Dgtal Sgnal Processng Dscrete-tme System Analyss Manar Mohasen Offce: F8 Emal: manar.subh@ut.ac.r School of IT Engneerng
Revew of Precedent Class Contnuous Sgnal The value of the sgnal s avalable over a contnuum of tme. Dscrete-tme Sgnal The sgnal s avalable at dscrete nstants of tme. Dgtal Sgnal The sgnal s dscrete-tme and has a dscrete number of levels. Contnuous sgnal Dscrete-tme sgnal Dgtal sgnal
Dscrete-tme System Analyss Assumpton Input x() and output y() are consdered to be causal dscrete-tme sgnals. Learnng Objectves Z-transform and ts propertes, and nverse Z-transform System representaton Dfference equatons Transfer functon, ero-state response, eros poles and modes, etc. Sgnal flow graph Impulse response and convoluton FIR and IIR systems System Stablty and the Jury test Frequency response Steady-state response
Z-transform Pars Z-transform The Z-transform of a causal dscrete-tme sgnal x() s functon X() of a complex varable defned as { } X () Z x() x() The Z-transform can be expressed as a rato of two polynomals b ( )( ) L( ) () m X ( p)( p) L( p ) The roots of the numerator are called the eros of X(). The roots of the denomnator are called the poles of X(). m Regon of Convergence (ROC) The set of values n the complex plane (C)over whch the power seres n - converges.
Z-transform Pars contd. Regon of Convergence (ROC) { C R } ROC > x R x max,..., n p Regon of convergence (shaded) ROC Example. x {, -3, 7, 4,,, } X() 3 + 7 + 4 3 3 3 + 7 + 4, > 3 ROC s the entre complex plane except the orgn. Example.: Unt mpulse δ() )defned edas follows. o δ ( ),, Z { } δ ( ). ROC s the entre complex plane. Im() Re() R x
Z-transform: Generc Seres Practcal Sgnals Most of practcal sgnals have nfnte number of nonero samples. It s, therefore, helpful l to use the followng result called geometrc seres. Unt step Example: Unt Step m f () f (), f() <, m f ( ) m The unt step functon s defned as followng:,, < u ( ), < Therefore, () U, > x() Im().5.5 -.5-5 5 Pole-ero pattern - - - Re() O X
Z-transform: Generc Seres contd. Example: Causal Exponental x() a u() Then, () ( a ) X, a < a, > a a x().5 Causal exponentals Im() Pole-ero patterns O a.8 5 5 - - 5 a. Zero at the orgn. Pole at a. If a, the sgnals reduces to the unt step. 5 5 x() 4 3 - Im() - O - - Re() X X
Z-transform Example Causal exponental damped cosne x a bu ( ) cos( ) ( ) Note that cos( b ) exp( jb )/ + exp( jb )/ Then, Z () x + ( aexp( jb ) ) ( aexp( jb ) ) + ( aexp( jb)) ( aexp( jb)) ( aexp( jb)) + ( aexp( jb)) ( aexp( jb ))( aexp( jb )) ( acos( b)) ( acos( b)) ( acos( b) + a) ( acos( b) + a) Poles at ( a exp(± jb) ). Zeros at and ( a cos(b) ). rt Imagnary Pa ) (.5 -.5 -.5 -.5 - -.5.5 Real Part x() a cos(b) u() a.8, b.5 5 5
Z-transform Propertes Lnearty Property Z-transform s a lnear operator. { } Z ax( ) + by( ) ax( ) + by( ) Example: Gven that Then, { } a x( ) + b y( ) ax() + by() x ( ) (.5) u ( ) + 5 δ ( ) Zx ( ) Z(.5) u ( ) 5 δ ( ) { δ } Z (.5) u ( ) + 5 Z ( ) X() has a pole at.5 and a ero at.5/7. + 5 7+.5.5.5
Z-transform Propertes contd. Delay Property Delayng the sgnal by M samples s equvalent to multplyng ts Z-tra nsform by -M. { } Z x ( M) x ( M) Let M. Then + M and Example: Pulse ( + M) { ( )} ( ) Z x M x M M M M x () X() x ( ) au ( ) u ( M ) X() a U() MU() a( M) U() M M ( ) M ( ) a( ) a( ). If a and M, x() δ() and X().
