Find the inverse Z-transform of 2 z 2 + 2 z G ( z ) = z 2 + 2 z 3 G - PowerPoint PPT Presentation

Inverse Z-transform - Partial Fraction 1. Find the inverse Z-transform of 2 z 2 + 2 z G ( z ) = z 2 + 2 z − 3 G ( z ) 2 z + 2 = z ( z + 3)( z − 1) A B = z + 3 + z − 1 Multiply throughout by z +3 and let z = − 3 to get A = 2 z + 2 � = − 4 � − 4 = 1 � z − 1 � z = − 3 1 Digital Control Kannan M. Moudgalya, Autumn 2007

Inverse Z-transform - Partial Fraction 2. G ( z ) A B = z + 3 + z z − 1 Multiply throughout by z − 1 and let z = 1 to get B = 4 4 = 1 G ( z ) 1 1 = z + 3 + | z | > 3 z z − 1 z z G ( z ) = z + 3 + | z | > 3 z − 1 ↔ ( − 3) n 1( n ) + 1( n ) 2 Digital Control Kannan M. Moudgalya, Autumn 2007

Partial Fraction - Repeated Poles 3. N ( z ) G ( z ) = ( z − α ) p D 1 ( z ) α not a root of N ( z ) and D 1 ( z ) A 1 A 2 A p G ( z ) = z − α + ( z − α ) 2 + · · · + ( z − α ) p + G 1 ( z ) G 1 ( z ) has poles corresponding to those of D 1 ( z ) . Multiply by ( z − α ) p ( z − α ) p G ( z ) = A 1 ( z − α ) p − 1 + A 2 ( z − α ) p − 2 + · · · + A p − 1 ( z − α ) + A p + G 1 ( z )( z − α ) p 3 Digital Control Kannan M. Moudgalya, Autumn 2007

Partial Fraction - Repeated Poles 4. ( z − α ) p G ( z ) = A 1 ( z − α ) p − 1 + A 2 ( z − α ) p − 2 + · · · + A p − 1 ( z − α ) + A p + G 1 ( z )( z − α ) p Substituting z = α , A p = ( z − α ) p G ( z ) | z = α Differentiate and let z = α : A p − 1 = d dz ( z − α ) p G ( z ) | z = α d p − 1 1 dz p − 1 ( z − α ) p G ( z ) | z = α Continuing, A 1 = ( p − 1)! 4 Digital Control Kannan M. Moudgalya, Autumn 2007

Repeated Poles - an Example 5. G ( z ) = 11 z 2 − 15 z + 6 A 1 A 2 B ( z − 2)( z − 1) 2 = z − 1 + ( z − 1) 2 + z − 2 × ( z − 2) , let z = 2 , to get B = 20 . × ( z − 1) 2 , 11 z 2 − 15 z + 6 = A 1 ( z − 1) + A 2 + B ( z − 1) 2 z − 2 z − 2 With z = 1 , get A 2 = − 2 . Differentiating with respect to z and with z = 1 , A 1 = ( z − 2)(22 z − 15) − (11 z 2 − 15 z + 6) � � = − 9 � ( z − 2) 2 � z =1 9 2 20 G ( z ) = − z − 1 − ( z − 1) 2 + z − 2 5 Digital Control Kannan M. Moudgalya, Autumn 2007

Important Result from Differentiation 6. Problem 4.9 in Text: Consider ∞ z 1( n ) a n ↔ � � < 1 a n z − n , � � az − 1 � z − a = n =0 Differentiating with respect to a , ∞ z z � na n − 1 z − n , na n − 1 1( n ) ↔ ( z − a ) 2 = ( z − a ) 2 n =0 Substituting a = 1 , can obtain the Z-transform of n : z n 1( n ) ↔ ( z − 1) 2 Through further differentiation, can derive Z-transforms of n 2 , n 3 , etc. 6 Digital Control Kannan M. Moudgalya, Autumn 2007

Repeated Poles - an Example 7. 9 2 20 G ( z ) = − z − 1 − ( z − 1) 2 + z − 2 zG ( z ) = − 9 z ( z − 1) 2 + 20 z 2 z z − 1 − z − 2 ↔ ( − 9 − 2 n + 20 × 2 n )1( n ) Use shifting theorem: G ( z ) ↔ ( − 9 − 2( n − 1) + 20 × 2 n − 1 )1( n − 1) 7 Digital Control Kannan M. Moudgalya, Autumn 2007

