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0132130084_ism06 - CHAPTER 6 Exercises E6.1(a The frequency...

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221 CHAPTER 6 Exercises E6.1 (a) The frequency of ) 2000 2 cos( 2 ) ( in t t v = π is 2000 Hz. For this frequency . 60 2 ) ( o = f H Thus, o o o 60 4 0 2 60 2 ) ( in out = × = = V V f H and we have ). 60 2000 2 cos( 4 ) ( out o + = t t v π (b) The frequency of ) 20 3000 2 cos( ) ( in o = t t v π is 3000 Hz. For this frequency . 0 ) ( = f H Thus, 0 0 2 0 ) ( in out = × = = o V V f H and we have . 0 ) ( out = t v E6.2 The input signal ) 1500 2 cos( 3 ) 20 500 2 cos( 2 ) ( t t t v + + = π π o has two components with frequencies of 500 Hz and 1500 Hz. For the 500-Hz component we have: o o o 35 7 20 2 15 5 . 3 ) 500 ( in out,1 = × = = V V H ) 35 500 2 cos( 7 ) ( out,1 o + = t t v π For the 1500-Hz component: o o o 45 5 . 7 0 3 45 5 . 2 ) 1500 ( in out,2 = × = = V V H ) 45 1500 2 cos( 5 . 7 ) ( out,2 o + = t t v π Thus the output for both components is ) 45 1500 2 cos( 5 . 7 ) 35 500 2 cos( 7 ) ( out o o + + + = t t t v π π E6.3 The input signal ) 3000 2 cos( 3 ) 1000 2 cos( 2 1 ) ( t t t v π + π + = has three components with frequencies of 0, 1000 Hz and 3000 Hz. For the dc component, we have 4 1 4 ) ( ) 0 ( ) ( 1 , out,1 = × = × = t v H t v in For the 1000-Hz component, we have: o o o 30 6 0 2 30 3 ) 1000 ( in,2 out,2 = × = = V V H ) 30 1000 2 cos( 6 ) ( out,1 o + = t t v π For the 3000-Hz component: 0 0 3 0 ) 3000 ( in,3 out,3 = × = = o V V H 0 ) ( out,3 = t v Thus, the output for all three components is ) 30 1000 2 cos( 6 4 ) ( out o + π + = t t v
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222 E6.4 Using the voltage-division principle, we have: fL j R R π 2 in out + × = V V Then the transfer function is: B f jf R fL j fL j R R f H / 1 1 / 2 1 1 2 ) ( in out + = + = + = = π π V V E6.5 From Equation 6.9, we have Hz 200 ) 2 /( 1 = = RC f B π , and from Equation 6.9, we have . / 1 1 ) ( in out B f jf f H + = = V V For the first component of the input, the frequency is 20 Hz, , 71 . 5 995 . 0 ) ( o = f H o 0 10 in = V , and o 71 . 5 95 . 9 ) ( in out = = V V f H Thus the first component of the output is ) 71 . 5 40 cos( 95 . 9 ) ( out,1 o = t t v π For the second component of the input, the frequency is 500 Hz, , 2 . 68 371 . 0 ) ( o = f H o 0 5 in = V , and o 2 . 68 86 . 1 ) ( in out = = V V f H Thus the second component of the output is ) 2 . 68 40 cos( 86 . 1 ) ( out,2 o = t t v π For the third component of the input, the frequency is 10 kHz, , 9 . 88 020 . 0 ) ( o = f H o 0 5 in = V , and o 9 . 88 100 . 0 ) ( in out = = V V f H Thus the third component of the output is ) 9 . 88 10 2 cos( 100 . 0 ) ( 4 out,2 o × = t t v π Finally, the output with for all three components is: ) 9 . 88 10 2 cos( 100 . 0 ) 2 . 68 40 cos( 86 . 1 ) 71 . 5 40 cos( 95 . 9 ) ( 4 out o o o × + + = t t t t v π π π E6.6 dB 98 . 33 ) 50 log( 20 ) ( log 20 ) ( dB = = = f H f H E6.7 (a) dB 15 ) ( log 20 ) ( dB = = f H f H 0.75 /20 15 ) ( log = = f H 623 . 5 10 ) ( 0.75 = = f H
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223 (b) dB 30 ) ( log 20 ) ( dB = = f H f H 1.5 /20 30 ) ( log = = f H 62 . 31 10 ) ( 1.5 = = f H E6.8 (a) Hz 4000 2 1000 2 = × is two octaves higher than 1000 Hz. (b) 125 2 / 1000 3 = Hz is three octaves lower than 1000 Hz. (c) 100 10 1000 2 = × kHz is two decades higher than 1000 Hz. (d) 100 10 / 1000 = Hz is one decade lower than 1000 Hz. E6.9 (a) To find the frequency halfway between two frequencies on a logarithmic scale, we take the logarithm of each frequency, average the logarithms, and then take the antilogarithm. Thus 2 . 316 10 10 5 . 2 2 / )] 1000 log( ) 100 [log( = = = + f Hz is half way between 100 Hz and 1000 Hz on a logarithmic scale.
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