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# z – TRANSFORM

Module by: Nguyen Huu Phuong. E-mail the author

Summary: This section is an introduction to the z-transform. It comprises of some basic but very useful contents. The userfulness lies in the fact that the z-transform applied for both discrete-time signals and systems, and it has many properties.

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

X(z)=n=0x(n)znX(z)=n=0x(n)zn size 12{X $$z$$ = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { size 11{x $$n$$ }} z rSup { size 8{ - n} } } {}
(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 z1z1 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)=n=x(n)zn=x(0)+x(1)z1+x(2)z2+...X(z)=n=x(n)zn=x(0)+x(1)z1+x(2)z2+... size 12{X $$z$$ = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x $$n$$ z rSup { size 8{ - n} } =x $$0$$ } +x $$1$$ z rSup { size 8{ - 1} } +x $$2$$ z rSup { size 8{ - 2} } + "." "." "." } {}
(2)

In general we write

X(z)=Z[x(n)]X(z)=Z[x(n)] size 12{X $$z$$ =Z $x $$n$$$ } {}
(3)

In Equation 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)=n=h(n)zn=...x(2)z2+x(1)z+x(0)+x(1)z1+x(2)z2+...H(z)=n=h(n)zn=...x(2)z2+x(1)z+x(0)+x(1)z1+x(2)z2+...alignl { stack { size 12{H $$z$$ = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h $$n$$ z rSup { size 8{ - n} } } } {} # matrix { {} # {} # {} } = "." "." "." 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)

Because X(z)X(z) size 12{X $$z$$ } {} is an infinite power series of z1z1 size 12{z rSup { size 8{ - 1} } } {} , the transform only exists at values where the series converges (i.e. goes to zero as nn 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 1

Find the z–transform of the two signals of Figure 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 n0n0 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)

X ( z ) = n = 0 x ( n ) z n = 1 + 0 . 8z 1 + 0 . 64 z 2 + 0 . 512 z 3 + . . . = 1 + ( 0 . 8z 1 ) + ( 0 . 8z 1 ) 2 + ( 0 . 8z 1 ) 3 + . . . X ( z ) = n = 0 x ( n ) z n = 1 + 0 . 8z 1 + 0 . 64 z 2 + 0 . 512 z 3 + . . . = 1 + ( 0 . 8z 1 ) + ( 0 . 8z 1 ) 2 + ( 0 . 8z 1 ) 3 + . . . alignl { stack { size 12{X $$z$$ = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {x $$n$$ z rSup { size 8{ - n} } } } {} # matrix { {} # {} # {} } =1+0 "." 8z rSup { size 8{ - 1} } +0 "." "64"z rSup { size 8{ - 2} } +0 "." "512"z rSup { size 8{ - 3} } + "." "." "." {} # matrix { {} # {} # {} } =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+...=n=0an=11aa<11+a+a2+a3+...=n=0an=11aa<1 size 12{1+a+a rSup { size 8{2} } +a rSup { size 8{3} } + "." "." "." = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {a rSup { size 8{n} } } = { {1} over {1 - a} } matrix { {} # {} # {} } \lline a \lline <1} {}
(5)

to obtain

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

X ( z ) = n = 0 x ( n ) z n = 0 + 1 . 0 ( z 1 ) 1 . 2 ( z 1 ) 2 + 1 . 44 ( z 1 ) 3 1 . 718 ( z 1 ) 4 + . . . = z 1 [ 1 + ( 1 . 2z 1 ) + ( 1 . 2z 1 ) 2 + ( 1 . 2z 1 ) 3 + . . . ] = z 1 1 1 + 1 . 2z 1 = z 1 1 + 1 . 2z 1 = 1 z + 1 . 2 X ( z ) = n = 0 x ( n ) z n = 0 + 1 . 0 ( z 1 ) 1 . 2 ( z 1 ) 2 + 1 . 44 ( z 1 ) 3 1 . 718 ( z 1 ) 4 + . . . = z 1 [ 1 + ( 1 . 2z 1 ) + ( 1 . 2z 1 ) 2 + ( 1 . 2z 1 ) 3 + . . . ] = z 1 1 1 + 1 . 2z 1 = z 1 1 + 1 . 2z 1 = 1 z + 1 . 2 alignl { stack { size 12{X $$z$$ = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {x $$n$$ z rSup { size 8{ - n} } } } {} # matrix { {} # {} # {} } =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} } + "." "." "." {} # matrix { {} # {} # {} } =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} } + "." "." "."$ {} # matrix { {} # {} # {} } =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} } {} } } {}

