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1.1 definition: the z-transform x(z) of a causal discrete – time signal x(n) is defined as

X ( z ) = size 12{X \( z \) ={}} {} n = 0 x ( n ) z n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { size 11{x \( n \) }} z rSup { size 8{ - n} } } {} ( 4 . 1 ) size 12{ \( 4 "." 1 \) } {}

z is a complex variable of the transform domain and can be considered as the complex frequency. Remember index n can be time or space or some other thing, but is usually taken as time. As defined above , X ( z ) size 12{X \( z \) } {} is an integer power series of z 1 size 12{z rSup { size 8{ - 1} } } {} with corresponding x ( n ) size 12{x \( n \) } {} as coefficients. Let’s expand X ( z ) size 12{X \( z \) } {} :

X ( z ) = size 12{X \( z \) ={}} {} n = x ( n ) z n size 12{ Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \( n \) z rSup { size 8{ - n} } } } {} = size 12{ {}={}} {} x ( 0 ) + x ( 1 ) z 1 + x ( 2 ) z 2 + . . . size 12{x \( 0 \) +x \( 1 \) z rSup { size 8{ - 1} } +x \( 2 \) z rSup { size 8{ - 2} } + "." "." "." } {} (4.2)

In general one writes

X ( z ) = size 12{X \( z \) ={}} {} Z [ x ( n ) ] size 12{Z \[ x \( n \) \] } {} (4.3)

In Eq.(4.1) the summation is taken from n = 0 size 12{n=0} {} to size 12{ infinity } {} , ie , X ( z ) size 12{X \( z \) } {} is not at all related to the past history of x ( n ) size 12{x \( n \) } {} . This is one–sided or unilateral z-transform . Sometime the one–sided z-transform has to take into account the initial conditions of x ( n ) size 12{x \( n \) } {} (see section 4.7).

In general , signals exist at all time , and the two-sided or bilateral z–transform is defined as

H ( z ) = size 12{H \( z \) ={}} {} n = h ( n ) z n size 12{ Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \( n \) z rSup { size 8{ - n} } } } {}

= x ( 2 ) z 2 + x ( 1 ) z + x ( 0 ) + x ( 1 ) z 1 + x ( 2 ) z 2 + . . . size 12{ {}=x \( - 2 \) z rSup { size 8{2} } +x \( - 1 \) z+x \( 0 \) +x \( 1 \) z rSup { size 8{ - 1} } +x \( 2 \) z rSup { size 8{ - 2} } + "." "." "." } {} (4.4)

Because X ( z ) size 12{X \( z \) } {} is an infinite power series of z 1 size 12{z rSup { size 8{ - 1} } } {} , the transform only exists at values where the series converges (i.e. goes to zero as n size 12{n rightarrow infinity } {} or - size 12{ infinity } {} ). Thus the z-transform is accompanied with its region of convergence (ROC) where it is finite (see section 4.4).

A number of authors denote X + ( z ) size 12{X rSup { size 8{+{}} } \( z \) } {} for one-side z-transform.

Example 4.1.1

Find the z–transform of the two signals of Fig.4.1

Solution

(a) Notice the signal is causal and monotically decreasing and its value is just 0 . 8 n size 12{0 "." 8 rSup { size 8{n} } } {} for n 0 size 12{n>= 0} {} . So we write

x ( n ) = 0 . 8 n u ( n ) size 12{x \( n \) =0 "." 8 rSup { size 8{n} } u \( n \) } {}

and use the transform ( 4 . 1 ) size 12{ \( 4 "." 1 \) } {}

X ( z ) = size 12{X \( z \) ={}} {} n = 0 x ( n ) z n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {x \( n \) z rSup { size 8{ - n} } } } {}

= 1 + 0 . 8z 1 + 0 . 64 z 2 + 0 . 512 z 3 + . . . size 12{ {}=1+0 "." 8z rSup { size 8{ - 1} } +0 "." "64"z rSup { size 8{ - 2} } +0 "." "512"z rSup { size 8{ - 3} } + "." "." "." } {}