Z-transform Propertes contd. Z-scale Property Multplyng a tme sgnal by a s equvalent to scalng the Z-transform varable by /a. x ( ) a X a Z a x( ) a x( )
Z-transform Propertes contd. Tme Multplcaton Property Multplyng a dscrete tme sgnal by the tme varable s equvalent to tang the dervatve of the Z-transform and scalng by. dx () d d x ( ) d x( ) x( ) Then dx () x( ) d Example: Unt Ramp, r() u() R () du() d ( )
Z-transform Propertes contd. Intal and Fnal Value Theorems Intal value theorem Fnal value theorem Example: x () lm X () x ( ) lm( ) X( ) X () a ( M ) M ( ) x() lm a x( ) lm a ( M ) M ( ) a ( M )( ) M ( )
Basc Z-transform Propertes - Summary Lnearty { } Delay Z { x ( M) M } X( ) Z-scale Z { ax ( )} X ( / a) Tme Multplcaton Zx () { ( ) dx } d Intal Value Z ax( ) + by( ) ax ( ) + by( ) x() lm X( ) Fnal Value x( ) lm( ) X( ) Convoluton Zh ( ) x ( ) H ( ) X ( ) { } School of IT Engneerng
Inverse Z-transform Inverse Z-transform Sometmes, the Z-transform of a sgnal s easy to obtan, to fnd the actual sgnal n tme doman, we should perform the nverse Z- transform. x ( ) Z X( ) { } Recall that X() can be represented as the rato of two polynomals n. bm+ bm + L+ b () m X n + a n + L + a n The denomnator s normaled so that the leadng coeffcent equals. If x() s a causal sgnal X() s a proper ratonal polynomal wth m n. That s, the number of eros s less than or equal to the number of poles.
Inverse Z-transform contd. Synthetc Dvson Method Ths method s used when a fnte number of samples of x() s requred. STEPS: STEP I: Wrte X() as a normaled ratonal polynomal» r n m. r( b + b m + L+ bm ) + n + L+ n X() b ( ) a ( ) a a STEP II: Perform a long dvson of b( - )bya( - ) to obtan For r, x() q() X() q() + q() + q() +L r In the general case where r,, < r x ( ) q ( r ), r <
Inverse Z-transform contd. Synthetc Dvson Method Example ( ) ( X() + + + ) 3 3 + ( 3+ 3) 3 + 3 r m n. Long Dvson of b( - ) ( + - ) by a( - ) ( - 3 - + 3 - ) Snce r, 4 9 5 3 + 3 + 3 + 3 + + + 3+ 4 3 4 + L 3 3 { L} x ( ),, 4, 9,5, 9 9 7 + 7 3 4 3 4 5 7
Inverse Z-transform contd. Advantage of the Synthetc Dvson Method: It can be automated. Drawbac:. Wrte X() as a ratonal polynomal n -. Set q() ) b() ( ) r( b + b m + L+ bm ) + n + L+ n b ( ) X() a ( ) a a 3. For to p compute { q( ) b q ( ) n ( ), b } a q ( ) b, n q b n< p 4. Compute x() from q() by shftng the later r samples. a closed-form expresson can t be obtaned.
