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ch08 - CHAPTER 8 Additional Problems 8.16 Provided that the...

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1 CHAPTER 8 Additional Problems 8.16 Provided that the channel bandwidth is not smaller than the reciprocal of the transmitted pulse duration T , the received pulse is recognizable at the channel output. With T = 1 μ s, a small enough value for the channel bandwidth is (1/ T) = 10 6 Hz = 1 MHz. 8.17 For a low-pass filter of the Butterworth type, the squared magnitude response is defined by (1) At the edge of the passband, ϖ = ϖ p , we have (by definition) We may therefore write (2) Define Then solving Eq. (2) for ϖ p : Next, by definition, at the edge of the stopband, ϖ = ϖ s , we have Hence (3) Define Hence, solving Eq. (3) for ϖ s : H j ϖ ( 29 2 1 1 ϖ ϖ c ( 29 2 N + ---------------------------------- = H j ϖ p ( 29 1 - = (1 - ) 2 1 1 ϖ p ϖ c ( 29 2 N + ------------------------------------- = 0 1 (1 - ) 2 2 - 2 = = ϖ p 0 1 - 0 ----------------- ϖ c = H j ϖ s ( 29 δ = δ 2 1 1 ϖ s ϖ c ( 29 2 N + ------------------------------------ = δ 0 δ 2 = ϖ s 1 δ 0 δ 0 ------------- ϖ c =

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2 8.18 We start with the relation For a Butterworth low-pass filter of order 5, the 10 poles of H ( s ) H (- s ) are uniformly distributed around the unit circle in the s -plane as shown in Fig. 1. Let D ( s ) D (- s ) denote the denominator polynomial of H ( s ) H (- s ). Hence Identifying the zeros of D ( s ) D (- s ) in the left-half plane with D ( s ) and those in the right- half plane with D (- s ), we may express D(s) as Hence, 8.19 (a) For filter order N that is odd, the transfer function H ( s ) of the filter must have a real pole in the left-half plane. Let this pole be s = - a where a > 0. We may then write H j ϖ ( 29 j ϖ = s 2 H s ( 29 H s ( 29 = Figure 1 x x x x x x x x x x σ j ϖ s- plane D s ( 29 D s ( 29 s 1 + ( 29 ( s 144 ° j 144 ° ) sin + s 144 ° j 144 ° sin cos + ( 29 cos + = ( s 108 ° j 108 ° ) sin + s 108 ° j 108 ° sin cos + ( 29 cos + × s 1 ( 29 ( s 36 ° j 36 ° ) sin + s 36 ° j 36 ° sin cos ( 29 cos × ( s 72 ° j 72 ° ) sin + s 72 ° j 72 ° sin cos ( 29 cos × D s ( 29 s 1 + ( 29 ( s 144 ° j 144 ° ) sin + s 144 ° j 144 ° sin cos + ( 29 cos + = ( s 108 ° j 108 ° ) sin + s 108 ° j 108 ° sin cos + ( 29 cos + × s 5 3.2361 s 4 5.2361 s 3 5.2361 s 2 3.2361 s 1 + + + + + = H s ( 29 1 D s ( 29 ----------- = 1 s 5 3.2361 s 4 5.2361 s 3 5.2361 s 2 3.2361 s 1 + + + + + --------------------------------------------------------------------------------------------------------------------------- = H s ( 29 1 s a + ( 29 D s ( 29 ------------------------------ =
3 where is the remainder of the denominator polynomial. For a Butterworth low- pass filter of cutoff frequency ϖ c , all the poles of H ( s ) lie on a circle of radius ϖ c in the left-half plane. Hence, we must have a = - ϖ c . (b) For a Butterworth low-pass filter of even order N , all the poles of the transfer function H ( s ) are complex. They all lie on a circle of radius ϖ c in the left-half plane. Let s = - a - jb , with a > 0 and b > 0, denote a complex pole of H ( s ). All the coefficients of H ( s ) are real. This condition can only be satisfied if we have a complex conjugate pole at s = - a + jb . We may then express the contribution of this pair of poles as whose coefficients are all real. We therefore conclude that for even filter order N , all the poles of H ( s ) occur in complex-conjugate pairs. 8.20 The transfer function of a Butterworth low-pass filter of order 5 is (1) The low-pass to high-pass transformation is defined by where it is assumed that the cutoff frequency of the high-pass filter is unity. Hence replacing s with 1/ s in Eq. (1), we find that the transfer function of a Butterworth high-pass

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ch08 - CHAPTER 8 Additional Problems 8.16 Provided that the...

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