= 1 + ( 0 . 8z 1 ) + ( 0 . 8z 1 ) 2 + ( 0 . 8z 1 ) 3 + . . . size 12{ {}=1+ \( 0 "." 8z rSup { size 8{ - 1} } \) + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." } {}

Applying the formula of infinite geometric series which is repeated here

1 + a + a 2 + a 3 + . . . = size 12{1+a+a rSup { size 8{2} } +a rSup { size 8{3} } + "." "." "." ={}} {} n = 0 a n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {a rSup { size 8{n} } } } {} = 1 1 a size 12{ {}= { {1} over {1 - a} } } {} a < 1 size 12{ \lline a \lline<1} {} (4.5)

to obtain

X ( z ) = size 12{X \( z \) ={}} {} 1 1 0 . 8z 1 size 12{ { {1} over {1 - 0 "." 8z rSup { size 8{ - 1} } } } } {} = z z 0 . 8 size 12{ {}= { {z} over {z - 0 "." 8} } } {}

The result can be left in either of the two forms .

(b) The signal is alternatively positive and negative with increasing value .The signal is divergent . We can put the signal in the form

x ( n ) = ( 1 . 2 ) n 1 u ( n 1 ) size 12{x \( n \) = \( - 1 "." 2 \) rSup { size 8{n - 1} } u \( n - 1 \) } {}

which is ( 1 . 2 ) n u ( n ) size 12{ \( - 1 "." 2 \) rSup { size 8{n} } u \( n \) } {} delayed one index(sample) . Let’s use the transform ( 4 . 1 ) size 12{ \( 4 "." 1 \) } {}

X ( z ) = size 12{X \( z \) ={}} {} n = 0 x ( n ) z n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {x \( n \) z rSup { size 8{ - n} } } } {}

= 0 + 1 . 0 ( z 1 ) 1 . 2 ( z 1 ) 2 + 1 . 44 ( z 1 ) 3 1 . 718 ( z 1 ) 4 + . . . size 12{ {}=0+1 "." 0 \( z rSup { size 8{ - 1} } \) - 1 "." 2 \( z rSup { size 8{ - 1} } \) rSup { size 8{2} } +1 "." "44" \( z rSup { size 8{ - 1} } \) rSup { size 8{3} } - 1 "." "718" \( z rSup { size 8{ - 1} } \) rSup { size 8{4} } + "." "." "." } {}

= z 1 [ 1 + ( 1 . 2z 1 ) + ( 1 . 2z 1 ) 2 + ( 1 . 2z 1 ) 3 + . . . ] size 12{ {}=z rSup { size 8{ - 1} } \[ 1+ \( - 1 "." 2z rSup { size 8{ - 1} } \) + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." \] } {}

= z 1 1 1 + 1 . 2z 1 = z 1 1 + 1 . 2z 1 = 1 z + 1 . 2 size 12{ {}=z rSup { size 8{ - 1} } { {1} over {1+1 "." 2z rSup { size 8{ - 1} } } } = { {z rSup { size 8{ - 1} } } over {1+1 "." 2z rSup { size 8{ - 1} } } } = { {1} over {z+1 "." 2} } } {}

1.2 the inverse z-transform

The inverse z-transform is denoted by Z 1 size 12{Z rSup { size 8{ - 1} } } {} :

x ( n ) = Z 1 [ X ( z ) ] size 12{x \( n \) =Z rSup { size 8{ - 1} } \[ X \( z \) \] } {} (4.6)

The signal x ( n ) size 12{x \( n \) } {} and its transform constitutes a transform pair

X ( n ) Z ( z ) size 12{X \( n \) ↔Z \( z \) } {} (4.7)

One way to find the inverse transform , whenever possible , is to utilize just the z-transform definition. General methods of the inverse z-transform are discursed in section 4.5 and 4.6