Inverse Z-transform contd. Partal Fracton Method To fnd an explct expresson for x(). X() ) should have n dstnct t poles. X() can be expressed as X() b () ( p )( p ) L ( p n ) Then, express X()/ usng partal fracton [Why we dvde by?] X() n R R R Rn + + L + p p p p Here, R s called the resdue of X() at pole p. Multply both sdes of the prevous equaton by ( - p ) and evaluatng the result at p. We fnd that: ( p ) X( ) R, n p Fnally, multplyng both sdes by and the nverse Z-transform s gven by: x n ( ) ( ) R p u ( ) n
Inverse Z-transform contd. Partal Fracton Method contd. Example. Consder Then and () ( + 4) R R R 3 X + + + ( )( + )( 3) ( ) ( + ) ( 3) R R R 3 ( ) X( ) ( + 4) 8 ( + )( 3) ( + ) X( ) ( + 4) 5 ( 3)( ) 6 ( 3) X( ) ( + 4) 65 ( )( + ) 3 3 X ( ) ( 8/3) + (5/6) + (65/) ( ) ( + ) ( 3) X( ) ( 8/3) + (5/6) + (65/) ( ) ( + ) ( 3) 8() ( ) 5( ) 65(3) x + + u ( ) 3 6 Show that x() s consstent wth the ntal value property. 3
Inverse Z-transform contd. Partal Fracton Method contd. P. (b) X (), > Fnd x() usng the partal fracton method. X () R + R ( )( + ) + R R ( ) X( ) ( ) ( + ) X( ) ( ) x ( ) ( ) u ( ) +.5 x() 3.5.5 4 6 8
Inverse Z-transform contd. Resdue Method Suppose that X() s factored as follows. X() b () r m m ( ) ( ) m p + p + L + ( p ) q Note that f m, we call p as a smple pole, otherwse we say that p has a multplcty l t m. Does not use tables to do the nverse Z-transform. Uses the Z-transform propertes and Cauchy s resdue theorem. The resdue theorem states that x() s the sum of resdues of X() (-) evaluated at the poles of X() (-). ( ) q x Res( p ) Resdue for a smple pole Res( p ) ( p) V( ), V( ) r X( ) p Resdue for poles havng multplcty m Res( ) d m p ( ) p ( ) ( )! m V m d q
Inverse Z-transform contd. Resdue Method contd.. Wrte X() n the followng form X() b () m ( ) m ( ) m r p + p + L + ( p ) q. Remove the r poles at by lettng V() r X(). 3. Compute the ntal value 4. For to q do { f p s a smple pole } 5. Set x() lm X( ) Res( ) ( ) ( ) p p V p else p s a multple pole and 6. x() v( r). m d m { m } d Res( p) ( p) V( ) ( m )! v ( ) x() δ ( ) + Res( p) u ( ) q q
Inverse Z-transform contd. Resdue Method Example. Fnd the nverse Z-transform of X () ( a)( b) Intal condton st Resdue nd Resdue x() lm X( ) + a Res( a) ( a) X( ) a a b b + b X b Res( b) ( ) ( ) b a Thus x ( ) x () δ ( ) + Res( s ) + Res( b ) u ( ) + b+ a b + b+ a b δ ( ) + a u( ) a u ( )
Inverse Z-transform contd. Resdue Method Example Fnd the nverse Z-transform of X() 5( ) ( ) V() s gven as follows V() 5 X() ( ) ( ) ( )( ) Res() d ( ) ( ) d V d d ( ) st Resdue { } nd Resdue { V } Res() ( ) ( ) ( ) Then v ( ) x() δ ( ) + Res() + Res() u ( ) + u ( ) Fnally, ( ) ( ) 5 x vn + + 5 u ( 6) n 5
Transfer Functons Dfference Equaton The followng s a generc LTI dfference equaton. n y( ) + a y( ) b x( ) m Thus, the soluton depends on the causal nput x() and the ntal condton {y(-), y(-), (, y(-n)}. ( Therefore, the output t can be decomposed d nto the sum of two parts. y( ) y ( ) y( ) + 443 443 ero nput ero state - Zero-nput response - Zero-state t response - Part of the output that s generated by the ntal condtons (x() ). - For a stable system y () as. - Part of the output that s generated by the nput sgnal. - For a stable system y () y() as.
