# ch06 - Chapter 6 Noise in Modulation Systems 6.1 Problems...

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Chapter 6 Noise in Modulation Systems 6.1 Problems Problem 6.1 The signal power at the output of the lowpasss …lter is P T . The noise power is N 0 B N , where B N is the noise-equivalent bandwidth of the …lter. From (5.116), we know that the noise-equivalent bandwidth of an n th order Butterworth …lter with 3 dB bandwidth W is B n ( n )= ¼W= 2 n sin( ¼= 2 n ) Thus, the signal-to-noise ratio at the …lter output is SNR = P T N 0 B N = ¼= 2 n ) 2 n P T N 0 W so that f ( n ) is given by f ( n 2 n ) 2 n We can see from the form of f ( n ) that since x ) ¼ 1 for x ¿ 1 , f ( 1 ) = 1 . Thus for large n , the SNR at the …lter output is close to P T =N 0 W . The plot is shown In Figure 6.1. Problem 6.2 We express n ( t ) as n ( t ) = n c ( t )cos · ! c t § 1 2 (2 ¼W ) t + μ ¸ + n s ( t )sin · ! c t § 1 2 (2 ) t + μ ¸ where we use the plus sign for the USB and the minus sign for the LSB . The received signal is x r ( t ) = A c [ m ( t )cos( ! c t + μ ) ¨ b m ( t )sin( ! c t + μ )] 1

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2 CHAPTER 6. NOISE IN MODULATION SYSTEMS 1 2 3 4 5 6 7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 n f(n) Figure 6.1: Multiplying x r ( t )+ n ( t ) by 2cos( ! c t + μ ) and lowpass …ltering yields y D ( t ) = A c m ( t n c ( t )cos( ¼Wt ) § n s ( t )sin( ) From this expression, we can show that the postdetection signal and noise powers are given S D = A 2 c m 2 N D = N 0 W S T = A c m 2 N T = N 0 W This gives a detection gain of one. The power spectral densities of n c ( t ) and n s ( t ) are illustrated in Figure 6.2. Problem 6.3 The received signal and noise are given by x r ( t ) = A c m ( t )cos ( ! c t + μ n ( t ) At the output of the predetection …lter, the signal and noise powers are given by S T = 1 2 A 2 c m 2 N T = n 2 = N 0 B T The predetection SNR is ( SNR ) T = A 2 c m 2 2 N 0 B T
6.1. PROBLEMS 3 1 2 W 1 2 W - ( 29 ( 29 , cs nn SfSf f 0 N Figure 6.2: 1 2 T B 1 2 T B - ( 29 c n Sf 0 f 0 N Figure 6.3: If the postdetection …lter passes all of the n c ( t ) component, y D ( t ) is y D ( t ) = A c m ( t )+ n c ( t ) The output signal power is A 2 c m 2 and the output noise PSD is shown in Figure 6.3. Case I: B D > 1 2 B T For this case, all of the noise, n c ( t ) , is passed by the postdetection …lter, and the output noise power is N D = Z 1 2 B T ¡ 1 2 B D N 0 df =2 N 0 B T This yields the detection gain ( SNR ) D ( ) T = A 2 c m 2 =N 0 B T A 2 c m 2 = 2 N 0 B T

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4 CHAPTER 6. NOISE IN MODULATION SYSTEMS Case II: B D < 1 2 B T For this case, the postdetection …lter limits the output noise and the output noise power is N D = Z B D ¡ B D N 0 df = 2 N 0 B D This case gives the detection gain ( SNR ) D ( ) T = A 2 c m 2 = 2 N 0 B D A 2 c m 2 = 2 N 0 B T = B T B D Problem 6.4 This problem is identical to Problem 6.3 except that the predetection signal power is S T = 1 2 A 2 c h 1+ a 2 m 2 n i and the postdetection signal power is S D = A 2 c a 2 m 2 n The noise powers do not change. Case I: B D > 1 2 B T ( ) D ( ) T = A 2 c a 2 m 2 n =N 0 B T A 2 c h a 2 m 2 n i = 2 N 0 B T = 2 a 2 m 2 n a 2 m 2 n Case II: B D < 1 2 B T ( ) D ( ) T = A 2 c a 2 m 2 n = 2 N 0 B D A 2 c h 1 + a 2 m 2 n i = 2 N 0 B T = a 2 m 2 n a 2 m 2 n = B T B D Problem 6.5 Since the message signal is sinusoidal m n ( t ) = cos(8 ¼t )
6.1.

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ch06 - Chapter 6 Noise in Modulation Systems 6.1 Problems...

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