Example 4.1.2

Find the inverse z-transform of the following

  1. X ( z ) = size 12{X \( z \) ={}} {} z z 0 . 8 size 12{ { {z} over {z - 0 "." 8} } } {}
  2. 1 z + 1 . 2 size 12{ { {1} over {z+1 "." 2} } } {}

Solution

(a) Let’s write

X ( z ) = size 12{X \( z \) ={}} {} z z- 0 . 8 size 12{ { {z} over {"z-"0 "." 8} } } {} = size 12{ {}={}} {} 1 1 0 . 8z 1 size 12{ { {1} over {1-0 "." 8z rSup { size 8{-1} } } } } {}

= 1 + ( 0 . 8z 1 ) + ( 0 . 8z 1 ) 2 + ( 0 . 8z 1 ) 3 + . . . size 12{ {}=1+ \( 0 "." 8z rSup { size 8{ - 1} } \) + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." } {}

= 1 + 0 . 8z 1 + 0 . 64 z 2 + 0 . 512 z 3 + . . . size 12{ {}=1+0 "." 8z rSup { size 8{ - 1} } +0 "." "64"z rSup { size 8{ - 2} } +0 "." "512"z rSup { size 8{ - 3} } + "." "." "." } {}

By comparing term by term with Equation ( 4 . 2 ) size 12{ \( 4 "." 2 \) } {} we get

x ( n ) = [ 1,0 . 8,0 . 64 , 0 . 512 ; . . . ] size 12{x \( n \) = \[ 1,0 "." 8,0 "." "64",0 "." "512"; "." "." "." \] } {}

or

x ( n ) = size 12{x \( n \) ={}} {} 0 . 8 n u ( n ) size 12{0 "." 8 rSup { size 8{n} } u \( n \) } {}

(b) Let’s write

X ( z ) = size 12{X \( z \) ={}} {} 1 z + 1,2 size 12{ { {1} over {z+1,2} } } {} = z 1 1 + 1,2 z 1 size 12{ {}= { {z rSup { size 8{ - 1} } } over {1+1,2z rSup { size 8{ - 1} } } } } {} = z 1 1 1 + 1,2 z 1 size 12{ {}=z rSup { size 8{ - 1} } { {1} over {1+1,2z rSup { size 8{ - 1} } } } } {}

Next , let’s expand X ( z ) size 12{X \( z \) } {} :

X ( z ) size 12{X \( z \) } {} = z 1 [ 1 + ( 1 . 2z 1 ) + ( 1 . 2z 1 ) 2 + ( 1 . 2z 1 ) 3 + . . . ] size 12{ {}=z rSup { size 8{ - 1} } \[ 1+ \( - 1 "." 2z rSup { size 8{ - 1} } \) + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." \] } {}

= 0 + 1 . 0z 1 1 . 2z 2 + 1 . 44 z 3 1 . 728 z 4 + . . . size 12{ {}=0+1 "." 0z rSup { size 8{ - 1} } - 1 "." 2z rSup { size 8{ - 2} } +1 "." "44"z rSup { size 8{ - 3} } - 1 "." "728"z rSup { size 8{ - 4} } + "." "." "." } {}

Thus

x ( n ) = [ 0,1 . 0, 1 . 2,1 . 44 , 1 . 728 , . . . ] size 12{x \( n \) = \[ 0,1 "." 0, - 1 "." 2,1 "." "44", - 1 "." "728", "." "." "." \] } {}

or

x ( n ) = ( 1 . 2 ) n 1 u ( n 1 ) size 12{x \( n \) = \( - 1 "." 2 \) rSup { size 8{n - 1} } u \( n - 1 \) } {}

That is

x ( n ) = 0 size 12{x \( n \) =0} {} n 0 size 12{n<= 0} {}

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Source:  OpenStax, Z-transform. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10798/1.1
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