στο Αριστοτέλειο υλικού.

Σήματα Συστήματα Μετασχηματισμός aplace Λυμένες ασκήσεις Κωνσταντίνος Κοτρόπουλος Τμήμα Πληροφορικής Θεσσαλονίκη, Ιούνιος 03

Άδειες Χρήσης Το παρόν εκπαιδευτικό υλικό υπόκειται σε άδειες χρήσης Creative Commons. Για εκπαιδευτικό υλικό, όπως εικόνες, που υπόκειται σε άλλου τύπου άδειας χρήσης, η άδεια χρήσης αναφέρεται ρητώς. Χρηματοδότηση Το παρόν εκπαιδευτικό υλικό έχει αναπτυχθεί στα πλαίσια του εκπαιδευτικού έργου του διδάσκοντα. Το έργο «Ανοικτά Ακαδημαϊκά Μαθήματα στο Αριστοτέλειο Πανεπιστήμιο Θεσσαλονίκης»» έχει χρηματοδοτήσει μόνο τη αναδιαμόρφωση του εκπαιδευτικού υλικού. Το έργο υλοποιείται στο πλαίσιο του Επιχειρησιακού Προγράμματος «Εκπαίδευση και Δια Βίου Μάθηση»» και συγχρηματοδοτείται από την Ευρωπαϊκή Ένωση (Ευρωπαϊκό Κοινωνικό Ταμείο) και από εθνικούς πόρους. Θεσσαλονίκη, Ιούνιος 03

Kef laio 6 Metasqhmatismìc aplace 6.7 umènec ask seic 6.7.. Na upologistoôn oi metasqhmatismoð aplace twn shm twn (e t + t)u(t), u(t ) kai e t u( t). Gia to s ma (e t + t)u(t) o dðpleuroc kai o monìpleuroc metasqhmatismìc aplace tautðzontai: {(e t + t)u(t)} = {e t u(t)} + {tu(t)} = s + + s = s + s +, Re{s} > 0. s (6.7.) (s +) Gia to s ma u(t ) o dðpleuroc kai o monìpleuroc metasqhmatismìc aplace tautðzontai: {u(t )} = e s {u(t)} = e s, Re{s} > 0. (6.7.) s Gia to s ma e t u( t) o dðpleuroc kai o monìpleuroc metasqhmatismìc aplace eðnai diaforetikoð. Gia monìpleuro metasqhmatismì aplace to s ma eðnai e t u(t) u( t) =e t gia 0 t. 'Ara h perioq sôgklishc eðnai ìlo to s epðpedo. Pr gmati me th qr sh tou orismoô: X (s) = = e t e st dt = e t e st dt = e (s+)t dt 0 0 0 [ ] s + e (s+)t = e (s+). (6.7.3) 0 s +

K. Kotrìpouloc: S mata-sust mata Epeid to s = den eðnai pìloc. e (s+) lim s s + Gia dðpleuro metasqhmatismì aplace èqoume ( )e (s+) = lim s = (6.7.4) X(s) = e t u( t) e st dt = e t e st dt = e (s+)t dt = e (s+)t = ] [e (s+) e (s+)() (s +) s + = s + e (s+), Re{s} <. (6.7.5) Sto Ðdio apotèlesma katal goume me qr sh twn idiot twn: e t u( t) e t u( t) e (t ) u( (t )) ee t u( t) e t u( t) s +, s +, e s s +, e s s +, e s e s + Re{s} < Re{s} < Re{s} < Re{s} < e (s+) =, Re{s} <. (6.7.6) s + 6.7.. DÐnetai ìti o metasqhmatismìc aplace thc tu(t) eðnai s. Na upologðsete touc metasqhmatismoôc aplace twn shm twn (t )u(t) kai (at b)u(t). {(t )u(t)} = {tu(t)} {u(t)} = s s = s s, Re{s} > 0. (6.7.7) {(at + b)u(t)} = a s b s = a bs s, Re{s} > 0. (6.7.8) 6.7.3. Na upologisteð o monìpleuroc metasqhmatismìc aplace thc parag gou tou s matoc x(t) =te t u(t). ìpou o monìpleuroc metasqhmatismìc aplace tou x(t) eðnai U{ d dt x(t)} = sx (s) x(0 ) (6.7.9) X (s) =U{te t u(t)} =, Re{s} >. (6.7.0) (s +)

