j(.5 5) A (.5 5) ; Θ ta.5 5

22 v (t) A si(t Θ Chapter 7, Sluti. Fr ay, V [/ ] (/), ω. H becmes jω L j ad.5 F becmes /(jω C) j/ Ω j ) V V j/ V V {j(/)/[ j j(/)]}v {j/[ j( )]}[(/ ) (/)] (( / ) / ) ( ) ta (( [( / ) ( / ) ta ) / ) (( ) / )] v (t) Chapter 7, Sluti 5. cs t ta If v s i the circuit f Fig. 7.7 is the same as fucti f (t) i Fig. 7.57(b), determie the dc cmpet ad the first three zer harmics f v (t). Ω H v S F Ω v Figure 7.7 Fr Prb. 7.5

23 f (t) t Figure 7.57(b) Fr Prb. 7.5 he sigal is eve, hece, b. I additi,, ω /. v s (t) fr all < t < fr all < t <.5 a. 5 dt dt a cs(t / )dt.5 cs(t / )dt 6. 5 si( t / ) si( t / ) si( / ) v s (t) si( / ) cs(t / ) Nw csider this circuit, Ω j/ v S -j/() Ω v Let Z [-j/()]()/( j/()) -j/( - j) herefre, v Zv s /(Z j/). Simplifyig, we get v j9v s j( 8)

24 Fr the dc case, ad v s ¾ V ad v v s / /8 V. We ca w slve fr v (t) v (t) 8 A t cs Θ vlts where A 6 6 si( / ) 6 ad Θ 9 ta where we ca further simplify A t this, A 9si( / ) 8 Chapter 7, Sluti 6. v s (t) A cs(t θ ) dd where θ ta [(/())/(/())] ta ().5 A 9 si 9 si ω ad H becmes jω L j Let Z j j/( j) If V is the vltage at the -referece de r acrss the -H iductr. V ZV s /( Z) [j/( j)]v s /{ [j/( j)]} jv s /( j) But V s A θ V j A θ /( j)

25 I V /j [ A θ ]/ 6 ta 9 si 6.5 ta Sice si(/) () ()/ fr dd, si (/) I.5 6 ta i (t) cs(t.5 ta ) 6 dd Chapter 7, Sluti 7. Frm Example 5., v s (t) 5 si(t), k k Fr the DC cmpet, the capacitr acts like a pe circuit. Fr the th harmic, V 5 V s [/()] mf becmes /(jω C) j/(xx ) j/() v j Vs j j j 9 ta v (t) si(t 9 ta ) 5

26 Chapter 7, Sluti 8. v s (t) si t, k k V jω Vs, ω jω Fr dc, ω, V.5, V s Fr th harmic, Vs 9 V 9 9 ta ta v (t) k cs(t ta ), k Chapter 7, Sluti 9. Cmparig v s (t) with f(t) i Figure 5., v s is shifted by.5 ad the magitude is 5 times that f f(t). Hece v s (t) 5 si(t), k k, ω //, ω ω Fr the DC cmpet, i 5/( ) / Fr the kth harmic, V s (/()) mh becmes jω L jx. j. 5 mf becmes /(jω C) j/() I Ω Ω I V S j/() j. Z

27 Let Z j/() ( j.) j ( j.) j j. j( j. j j. j8 j(. ) Z i Z 8 j( ) j(. ) I Vs Z i [8 j( ) j( )] I j j I ( j.) ji j(. ) j [8 j( )] hus 9 ta (8) {( ( i (t) k ) /(8)} ) I si(t θ ), k where θ 9 ta 8 I (8) ( )

28 Chapter 7, Sluti., ω / t a v(t)dt ( t)dt t / a v(t) cs(t)dt ( t) cs(t)dt si(t) t cs( t) si( t) ( cs ),, eve dd ( ) b v(t) si(t)dt ( t) si(t)dt cs(t) si(t) t cs(t) v s (t) A cs(t ϕ ) ( ) 6 where φ ta, A ( ) Fr the DC cmpet, v s /. As shw i Figure (a), the capacitr acts like a pe circuit. Ω V x V x V.5V i V x Ω V (a)

29 Ω V x V x V V S (/)F Ω V (b) Applyig KVL t the circuit i Figure (a) gives.5 V x i () But.5 i V x r V x i () Addig () ad (),.5 6i r i.5 V i.75 Fr the th harmic, we csider the circuit i Figure (b). ω, V s A φ, /(jω C) j/() At the superde, (V s V x )/ [/(j)]v x V / V s [ j/]v x V / () But V x V x V r V V x Substitutig this it (), V s [ j/]v x V x [ j/]v x (/)[ j/]v (/)[8 j]v V V s /(8 j) 6 A φ ta ( / 8) 6 V [ta ( / 8) ta ( ( ) /())] 6 ( )

30 hus v (t) V cs(t θ ) 6 where V ) 6 ( θ ta (/8) ta (( )/()) Chapter 7, Sluti. Fr the full wave rectifier,, ω /, ω ω Hece t v i (t) cs( ) Fr the DC cmpet, V i / he iductr acts like a shrt-circuit, while the capacitr acts like a pe circuit. V V i / Fr the th harmic, V i [ /(( ))] H becmes jω L j. F becmes /(jω C) j5/ Z ( j5/) j/( j) V [Z/(Z j)]v i jv i /( j(8 )) j j(8 ) ( )

31 {9 ta ( ) (.5)} 6 (8 ) Hece v (t) where A ( ) 6 A cs(t θ ) 9 θ 9 ta (.5) Chapter 7, Sluti. v s 5 si t, k - k Vs R jωc( V ) V j Vs, ωrc ω ω Fr (dc cmpet), V. Fr the th harmic, Hece, 9 V 9 RC 9 x xx v 5 (t) cs t, k - k 5 Alteratively, we tice that this is a itegratr s that v 5 (t) vsdt cs t, k - RC k

32 Chapter 7, Sluti. (a) V rms a (a b ) ( ).9 V (b) I rms 6 ( ) 6.78 A (c) P V dc I dc V I cs(θ Φ ) x6.5[xcs(5 - ) xcs(-5 6 )] W Chapter 7, Sluti. (a) p vi [ 6cs 5 cs 5 ] W (b) he pwer spectrum is shw belw. p ω Chapter 7, Sluti 5. ω jω L jxx j /(jω C) j/(xx 6 ) j5/

33 Z R jω L /(jω C) j j5/ I V/Z Fr, V, Z j j5 j I /( j) Fr, V 5, Z j j.5 j8.5 I 5/( j8.5).8.6 Fr, V 5, Z j6 j5/ j. I 5/( j.).5. I rms A p V DC I DC V cs( θ φ I ).5[x.987cs(7.89 ) 5x.8cs(.6 ) 5x.5cs(. )].5[ ] 57.5 watts Chapter 7, Sluti 6. (a) his is a eve fucti I rms f (t)dt / f (t)dt f(t) t,, < t < < t <, ω / / I rms ( t) dt (t t t / )

34 ( /) / r I rms.865 A (b) Frm Prblem 6., a [8/( )][ cs(/)], a.5 a 8/, a /, a 8/(9 ), a, a 5 9/(5 ), a 6 /(9 ) I rms a A I rms.86 A. 666 Chapter 7, Sluti 7. Let I I DC I I Fr the DC cmpet I DC [5/(5 )]() A I j8 I s 5 Ω Ω Fr AC, ω jωl jx8x j8 I 5I s /(5 j8) Fr I s.5 6 I 6 /(5 j8) r I / Fr I s.5 I.5 /(5 j8) r I.5/ p (I DC I / I /) ( [/(x89)] [6.5/(x89)])x p.88 watts

