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)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) 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 0.8n 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+a2+a3+...= size 12{1+a+a rSup { size 8{2} } +a rSup { size 8{3} } + "." "." "." ={}} {}∑n=0∞an size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {a rSup { size 8{n} } } } {}=11−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)nu(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 X ( z ) = size 12{X \( z \) ={}} {} z z − 0 . 8 size 12{ { {z} over {z - 0 "." 8} } } {} 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} {} = ( − 1 . 2 ) n size 12{ {}= \( - 1 "." 2 \) rSup { size 8{n} } } {} n > 0 size 12{n>0} {}

1.3 z–transform pairs Table 4.1 gives many useful z-transform pairs . All signals are causal (right-sided), except two which are anticausal (left-sided). Notice that a transform can be expresed equivalently as a function of z−1 size 12{z rSup { size 8{ - 1} } } {} or z , for example Table 4.1 : Common z-transform pairs u(n)→X(z)= size 12{u \( n \) rightarrow X \( z \) ={}} {}11−z−1 size 12{ { {1} over {1 - z rSup { size 8{ - 1} } } } } {} or zz−1 size 12{ { {z} over {z - 1} } } {} anu(n)→X(z)= size 12{a rSup { size 8{n} } u \( n \) rightarrow X \( z \) ={}} {}11−az−1 size 12{ { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } } {} or zz−a size 12{ { {z} over {z - a} } } {} (cosnΩ0)u(n)→X(z)= size 12{ \( "cos"n %OMEGA rSub { size 8{0} } \) u \( n \) rightarrow X \( z \) ={}} {}1−z−1cosΩ01−2z−1cosΩ0+z−2 size 12{ { {1 - z rSup { size 8{ - 1} } "cos" %OMEGA rSub { size 8{0} } } over {1 - 2z rSup { size 8{ - 1} } "cos" %OMEGA rSub { size 8{0} } +z rSup { size 8{ - 2} } } } } {} or z(z−cosΩ0)z2−2zcosΩ0+1 size 12{ { {z \( z - "cos" %OMEGA rSub { size 8{0} } \) } over {z rSup { size 8{2} } - 2z"cos" %OMEGA rSub { size 8{0} } +1} } } {} Which form is more appropriate depending on what we would like to do with the transform. (see sections 4.1.6 , 4.3 and 4.6).

1.4 z–transform for systems The z-transform applies to systems as well as signals because systems are represented by their impulse responses which are functions of index n (time or space …) just like signals. Remember that many other transforms (Laplace, Fourier …) have the same property, and due to this property that the transforms are useful in the analysis and design of systems because signals and systems interact. Specifically , the z–transform of impulse response h(n) is H(z)= size 12{H \( z \) ={}} {}∑n=0∞h(n)z−n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {h \( n \) z rSup { size 8{ - n} } } } {} (one–sided transform ) (4.8) or 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} } } } {} (two–sided transform )(4.9) depending on whether the system is causal or noncausal . H(z) size 12{H \( z \) } {}, the z–transform of h(n) size 12{h \( n \) } {}, is called transfer function or system function of the system . Example 4.1.3 A system has impulse respone h(n) = [1 , 2 , 3 , 4 , 5 , 6] Find the transfer function . Solution The system is of noncausal FIR type . Its transfer function is 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} } } } {} = ∑ n = − 2 3 h ( n ) z − n size 12{ {}= Sum cSub { size 8{n= - 2} } cSup { size 8{3} } {h \( n \) z rSup { size 8{ - n} } } } {} = h ( − 2 ) z 2 + h ( − 1 ) z − 1 + h ( 0 ) z 0 + h ( 1 ) z − 1 + h ( 2 ) z − 2 + h ( 3 ) z − 3 size 12{ {}=h \( - 2 \) z rSup { size 8{2} } +h \( - 1 \) z rSup { size 8{ - 1} } +h \( 0 \) z rSup { size 8{0} } +h \( 1 \) z rSup { size 8{ - 1} } +h \( 2 \) z rSup { size 8{ - 2} } +h \( 3 \) z rSup { size 8{ - 3} } } {} = z − 2 + 2z − 1 + 3 + 4z − 1 + 5z − 2 + 6z − 3 size 12{ {}=z rSup { size 8{ - 2} } +2z rSup { size 8{ - 1} } +3+4z rSup { size 8{ - 1} } +5z rSup { size 8{ - 2} } +6z rSup { size 8{ - 3} } } {} On the contrary, if H(z) size 12{H \( z \) } {} is known as above we can easily obtain h(n) size 12{h \( n \) } {} .

1.5 Eigen-function and eigen-value We know that if the frequency response of a system is H(ω) size 12{H \( ω \) } {} then for input x(n)= size 12{x \( n \) ={}} {}ejnω size 12{e rSup { size 8{ ital "jn"ω} } } {} , the output is y(n)= size 12{y \( n \) ={}} {}ejnω size 12{e rSup { size 8{ ital "jn"ω} } } {}H(ω) size 12{H \( ω \) } {}. Because of this , ejnω size 12{e rSup { size 8{ ital "jn"ω} } } {} is the eigen-function , and H(ω) size 12{H \( ω \) } {} the eigen-value of the system. Now , for input x(n)=zn size 12{x \( n \) =z rSup { size 8{n} } } {} (4.10) the system output is y ( n ) = h ( n ) ∗ x ( n ) = size 12{y \( n \) =h \( n \) *x \( n \) ={}} {} ∑ k = 0 ∞ h ( k ) z n − k size 12{ Sum cSub { size 8{k=0} } cSup { size 8{ infinity } } {h \( k \) z rSup { size 8{n - k} } } } {} = z n ∑ k = 0 ∞ h (