K. Kotrìpouloc: S mata-sust mata 3 Epeid x(0 )=0paÐrnoume: U{ d s x(t)} = sx (s) = dt, Re{s} >. (6.7.) (s +) Sto Ðdio apotèlesma ja katal game kai me efarmog tou dðpleurou metasqhmatismoô aplace sto Ðdio s ma, epeid to s ma eðnai aitiatì. 6.7.4. Me th qr sh twn idiot twn na upologðsete to metasqhmatismì aplace tou s matoc x(t) = 0 λu(t λ) dλ. Parathr ìti x(t) =tu(t) u(t). Epomènwc { 0 } λu(t λ)dλ = {tu(t) u(t)} = s s =, Re{s} > 0. (6.7.) s3 6.7.5. Na upologisteð o monìpleuroc metasqhmatismìc aplace tou hmi-periodikoô s matoc pou èqei pr th perðodo: x T (t) = { u(t) u(t 0.5), an t 0.5 0, an 0.5 <t<. (6.7.3) U{x T (t)} = X T (s) = s e s/ s X (s) = = e s/, ROC: ìlo to s-epðpedo s e stx T (s) = e s/, Re{s} > 0. ( e s (6.7.4) )s 'Ontwc h perioq sôgklishc eðnai Re{s} > 0, epeid lim s 0 e s/ ( e s )s = ( /)e s/ [ ( )e s ]s +[ e s ] = lim s 0 'Ara prèpei na exairejeð to s =0. ( s)e s = 0. (6.7.5) 6.7.6. Na antistrèyete touc ex c metasqhmatismoôc aplace: X(s) =, Re{s} > s (6.7.6) +4s +4 X(s) = s3 +3s +, Re{s} > (s +) 3 (6.7.7) X(s) = e 4s, Re{s} > 0. 3s 3 (6.7.8) +s

4 K. Kotrìpouloc: S mata-sust mata a. X(s) = s +4s +4 = (s +) x(t) =te t u(t). (6.7.9) b. Prosjètontac kai afair ntac 3s ston arijmht paðrnoume X(s) = s3 +3s + = s3 +3s +3s + 3s + (s +) 3 (s +) 3 = + 3s (s +) 3. (6.7.0) Opìte prèpei na epekteðnoume to deôtero ìro sthn (6.7.0) se jroisma merik n klasm twn: 3s A (s +) 3 s + + B (s +) + C (s +) 3 3s A(s +) + B(s +)+C 3s As +(A + B)s + A + B + C. (6.7.) Apì ta ek tautìthtac Ðsa polu numa sthn (6.7.) prokôptei èna sôsthma tri n e- xis sewn me treic agn stouc apì to opoðo mporoôme na upologðsoume ta A, B kai C: A = 3 (6.7.) A + B = 0 B = A =6 (6.7.3) A + B + C = C = A B =+3 6=. (6.7.4) Epomènwc apì ton PÐnaka basik n dðpleurwn metasqhmatism n aplace èqoume δ(t) (6.7.5) 3 s + 6 (s +) 6 te t u(t) (6.7.7) 3 e t u(t) (6.7.6) (s +) 3 e t t u(t). (6.7.8) SunoyÐzoume { s 3 +3s + } (s +) 3 = δ(t) 3 e t u(t)+6te t u(t) t e t u(t) = δ(t) (t 6t +3)e t u(t). (6.7.9)