35 Chapter 7, Sluti 8. (a) Fr the DC cmpet, i(t) ma. he capacitr acts like a pe circuit s that v Ri(t) x xx Fr the AC cmpet, ω,, /(jω C) j/(xx 6 ) ( j/) kω Z ( j/) ( j/)/( j/) j/( j) V ZI [ j/( j)]i Fr, Fr, V [ j/( j)] mv V [ j/( j)] mv v(t).cs(t 8. ).58cs(t 5.96 ) V (b) p V DC I DC V cs( θ φ I ) x.5xx.cs(5 8. ).5xx.58cs( ) 8. mw Chapter 7, Sluti 9. (a) Z rms z (t)dt dt dt (5). 5 (b) Z rms Z rms.58 6 a (a b ) Z rms (c ) %errr x

36 Chapter 7, Sluti 5. jω t c f (t)e dt, ω t te j dt Usig itegrati by parts, u t ad du dt dv e jt dt which leads t v [/(j)]e jt t jt jt c e e dt j j j j jt [ e e ] ( j) e jt [j/()]cs() [/( )](e j e j ) hus c j( ) j si() j( ) f(t) jωt c e ( ) j e jt Chapter 7, Sluti 5., ω / ( t jt ) jt jωt jt e f (t)e dt t e dt ( j) c c j ( j) ( j) f (t) ( j)e jt

37 Chapter 7, Sluti 5. jω t c f (t)e dt, ω t te j dt Usig itegrati by parts, u t ad du dt dv e jt dt which leads t v [/(j)]e jt t jt jt c e e dt j j j j jt [ e e ] ( j) e jt [j/()]cs() [/( )](e j e j ) hus c j( ) j si() j( ) f(t) jωt c e ( ) j e jt Chapter 7, Sluti 5. ω / c e t e jωt dt e ( jω )t ( j)t ( j) e [ e ] j dt j [/(j)][ e (cs() jsi())] ( e )/( j).6/( j f(t).6e j jt

38 Chapter 7, Sluti 5., ω / / c f (t)e jωt dt e jt / dt jt / e dt e jt / dt j j / j j / j j [ e e e e e ] j j / j [ e e ] f(t) c e jωt Chapter 7, Sluti 55., ω / c i(t)e jωt dt But i(t) si(t),, < t < < t < c jt si(t)e dt j (e jt e jt )e jt dt j jt( ) e j( ) jt( ) e j( ) e j ( ) e j ( )

39 ( ) j( ) j( ) j( ) j( [ e e e e ] ) But e j cs() jsi() e j hus j j j j c [ e e e e ] ( ) i(t) j e ( e ) jt j e ( ) Chapter 7, Sluti 56. c a, ω c (a jb )/ ( j)/[( )] f(t) ( j) ( ) jt e Chapter 7, Sluti 57. a (6/ ) c c.5(a jb ) a / /( ) f(t) j5t e Chapter 7, Sluti 58. c (a jb )/, ( ) cs(), ω / c [(cs() )/( )] j cs()/() hus f(t) cs() cs() j jt e

40 Chapter 7, Sluti 59. Fr f(t),, ω /. a DC cmpet (x )/.5 Fr h(t),, ω /. a (x x)/.5 hus by replacig ω with ω ad multiplyig the magitude by five, we btai j( ) t j5e h(t) ( ) Chapter 7, Sluti 6. Frm Prblem 6.7, a a, b [/()][ cs()], c c (a jb )/ [j/()][ cs() ],. Chapter 7, Sluti 6. (a) ω. f(t) a A cs(ω t φ ) 6 cs(t 5 ) cs(t 5 ) cs(t 5 ).5cs(t ) 6 cs(t)cs(5 ) si(t)si(5 ) cs(t)cs(5 ) si(t)si(5 ) cs(t)cs(5 ) si(t)si(5 ).5cs(t)cs( ).5si(t)si( ) 6.57cs(t).7si(t).65cs(t).7si(t).96cs(t).si(t).7cs(t).7si(t)

41 (b) f rms a A f rms 6.5[ (.5) ] 6.65 f rms 6.88 Chapter 7, Sluti 6. (a) ω /.s (b) f (t) a A cs(ωt φ ) cs(t 9 ) 5.cs(t 9 )... f (t) si t 5.si t.7si 6t.8si 8t... Chapter 7, Sluti 6. his is a eve fucti., ω /, b. f(t),, < t < < t <.5 a / f (t)dt dt.5 dt (/)[ ] / a / f (t)cs(ω t)dt t si cs(t / )dt 6 t si.5.5 cs(t / ) dt [ /()]si(/)

42 f (t) t si cs a /., ω /, a [/()]si(t/) A a b si A.55, A.757, A, A.75, A 5. he amplitude spectra are shw belw.. A Chapter 7, Sluti 6. he amplitude ad phase spectra are shw belw.

43 A ω φ 6 ω -8 Chapter 7, Sluti 65. a /( ), b /(), ω 9 A a b.,,, 5, 7, 9, etc. A

44 φ ta (b /a ) ta {[ /()][ /]} ta ( x.7) φ ω 5. A φ ω Chapter 7, Sluti 66. he schematic is shw belw. he wavefrm is iputted usig the attributes f VPULSE. I the rasiet dialg bx, we eter Prit Step.5, Fial ime, Ceter Frequecy.5, Output Vars V() ad click eable Furier. After simulati, the utput plt is shw belw. he utput file icludes the fllwig Furier cmpets.

45 DC COMPONEN 5.995E FOURIER COMPONENS OF RANSIEN RESPONSE V() HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONEN COMPONEN (DEG) PHASE (DEG) 5.E-.8E.E.78E.E.E.59E 5.E-.56E.78E.5E.6E.8E- 5.7E.56E.E 7.978E-.56E- 7.9E 5.7E 5.5E 6.9E-.8E- 8.9E 7.9E 6.E 5.6E-.676E-.69E 8.9E 7.5E.58E-.E-.8E.69E 8.E.E-.6E-.6E.8E 9.5E.58E-.6E-.6E.6E OAL HARMONIC DISORION 7.66E PERCEN Chapter 7, Sluti 67. he Schematic is shw belw. I the rasiet dialg bx, we type Prit step.s, Fial time 6s, Ceter frequecy.667, Output vars v(), ad click Eable Furier. After simulati, the utput file icludes the fllwig Furier cmpets,

46 FOURIER COMPONENS OF RANSIEN RESPONSE V() DC COMPONEN.96E HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONEN COMPONEN (DEG) PHASE (DEG).667E-.E.E E.E.E E-.75E- -8.9E 6.67E- 5.E- 5.E-.E- 9.E.8E 6.668E-.E-.75E- 9.E.8E 5 8.5E- 9.76E-.996E E.E- 6.E 7.8E-6.76E-6-9.E -.58E- 7.67E.968E-.E E.7E- 8.E.6E- 6.6E-5-8.7E.78E 9.5E 6.E-.68E- 9.E.8E OAL HARMON IC DISORION.865E PERCEN

47 Chapter 7, Sluti 68. he schematic is shw belw. We set the fial time 6s ad the ceter frequecy /.5. Whe the schematic is saved ad ru, we btai the Furier series frm the utput file as shw belw. FOURIER COMPONENS OF RANSIEN RESPONSE V() DC COMPONEN.99E HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (DEG) (HZ) COMPONEN COMPONEN (DEG) PHASE 5.E-.7E.E 9.E-.E.E 6.67E- 5.E- -.78E.79E.5E.6E-.E-.7E.8E.E.85E-.5E- -.76E -.77E 5.5E.59E-.E-.5E.6E 6.E.5E-.669E- -.76E -.755E 7.5E.8E-.E- 6.E 5.E 8.E.596E-.5E- -.78E -.77E 9.5E.9E-.5E- 8.E 7.E Chapter 7, Sluti 69.

