Let z1,z2,z3...z1,z2,z3... size 12{z rSub { size 8{1} } ,z rSub { size 8{2} } ,z rSub { size 8{3} } "." "." "." } {} are roots of N(z)N(z) size 12{N \( z \) } {}, and p1,p2,p3...p1,p2,p3... size 12{p rSub { size 8{1} } ,p rSub { size 8{2} } ,p rSub { size 8{3} } "." "." "." } {} those of D(z)D(z) size 12{D \( z \) } {}, then the z-transform can be put in the form

where G is a gain factor ; z1,z2,z3...z1,z2,z3... size 12{z rSub { size 8{1} } ,z rSub { size 8{2} } ,z rSub { size 8{3} } "." "." "." } {} are the zeros which make X(z)X(z) size 12{X \( z \) } {} go to zero; and p1,p2,p3...p1,p2,p3... size 12{p rSub { size 8{1} } ,p rSub { size 8{2} } ,p rSub { size 8{3} } "." "." "." } {}the poles which make X(z)X(z) size 12{X \( z \) } {}go to infinity , L is the order of mumerator , M of denominator . X(z)X(z) size 12{X \( z \) } {} is only a proper rational polynimomial when L≤ML≤M size 12{L <= M} {} (the order of numerator is less than the order of denominator). Above are finite zeros and poles . Besides ,when variable z in the demoninator goes to infinity X(z)X(z) size 12{X \( z \) } {} may go to zero , this is an infinite zero . Similarly , when z in the numerator goes to infinity X(z)X(z) size 12{X \( z \) } {} may go to infinity, this is an infinite pole . When the order M of numerator is smaller than the order N of the demominator there will be an infinite zero of order M−NM−N size 12{M - N} {}, when M−NM−N size 12{M - N} {} there will be an infinity pole of order M−NM−N size 12{M - N} {} . For most of the cases we just ignore the infinite poles and zeros .

Distribution of poles and zeros of X(z)X(z) size 12{X \( z \) } {} or H(z)H(z) size 12{H \( z \) } {} in the z-plane is called pole – zero plot . Figure 1 shows the pole – zero plot for sereval simple signals. Notice the unit sample δnδn size 12{δ rSup { size 8{n} } } {} is very special in that it is the only function which doesn’t have any pole and zero .

The pole – zero plot is very useful in the analysis and design of digital filters , or systems in general . Figure 2 shows the relation of pole – zero plot and characteristic of signals. Actually only the pole location affects the nature of signals .

It’s worth remembering that when the signal x(n)x(n) size 12{x \( n \) } {} or impulse response h(n)h(n) size 12{h \( n \) } {} are real – valued , their poles and zeros are either real or occur in complex conjugate pairs .

Multiplying both numerator and denominator by z4z4 size 12{z rSup { size 8{4} } } {}, we get

H ( z ) = z 2 + z z 4 − 3,6 z 3 + 4, 59 z 2 − 2, 38 z + 0, 39 = N ( z ) D ( z ) H ( z ) = z 2 + z z 4 − 3,6 z 3 + 4, 59 z 2 − 2, 38 z + 0, 39 = N ( z ) D ( z ) size 12{H \( z \) = { {z rSup { size 8{2} } +z} over {z rSup { size 8{4} } - 3,6z rSup { size 8{3} } +4,"59"z rSup { size 8{2} } - 2,"38"z+0,"39"} } = { {N \( z \) } over {D \( z \) } } } {}

Factorize out the numerator and denominator :

N ( z ) = z ( z + 1 ) D ( z ) = ( z − 1 ) 2 ( z 2 − 1 . 6z + 0 . 39 ) N ( z ) = z ( z + 1 ) D ( z ) = ( z − 1 ) 2 ( z 2 − 1 . 6z + 0 . 39 ) alignl { stack { size 12{N \( z \) =z \( z+1 \) } {} # size 12{D \( z \) = \( z - 1 \) rSup { size 8{2} } \( z rSup { size 8{2} } - 1 "." 6z+0 "." "39" \) } {} } } {}

Thus the zeros of the system are

z ( z + 1 ) = 0 ⇒ z = 0, z = − 1 z ( z + 1 ) = 0 ⇒ z = 0, z = − 1 size 12{z \( z+1 \) =0 drarrow z=0, matrix { {} # {} } z= - 1} {}

and the poles are

( z − 1 ) 2 ( z 2 − 1 . 6z + 0 . 39 ) = 0 ⇒ z = 1 ( double ) , z = 0 . 8 + j0 . 5, z = 0 . 8 − j0 . 5 ( z − 1 ) 2 ( z 2 − 1 . 6z + 0 . 39 ) = 0 ⇒ z = 1 ( double ) , z = 0 . 8 + j0 . 5, z = 0 . 8 − j0 . 5 size 12{ \( z - 1 \) rSup { size 8{2} } \( z rSup { size 8{2} } - 1 "." 6z+0 "." "39" \) =0 matrix { {} # {} } drarrow matrix { {} # {} } z=1 \( ital "double" \) , matrix { {} # {} } z=0 "." 8+j0 "." 5, matrix { {} # {} } z=0 "." 8 - j0 "." 5} {}

Figure 3 is the pole – zero plot of the system .