Transfer Functons contd. Dfference Equaton contd. Let x() be a nonero nput to an LTI dscrete-tme system. Let y() be the output t wth a ero ntal t condton. Then, the transfer functon of the system s gven by: n y ( ) + a y ( ) b x ( ) m H () Y () X () n Y () + a Y () b X () m + n m + a Y ( ) b X ( ) b b b H () + + L+ + a + + m m L a n n
Transfer Functons contd. Example.4 Gven y( ). y( ).3 y( ) + x( ) + 6 x( ) Then Y ( ). Y ( ).3 Y ( ) + X ( ) + 6 X ( ) + 6 + 6 H (). +.3.+.3
Transfer Functons contd. Zero-state Response The ero-state response can be drectly obtaned from the transfer functon. y( ) Z H( ) X( ) { } Example.5 H() and X() are gven as follows. H () + 6 + 6.+.3 (.8)(.6) Usng the Resdue Method y ( ) Z H( ) X( ) { } Z (+ 6) (.8)(.6)( ) 5 + y() 33.3 75(.8) + 4.7(.4) u ( ) 5 X() 3 4 5
Transfer Functons contd. Poles, Zeros, and Modes Consder the transfer functon to be factored n the followng form H () b ( )( ) L ( ) m m n( p)( p ) L( p ) The roots of the numerator are the eros of the system. The roots of the denomnator are the poles of the system. If m > n, we wll have (m n) poles at ero. If m < n, we wll have (n m) eros at ero. Example: y( ) y( ) y( ) + x( ) + x( ) H () Y j j X() + + ( )( + ) () + + ( )( + ) Two poles at + and -. Two eros at +j and j. n
Transfer Functons contd. Poles, Zeros, and Modes contd. The output y() can be decomposed nto terms called modes. y( ) natural modes + forced modes - Each term s generated by a - Each term s generated by a pole of H(). pole of X(); the forcng functon. Each mode taes the followng form mode ( c + c + L + c r ) p, r r Where c s a constant and r s the multplcty of the pole)
Transfer Functons contd. Poles, Zeros, and Modes contd. Example.7 Let H () ( +.6) (.8)(.4) X() (.4) +.6 Then, { } y ( ) Z H( ) X( ) Z (.8) u ( ).8 Note on Example.7 A natural mode and a forced mode dd not appear n the ero-state output y(). [Can ths be useful?]
Transfer Functons contd. DC Gan If all the poles of the system strctly resde nsde the unt crcle, then all the natural modes of the system decay to ero wth ncreasng tme and the system s sad to be stable. Then, the amount by whch a DC nput (au()) s scaled, s called the DC gan of the system. From the fnal value theorem Therefore, lm y ( ) lm( ) Y ( ) lm( ) H ( ) X ( ) lm ( ) H( ) a lm Ha ( ) H() a DC gan H()
Summary and Dscusson Assumpton Input x() and output y() are consdered to be causal dscrete-tme sgnals. Learnng Objectves Z-transform and ts propertes, and nverse Z-transform System representaton Dfference equatons Transfer functon, ero-state response, eros poles and modes, etc.
Contnuous, Dscrete, and Dgtal Contnuous Sgnal The value of the sgnal s avalable over a contnuum of tme. Dscrete-tme Sgnal The sgnal s avalable at dscrete nstants of tme. Dgtal Sgnal The sgnal s dscrete-tme and has a dscrete number of levels. Contnuous sgnal Dscrete-tme sgnal Dgtal sgnal
Dscrete-tme System Analyss Assumpton Input x() and output y() are consdered to be causal dscrete-tme sgnals. Learnng Objectves Z-transform and ts propertes, and nverse Z-transform System representaton Dfference equatons Transfer functon, ero-state response, eros poles and modes, etc. Sgnal flow graph Impulse response and convoluton FIR and IIR systems System Stablty Frequency response Steady-state response
Sgnal Flow Graph Sgnal Flow Chart Dscrete-tme systems can be represented effcently usng the sgnal flow dagrams. Example: y x au + bυ multplcaton υ output u a x b y addton
Sgnal Flow Graph contd. Transfer Functon Decomposton v() s output of the frst subsystem and the nput to the second subsystem. Ha( ) H () b x ( ) v ( ) b + b + L+ bm L + a n + + an m y ( )
Sgnal Flow Graph contd. Example H ().4 +.6.8 +.9 v x.8 4.4 y.9.6 x v v y.8.4.9.6
The Impulse Response and Convoluton Impulse Response The output of the system produced by the unt mpulse nput. School of IT Engneerng Convoluton Input s x() ( ) Output s y() x ( ) y( ) h ( ) x ( ) x ( ) δ ( ), y ( ) x ( ) h ( ) x() δ ( ) + x() δ ( ) + x() δ ( ) LL xn () δ ( n) h ( ) x() h( ) + x() h( ) + x() h( ) LL xnh ()( n)
Convoluton contd. Example 3 3 () a( () b( 3 4 5 6 Step I: Flp one of the sgnals Step II: Move and compute a() 3 3 4 5 6 b(-) 3 3 4 5 6 Move and compute 3 4 5 6 5 y() 5 4 6 8
FIR & IIR Systems Fnte-mpulse Response (FIR) System Iff the mpulse response has a fnte number of nonero samples. Every FIR system has an equal number of poles and eros. The poles of an FIR system are all located at the orgn. b, m h ( ), m< < Example: m m + y( ) x( ) Infnte-mpulse mpulse Response (IIR) System The mpulse response has an nfnte number of nonero samples.