48 he schematic is shw belw. I the rasiet dialg bx, set Prit Step.5 s, Fial ime, Ceter Frequecy.5, Output Vars V() ad click eable Furier. After simulati, we btai V() as shw belw. We als btai a utput file which icludes the fllwig Furier cmpets. FOURIER COMPONENS OF RANSIEN RESPONSE V() DC COMPONEN 5.85E- HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONEN COMPONEN (DEG) PHASE (DEG) 5.E-.56E-.E -9.9E.E.E.977E- 7.E E.8E.5E.5E-.7E E -.76E.E.969E- 7.E- -8.E 6.757E 5.5E.68E-.6E- -9.E -.7E 6.E.955E- 7.85E- -8.E 9.659E 7.5E 8.55E-.E E -.9E 8.E.95E- 7.8E E.5E 9.5E 5.58E-.96E- -9.7E -6.97E OAL HARMONIC DISORION.85E PERCEN

49 Chapter 7, Sluti 7. he schematic is shw belw. I the rasiet dialg bx, we set Prit Step. s, Fial Step s, Ceter Frequecy.5, Output Vars V() ad V(), ad click eable Furier. After simulati, we cmpare the utput ad utput wavefrms as shw. he utput icludes the fllwig Furier cmpets.

50 FOURIER COMPONENS OF RANSIEN RESPONSE V() DC COMPONEN E- HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONEN COMPONEN (DEG) PHASE (DEG) 5.E-.7E.E.E.E.E.758E-.5E- -.9E -.98E.5E.E-.97E E -.99E.E.7E-.66E E -6.87E 5.5E 8.58E- 7.98E E E 6.E 6.9E- 5.78E E E 7.5E.7E-.E- -6.5E -7.5E 8.E.7E-.69E- -7.E -8.6E 9.5E.997E-.8E E -8.9E OAL HARMONIC DISORION.5895E PERCEN

51 Chapter 7, Sluti 7. he schematic is shw belw. We set Prit Step.5, Fial ime s, Ceter Frequecy.5, Output Vars I(), ad click eable Furier i the rasiet dialg bx. After simulati, the utput wavefrm is as shw. he utput file icludes the fllwig Furier cmpets.

52 FOURIER COMPONENS OF RANSIEN RESPONSE I(L_L) DC COMPONEN E- HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONEN COMPONEN (DEG) PHASE (DEG) 5.E-.87E-.E -6.79E.E.E.89E- 8.68E- 8.7E 7.566E.5E.78E-.E E -.E.E 9.58E-5.9E- -.8E 6.565E 5.5E.7E-.6E E -.76E 6.E 6.66E-5.78E- -7.8E 6.8E 7.5E 5.97E-.596E E -.8E 8.E 6.59E-5.69E- -.88E.9E 9.5E.E- 9.E- -.E -5.87E OAL HARMONIC DISORION.8E PERCEN Chapter 7, Sluti 7. 5, ω / /5 f(t) is a dd fucti. a a / b f (t)si(ω t)dt si(. 5 t) dt 8x5 cs(.t) [ cs(.)] f(t) [ cs(.)]si(.t) Chapter 7, Sluti 7. p V DC R V R.5[( )/] mw

53 Chapter 7, Sluti 7. (a) A A A a b, φ ta (b /a ) 6 8, φ ta (6/8) , φ ta (/) 6.87 i(t) { cs(t 6.87 ) 5cs(t 6.87 )} A (b) p I DCR.5 I R [.5( 5 )] 57 W Chapter 7, Sluti 75. he lwpass filter is shw belw. R v s C v - - v s Aτ A τ si cs ωt V jωc V R jω C s jω V RC s, ω ω / Fr, (dc cmpet), V Aτ V s ()

54 Fr the th harmic, A τ V si 9 ω R C ta ω RC Whe, A τ V si () R C Frm () ad (), Aτ A 5x si R C R C.9 τ.9x R C C 5 R x.9x x.59 mf Chapter 7, Sluti 76. v s (t) is the same as f(t) i Figure 6. except that the magitude is multiplied by. Hece v (t) 5 si(t), k k, ω /, ω ω jω L j; Z R R/( R) V ZV s /(Z j) [R/(R j( R))]V s V R ta {( / 5R)( R)} V R ( R) s V s [/()] he surce curret I s is ( R) Vs Vs I s Z j R R j( R) j R

55 ( R) ta {( / )( R)} R ( R) p s V DC I DC Vs Is cs(θ φ ) Fr the DC case, L acts like a shrt-circuit. I s 5 R R 5( R), V s 5 V R p s 5( R) R ( R)cs ta R ( R) 5 ( R) ( R) cs ta ( R) 5 R 6 ( R) V p s DC V R R 5 R R R R ( R) R ( R) We wat p (7/)p s.7p s. Due t the cmplexity f the terms, we csider ly the DC cmpet as a apprximati. I fact the DC cmpet has the latgest share f the pwer fr bth iput ad utput sigals ( R) x R R 7 7R which leads t R /7.86 Ω

56 Chapter 7, Sluti 77. (a) Fr the first tw AC terms, the frequecy rati is 6/.5 s that the highest cmm factr is. Hece ω. /ω / (b) he average value is the DC cmpet (c) V rms a (a b ) V rms ( ) ( 8 6 ).5 V rms. V Chapter 7, Sluti 78. V DC DC (a) p R V R V R V,rms ( /5) ( /5) ( /5) W (b) 5% icrease (5/) DC p DC W which leads t VDC R 5 R V DC.5 V R V Chapter 7, Sluti 79. Frm able 7., it is evidet that a, b A/[( )], A. A Frtra prgram t calculate b is shw belw. he result is als shw.

57 C FOR PROBLEM 7.79 DIMENSION B() A PIE. C.*A/PIE DO N, B(N) C/(.*FLOA(N).) PRIN *, N, B(N) CONINUE SOP END b Chapter 7, Sluti 8. Frm Prblem 7.55, c [ e j ]/[( )] his is calculated usig the Frtra prgram shw belw. he results are als shw. C FOR PROBLEM 7.8 COMPLEX X, C(:) PIE.597 A.*PIE DO N, IF(N.EQ.) GO O X CMPLX(, PIE*FLOA(N)) C(N) (. CEXP( X))/(A*( FLOA(N*N))) PRIN *, N, C(N) CONINUE SOP END

58 c.88 j.6 j.x j x j x j 9.5x j Chapter 7, Sluti 8. (a) A A A f(t) cs(ω t) he ttal average pwer is p avg F rms R F rms sice R hm. (b) P avg F rms f Frm the Furier series abve (t)dt c A/, c A/[( )].5A

59 ω c c % pwer A/ A /( ) 8.% ω A/() 8A /(9 ) 8.% ω A/(5) A /(5 ).7% 6ω A/(5) 8A /(5 ).% 8ω A/(6) 8A /(969 ).% (c) 8.% (d).7% Chapter 7, Sluti 8. P VDC V R R Assumig V is a amplitude-phase frm f Furier series. But A C, c a A C Hece, P c c R R Alteratively, where V rms a P V rms R A c c c ( ) x P 89./ 7. W

The Excel FFT Function v1.1 P. T. Debevec February 12, The discrete Fourier transform may be used to identify periodic structures in time ht.