## The inverse z-transform

The inverse z-transform is denoted by Z1Z1 size 12{Z rSup { size 8{ - 1} } } {}:

x(n)=Z1[X(z)]x(n)=Z1[X(z)] size 12{x $$n$$ =Z rSup { size 8{ - 1} } $X $$z$$$ } {}
(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 $$n$$ widevec {} Z $$z$$ } {}
(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 2

Find the inverse z-transform of the following

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

Solution

(a) Let’s write

X ( z ) = z z- 0 . 8 = 1 1 0 . 8z 1 = 1 + ( 0 . 8z 1 ) + ( 0 . 8z 1 ) 2 + ( 0 . 8z 1 ) 3 + . . . = 1 + 0 . 8z 1 + 0 . 64 z 2 + 0 . 512 z 3 + . . . X ( z ) = z z- 0 . 8 = 1 1 0 . 8z 1 = 1 + ( 0 . 8z 1 ) + ( 0 . 8z 1 ) 2 + ( 0 . 8z 1 ) 3 + . . . = 1 + 0 . 8z 1 + 0 . 64 z 2 + 0 . 512 z 3 + . . . alignl { stack { size 12{X $$z$$ = { {z} over {"z-"0 "." 8} } = { {1} over {1-0 "." 8z rSup { size 8{-1} } } } } {} # matrix { {} # {} # {} } =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} } + "." "." "." {} # matrix { {} # {} # {} } =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 2 we get

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

or

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 $$n$$ } {}

(b) Let’s write

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

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

X ( z ) = z 1 [ 1 + ( 1 . 2z 1 ) + ( 1 . 2z 1 ) 2 + ( 1 . 2z 1 ) 3 + . . . ] = 0 + 1 . 0z 1 1 . 2z 2 + 1 . 44 z 3 1 . 728 z 4 + . . . X ( z ) = z 1 [ 1 + ( 1 . 2z 1 ) + ( 1 . 2z 1 ) 2 + ( 1 . 2z 1 ) 3 + . . . ] = 0 + 1 . 0z 1 1 . 2z 2 + 1 . 44 z 3 1 . 728 z 4 + . . . alignl { stack { size 12{X $$z$$ =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} } + "." "." "."$ } {} # matrix { {} # {} # {} } =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 , . . . ] 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 ) 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 n 0 = ( 1 . 2 ) n n > 0 x ( n ) = 0 n 0 = ( 1 . 2 ) n n > 0 alignl { stack { size 12{x $$n$$ =0 matrix { {} # {} # {} } n <= 0} {} # size 12{ matrix { {} # {} # {} } = $$- 1 "." 2$$ rSup { size 8{n} } matrix { {} # {} } n>0} {} } } {}

## z–transform pairs

Figure 2 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 expressed equivalently as a function of z1z1 size 12{z rSup { size 8{ - 1} } } {} or z , for example

