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= ( 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 \( z \) ={}} {} 1 1 z 1 size 12{ { {1} over {1 - z rSup { size 8{ - 1} } } } } {} or z z 1 size 12{ { {z} over {z - 1} } } {}

a n u ( n ) X ( z ) = size 12{a rSup { size 8{n} } u \( n \) rightarrow X \( z \) ={}} {} 1 1 az 1 size 12{ { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } } {} or z z a size 12{ { {z} over {z - a} } } {}

( cos n Ω 0 ) u ( n ) X ( z ) = size 12{ \( "cos"n %OMEGA rSub { size 8{0} } \) u \( n \) rightarrow X \( z \) ={}} {} 1 z 1 cos Ω 0 1 2z 1 cos Ω 0 + 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 cos Ω 0 ) z 2 2z cos Ω 0 + 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 \( n \) ={}} {} e jn ω size 12{e rSup { size 8{ ital "jn"ω} } } {} , the output is y ( n ) = size 12{y \( n \) ={}} {} e jn ω size 12{e rSup { size 8{ ital "jn"ω} } } {} H ( ω ) size 12{H \( ω \) } {} . Because of this , e jn ω 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 ) = z n 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 ( k ) z k size 12{ {}=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) , thus

{} y ( n ) = z n H ( z ) size 12{y \( n \) =z rSup { size 8{n} } H \( z \) } {} (4.11)

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

1.6 transfer function in terms of filter coefficients

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

y ( n ) = size 12{y \( n \) ={}} {} k = 1 M a k y ( n k ) size 12{ Sum cSub { size 8{k=1} } cSup { size 8{M} } {a rSub { size 8{k} } y \( n - k \) } } {} + k = N N b k x ( n k ) size 12{+ Sum cSub { size 8{k= - N} } cSup { size 8{N} } {b rSub { size 8{k} } x \( n - k \) } } {} (4.12)

where a k size 12{a rSub { size 8{k} } } {} and b k 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 ) = z n size 12{x \( n \) =z rSup { size 8{n} } } {} and y ( n ) = z n size 12{y \( n \) =z rSup { size 8{n} } } {} H ( z ) size 12{H \( z \) } {} to get

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

H ( z ) = size 12{H \( z \) ={}} {} k = N N b k z k 1 k = 1 M a k z k size 12{ { { 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} } } } } } {} (4.13)

It is worthwhile to notice that the above transfer function is resulted from the filter equation (4.12) . 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 ) size 12{H \( z \) } {} . For nonrecursive fillters the denominator is just 1 and 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 ) size 12{H \( z \) } {} without the need to take the z-transform. Vice versa, if we know H ( z ) size 12{H \( z \) } {} that means we know the filler coefficients , hence the filter equation.

Example 4.1.4

Given

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

(b) H ( z ) = size 12{H \( z \) ={}} {} 20 z 2 + 5z 10 z 3 + 5z 2 8z 1 size 12{ { {-"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 ) size 12{H \( z \) } {} as a function of z 1 size 12{z rSup { size 8{ - 1} } } {} by multiplying the numerator and denominator with z 2 size 12{z rSup { size 8{ - 2} } } {} :

H ( z ) = size 12{H \( z \) ={}} {} 2 3z 1 1 + 0 . 5z 1 0 . 8z 2 = 2 3z 1 1 ( 0 . 5z 1 + 0 . 8z 2 ) size 12{ { {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 size 12{a rSub { size 8{0} } =2} {} a 1 = 3 size 12{a rSub { size 8{1} } = - 3} {} b 1 = 0 . 5 size 12{b rSub { size 8{1} } = - 0 "." 5} {} b 2 = 0 . 8 size 12{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 ) 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 . 1z 3 size 12{0 "." 1z rSup { size 8{ - 3} } } {} to make 10 z 3 size 12{"10"z rSup { size 8{ - 3} } } {} in the denominator equal to 1 :

H ( z ) = size 12{H \( z \) ={}} {} 2z 1 + 5z 2 1 + 0 . 5z 1 0 . 8z 2 0 . 1z 3 size 12{ { { - 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 size 12{a rSub { size 8{1} } = - 2} {} a 2 = 5 size 12{a rSub { size 8{2} } =5} {} b 1 = 0 . 5 size 12{b rSub { size 8{1} } = - 0 "." 5} {} b 2 = 0 . 8 size 12{b rSub { size 8{2} } =0 "." 8} {} b 3 = 0 . 1 size 12{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 ) size 12{y \( n \) = - 0 "." 5y \( n - 1 \) +0 "." 8y \( n - 2 \) +0 "." 1y \( n - 3 \) - 2x \( n - 1 \) +5x \( n - 2 \) } {} y(n)

Questions & Answers

what are the products of Nano chemistry?
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Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
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industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
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LITNING Reply
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Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
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Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
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Bob Reply
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The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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biomolecules are e building blocks of every organics and inorganic materials.
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research.net
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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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