The Excel FFT Function v1.1 P. T. Debevec February 12, The discrete Fourier transform may be used to identify periodic structures in time ht. The Excel FFT Fucti v P T Debevec February 2, 26 The discrete Furier trasfrm may be used t idetify peridic structures i time ht series data Suppse that a physical prcess is represeted by the fucti f time,

More information

Chapter 15: Fourier Series

Chapter 15: Fourier Series Chapter 5: Fourier Series Ex. 5.3- Ex. 5.3- Ex. 5.- f(t) K is a Fourier Series. he coefficiets are a K; a b for. f(t) AcosZ t is a Fourier Series. a A ad all other coefficiets are zero. Set origi at t,

More information

2. Find i, v, and the power dissipated in the 6-Ω resistor in the following figure.

2. Find i, v, and the power dissipated in the 6-Ω resistor in the following figure. CSC Class exercise DC Circuit analysis. Fr the ladder netwrk in the fllwing figure, find I and R eq. Slutin Req 4 ( 6 ) 5Ω 0 0 I Re q 5 A. Find i, v, and the pwer dissipated in the 6-Ω resistr in the fllwing

More information

D.S.G. POLLOCK: TOPICS IN TIME-SERIES ANALYSIS STATISTICAL FOURIER ANALYSIS

D.S.G. POLLOCK: TOPICS IN TIME-SERIES ANALYSIS STATISTICAL FOURIER ANALYSIS STATISTICAL FOURIER ANALYSIS The Furier Represetati f a Sequece Accrdig t the basic result f Furier aalysis, it is always pssible t apprximate a arbitrary aalytic fucti defied ver a fiite iterval f the

More information

MATH Midterm Examination Victor Matveev October 26, 2016

MATH Midterm Examination Victor Matveev October 26, 2016 MATH 33- Midterm Examiati Victr Matveev Octber 6, 6. (5pts, mi) Suppse f(x) equals si x the iterval < x < (=), ad is a eve peridic extesi f this fucti t the rest f the real lie. Fid the csie series fr

More information

Sinusoidal Steady-state Analysis

Sinusoidal Steady-state Analysis Siusoidal Steady-state Aalysis Complex umber reviews Phasors ad ordiary differetial equatios Complete respose ad siusoidal steady-state respose Cocepts of impedace ad admittace Siusoidal steady-state aalysis

More information

Quantum Mechanics for Scientists and Engineers. David Miller

Quantum Mechanics for Scientists and Engineers. David Miller Quatum Mechaics fr Scietists ad Egieers David Miller Time-depedet perturbati thery Time-depedet perturbati thery Time-depedet perturbati basics Time-depedet perturbati thery Fr time-depedet prblems csider

More information

CHAPTER 5. Solutions for Exercises

CHAPTER 5. Solutions for Exercises HAPTE 5 Slutins fr Exercises E5. (a We are given v ( t 50 cs(00π t 30. The angular frequency is the cefficient f t s we have ω 00π radian/s. Then f ω / π 00 Hz T / f 0 ms m / 50 / 06. Furthermre, v(t attains

More information

ZVS Boost Converter. (a) (b) Fig 6.29 (a) Quasi-resonant boost converter with M-type switch. (b) Equivalent circuit.

ZVS Boost Converter. (a) (b) Fig 6.29 (a) Quasi-resonant boost converter with M-type switch. (b) Equivalent circuit. EEL6246 Pwer Electrnics II Chapter 6 Lecture 6 Dr. Sam Abdel-Rahman ZVS Bst Cnverter The quasi-resnant bst cnverter by using the M-type switch as shwn in Fig. 6.29(a) with its simplified circuit shwn in

More information

are specified , are linearly independent Otherwise, they are linearly dependent, and one is expressed by a linear combination of the others

are specified , are linearly independent Otherwise, they are linearly dependent, and one is expressed by a linear combination of the others Chater 3. Higher Order Liear ODEs Kreyszig by YHLee;4; 3-3. Hmgeeus Liear ODEs The stadard frm f the th rder liear ODE ( ) ( ) = : hmgeeus if r( ) = y y y y r Hmgeeus Liear ODE: Suersiti Pricile, Geeral

More information

ECE 2100 Circuit Analysis

ECE 2100 Circuit Analysis ECE 00 Circuit Analysis Lessn 6 Chapter 4 Sec 4., 4.5, 4.7 Series LC Circuit C Lw Pass Filter Daniel M. Litynski, Ph.D. http://hmepages.wmich.edu/~dlitynsk/ ECE 00 Circuit Analysis Lessn 5 Chapter 9 &

More information

Solutions to Midterm II. of the following equation consistent with the boundary condition stated u. y u x y

Solutions to Midterm II. of the following equation consistent with the boundary condition stated u. y u x y Sltis t Midterm II Prblem : (pts) Fid the mst geeral slti ( f the fllwig eqati csistet with the bdary cditi stated y 3 y the lie y () Slti : Sice the system () is liear the slti is give as a sperpsiti

More information

Revision: August 19, E Main Suite D Pullman, WA (509) Voice and Fax

Revision: August 19, E Main Suite D Pullman, WA (509) Voice and Fax .7.4: Direct frequency dmain circuit analysis Revisin: August 9, 00 5 E Main Suite D Pullman, WA 9963 (509) 334 6306 ice and Fax Overview n chapter.7., we determined the steadystate respnse f electrical

More information

Dr. Kasra Etemadi February 27, 2007

Dr. Kasra Etemadi February 27, 2007 Dr. Kasra Eteadi February 7, 7 Chapter 4:Transients Chapter 5: Sinusidal Surces Chapter 6: nnsinusidal surces Furier Trasr Transer Functin Filters Lwpass Filters Highpass Filters andpass Filters Surce

More information

Solutions. Definitions pertaining to solutions

Solutions. Definitions pertaining to solutions Slutis Defiitis pertaiig t slutis Slute is the substace that is disslved. It is usually preset i the smaller amut. Slvet is the substace that des the disslvig. It is usually preset i the larger amut. Slubility

More information

Fourier Series & Fourier Transforms

Fourier Series & Fourier Transforms Experimet 1 Furier Series & Furier Trasfrms MATLAB Simulati Objectives Furier aalysis plays a imprtat rle i cmmuicati thery. The mai bjectives f this experimet are: 1) T gai a gd uderstadig ad practice

More information

Signal Processing in Mechatronics. Lecture 3, Convolution, Fourier Series and Fourier Transform

Signal Processing in Mechatronics. Lecture 3, Convolution, Fourier Series and Fourier Transform Sigal Processig i Mechatroics Summer semester, 1 Lecture 3, Covolutio, Fourier Series ad Fourier rasform Dr. Zhu K.P. AIS, UM 1 1. Covolutio Covolutio Descriptio of LI Systems he mai premise is that the

More information

Frequency-Domain Study of Lock Range of Injection-Locked Non- Harmonic Oscillators

Frequency-Domain Study of Lock Range of Injection-Locked Non- Harmonic Oscillators 0 teratial Cferece mage Visi ad Cmputig CVC 0 PCST vl. 50 0 0 ACST Press Sigapre DO: 0.776/PCST.0.V50.6 Frequecy-Dmai Study f Lck Rage f jecti-lcked N- armic Oscillatrs Yushi Zhu ad Fei Yua Departmet f

More information

Supplementary Course Notes Adding and Subtracting AC Voltages and Currents

Supplementary Course Notes Adding and Subtracting AC Voltages and Currents Supplementary Curse Ntes Adding and Subtracting AC Vltages and Currents As mentined previusly, when cmbining DC vltages r currents, we nly need t knw the plarity (vltage) and directin (current). In the