Notice that for the purpose of finding the zeros and poles we first turn H(z)H(z) size 12{H \( z \) } {}as a function of z−1z−1 size 12{z rSup { size 8{ - 1} } } {} into a function of z .

There are functions in Matlab which allow us to plot in three dimensions the variation of the magnitude ∣X(z)∣∣X(z)∣ size 12{ \lline X \( z \) \lline } {} or ∣H(z)∣∣H(z)∣ size 12{ \lline H \( z \) \lline } {} with respect to the coordinates Real (z)(z) size 12{ \( z \) } {} and Imaginary (z)(z) size 12{ \( z \) } {} . Figure 4 is an example of the plot of a transfer function having a zero and a pole.

Poles and zeros of a system do not affect its magnitude and phase response but do affect timing of its response, i.e the response comes earlier or later compared with the time when the input is applied . Specifically , a zero at origin will delay the response a sample , whereas a pole at origin will advance the response a sample . We are free to add zeros and poles to the system , by adding appropriate factors in the numerator and in the denominator to make the system response immediately or become causal .

For example, consider the transfer function

H ( z ) = 1 z ( z − 1 ) ( z − 2 ) H ( z ) = 1 z ( z − 1 ) ( z − 2 ) size 12{H \( z \) = { {1} over {z \( z - 1 \) \( z - 2 \) } } } {}

having three poles at z=0.1z=0.1 size 12{z=0 "." 1} {} and 2 . The corresponding difference aquation can be found to be

y ( n ) = y ( n − 1 ) − y ( n − 2 ) + x ( n − 3 ) y ( n ) = y ( n − 1 ) − y ( n − 2 ) + x ( n − 3 ) size 12{y \( n \) =y \( n - 1 \) - y \( n - 2 \) +x \( n - 3 \) } {}

which shows that the present output signal depends on the input signal of 3 indices (samples) before . To make the system response immediately , we advance the system 3 indices by adding 3 zeros at origin . The transfer function becomes

H ( z ) = z 3 z ( z − 1 ) ( 2z − 1 ) H ( z ) = z 3 z ( z − 1 ) ( 2z − 1 ) size 12{H \( z \) = { {z rSup { size 8{3} } } over {z \( z - 1 \) \( 2z - 1 \) } } } {}

corresponding to the difference equation

y ( n ) = y ( n − 1 ) − y ( n − 2 ) + x ( n ) y ( n ) = y ( n − 1 ) − y ( n − 2 ) + x ( n ) size 12{y \( n \) =y \( n - 1 \) - y \( n - 2 \) +x \( n \) } {}

Thus for a system to response immediately to the input , its transfer function must have equal number of poles and zeros or , in other word , the orders of numerator polynomial and denominator polynomial must be equal .

In the rational polynomial of the z – transform if a zero coincides with a pole, this pole – zero pair is canceled out , thus reducing the orders of the polynomials, and consequently simplifying the conresponding difference equation .

The pole – zero cancelation technique is sometimes used in digital signal processing and design of control systems .The output is an interaction between the input signal and the system , thus we can choose the system so to cancell a pole or a zero of the input signal. The pole – zero cancellation can occur within the system . The problem is that when the pole – zero cancellation is not perfect due to finite word length effect and other reasons , the designed system may become unstable .

We know that the z – transform when z is restricted on the unit circle is just the DTFT. Hence the frequency response can be obtained approximately by geometrical (graphical) method. Let’s consider a simple example . The transfer function is

H ( z ) = z − 0,8 z + 0,8 H ( z ) = z − 0,8 z + 0,8 size 12{H \( z \) = { {z - 0,8} over {z+0,8} } } {}

having a zero at z=0.8z=0.8 size 12{z=0 "." 8} {} and a pole at p=−0.8p=−0.8 size 12{p= - 0 "." 8} {} . For frequency response we replace z=ejωz=ejω size 12{z=e rSup { size 8{jω} } } {}:

H ( ω ) = e jω − 0,8 e jω + 0,8 H ( ω ) = e jω − 0,8 e jω + 0,8 size 12{H \( ω \) = { {e rSup { size 8{ ital "jω"} } - 0,8} over {e rSup { size 8{ ital "jω"} } +0,8} } } {}

The numerator is represented by vector Z from the zero z to a point z=ejωz=ejω size 12{z=e rSup { size 8{jω} } } {} on the unit circle , and the denominator by a vector P from the pole p to the same point z (Figure 5). We adopt the notation ∠∠ size 12{∠} {}for phase angle . Thus

Notice that the phase response is the angle seen from the point z on the unit circle to pole p and zero z.

Hence the magnitude and phase responses are respectively

∣ H ( ω ) ∣ = ∣ Z ∣ ∣ P ∣ ∣ H ( ω ) ∣ = ∣ Z ∣ ∣ P ∣ size 12{ \lline H \( ω \) \lline = { { lline Z rline } over { lline P rline } } } {}

Φ H ( ω ) =∠ ( Z − P ) Φ H ( ω ) =∠ ( Z − P ) size 12{Φ rSub { size 8{H} } \( ω \) "=∠" \( Z - P \) } {}

Let’s consider some special cases where the evaluation is obvious :