K. Kotrìpouloc: S mata-sust mata 5 g. ìpou y(t) X(s) = e 4s 3s 3 +s Y (s) = x(t) =y(t) y(t 4) (6.7.30) 3s 3 +s = s (3s +) = 3s (s + 3 ) = A s + B s + C s +. (6.7.3) 3 Apì ta ek tautìthtac Ðsa polu numa paðrnoume 3s(s + 3 )A +3(s + 3 )B + C3s 3As +As +3Bs +B +3Cs 3(A + C)s +(A +3B)s +B. (6.7.3) opìte B = A +3B = 0 A = 3 B = 3 = 3 4 (6.7.33) (6.7.34) C = A C = 3 4. (6.7.35) Epomènwc Y (s) = y(t) = 3s (s + ) = 3/4 + / s s + 3/4 3 s +/3 ( 3 4 + t + 3 ) 4 e t/3 u(t). (6.7.36) Antikajist ntac sthn (6.7.30) prokôptei to zhtoômeno apotèlesma. 6.7.7. 'Ena sôsthma perigr fetai apì th diaforik exðswsh: y +3y + y = x + y (6.7.37) me arqikèc sunj kec y(0) = 0 kai y (0) = 0. Na upologðsete thn kroustik apìkrish tou sust matoc kai th sun rthsh metafor c tou. Poi eðnai h bhmatik apìkrish tou sust matoc? Sthn skhsh monìpleuroi kai dðpleuroi metasqhmatismoð tautðzontai, epeid to s ma y(t) eðnai aitiatì kai oi arqikèc sunj kec eðnai mhdenikèc. Skìpima ja k noume qr sh thc idiìthtac

6 K. Kotrìpouloc: S mata-sust mata thc parag gishc tou monìpleurou metasqhmatismoô aplace gia na upenjum soume ston anagn sth th qrhsimìthta tou monìpleurou metasqhmatismoô aplace sth melèth dunamik n susthm twn pou perigr fontai apì diaforikèc exis seic me mh-mhdenikèc arqikèc sunj kec. 'Eqoume { d y(t) } U dt { dy(t) } 3 U dt = s Y (s) sy (0) (0) y () (0) (6.7.38) = 3sY (s) 3y (0) (0) (6.7.39) ìpou { } U y(t) { } U x(t) = Y(s) =Y (s) (6.7.40) = X (s) =X(s). (6.7.4) PaÐrnontac to metasqhmatismì aplace amfotèrwn twn mel n thc (6.7.37) prokôptei (s +s +)Y (s) =X(s) H(s) = Y (s) X(s) =, Re{s} >. (6.7.4) (s +) Epomènwc h kroustik apìkrish eðnai h(t) = {H(s)} = te t u(t). (6.7.43) H bhmatik apìkrish ja prokôyei an x(t) =u(t). Tìte X(s) = s, Re{s} > 0, opìte Y (s) = (s +) s = Apì ta ek tautìthtac Ðsa polu numa ja èqoume A s + + As(s +)+Bs +Γ(s +) B (s +) + Γ s. (6.7.44) (A +Γ)s +(A + B +Γ)s +Γ (6.7.45) Γ = A = A + B +Γ = 0 B =. (6.7.46) 'Ara Y (s) = s + + (s +) + s y(t) = ( e t te t )u(t). (6.7.47)

K. Kotrìpouloc: S mata-sust mata 7 6.7.8. To sôsthma tou Sq matoc 6.7. lègetai bajmðda sugkr thshc (holding system). To olokl rwma pou upologðzei h sqetik bajmðda eðnai t apìkrish kai h apìkrish suqnìthtac tou sust matoc. dτ. Na upologisteð h kroustik x(t) t t 0 + Σ t dτ y(t) Sq ma 6.7.: BajmÐda sugkr thshc. H èxodoc thc bajmðdac sugkr thshc y(t) eðnai y(t) =[x(t) x(t t 0 )] h(t) = opìte o metasqhmatismìc Fourier thc exìdou eðnai All f(t) =x(t) x(t t 0 ), opìte t [x(τ) x(τ t 0 )] dτ (6.7.48) }{{} f(τ) Y (jω)=πf(0)δ(ω)+ F (jω) jω. (6.7.49) F (jω)=[ e jωt 0 ]X(jω) (6.7.50) kai F (0) = 0. Antikajist ntac sthn (6.7.49) prokôptei Y (jω) = F (jω) jω = [ e jωt0 ]X(jω) jω H(jω) = Y (jω) X(jω) = e jωt0 jω Sunep c h(t) =u(t) u(t t 0 ). (6.7.5) F h(t) = sgn(t) sgn(t t 0). (6.7.5) 6.7.9. DÐnetai to sôsthma tou Sq matoc 6.7. ìpou h (t) = d [ ] sin ω0 t dt π t (6.7.53) H (ω) = exp{ jπ ω } ω 0 (6.7.54) h 3 (t) = sin 3ω 0t πt (6.7.55) h 4 (t) = u(t). (6.7.56)