More information

Power Flow in Electromagnetic Waves. The time-dependent power flow density of an electromagnetic wave is given by the instantaneous Poynting vector

Power Flow in Electromagnetic Waves. The time-dependent power flow density of an electromagnetic wave is given by the instantaneous Poynting vector Pwer Flw in Electrmagnetic Waves Electrmagnetic Fields The time-dependent pwer flw density f an electrmagnetic wave is given by the instantaneus Pynting vectr P t E t H t ( ) = ( ) ( ) Fr time-varying

More information

ECE 2100 Circuit Analysis

ECE 2100 Circuit Analysis ECE 2100 Circuit Analysis Lessn 25 Chapter 9 & App B: Passive circuit elements in the phasr representatin Daniel M. Litynski, Ph.D. http://hmepages.wmich.edu/~dlitynsk/ ECE 2100 Circuit Analysis Lessn

More information

Supplementary Course Notes Adding and Subtracting AC Voltages and Currents

Supplementary Course Notes Adding and Subtracting AC Voltages and Currents Supplementary Curse Ntes Adding and Subtracting AC Vltages and Currents As mentined previusly, when cmbining DC vltages r currents, we nly need t knw the plarity (vltage) and directin (current). In the

More information

Physics 2B Chapter 23 Notes - Faraday s Law & Inductors Spring 2018

Physics 2B Chapter 23 Notes - Faraday s Law & Inductors Spring 2018 Michael Faraday lived in the Lndn area frm 1791 t 1867. He was 29 years ld when Hand Oersted, in 1820, accidentally discvered that electric current creates magnetic field. Thrugh empirical bservatin and

More information

Chapter 4 : Laplace Transform

Chapter 4 : Laplace Transform 4. Itroductio Laplace trasform is a alterative to solve the differetial equatio by the complex frequecy domai ( s = σ + jω), istead of the usual time domai. The DE ca be easily trasformed ito a algebraic

More information

6.003 Homework #12 Solutions

6.003 Homework #12 Solutions 6.003 Homework # Solutios Problems. Which are rue? For each of the D sigals x [] through x 4 [] below), determie whether the coditios listed i the followig table are satisfied, ad aswer for true or F for

More information

ANALOG FILTERS. C. Sauriol. Algonquin College Ottawa, Ontario

ANALOG FILTERS. C. Sauriol. Algonquin College Ottawa, Ontario LOG ILT By. auril lgqui llege Ottawa, Otari ev. March 4, 003 TBL O OTT alg ilters TIO PI ILT. irst-rder lw-pass filter- -4. irst-rder high-pass filter- 4-6 3. ecd-rder lw-pass filter- 6-4. ecd-rder bad-pass

More information

x 2 x 3 x b 0, then a, b, c log x 1 log z log x log y 1 logb log a dy 4. dx As tangent is perpendicular to the x axis, slope

x 2 x 3 x b 0, then a, b, c log x 1 log z log x log y 1 logb log a dy 4. dx As tangent is perpendicular to the x axis, slope The agle betwee the tagets draw t the parabla y = frm the pit (-,) 5 9 6 Here give pit lies the directri, hece the agle betwee the tagets frm that pit right agle Ratig :EASY The umber f values f c such

More information

6.003 Homework #12 Solutions

6.003 Homework #12 Solutions 6.003 Homework # Solutios Problems. Which are rue? For each of the D sigals x [] through x 4 [] (below), determie whether the coditios listed i the followig table are satisfied, ad aswer for true or F

More information

Sinusoidal Steady-State Analysis

Sinusoidal Steady-State Analysis Chapter 4 Sinusoidal Steady-State Analysis In this unit, we consider circuits in which the sources are sinusoidal in nature. The review section of this unit covers most of section 9.1 9.9 of the text.

More information

Synchronous Motor V-Curves

Synchronous Motor V-Curves Synchrnus Mtr V-Curves 1 Synchrnus Mtr V-Curves Intrductin Synchrnus mtrs are used in applicatins such as textile mills where cnstant speed peratin is critical. Mst small synchrnus mtrs cntain squirrel

More information

BASIC DIRECT-CURRENT MEASUREMENTS

BASIC DIRECT-CURRENT MEASUREMENTS Brwn University Physics 0040 Intrductin BASIC DIRECT-CURRENT MEASUREMENTS The measurements described here illustrate the peratin f resistrs and capacitrs in electric circuits, and the use f sme standard

More information

Examination No. 3 - Tuesday, Nov. 15

Examination No. 3 - Tuesday, Nov. 15 NAME (lease rit) SOLUTIONS ECE 35 - DEVICE ELECTRONICS Fall Semester 005 Examiati N 3 - Tuesday, Nv 5 3 4 5 The time fr examiati is hr 5 mi Studets are allwed t use 3 sheets f tes Please shw yur wrk, artial

More information

Chapter 3.1: Polynomial Functions

Chapter 3.1: Polynomial Functions Ntes 3.1: Ply Fucs Chapter 3.1: Plymial Fuctis I Algebra I ad Algebra II, yu ecutered sme very famus plymial fuctis. I this secti, yu will meet may ther members f the plymial family, what sets them apart

More information

POWER AMPLIFIERS. 1. Explain what are classes A, B, AB and C amplifiers in terms of DC biasing using a MOSFET drain characteristic.

POWER AMPLIFIERS. 1. Explain what are classes A, B, AB and C amplifiers in terms of DC biasing using a MOSFET drain characteristic. CTONIC 3 XCI OW AMII. xpla what are classes A, B, AB and C amplifiers terms f DC biasg usg a MOT dra characteristic.. efer t the graphs f page and the table at the tp f page 3 f the thery ntes t answer

More information

y[ n] = sin(2" # 3 # n) 50

y[ n] = sin(2 # 3 # n) 50 Period of a Discrete Siusoid y[ ] si( ) 5 T5 samples y[ ] y[ + 5] si() si() [ ] si( 3 ) 5 y[ ] y[ + T] T?? samples [iteger] 5/3 iteger y irratioal frequecy ysi(pisqrt()/5) - - TextEd si( t) T sec cotiuous

More information

Series and Parallel Resonances

Series and Parallel Resonances Series and Parallel esnances Series esnance Cnsider the series circuit shwn in the frequency dmain. The input impedance is Z Vs jl jl I jc C H s esnance ccurs when the imaginary part f the transfer functin

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science. BACKGROUND EXAM September 30, 2004.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science. BACKGROUND EXAM September 30, 2004. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Departmet of Electrical Egieerig ad Computer Sciece 6.34 Discrete Time Sigal Processig Fall 24 BACKGROUND EXAM September 3, 24. Full Name: Note: This exam is closed

More information

Intermediate Division Solutions

Intermediate Division Solutions Itermediate Divisi Slutis 1. Cmpute the largest 4-digit umber f the frm ABBA which is exactly divisible by 7. Sluti ABBA 1000A + 100B +10B+A 1001A + 110B 1001 is divisible by 7 (1001 7 143), s 1001A is

More information

Chapter 9 Frequency-Domain Analysis of Dynamic Systems

Chapter 9 Frequency-Domain Analysis of Dynamic Systems ME 43 Systes Dyaics & Ctrl Chapter 9: Frequecy Dai Aalyis f Dyaic Systes Systes Chapter 9 Frequecy-Dai Aalysis f Dyaic Systes 9. INTRODUCTION A. Bazue The ter Frequecy Respse refers t the steady state

