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(b) Because the order of this filter is N = 14, the system function has 14 zeros. For a linear phase filter, we know that the zeros of the system function may lie on the unit circle, or they may occur in conjugate reciprocal pairs. From the plot of the frequency response magnitude, we see that I H ( e J ' " ) (= 0 at w l w 0.175n, w2 w 0.3n, . and w3 0 . 3 9 ~ Therefore, there are three zeros on the unit circle at these frequencies. Because there must also be zeros at the conjugate positions, z = e-'"'1 for i = 1, 2,3, these unit circle zeros account for six of the fourteen zeros. In addition to these. there must be a conjugate pair of zeros at z = re*'", where w4 0.71~. These zeros account for the dip in I H ( e l " ) I at w = 0.71~. Because the filter has linear phase, in addition to this pair of complex zeros, there must be a pair at the reciprocal locations, z - ' = re*jW4. For the same reason, there will be zeros on the real axis at z = a land z = I/cul,as well as zeros on the real axis at z = -a2 and z = -I/a2,where a , and a2 are positive real numbers that are less than I. These four zeros account for the minima in I H ( e J w at w = 0 and w = n . A plot showing the actual positions of the 14 zeros of H ( z ) is given ) below.

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EAN - 128 (also known as: EAN - 128 , UCC- 128 , USS- 128 , UCC. EAN - 128 , and GTIN- 128 ) is developed to provide a worldwide format and standard for exchanging common data between companies. It is a variable-length linear barcode with high density.
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1 2 x

Reasons Given A diameter cuts a circle into two equal semicircles. Given Parallel lines intercept > arcs on a circle. If equals are subtracted from equals, the differences are equal. Definition of > arcs. 6. In a circle, equal arcs have chords which are equal in length.

With the frequency sampling method, the frequency samples match the ideal frequency response exactly. Derive an interpolation formula that shows how the frequency samples H (k) are interpolated. The frequency response of an FIR filter of length N is

6.1. Provide the proofs requested in Fig. 6-44. (a) Given: AB DE AC DF To Prove: jB > jE (b) Given: Circle O, AB BC Diameter BD To Prove: BD bisects jAOC. (c) Given: (6.3) Circle O AB CD To Prove: jAOC > jBOD

H ( k )=

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h(n)e-"n"klN - ~ , , ( e ~ ' " ~ / k = 0 , I . ~)

(b) Fig. 6-44

Because these frequency samples correspond to the N-point DFT of h ( n ) ,the unit sample response may be expressed in terms of these samvles as follows:

=

Provide the proofs requested in Fig. 6-45. Please refer to figure 6-45(a) for problems 6.2(a) and (b); to figure 6-45(b) for problems 6.2(c) and (d); and figure 6-45(c) for problems 6.2(e) and (f). (6.3) (a) Given: AB AC To Prove: ABC > ACB (b) Given: ABC > ACB To Prove: AB AC (c) Given: Circle O, AB AD Diameter AC To Prove: BC CD (d) Given: Circle O AB AD, BC CD To Prove: AC is a diameter. (e) Given: AD BC To Prove: AC BD (f) Given: AC BD To Prove: AD BC

Substituting this into the expression above for H ( e J w ) we have ,

Fig. 6-45

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Given a low-pass filter that has been designed and implemented, either in hardware or software, it may be of interest to try to improve the frequency response characteristics by repetitive use of the filter. Suppose that h ( n ) is the unit sample response of a zero phase FIR filter with a frequency response that satisfies the following specifications:

Prove each of the following: (a) If a radius bisects a chord, then it bisects its arcs. (b) If a diameter bisects the major arc of a chord, then it is perpendicular to the chord. (c) If a diameter is perpendicular to a chord, it bisects the chord and its arcs.

(Note that H ( e J w ) having zero phase implies that H ( e J W is real-valued for all LO). )

Sometimes the value f (x) will get larger without bound or smaller without bound as x approaches a from one side (x a or x a+ ).

( a ) If a new filter is formed by cascading h ( n ) with itself,

(6.4)

Find A . B. C , and D in terms of 6, and 6 , of the low-pass filter h ( n ) .

Prove each of the following: (a) A radius through the point of intersection of two congruent chords bisects an angle formed by them.

(b) If 6,

(6.4)

CHAP. 9]

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