Properties of Cont. Time Sinusoidal Signals 8. u a ( t ) = A cos (Ω t + θ ) u a ( t ) = A cos (2 πF t + θ ) , − ∞ < t < ∞ A : amplitude Ω : frequency in rad/s θ : phase in rad F : frequency in cycles/s or Hertz Ω = 2 πF T p = 1 F 8 Digital Control Kannan M. Moudgalya, Autumn 2007

Properties of Cont. Time Sinusoidal Signals 9. u a ( t ) = A cos (Ω t + θ ) u a ( t ) = A cos (2 πF t + θ ) , − ∞ < t < ∞ With F fixed, periodic with period T p u a [ t + T p ] = A cos (2 πF ( t + 1 F ) + θ ) = A cos (2 π + 2 πF t + θ ) = A cos (2 πF t + θ ) = u a [ t ] 9 Digital Control Kannan M. Moudgalya, Autumn 2007

Properties of Cont. Time Sinusoidal Signals 10. Cont. signals with different frequencies are different 2 π t 2 π 7 t � � � � u 1 = cos , u 2 = cos 8 8 1 0.8 0.6 0.4 0.2 coswt 0 −0.2 −0.4 −0.6 −0.8 −1 0 0.5 1 1.5 2 2.5 3 3.5 4 t Different wave forms for different frequencies. Can increase f all the way to ∞ or decrease to 0. 10 Digital Control Kannan M. Moudgalya, Autumn 2007

Discrete Time Sinusoidal Signals 11. u ( n ) = A cos ( wn + θ ) , −∞ < n < ∞ w = 2 πf n integer variable, sample number A amplitude of the sinusoid w frequency in radians per sample θ phase in radians. f normalized frequency, cycles/sample 11 Digital Control Kannan M. Moudgalya, Autumn 2007

Discrete Time Sinusoids - Identical Signals 12. Discrete time sinusoids whose frequencies are sepa- rated by integer multiple of 2 π are identical, i.e., cos (( w 0 + 2 π ) n + θ ) = cos ( w 0 n + θ ) , ∀ n • All sinusoidal sequences of the form, u k ( n ) = A cos ( w k n + θ ) , w k = w 0 + 2 kπ , − π < w 0 < π , are indistinguishable or identical. • Only sinusoids in − π < w 0 < π are different − π < w 0 < π or − 1 2 < f 0 < 1 2 12 Digital Control Kannan M. Moudgalya, Autumn 2007

Sampling a Cont. Time Signal - Preliminaries 13. Let analog signal u a [ t ] have a frequency of F Hz u a [ t ] = A cos(2 πF t + θ ) Uniform sampling rate ( T s s) or frequency ( F s Hz) t = nT s = n F s u ( n ) = u a [ nT s ] − ∞ < n < ∞ = A cos (2 πF T s n + θ ) 2 π F � � = A cos n + θ F s △ = A cos(2 πfn + θ ) 13 Digital Control Kannan M. Moudgalya, Autumn 2007

Sampling a Cont. Time Signal - Preliminaries 14. In our standard notation, u ( n ) = A cos (2 πfn + θ ) It follows that f = F w = Ω = Ω T s F s F s Reason to call f normalized frequency, cycles/sample. Apply the uniqueness condition for sampled signals − 1 2 < f < 1 − π < w < π 2 F max = F s Ω max = 2 πF max = πF s = π 2 T s 14 Digital Control Kannan M. Moudgalya, Autumn 2007

Properties of Discrete Time Sinusoids - Alias 15. • As w 0 ↑ , freq. of oscillation ↑ , reaches maximum at w 0 = π • What if w 0 > π ? w 1 = w 0 w 2 = 2 π − w 0 u 1 ( n ) = A cos w 1 n = A cos w 0 n u 2 ( n ) = A cos w 2 n = A cos(2 π − w 0 ) n = A cos w 0 n = u 1 ( n ) w 2 is an alias of w 1 15 Digital Control Kannan M. Moudgalya, Autumn 2007