8 K. Kotrìpouloc: S mata-sust mata x(t) + h h + + h 3 h 4 y(t) Sq ma 6.7.: SÔsthma tou Probl matoc 6.7.9. Na upologisjeð h kroustik apìkrish tou sust matoc. 'Estw x (t) h èxodoc thc bajmðdac me kroustik apìkrish h (t). Tìte x (t) =(h x)(t). An x (t) eðnai h diègersh thc bajmðdac me kroustik apìkrish h 3 (t), ja èqoume x (t) =(h x)(t)+h (h x)(t) (6.7.57) kai h apìkrish tou sust matoc eðnai: y(t) = (h 3 h 4 ) x (t) =(h 3 h 4 ) h x(t)+(h 3 h 4 ) (h h ) x(t) = [h 3 h 4 h + h 3 h 4 h h ] x(t) [ ] H 3 (s)h 4 (s)h (s)+h 3 (s)h 4 (s)h (s)h (s) X(s). (6.7.58) Epeid isqôei: ja eðnai: h (t) = d dt sin ω 0 t πt ( sin ω0 t ) πt h 3 (t) = sin 3ω 0t πt h 4 (t) =u(t) F {, an ω <ω 0 0, an ω >ω 0 (6.7.59) { jω F H (ω) =, an ω <ω 0 (6.7.60) 0, an ω >ω 0 { F, an ω < 3ω0 H 3 (ω) = (6.7.6) 0, alloô F H 4 (ω) = + πδ(ω). (6.7.6) jω Opìte H 3 (ω)h 4 (ω)h (ω) = { [ ] + πδ(ω) jω jω, an ω <ω 0 0, an ω >ω 0 = H 3 (ω)h 4 (ω)h (ω)h (ω) = {, an ω <ω 0 0, an ω >ω 0. { jπω ω e( ) 0, an ω <ω 0 0, alloô. (6.7.63) (6.7.64)

K. Kotrìpouloc: S mata-sust mata 9 Sunep c { H(ω) = + πω ω e j 0, an ω <ω 0 0 alloô. H apìkrish suqnìthtac H(ω) mporeð na grafeð wc [ ] F H(ω) =G(ω) u(ω + ω 0 ) u(ω ω 0 ) h(t) =g(t) sin ω 0t πt (6.7.65) (6.7.66) ìpou Opìte telik G(ω) = + πω ω e j 0 h(t) = sin ω 0 t + sin ω 0 (t π ω 0 ) πt π(t π = sin ω 0 t πt F δ(t)+ π δ(t ). (6.7.67) ω 0 ω 0 ) + sin ω 0 t π(t π ω 0 ) = [ ] π tω0 sin ω 0 t. (6.7.68) πt π tω 0 6.7.0. Aitiatì kai eustajèc G.Q.A. sôsthma S èqei kroustik apìkrish h(t) kai rht sun rthsh sust matoc H(s). GnwrÐzoume ìti (a) H() = 0.. (b) 'Otan to sôsthma S diegeðretai me u(t), h èxodoc eðnai apolôtwc oloklhr simh. (g) 'Otan to sôsthma S diegeðretai me tu(t), h èxodoc den eðnai apolôtwc oloklhr simh. (d) To s ma d h(t) dt + dh(t) dt +h(t) eðnai peperasmènhc di rkeiac. (e) H sun rthsh sust matoc H(s) èqei èna dhlo mhdenikì sto peiro. Na prosdiorðste thn H(s) kai thn perioq sôgklis c thc.. To dedomèno (b) sunep getai ìti to sôsthma S gia diègersh x (t) =u(t) X (s) =, Re{s} > 0 (6.7.69) s par gei apìkrish y (t) =T[x (t)] pou diajètei metasqhmatismì Fourier. Epeid Y (s) =H(s)X (s) (6.7.70) toôto sumbaðnei ìtan h H(s) prosfèrei èna mhdenikì gia s =0, pou akur nei ton pìlo thc X (s) sto s =0. Dhlad, to s =0prèpei na an kei sth ROC thc sun rthshc sust matoc H(s), ra na mhn eðnai pìloc.

