Z-Transform - ver 1

# Z-Transform - ver 1 - Z-Transform Linear Systems and...

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Z-Transform Linear Systems and Signals – Lectures 21-24 Dr. J. K. Aggarwal The University of Texas at Austin

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Z-Transform Analysis ± The Z-transform of a discrete-time signal x [ n ] is where z is a complex variable ± The inverse z-transform is defined by but we shall use tables ± 0 [] n n Xz x n z = = 1 1 2 n xn X zz d z j π = v { } {} 1 [ [[ ] ] ] X z Zx n xn Z or = =
Linearity ± Like Laplace Transform, Z-transform is a linear operator ² We shall not worry about the existence of z-transform 11 22 [] xn X z X z If then 2 2 1 1 ax n ax n aX z + ⇔+

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An Example ± Let us compute Z-transform of some simple discrete-time signals: ² Step function This is the geometric series { 1, 0 [] 0, 0 n un n = < { } 123 [] 1 Zu n z z z −−− = ++++ 23 1 () 1 () n n nn n Gx x x x x x x x x x x + =+ + + + + + + + + 1 1 (1 ) ( ) 1 1 1 n n n n xG x x x x + + −= ⇒= Subtracting,
An Example ...contd 2 1, , 1 () 1 1 .. 1 , 1 1 n If x then as n Gx x ie x x if x x <→ = + ++= < {} 12 1 1 1 [] , 1 ,1 1 z z o z n z S Zu −− +++ = < ⎩− Notice the strict inequality

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Another Example ± {} 0 0 2 2 [] 1 1 ,1 1 , n n nn n n z Z run z r z xn run Zx n r u n z rz rr zz r z r r z = = => = = = =+ + + = <
More examples ± Z-transform of ± Z-transform of Recall [] n δ { } 1 Zn = cos [ ] nun β cos 2 jn ee n + = {} cos [ ] [ ] 2 22 u n Z u n Z un Z ββ ⎧⎫ + = ⎨⎬ ⎩⎭ ⎫⎧ =+ ⎬⎨

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More examples ...contd {} () ( ) 1 2 2 1 11 1 1 cos [ ] 21 ,1 22 cos sin cos cos cos [ sin 1 2 2c o s 1 2 ] 2o o c s s jj z Zn u n ze zz z zj z z z z z z z nu z ββ β −− = =+ > + −+ = ⎩⎭ = 1 1
Finding the Inverse Z-Transform ± As in the case of Laplace, we shall avoid the complex contour integral – instead, we shall develop a table based upon what we have already learned ² ² ² {} 1 1 [] , 1 11 z Zu n z zz = => −− , n z Z run z r zr = > [] 1 Zn δ =

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Finding the Inverse Z-Transform …contd ± ± {} ( ) 2 cos cos [ ] 2c o s zz Zn u n β = +1 2 sin sin [ ] o s z u n = Prove using the same method as cos [ ] nu n
Example 1 ± Consider a simple function We expand into partial fractions () ( ) 0.5 [] 10 . 5 z Xz zz = −− X z z ( ) 0 . 5 . 5 11 . 5 . 5 [] ( 0 . 5 ) [] , 0 n z xn un un n = =−

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Example 2 ± Exercise 5.1-5b, Lathi 2 4 [] 56 z Xz zz = + () 4 32 21 1 33 23 3 3 2 3 nn z z z z nu n xn δ = −− =+− =− + ⎛⎞ +− ⎜⎟ ⎝⎠ =− Using partial fraction expansion
Example 3: Complex Poles ± Lathi, page 503 ( ) () ( ) ( ) 2 2.246 2.246 2.246 2.246 23 17 [] 16 2 5 23 17 13 4 3 4 21 . 6 1 . 6 4 3 4 [] 2 1 . 6 1 . 6 4 3 4 jj z Xz z zz z z j z j ee zzj zj z e e z + = −− + + = + =+ + + + + ( ) 2 1 7 2 5 z + = +

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Example 3: Complex Poles . ..contd 0.927 ,3 45 j Now j e += () ( ) ( ) ( ) 2.246 0.927 2.246 0.927 2.246 2.246 0.927 0.927 [] 1 . 6 5 . 6 5 [] 1.6 5 cos 0.927 2.246 sin 0.927 2.246 1.6 5 cos 0.927 2.246 sin 0.927 2.246 [] 2 1 . 6 1 . 6 15 5 2[] nn jj j j n n un
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Z-Transform - ver 1 - Z-Transform Linear Systems and...

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