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z – TRANSFORM

Module by: Nguyen Huu Phuong

1.1 Definition: The z-transform X(z) of a causal discrete – time signal x(n) is defined as

X ( z ) = X ( z ) = size 12{X \( z \) ={}} {} ∑ n = 0 ∞ x ( n ) z − n ∑ 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 ) ( 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)X(z) size 12{X \( z \) } {} is an integer power series of z−1z−1 size 12{z rSup { size 8{ - 1} } } {} with corresponding x(n)x(n) size 12{x \( n \) } {} as coefficients. Let’s expand X(z)X(z) size 12{X \( z \) } {}:
X(z)=X(z)= size 12{X \( z \) ={}} {}∑n=−∞∞x(n)z−n∑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+...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)=X(z)= size 12{X \( z \) ={}} {}Z[x(n)]Z[x(n)] size 12{Z \[ x \( n \) \] } {}(4.3)
In Eq.(4.1) the summation is taken from n=0n=0 size 12{n=0} {} to ∞∞ size 12{ infinity } {}, ie , X(z)X(z) size 12{X \( z \) } {} is not at all related to the past history of x(n)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)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 ) = H ( z ) = size 12{H \( z \) ={}} {} ∑ n = − ∞ ∞ h ( n ) z − n ∑ 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)z2+x(−1)z+x(0)+x(1)z−1+x(2)z−2+...=x(−2)z2+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)X(z) size 12{X \( z \) } {}is an infinite power series of z−1z−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→∞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)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 0.8n0.8n size 12{0 "." 8 rSup { size 8{n} } } {} for n≥0n≥0 size 12{n >= 0} {} . So we write
x ( n ) = 0 . 8 n u ( n ) 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)(4.1) size 12{ \( 4 "." 1 \) } {}
X ( z ) = X ( z ) = size 12{X \( z \) ={}} {} ∑ n = 0 ∞ x ( n ) z − n ∑ 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 + . . . = 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 + . . . = 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+a2+a3+...=1+a+a2+a3+...= size 12{1+a+a rSup { size 8{2} } +a rSup { size 8{3} } + "." "." "." ={}} {}∑n=0∞an∑n=0∞an size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {a rSup { size 8{n} } } } {}=11−a=11−a size 12{ {}= { {1} over {1 - a} } } {}∣a∣<1∣a∣<1 size 12{ \lline a \lline <1} {} (4.5)
to obtain
X ( z ) = X ( z ) = size 12{X \( z \) ={}} {} 1 1 − 0 . 8z − 1 1 1 − 0 . 8z − 1 size 12{ { {1} over {1 - 0 "." 8z rSup { size 8{ - 1} } } } } {} = z z − 0 . 8 = 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 ) 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)nu(n)(−1.2)nu(n) size 12{ \( - 1 "." 2 \) rSup { size 8{n} } u \( n \) } {} delayed one index(sample) . Let’s use the transform (4.1)(4.1) size 12{ \( 4 "." 1 \) } {}
X ( z ) = X ( z ) = size 12{X \( z \) ={}} {} ∑ n = 0 ∞ x ( n ) z − n ∑ 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 + . . . = 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 + . . . ] = 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 = 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−1Z−1 size 12{Z rSup { size 8{ - 1} } } {}:
x(n)=Z−1[X(z)]x(n)=Z−1[X(z)] size 12{x \( n \) =Z rSup { size 8{ - 1} } \[ X \( z \) \] } {}(4.6)
The signal x(n)x(n) size 12{x \( n \) } {} and its transform constitutes a transform pair
X(n)↔Z(z)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 ) = X ( z ) = size 12{X \( z \) ={}} {} z z − 0 . 8 z z − 0 . 8 size 12{ { {z} over {z - 0 "." 8} } } {}
  2. 1 z + 1 . 2 1 z + 1 . 2 size 12{ { {1} over {z+1 "." 2} } } {}

Solution

(a) Let’s write
X ( z ) = X ( z ) = size 12{X \( z \) ={}} {}