0 K. Kotrìpouloc: S mata-sust mata. To dedomèno (g) sunep getai ìti to sôsthma S gia diègersh x (t) =tu(t) X (s) =, Re{s} > 0 (6.7.7) s par gei apìkrish y (t) =T[x (t)] me metasqhmatismì aplace Y (s) =H(s)X (s) (6.7.7) pou gia na eðnai mh-apolôtwc oloklhr simh, prèpei o xonac jω na mhn an kei sthn ROC tou metasqhmatismoô aplace Y (s). Autì mporeð na sumbeð ìtan h sun rthsh sust matoc H(s) den èqei mhdenikì deôterhc t xhc sto s = 0. 'Ara h sun rthsh sust matoc H(s) diajètei mìno mhdenikì t xhc sto s =0, an sunektim soume kai to apotèlesma thc epexergasðac tou dedomènou (b). 3. 'Estw p(t) = d h(t) dt + dh(t) dt +h(t). Tìte P (s) =s H(s)+sH(s)+H(s) (6.7.73) epeid to sôsthma eðnai aitiatì, ra h(0 )=0. Opìte H(s) = P (s) s +s +. (6.7.74) SÔmfwna me to dedomèno (g) to p(t) eðnai peperasmènhc di rkeiac, kat sunèpeia den prèpei na èqei prìdhlouc pìlouc, ra P (s) =A N i= (s z i), opìte H(s) = A N i= (s z i) s +s +. (6.7.75) GnwrÐzoume ìti h H(s) èqei èna dhlo mhdenikì sto peiro, ra o arijmht c prèpei na eðnai polu numo to polô bajmoô, dhlad na tou leðpei èna mhdenikì. Sunep c H(s) = 4. Apì thn H(s) s= =0. èpetai ìti A =. 'Ara H(s) = As s +s +. (6.7.76) s s +s + me pìlouc s, = ± j. (6.7.77) Epeid h h(t) eðnai aitiat kai eustaj c sun rthsh, ìloi oi pìloi keðntai sto arnhtikì hmiepðpedo, opìte Re{s} >.

K. Kotrìpouloc: S mata-sust mata 6.7.. Na brejeð o metasqhmatismìc aplace tou s matoc x(t) ìtan: (a) To s ma x(t) eðnai pragmatikì rtiac summetrðac. (b) O metasqhmatismìc aplace X(s) èqei tèsseric prìdhlouc pìlouc kai kanèna prìdhlo mhdenikì. (g) O metasqhmatismìc aplace X(s) èqei pìlo sto s =( )ejπ/4. (d) x(τ)dτ =4. Apì to dedomèno (b) prokôptei X(s) = A (s a)(s b)(s c)(s d). (6.7.78) SÔmfwna me to dedomèno (a) to s ma x(t) eðnai pragmatikì s ma, ra oi pìloi tou metasqhmatismoô aplace sumbaðnoun se suzugeðc migadikèc jèseic sto epðpedo. 'Ara: b = a, d = c. (6.7.79) Efìson to s ma x(t) eðnai rtiac summetrðac sômfwna me to dedomèno (a), o metasqhmatismìc aplace ja prèpei na eðnai rtiac summetrðac epðshc. 'Ara oi pìloi sumbaðnoun se summetrikèc wc proc ton jω xona jèseic. ToÔto shmaðnei ìti: c = a. (6.7.80) 'Ara X(s) = GnwrÐzoume ìti a = ejπ/4. 'Ara: A (s a)(s a )(s + a )(s + a). (6.7.8) X(s) = A (s s + 4 )(s + s + (6.7.8) ). 4 All : opìte A = 4. x(τ) dτ = X(0) = 4 (6.7.83)