More information

5.1. Periodic Signals: A signal f(t) is periodic iff for some T 0 > 0,

5.1. Periodic Signals: A signal f(t) is periodic iff for some T 0 > 0, 5. Periodic Sigals: A sigal f(t) is periodic iff for some >, f () t = f ( t + ) i t he smallest value that satisfies the above coditios is called the period of f(t). Cosider a sigal examied over to 5 secods

More information

Coupled Inductors and Transformers

Coupled Inductors and Transformers Cupled nductrs and Transfrmers Self-nductance When current i flws thrugh the cil, a magnetic flux is prduced arund it. d d di di v= = = dt di dt dt nductance: = d di This inductance is cmmnly called self-inductance,

More information

Lab 11 LRC Circuits, Damped Forced Harmonic Motion

Lab 11 LRC Circuits, Damped Forced Harmonic Motion Physics 6 ab ab 11 ircuits, Damped Frced Harmnic Mtin What Yu Need T Knw: The Physics OK this is basically a recap f what yu ve dne s far with circuits and circuits. Nw we get t put everything tgether

More information

LECTURE 12 Sections Introduction to the Fourier series of periodic signals

LECTURE 12 Sections Introduction to the Fourier series of periodic signals Signals and Systems I Wednesday, February 11, 29 LECURE 12 Sections 3.1-3.3 Introduction to the Fourier series of periodic signals Chapter 3: Fourier Series of periodic signals 3. Introduction 3.1 Historical

More information

Three Phase Circuits

Three Phase Circuits Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/ OUTLINE Previously on ELCN102 Three Phase Circuits Balanced

More information

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense, 3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [

More information

Chapter 5. Root Locus Techniques

Chapter 5. Root Locus Techniques Chapter 5 Rt Lcu Techique Itrducti Sytem perfrmace ad tability dt determied dby cled-lp l ple Typical cled-lp feedback ctrl ytem G Ope-lp TF KG H Zer -, - Ple 0, -, - K Lcati f ple eaily fud Variati f

More information

Ch 4: The Continuous-Time Fourier Transform

Ch 4: The Continuous-Time Fourier Transform Ch 4: The Continuous-Time Fourier Transform Fourier Transform of x(t) Inverse Fourier Transform jt X ( j) x ( t ) e dt jt x ( t ) X ( j) e d 2 Ghulam Muhammad, King Saud University Continuous-time aperiodic

More information

Relationships Between Frequency, Capacitance, Inductance and Reactance.

Relationships Between Frequency, Capacitance, Inductance and Reactance. P Physics Relatinships between f,, and. Relatinships Between Frequency, apacitance, nductance and Reactance. Purpse: T experimentally verify the relatinships between f, and. The data cllected will lead

More information

Dynamic Response of Second Order Mechanical Systems with Viscous Dissipation forces

Dynamic Response of Second Order Mechanical Systems with Viscous Dissipation forces Hadut #a (pp. 1-39) Dyamic Respse f Secd Order Mechaical Systems with Viscus Dissipati frces d X d X + + = ext() t M D K X F dt dt Free Respse t iitial cditis ad F (t) = 0, Uderdamped, Critically Damped

More information

Chapter 5 Frequency Domain Analysis of Systems

Chapter 5 Frequency Domain Analysis of Systems Chapter 5 Frequency Domain Analysis of Systems CT, LTI Systems Consider the following CT LTI system: xt () ht () yt () Assumption: the impulse response h(t) is absolutely integrable, i.e., ht ( ) dt< (this

More information

Solutions to Problems in Chapter 4

Solutions to Problems in Chapter 4 Solutions to Problems in Chapter 4 Problems with Solutions Problem 4. Fourier Series of the Output Voltage of an Ideal Full-Wave Diode Bridge Rectifier he nonlinear circuit in Figure 4. is a full-wave

More information

Physics 102. Second Midterm Examination. Summer Term ( ) (Fundamental constants) (Coulomb constant)

Physics 102. Second Midterm Examination. Summer Term ( ) (Fundamental constants) (Coulomb constant) ε µ0 N mp T kg Kuwait University hysics Department hysics 0 Secnd Midterm Examinatin Summer Term (00-0) July 7, 0 Time: 6:00 7:0 M Name Student N Instructrs: Drs. bdel-karim, frusheh, Farhan, Kkaj, a,

More information

REACTANCE. By: Enzo Paterno Date: 03/2013

REACTANCE. By: Enzo Paterno Date: 03/2013 REACTANCE REACTANCE By: Enzo Paterno Date: 03/2013 5/2007 Enzo Paterno 1 RESISTANCE - R i R (t R A resistor for all practical purposes is unaffected by the frequency of the applied sinusoidal voltage or

More information

Solution of EECS 315 Final Examination F09

Solution of EECS 315 Final Examination F09 Solutio of EECS 315 Fial Examiatio F9 1. Fid the umerical value of δ ( t + 4ramp( tdt. δ ( t + 4ramp( tdt. Fid the umerical sigal eergy of x E x = x[ ] = δ 3 = 11 = ( = ramp( ( 4 = ramp( 8 = 8 [ ] = (

More information

Physical Chemistry Laboratory I CHEM 445 Experiment 2 Partial Molar Volume (Revised, 01/13/03)

Physical Chemistry Laboratory I CHEM 445 Experiment 2 Partial Molar Volume (Revised, 01/13/03) Physical Chemistry Labratry I CHEM 445 Experimet Partial Mlar lume (Revised, 0/3/03) lume is, t a gd apprximati, a additive prperty. Certaily this apprximati is used i preparig slutis whse ccetratis are

More information

Sections 15.1 to 15.12, 16.1 and 16.2 of the textbook (Robbins-Miller) cover the materials required for this topic.

Sections 15.1 to 15.12, 16.1 and 16.2 of the textbook (Robbins-Miller) cover the materials required for this topic. Tpic : AC Fundamentals, Sinusidal Wavefrm, and Phasrs Sectins 5. t 5., 6. and 6. f the textbk (Rbbins-Miller) cver the materials required fr this tpic.. Wavefrms in electrical systems are current r vltage

More information

1 1 2 = show that: over variables x and y. [2 marks] Write down necessary conditions involving first and second-order partial derivatives for ( x0, y

1 1 2 = show that: over variables x and y. [2 marks] Write down necessary conditions involving first and second-order partial derivatives for ( x0, y Questio (a) A square matrix A= A is called positive defiite if the quadratic form waw > 0 for every o-zero vector w [Note: Here (.) deotes the traspose of a matrix or a vector]. Let 0 A = 0 = show that:

More information

Ch. 1 Introduction to Estimation 1/15

Ch. 1 Introduction to Estimation 1/15 Ch. Itrducti t stimati /5 ample stimati Prblem: DSB R S f M f s f f f ; f, φ m tcsπf t + φ t f lectrics dds ise wt usually white BPF & mp t s t + w t st. lg. f & φ X udi mp cs π f + φ t Oscillatr w/ f

More information

Chapter 30. Inductance

Chapter 30. Inductance Chapter 30 nductance 30. Self-nductance Cnsider a lp f wire at rest. f we establish a current arund the lp, it will prduce a magnetic field. Sme f the magnetic field lines pass thrugh the lp. et! be the

More information

Part a: Writing the nodal equations and solving for v o gives the magnitude and phase response: tan ( 0.25 )

Part a: Writing the nodal equations and solving for v o gives the magnitude and phase response: tan ( 0.25 ) + - Hmewrk 0 Slutin ) In the circuit belw: a. Find the magnitude and phase respnse. b. What kind f filter is it? c. At what frequency is the respnse 0.707 if the generatr has a ltage f? d. What is the