Stablty Stable System A system s a BIBO stable ff every bounded nput produces a bounded output. Otherwse, the system s unstable. Frst Stablty Chec [Tme doman] y ( ) hx ( ) ( ) The bound of the output y ( ) hx ( ) ( ) B h () x ( ) Therefore, the system has a bounded output ff S h x h() h( ) <
Stablty contd. Example Consder the mpulse response h ( ) a. u ( ) Is the system BIBO stable? S h( ) au( ) h a, a< a BIBO stable ff a <. Note N b bn+ b
Stablty contd. Second Stablty Chec An LTI dscrete-tme system wth TF H() s BIBO stable ff All poles of fh( H() resde nsded the unt crcle. That s, p <, n.5 Regon of stable poles (shaded).5 Im() -.5 - -.5 -.5 - -.5.5.5 Re()
Frequency Response Frequency Response For an LTI and BIBO stable system, the frequency response s defned as: H( f) H( ), f exp( j π ft) If H() s generated by a dfference equaton wth real coeff. All the nformaton of H(f) s contaned n [, f s ] because H f H f ( ) * ( ) Then, the power spectrum s a symmetrc functon. If we have h(), how can I fnd H(f)? From the -transform H( f) h( ) exp( j π ft ) h ( )exp( j π ft) f s
Frequency Response contd. Ln wth the Fourer Seres H( f) h( )exp( j π ft) () Note I: H(f) H(f + fs) We stated ths remar earler when we addressed the samplng theory Note II: () s a complex Fourer seres representaton of H(f). Therefore, f s / f fs / h ( ) T H( f)exp( j π ft) df /
Frequency Response contd. Snusodal Inputs Consder an nput gven by x ( ) sn( π ft a ) sn( θ ) θ π Then, the output of the LTI system wth TF H() Y () H () sn( θ) [ exp( j θ )][ exp( j θ )] { a } y ( ) Im H( f )exp( jθ ) ss fat { φ θ } a { φ a θ } y ( ) Im A( f )exp( j ( f )exp( j ) ss a a Af ( )Im exp( j( f) + j) Af ( )sn( θ + φ( f )) a Result We can measure the response (magntude and phase) of the system at certan frequency a
Frequency Response contd. Example H () Then (θ πft) +.64 H( f) exp( jθ ) + cos( θ) + + jsn( θ) exp( j θ).64 cos( θ).64 + jsn( θ) Magntude Response Af ( ) H( f) Phase Response φ( f ) H( f) cos( π ft) sn + + ( π ft) π ft + π cos(4 ).64 sn (4 ft) tan sn( π ft ) sn(4 π ft ) tan cos( π ft) + cos(4 π ft).64
Summary & Dscusson Assumpton Input x() and output y() are consdered to be causal dscrete-tme sgnals. Learnng Objectves Z-transform and ts propertes, and nverse Z-transform System representaton Dfference equatons Transfer functon, ero-state response, eros poles and modes, etc. Sgnal flow graph Impulse response and convoluton FIR and IIR systems System Stablty Frequency response Steady-state response