Lecture 17 z-transforms 2

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1 Lecture 17 z-transforms 2 Fundamentals of Digital Signal Processing Spring, 2012 Wei-Ta Chu 2012/5/3 1

2 Factoring z-polynomials We can also factor z-transform polynomials to break down a large system into smaller modules. Since cascading systems is equivalent to multiplying their system functions, the factors of a high-order polynomial H(z) would represent component systems that make up H(z) in a cascade connection. 2

3 Example One of the roots is z=1, so is a factor of H(z) 3

4 Deconvolution Suppose that we have the cascade of two filters H 1 (z) and H 2 (z), and H 1 (z) is known. Is it possible to find H 2 (z) so that the overall system has its output equal to the input? If so, the z-transform analysis tells us that its system function would have to be H(z) = 1 The second filter tries to undo convolution called deconvolution or inverse filtering If, then is said to be the inverse of More related content will be described in Chapter 8 4

5 The z-domain and The -Domain Recall the frequency response formula There is a clear correspondence: The connection between and the z-transform 5

6 The z-domain and The -Domain To see why this relationship is the key, we need only recall that if the signal z n is the input to an LTI filter, the resulting output is y[n] = H(z)z n. If the value of z is where is obviously the same as what we have called the frequency response 6

7 The z-plane and The Unit Circle The frequency response is obtained from the system function by evaluating for a specific set of values of z. We see that the corresponding values of z all have unit magnitude and that the angle varies from to The values of lie on a unit circle. 7

8 The z-plane and The Unit Circle This figure gives us a convenient way of visualizing the relationship between the -domain and the z- domain. The periodicity of the frequency response is obvious Since frequency is equivalent to angle in the z-plane, radians in the z-plane correspond to an interval of radians of frequency. 8

9 The Zeros and Poles of H(z) The system function for an FIR system is essentially determined by its zeros. Consider the system function which can be expressed as If we multiply H(z) by, we obtain the equivalent forms: 9

10 The Zeros and Poles of H(z) Zeros of H(z) are at locations for Values of z for which H(z) is undefined (infinite) are called poles of H(z). In this case, we say that the term z 3 represents three poles at z=0 or that H(z) has a thirdorder pole at z = 0. 10

11 The Zeros and Poles of H(z) Pole-zero plot Three poles at z=0 are indicated by a single with a numeral 3 beside it. Each zero location is denoted by a small circle 11

12 Significance of the Zeros of H(z) The zeros of a polynomial system function are sufficient to determine H(z) except for a constant multiplier. The system function determines the difference equation of the filter. The difference equation is the direct link between an input x[n] and the corresponding output y[n] However, there are some inputs where knowledge of the zero locations is sufficient to make a precise statement about the output without actually computing it using the difference equation. 12

13 Significance of the Zeros of H(z) Such signals are of the form for all n, where the subscript signifies that z 0 is a particular complex number. The output is The quantity is a complex constant, which, through complex multiplication, causes a magnitude and phase change of the input signal If is one of the zeros of, then so the output will be zero. 13

14 Example When, the roots are The complex sinusoids will be zero with frequencies The output resulting from each of the following three signals will be zero. Complex sinusoids at those frequencies are blocked. (or nulled) 14

15 Nulling Filters If the zeros of H(z) lie on the unit circle, then certain sinusoidal input signals are removed or nulled by the filter. Therefore, it should be possible to use this result in designing an FIR filter that can null a particular sinusoidal input. Zeros in the z-plane can remove only signals that have the special form If we want to eliminate a sinusoidal input signal, then we actually have to remove two signals of the form 15

16 Nulling Filters Each complex exponential can be removed with a firstorder FIR filter, and then the two filters would be cascaded to form the second-order nulling filter that removes the cosine. The second-order FIR filter will have two zeros at and The signal will be nulled by a filter with system function because 16

17 Nulling Filters Similarly, will remove Thus the second-order nulling filter will be the product 17

18 Nulling Filters This figure shows the two zeros needed to remove components at The numerical values for the coefficients of H(z) are The nulling filter that will remove the signal from its input is the FIR filter whose difference equation is 18

19 Graphical Relation Between z and The frequency response is obtained by evaluating the system function on the unit circle of the z-plane. Evaluating the z-transform magnitude H(z) over a region of the z-plane that includes the unit circle as well as values both inside and outside the unit circle. 11-point running sum 19

20 Graphical Relation Between z and The zeros of the system function of this filter are on the unit circle at angles, k=1,2,,10. This means that the polynomial can be represented in the form Recall that each factor of the form represents a zero at and a pole at z = 0. Thus, this equation displays the 10 zeros of H(z) at, for k=1,2,,10 and the 10 poles at z=0. 20

21 Graphical Relation Between z and A plot of versus 10 zeros of H(z) at This example illustrates that intuitive picture of the frequency response of an LTI system can be visualized from a plot of the poles and zeros of the system function H(z) 21

22 The L-Point Running-Sum Filters The L-point running-sum filter has system function H(z) can be represented in the forms: Zeros: Since, an integer, each value is These L numbers are the roots of the L th -order equation. Because the values satisfy the equation, they are called the L th roots of unity. 22

23 The L-Point Running-Sum Filters The zeros of the denominator in, which are either z=0 (of order L-1) or z=1, would normally be the poles of H(z). However, since one of the L th roots of unity is z=1 that zero of the numerator cancels the corresponding zero of the denominator, so that only the term really causes a pole of H(z). Therefore, it follows that H(z) can be expressed in the factored form 23

24 Example For a 10-point running-sum filter (L=10), the system function is The factors of the numerator are the tenth roots of unity, and the zero at z = 1 is canceled by the corresponding term in the denominator. This explains why we have only nine zeros around the unit circle with the gap at z = 1. 24

25 Example The nine zeros show up as zeros along the axis at, and it is the gap at z = 1 that allows the frequency response to be larger at. The other zeros around the unit circle keep small, thereby creating the lowpass filter frequency response. 25

26 A Complex Bandpass Filter To get the filter coefficients, we compare two figures. Rotation of the z-plane representation will have the corresponding effect of shifting the frequency response along the -axis by the amount of the rotation. The desired rotation in this case is by the angle 26

27 A Complex Bandpass Filter The question is how to move the roots of a polynomial through a rotation. The answer is that we must multiply the kth filter coefficient b k by where is the desired angle of rotation. Consider the following general operation on a polynomial G(z): Every occurrence of the variable z in the polynomial G(z) is replaced by z/r. The effect of this substitution on the roots of G(z) is to multiply them by r and make these the roots of H(z). 27

28 A Complex Bandpass Filter E.g. The two roots of H(z) are now z=2r and z=r 28

29 A Complex Bandpass Filter In the case of the complex bandpass filter, G(z) is the running-sum system function and the parameter r is a complex exponential The filter coefficients of the complex bandpass filter are 29

30 A Complex Bandpass Filter Another way to determine the frequency response of the complex bandpass filter is to compute it directly The frequency response of the system is a shifted (by ) version of the frequency response of the L-point running-sum filter. 30

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