u ( n ) X ( z ) = 1 1 z 1 or z z 1 a n u ( n ) X ( z ) = 1 1 az 1 or z z a ( cos n Ω 0 ) u ( n ) X ( z ) = 1 z 1 cos Ω 0 1 2z 1 cos Ω 0 + z 2 or z ( z cos Ω 0 ) z 2 2z cos Ω 0 + 1 u ( n ) X ( z ) = 1 1 z 1 or z z 1 a n u ( n ) X ( z ) = 1 1 az 1 or z z a ( cos n Ω 0 ) u ( n ) X ( z ) = 1 z 1 cos Ω 0 1 2z 1 cos Ω 0 + z 2 or z ( z cos Ω 0 ) z 2 2z cos Ω 0 + 1 alignl { stack { size 12{u $$n$$ rightarrow X $$z$$ = { {1} over {1 - z rSup { size 8{ - 1} } } } matrix { {} # {} } ital "or" matrix { {} # {} } { {z} over {z - 1} } } {} # a rSup { size 8{n} } u $$n$$ rightarrow X $$z$$ = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix { {} # {} } ital "or" matrix { {} # {} } { {z} over {z - a} } {} # $$"cos"n %OMEGA rSub { size 8{0} }$$ u $$n$$ rightarrow X $$z$$ = { {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} } } } matrix { {} # {} } ital "or" matrix { {} # {} } { {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).

## 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)h(n) size 12{h $$n$$ } {} is

H(z)=n=0h(n)zn(onesidetransform)H(z)=n=0h(n)zn(onesidetransform) size 12{H $$z$$ = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {h $$n$$ z rSup { size 8{ - n} } } matrix { {} # {} # {} } $$ital "one" - ital "side" matrix { {} } ital "transform"$$ } {}
(8)

or

H(z)=n=h(n)zn(twosidetransform)H(z)=n=h(n)zn(twosidetransform) size 12{H $$z$$ = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h $$n$$ z rSup { size 8{ - n} } } matrix { {} # {} # {} } $$ital "two" - ital "side" matrix { {} } ital "transform"$$ } {}
(9)

depending on whether the system is causal or noncausal .

H(z)H(z) size 12{H $$z$$ } {}, the z–transform of h(n)h(n) size 12{h $$n$$ } {} , is called transfer function or system function of the system .

### Example 3

A system has impulse respone h ( n ) = [ 1,2,3,4,5,6 ] h ( n ) = [ 1,2,3,4,5,6 ] size 12{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 ) = n = h ( n ) z n = n = 2 3 h ( n ) z 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 = z 2 + 2z 1 + 3 + 4z 1 + 5z 2 + 6z 3 H ( z ) = n = h ( n ) z n = n = 2 3 h ( n ) z 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 = z 2 + 2z 1 + 3 + 4z 1 + 5z 2 + 6z 3 alignl { stack { size 12{H $$z$$ = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h $$n$$ z rSup { size 8{ - n} } } = Sum cSub { size 8{n= - 2} } cSup { size 8{3} } {h $$n$$ z rSup { size 8{ - n} } } } {} # matrix { {} # {} # {} } =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} } {} # matrix { {} # {} # {} } =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)H(z) size 12{H $$z$$ } {}is known as above we can easily obtain h(n)h(n) size 12{h $$n$$ } {} .

## Eigen-function and eigen-value

We know that if the frequency response of a system is H(ω)H(ω) size 12{H $$ω$$ } {}then for input x(n)=ejnωx(n)=ejnω size 12{x $$n$$ =e rSup { size 8{ ital "jn"ω} } } {} , the output is y(n)=ejnωH(ω)y(n)=ejnωH(ω) size 12{y $$n$$ =e rSup { size 8{ ital "jn"ω} } H $$ω$$ } {}. Because of this , ejnωejnω size 12{e rSup { size 8{ ital "jn"ω} } } {} is the eigen-function , and H(ω)H(ω) size 12{H $$ω$$ } {}the eigen-value of the system.

Now , for input

x(n)=znx(n)=zn size 12{x $$n$$ =z rSup { size 8{n} } } {}
(10)

the system output is

y ( n ) = h ( n ) x ( n ) = k = 0 h ( k ) z n k = z n k = 0 h ( k ) z k y ( n ) = h ( n ) x ( n ) = k = 0 h ( k ) z n k = z n k = 0 h ( k ) z k size 12{y $$n$$ =h $$n$$ *x $$n$$ = Sum cSub { size 8{k=0} } cSup { size 8{ infinity } } {h $$k$$ z rSup { size 8{n - k} } } =z rSup { size 8{n} } left [ Sum cSub { size 8{k=0} } cSup { size 8{ infinity } } {h $$k$$ z rSup { size 8{ - k} } } right ]} {}

In the brackets is just H(z)H(z) size 12{H $$z$$ } {} , thus

y(n)=znH(z)y(n)=znH(z) size 12{y $$n$$ =z rSup { size 8{n} } H $$z$$ } {}
(11)

Hence in the z–transform domain , znzn size 12{z rSup { size 8{n} } } {} is the eigen-function and H(z)H(z) size 12{H $$z$$ } {} is the eigen-value of the system.