More information

1. Nature of Impulse Response - Pole on Real Axis. z y(n) = r n. z r

1. Nature of Impulse Response - Pole on Real Axis. z y(n) = r n. z r . Nature of Impulse Respose - Pole o Real Axis Causal system trasfer fuctio: Hz) = z yz) = z r z z r y) = r r > : the respose grows mootoically > r > : y decays to zero mootoically r > : oscillatory, decayig

More information

MH1101 AY1617 Sem 2. Question 1. NOT TESTED THIS TIME

MH1101 AY1617 Sem 2. Question 1. NOT TESTED THIS TIME MH AY67 Sem Questio. NOT TESTED THIS TIME ( marks Let R be the regio bouded by the curve y 4x x 3 ad the x axis i the first quadrat (see figure below. Usig the cylidrical shell method, fid the volume of

More information

Dynamic Response of Linear Systems

Dynamic Response of Linear Systems Dyamic Respose of Liear Systems Liear System Respose Superpositio Priciple Resposes to Specific Iputs Dyamic Respose of st Order Systems Characteristic Equatio - Free Respose Stable st Order System Respose

More information

Quantum Theory Assignment 3

Quantum Theory Assignment 3 Quatum Theory Assigmet 3 Assigmet 3.1 1. Cosider a spi-1/ system i a magetic field i the z-directio. The Hamiltoia is give by: ) eb H = S z = ωs z. mc a) Fid the Heiseberg operators S x t), S y t), ad

More information

RMO Sample Paper 1 Solutions :

RMO Sample Paper 1 Solutions : RMO Sample Paper Slutis :. The umber f arragemets withut ay restricti = 9! 3!3!3! The umber f arragemets with ly e set f the csecutive 3 letters = The umber f arragemets with ly tw sets f the csecutive

More information

Axial Temperature Distribution in W-Tailored Optical Fibers

Axial Temperature Distribution in W-Tailored Optical Fibers Axial Temperature Distributi i W-Tailred Optical ibers Mhamed I. Shehata (m.ismail34@yah.cm), Mustafa H. Aly(drmsaly@gmail.cm) OSA Member, ad M. B. Saleh (Basheer@aast.edu) Arab Academy fr Sciece, Techlgy

More information

Sinusoids and Phasors

Sinusoids and Phasors CHAPTER 9 Sinusoids and Phasors We now begins the analysis of circuits in which the voltage or current sources are time-varying. In this chapter, we are particularly interested in sinusoidally time-varying

More information

Fall 2011 ME 2305 Network Analysis. Sinusoidal Steady State Analysis of RLC Circuits

Fall 2011 ME 2305 Network Analysis. Sinusoidal Steady State Analysis of RLC Circuits Fall 2011 ME 2305 Network Analysis Chapter 4 Sinusoidal Steady State Analysis of RLC Circuits Engr. Humera Rafique Assistant Professor humera.rafique@szabist.edu.pk Faculty of Engineering (Mechatronics)

More information

JAZAN University. Department: Electrical Engineering. Names & ID: Electronics LAB - 1/ / Electronics LAB - EngE : 314 G:...

JAZAN University. Department: Electrical Engineering. Names & ID: Electronics LAB - 1/ / Electronics LAB - EngE : 314 G:... Electrnics LAB - EngE : 314 G:... Electrnics LAB - 1/2-201. /201. Electrnics LAB - EngE : 314 G:... Electrnics LAB - 2/2-201. /201. Silicn JAZAN University Experiment 1: characteristics Electrnics LAB

More information

ACTIVE FILTERS EXPERIMENT 2 (EXPERIMENTAL)

ACTIVE FILTERS EXPERIMENT 2 (EXPERIMENTAL) EXPERIMENT ATIVE FILTERS (EXPERIMENTAL) OBJETIVE T desig secd-rder lw pass ilters usig the Salle & Key (iite psitive- gai) ad iiite-gai apliier dels. Oe circuit will exhibit a Butterwrth respse ad the

More information

ENGIN 211, Engineering Math. Fourier Series and Transform

ENGIN 211, Engineering Math. Fourier Series and Transform ENGIN 11, Engineering Math Fourier Series and ransform 1 Periodic Functions and Harmonics f(t) Period: a a+ t Frequency: f = 1 Angular velocity (or angular frequency): ω = ππ = π Such a periodic function

More information

E o and the equilibrium constant, K

E o and the equilibrium constant, K lectrchemical measuremets (Ch -5 t 6). T state the relati betwee ad K. (D x -b, -). Frm galvaic cell vltage measuremet (a) K sp (D xercise -8, -) (b) K sp ad γ (D xercise -9) (c) K a (D xercise -G, -6)

More information

radians A function f ( x ) is called periodic if it is defined for all real x and if there is some positive number P such that:

radians A function f ( x ) is called periodic if it is defined for all real x and if there is some positive number P such that: Fourier Series. Graph of y Asix ad y Acos x Amplitude A ; period 36 radias. Harmoics y y six is the first harmoic y y six is the th harmoics 3. Periodic fuctio A fuctio f ( x ) is called periodic if it

More information

Multi-objective Programming Approach for. Fuzzy Linear Programming Problems

Multi-objective Programming Approach for. Fuzzy Linear Programming Problems Applied Mathematical Scieces Vl. 7 03. 37 8-87 HIKARI Ltd www.m-hikari.cm Multi-bective Prgrammig Apprach fr Fuzzy Liear Prgrammig Prblems P. Padia Departmet f Mathematics Schl f Advaced Scieces VIT Uiversity

More information

Super-efficiency Models, Part II

Super-efficiency Models, Part II Super-efficiec Mdels, Part II Emilia Niskae The 4th f Nvember S steemiaalsi Ctets. Etesis t Variable Returs-t-Scale (0.4) S steemiaalsi Radial Super-efficiec Case Prblems with Radial Super-efficiec Case

More information

Lecture 20a. Circuit Topologies and Techniques: Opamps

Lecture 20a. Circuit Topologies and Techniques: Opamps Lecture a Circuit Tplgies and Techniques: Opamps In this lecture yu will learn: Sme circuit tplgies and techniques Intrductin t peratinal amplifiers Differential mplifier IBIS1 I BIS M VI1 vi1 Vi vi I

More information

Chapter 4 The Fourier Series and Fourier Transform

Chapter 4 The Fourier Series and Fourier Transform Chapter 4 The Fourier Series and Fourier Transform Representation of Signals in Terms of Frequency Components Consider the CT signal defined by N xt () = Acos( ω t+ θ ), t k = 1 k k k The frequencies `present

More information

Markov processes and the Kolmogorov equations

Markov processes and the Kolmogorov equations Chapter 6 Markv prcesses ad the Klmgrv equatis 6. Stchastic Differetial Equatis Csider the stchastic differetial equati: dx(t) =a(t X(t)) dt + (t X(t)) db(t): (SDE) Here a(t x) ad (t x) are give fuctis,

More information

Module B3. VLoad = = V S V LN

Module B3. VLoad = = V S V LN Mdule B Prblem The -hase lads are cnnected n arallel. One s a urely resste lad cnnected n wye. t cnsumes 00kW. The secnd s a urely nducte 00kR lad cnnected n wye. The thrd s a urely caacte 00kR lad cnnected

More information

Cop yri ht 2006, Barr Mabillard.