Properties of Discrete Time Sinusoids - Alias 16. u 1 [ t ] = cos (2 π t 8) , u 2 [ t ] = cos (2 π 7 t 8 ) , T s = 1 u 2 ( n ) = cos (2 π 7 n 8 ) = cos 2 π (1 − 1 8) n = cos (2 π − 2 π 8 ) n = cos (2 πn 8 ) = u 1 ( n ) 1 0.8 0.6 0.4 0.2 coswt 0 −0.2 −0.4 −0.6 −0.8 −1 0 0.5 1 1.5 2 2.5 3 3.5 4 t 16 Digital Control Kannan M. Moudgalya, Autumn 2007

Fourier Transform of Aperiodic Signals 17. � ∞ x ( t ) e − j 2 πF t dt X ( F ) = −∞ � ∞ X ( F ) e j 2 πF t dF x ( t ) = −∞ If we let radian frequency Ω = 2 πF � ∞ x ( t ) = 1 X [Ω] e j Ω t d Ω , 2 π −∞ � ∞ x ( t ) e − j Ω t dt X [Ω] = −∞ x ( t ) , X [Ω] are Fourier Transform Pair 17 Digital Control Kannan M. Moudgalya, Autumn 2007

Frequency Response 18. Apply u ( n ) = e jwn to g ( n ) and obtain output y : ∞ � y ( n ) = g ( n ) ∗ u ( n ) = g ( k ) u ( n − k ) k = −∞ ∞ ∞ g ( k ) e jw ( n − k ) = e jwn � � g ( k ) e − jwk = k = −∞ k = −∞ Define Discrete Time Fourier Transform � ∞ ∞ � △ g ( k ) e − jwk = � � G ( e jw ) g ( k ) z − k � = � � k = −∞ k = −∞ � z = e jw = G ( z ) | z = e jw Provided the sequence converges absolutely 18 Digital Control Kannan M. Moudgalya, Autumn 2007

Frequency Response 19. ∞ g ( k ) e − jwk = e jwn G ( e jw ) � y ( n ) = e jwn k = −∞ Write in polar coordinates: G ( e jw ) = | G ( e jw ) | e jϕ - ϕ is phase angle: y ( n ) = e jwn | G ( e jw ) | e jϕ = | G ( e jwn ) | e j ( wn + ϕ ) 1. Input is sinusoid ⇒ output also is a sinusoid with following properties: • Output amplitude gets multiplied by the magnitude of G ( e jw ) • Output sinusoid shifts by ϕ with respect to input 19 Digital Control Kannan M. Moudgalya, Autumn 2007

2. At ω where | G ( e jw ) | is large, the sinusoid gets amplified and at ω where it is small, the sinusoid gets attenuated. • The system with large gains at low frequencies and small gains at high frequencies are called low pass filters • Similarly high pass filters 20 Digital Control Kannan M. Moudgalya, Autumn 2007

Discrete Fourier Transform 20. Define Discrete Time Fourier Transform � ∞ ∞ � △ g ( k ) e − jwk = � � G ( e jw ) g ( k ) z − k � = � � k = −∞ k = −∞ � z = e jw = G ( z ) | z = e jw Provided, absolute convergence ∞ ∞ � � | g ( k ) e − jwk | < ∞ ⇒ | g ( k ) | < ∞ k = −∞ k = −∞ For causal systems, BIBO stability. Required for DTFT. 21 Digital Control Kannan M. Moudgalya, Autumn 2007

FT of Discrete Time Aperiodic Signals - Defi- 21. nition ∞ △ � U ( e jω ) u ( n ) e − jωn = n = −∞ � π u ( m ) = 1 U ( e jω ) e jωm dω 2 π − π � 1 / 2 U ( f ) e j 2 πfm d = f − 1 / 2 22 Digital Control Kannan M. Moudgalya, Autumn 2007

FT of a Moving Average Filter - Example 22. y ( n ) = 1 3[ u ( n + 1) + u ( n ) + u ( n − 1)] ∞ � y ( n ) = g ( k ) u ( n − k ) k = −∞ = g ( − 1) u ( n + 1) + g (0) u ( n ) + g (1) u ( n − 1) g ( − 1) = g (0) = g (1) = 1 3 ∞ � e jw � g ( n ) z − n | z = e jw � G = n = −∞ = 1 = 1 e jw + 1 + e − jw � � 3(1 + 2 cos w ) 3 23 Digital Control Kannan M. Moudgalya, Autumn 2007