## Transfer function in terms of filter coefficients

For , Let’s begin with the general filter difference equation which is repeated here

y(n)=k=1Maky(nk)+k=NNbkx(nk)y(n)=k=1Maky(nk)+k=NNbkx(nk) size 12{y $$n$$ = Sum cSub { size 8{k=1} } cSup { size 8{M} } {a rSub { size 8{k} } y $$n - k$$ } + Sum cSub { size 8{k= - N} } cSup { size 8{N} } {b rSub { size 8{k} } x $$n - k$$ } } {}
(12)

where akak size 12{a rSub { size 8{k} } } {} and bkbk size 12{b rSub { size 8{k} } } {} are the filter coefficients (constants) , and the limits M, N can be extended to infinity . Now we make the replacement x(n)=znx(n)=zn size 12{x $$n$$ =z rSup { size 8{n} } } {} and y(n)=znH(z)y(n)=znH(z) size 12{y $$n$$ =z rSup { size 8{n} } H $$z$$ } {} to get

z n H ( z ) = k = 1 M a k z n k H ( z ) + k = N N b k z n k z n H ( z ) = k = 1 M a k z n k H ( z ) + k = N N b k z n k size 12{z rSup { size 8{n} } H $$z$$ = Sum cSub { size 8{k=1} } cSup { size 8{M} } {a rSub { size 8{k} } z rSup { size 8{n - "k "} } H $$z$$ } + Sum cSub { size 8{k= - N} } cSup { size 8{N} } {b rSub { size 8{k} } z rSup { size 8{n - k} } } } {}

From this we derive the expression of H(z)H(z) size 12{H $$z$$ } {}

H(z)=k=NNbkzk1k=1MakzkH(z)=k=NNbkzk1k=1Makzk size 12{H $$z$$ = { { Sum cSub { size 8{k= - N} } cSup { size 8{N} } {b rSub { size 8{k} } z rSup { size 8{ - k} } } } over {1 - Sum cSub { size 8{k=1} } cSup { size 8{M} } {a rSub { size 8{k} } z rSup { size 8{ - k} } } } } } {}
(13)

It is worthwhile to notice that the above transfer function is resulted from the filter Equation 12. Other authors write the filter equation differently (for example, all the y terms are on the left side of the equation), leading to a slightly different expression of H(z)H(z) size 12{H $$z$$ } {}. For nonrecursive fillters the denominator is just 1 and H(z)H(z) size 12{H $$z$$ } {} becomes that of nonrecursive filters as we know .

The idea here is that when the filter equation is given, we collect its coefficients to place into the expression above of H(z)H(z) size 12{H $$z$$ } {} without the need to take the z-transform. Vice versa, if we know H(z)H(z) size 12{H $$z$$ } {}that means we know the filler coefficients , hence the filter equation.

### Example 4

Given (a) H(z)=2z23zz2+0.5z0.8H(z)=2z23zz2+0.5z0.8 size 12{H $$z$$ = { {2z rSup { size 8{2} } - 3z} over {z rSup { size 8{2 } } +0 "." 5z - 0 "." 8} } } {} (b) H(z)=20z2+5z10z3+5z28z1H(z)=20z2+5z10z3+5z28z1 size 12{H $$z$$ = { {-"20"z rSup { size 8{2} } +5z} over {"10"z rSup { size 8{3} } + 5z rSup { size 8{2} } -8z - 1} } } {} Find the filter difference equation.