Cop yri ht 2006, Barr Mabillard. Trignmetry II Cpyright Trignmetry II Standards 006, Test Barry ANSWERS Mabillard. 0 www.math0s.cm . If csα, where sinα > 0, and 5 cs α + β value f sin β, where tan β > 0, determine the exact 9 First determine

More information

MATHEMATICS 9740/01 Paper 1 14 Sep hours

MATHEMATICS 9740/01 Paper 1 14 Sep hours Cadidate Name: Class: JC PRELIMINARY EXAM Higher MATHEMATICS 9740/0 Paper 4 Sep 06 3 hurs Additial Materials: Cver page Aswer papers List f Frmulae (MF5) READ THESE INSTRUCTIONS FIRST Write yur full ame

More information

Fourier Method for Solving Transportation. Problems with Mixed Constraints

Fourier Method for Solving Transportation. Problems with Mixed Constraints It. J. Ctemp. Math. Scieces, Vl. 5, 200,. 28, 385-395 Furier Methd fr Slvig Trasprtati Prblems with Mixed Cstraits P. Padia ad G. Nataraja Departmet f Mathematics, Schl f Advaced Scieces V I T Uiversity,

More information

R-L-C Circuits and Resonant Circuits

R-L-C Circuits and Resonant Circuits P517/617 Lec4, P1 R-L-C Circuits and Resonant Circuits Consider the following RLC series circuit What's R? Simplest way to solve for is to use voltage divider equation in complex notation. X L X C in 0

More information

ELEG 635 Digital Communication Theory. Lecture 10

ELEG 635 Digital Communication Theory. Lecture 10 ELEG 635 Digital Cmmuiati hery Leture Ergdi Radm Presses CPM reeiver Sigal Parameter Estimati Carrier Syhrizati Ageda Ergdi Radm Presses - A radm press is said t be ergdi if time averagig is equivalet

More information

Electrostatics. . where,.(1.1) Maxwell Eqn. Total Charge. Two point charges r 12 distance apart in space

Electrostatics. . where,.(1.1) Maxwell Eqn. Total Charge. Two point charges r 12 distance apart in space Maxwell Eq. E ρ Electrstatics e. where,.(.) first term is the permittivity i vacuum 8.854x0 C /Nm secd term is electrical field stregth, frce/charge, v/m r N/C third term is the charge desity, C/m 3 E

More information

Review Problems 3. Four FIR Filter Types

Review Problems 3. Four FIR Filter Types Review Prblems 3 Fur FIR Filter Types Fur types f FIR linear phase digital filters have cefficients h(n fr 0 n M. They are defined as fllws: Type I: h(n = h(m-n and M even. Type II: h(n = h(m-n and M dd.

More information

ELEG 635 Digital Communication Theory. Lecture 11

ELEG 635 Digital Communication Theory. Lecture 11 EEG 635 Digital Cmmuiati Thery eture 11 110801 Ageda Carrier Syhrizati Phase k p (P) Deisi Direted ps N-Deisi Direted ps Timig Syhrizati Deisi Direted ps N-Deisi Direted ps frmati Thery Sha's aw iear Blk

More information

Analysis of a Positive Output Super-Lift Luo Boost Converter

Analysis of a Positive Output Super-Lift Luo Boost Converter Ausha eade et al. t. Jural f Egeerg esearch ad Applicats SSN: 8-96, l. 6, ssue, (Part - 5) February 06, pp.7-78 ESEACH ACE www.ijera.cm OPEN ACCESS Aalys f a Psitive Output Super-ift u Bst Cverter Ausha

More information

Sinusoidal Steady State Power Calculations

Sinusoidal Steady State Power Calculations 10 Sinusoidal Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/ 45 V, Therefore I = 20/15 A P = 1 (100)(20)cos[ 45 (15)] = 500W, 2 A B Q = 1000sin 60 = 866.03 VAR, B A [b] V = 100/

More information

ALE 26. Equilibria for Cell Reactions. What happens to the cell potential as the reaction proceeds over time?

ALE 26. Equilibria for Cell Reactions. What happens to the cell potential as the reaction proceeds over time? Name Chem 163 Secti: Team Number: AL 26. quilibria fr Cell Reactis (Referece: 21.4 Silberberg 5 th editi) What happes t the ptetial as the reacti prceeds ver time? The Mdel: Basis fr the Nerst quati Previusly,

More information

Project Components. MC34063 or equivalent. Bread Board. Energy Systems Research Laboratory, FIU

Project Components. MC34063 or equivalent. Bread Board. Energy Systems Research Laboratory, FIU Project Components MC34063 or equivalent Bread Board PSpice Software OrCAD designer Lite version http://www.cadence.com/products/orcad/pages/downloads.aspx#pspice More Details on the Introduction CONVERTER

More information

Frequency Response. Re ve jφ e jωt ( ) where v is the amplitude and φ is the phase of the sinusoidal signal v(t). ve jφ

Frequency Response. Re ve jφ e jωt ( ) where v is the amplitude and φ is the phase of the sinusoidal signal v(t). ve jφ 27 Frequency Response Before starting, review phasor analysis, Bode plots... Key concept: small-signal models for amplifiers are linear and therefore, cosines and sines are solutions of the linear differential

More information

Edexcel GCSE Physics

Edexcel GCSE Physics Edexcel GCSE Physics Tpic 10: Electricity and circuits Ntes (Cntent in bld is fr Higher Tier nly) www.pmt.educatin The Structure f the Atm Psitively charged nucleus surrunded by negatively charged electrns

More information

BC Calculus Review Sheet. converges. Use the integral: L 1

BC Calculus Review Sheet. converges. Use the integral: L 1 BC Clculus Review Sheet Whe yu see the wrds.. Fid the re f the uuded regi represeted y the itegrl (smetimes f ( ) clled hriztl imprper itegrl).. Fid the re f differet uuded regi uder f() frm (,], where

More information

Electric Circuit Theory

Electric Circuit Theory Electric Circuit Theory Nam Ki Min nkmin@korea.ac.kr 010-9419-2320 Chapter 11 Sinusoidal Steady-State Analysis Nam Ki Min nkmin@korea.ac.kr 010-9419-2320 Contents and Objectives 3 Chapter Contents 11.1

More information

GATE EE Topic wise Questions SIGNALS & SYSTEMS

GATE EE Topic wise Questions SIGNALS & SYSTEMS www.gatehelp.com GATE EE Topic wise Questions YEAR 010 ONE MARK Question. 1 For the system /( s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) s (C)

More information

B U Department of Mathematics Math 101 Calculus I

B U Department of Mathematics Math 101 Calculus I B U Departmet of Mathematics Math Calculus I Sprig 5 Fial Exam Calculus archive is a property of Boğaziçi Uiversity Mathematics Departmet. The purpose of this archive is to orgaise ad cetralise the distributio

More information

[1 & α(t & T 1. ' ρ 1

[1 & α(t & T 1. ' ρ 1 NAME 89.304 - IGNEOUS & METAMORPHIC PETROLOGY DENSITY & VISCOSITY OF MAGMAS I. Desity The desity (mass/vlume) f a magma is a imprtat parameter which plays a rle i a umber f aspects f magma behavir ad evluti.

More information

Fourier Series and their Applications

Fourier Series and their Applications Fourier Series ad their Applicatios The fuctios, cos x, si x, cos x, si x, are orthogoal over (, ). m cos mx cos xdx = m = m = = cos mx si xdx = for all m, { m si mx si xdx = m = I fact the fuctios satisfy

More information

EE 221 Practice Problems for the Final Exam

EE 221 Practice Problems for the Final Exam EE 1 Practce Prblems fr the Fnal Exam 1. The netwrk functn f a crcut s 1.5 H. ω 1+ j 500 Ths table recrds frequency respnse data fr ths crcut. Fll n the blanks n the table:. The netwrk functn f a crcut

More information