Solution

(a) Write H(z)H(z) size 12{H $$z$$ } {} as a function of z1z1 size 12{z rSup { size 8{ - 1} } } {} by multiplying the numerator and denominator with z2z2 size 12{z rSup { size 8{ - 2} } } {} :

H ( z ) = 2 3z 1 1 + 0 . 5z 1 0 . 8z 2 = 2 3z 1 1 ( 0 . 5z 1 + 0 . 8z 2 ) H ( z ) = 2 3z 1 1 + 0 . 5z 1 0 . 8z 2 = 2 3z 1 1 ( 0 . 5z 1 + 0 . 8z 2 ) size 12{H $$z$$ = { {2-3z rSup { size 8{ - 1} } } over {1+0 "." 5z rSup { size 8{ - 1} } - 0 "." 8z rSup { size 8{ - 2} } } } = { {2 - 3z rSup { size 8{ - 1} } } over {1 - $$- 0 "." 5z rSup { size 8{ - 1} } +0 "." 8z rSup { size 8{ - 2} }$$ } } } {}

The coefficients are

a 0 = 2 a 1 = 3 b 1 = 0 . 5 b 2 = 0 . 8 a 0 = 2 a 1 = 3 b 1 = 0 . 5 b 2 = 0 . 8 size 12{a rSub { size 8{0} } =2 matrix { {} # {} # {} } a rSub { size 8{1} } = - 3 matrix { {} # {} # {} } b rSub { size 8{1} } = - 0 "." 5 matrix { {} # {} # {} } b rSub { size 8{2} } =0 "." 8} {}

Thus the filler equation is

y ( n ) = 0 . 5y ( n 1 ) 0 . 8y ( n 2 ) + 2x ( n ) 3x ( n 1 ) y ( n ) = 0 . 5y ( n 1 ) 0 . 8y ( n 2 ) + 2x ( n ) 3x ( n 1 ) size 12{y $$n$$ = - 0 "." 5y $$n - 1$$ - 0 "." 8y $$n - 2$$ +2x $$n$$ - 3x $$n - 1$$ } {}

(b) Multiply the numerator and denominator with 0.1z30.1z3 size 12{0 "." 1z rSup { size 8{ - 3} } } {} to make 10z310z3 size 12{"10"z rSup { size 8{ - 3} } } {} in the denominator equal to 1 :

H ( z ) = 2z 1 + 5z 2 1 + 0 . 5z 1 0 . 8z 2 0 . 1z 3 H ( z ) = 2z 1 + 5z 2 1 + 0 . 5z 1 0 . 8z 2 0 . 1z 3 size 12{H $$z$$ = { { - 2z rSup { size 8{ - 1} } +5z rSup { size 8{ - 2} } } over {1+0 "." 5z rSup { size 8{ - 1} } - 0 "." 8z rSup { size 8{ - 2} } - 0 "." 1z rSup { size 8{ - 3} } } } } {}

Collect the coefficients:

a 1 = 2 a 2 = 5 b 1 = 0 . 5 b 2 = 0 . 8 b 3 = 0 . 1 a 1 = 2 a 2 = 5 b 1 = 0 . 5 b 2 = 0 . 8 b 3 = 0 . 1 size 12{a rSub { size 8{1} } = - 2 matrix { {} # {} # {} } a rSub { size 8{2} } =5 matrix { {} # {} # {} } b rSub { size 8{1} } = - 0 "." 5 matrix { {} # {} # {} } b rSub { size 8{2} } =0 "." 8 matrix { {} # {} # {} } b rSub { size 8{3} } =0 "." 1} {}

Thus

y ( n ) = 0 . 5y ( n 1 ) + 0 . 8y ( n 2 ) + 0 . 1y ( n 3 ) 2x ( n 1 ) + 5x ( n 2 ) y ( n ) = 0 . 5y ( n 1 ) + 0 . 8y ( n 2 ) + 0 . 1y ( n 3 ) 2x ( n 1 ) + 5x ( n 2 ) size 12{y $$n$$ = - 0 "." 5y $$n - 1$$ +0 "." 8y $$n - 2$$ +0 "." 1y $$n - 3$$ - 2x $$n - 1$$ +5x $$n - 2$$ } {}

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