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Unformatted text preview: 1 CHAPTER 6 Exercises E6.1 (a) The frequency of ) 2000 2 cos( 2 ) ( in t v ⋅ = π is 2000 Hz. For this frequency . 60 2 ) ( o ∠ = f H Thus, o o o 60 4 2 60 2 ) ( in out ∠ = ∠ × ∠ = = V V and we have ). 60 2000 2 cos( 4 ) ( out o + ⋅ = (b) The frequency of ) 20 3000 2 cos( ) ( in o − ⋅ = is 3000 Hz. For this frequency . ) ( = Thus, 2 ) ( in out = ∠ × = = o V V and we have . ) ( out = E6.2 The input signal ) 1500 2 cos( 3 ) 20 500 2 cos( 2 ) ( ⋅ + + ⋅ = o has two components with frequencies of 500 Hz and 1500 Hz. For the 500Hz component we have: o o o 35 7 20 2 15 5 . 3 ) 500 ( in out,1 ∠ = ∠ × ∠ = = V V ) 35 500 2 cos( 7 ) ( out,1 o + ⋅ = For the 1500Hz component: o o o 45 5 . 7 3 45 5 . 2 ) 1500 ( in out,2 ∠ = ∠ × ∠ = = V V ) 45 1500 2 cos( 5 . 7 ) ( out,2 o + ⋅ = Thus the output for both components is ) 45 1500 2 cos( 5 . 7 ) 35 500 2 cos( 7 ) ( out o o + ⋅ + + ⋅ = E6.3 The input signal ) 3000 2 cos( 3 ) 1000 2 cos( 2 1 ) ( ⋅ π + ⋅ π + = has three components with frequencies of 0, 1000 Hz and 3000 Hz. For the dc component, we have 4 1 4 ) ( ) ( ) ( 1 , out,1 = × = × = in For the 1000Hz component, we have: o o o 30 6 2 30 3 ) 1000 ( in,2 out,2 ∠ = ∠ × ∠ = = V V ) 30 1000 2 cos( 6 ) ( out,1 o + ⋅ = For the 3000Hz component: 3 ) 3000 ( in,3 out,3 = ∠ × = = o V V ) ( out,3 = Thus, the output for all three components is ) 30 1000 2 cos( 6 4 ) ( out o + ⋅ π + = 2 E6.4 Using the voltagedivision principle, we have: fL j R π 2 in out + × = V V Then the transfer function is: B f jf 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 , and from Equation 6.9, we have . / 1 1 ) ( in out + = = V V For the first component of the input, the frequency is 20 Hz, , 71 . 5 995 . ) ( o − ∠ = o 10 in ∠ = V , and o 71 . 5 95 . 9 ) ( in out − ∠ = = V V Thus the first component of the output is ) 71 . 5 40 cos( 95 . 9 ) ( out,1 o − = t v π For the second component of the input, the frequency is 500 Hz, , 2 . 68 371 . ) ( o − ∠ = o 5 in ∠ = V , and o 2 . 68 86 . 1 ) ( in out − ∠ = = V V Thus the second component of the output is ) 2 . 68 40 cos( 86 . 1 ) ( out,2 o − = π For the third component of the input, the frequency is 10 kHz, , 9 . 88 020 . ) ( o − ∠ = o 5 in ∠ = V , and o 9 . 88 100 . ) ( in out − ∠ = = V V Thus the third component of the output is ) 9 . 88 10 2 cos( 100 . ) ( 4 out,2 o − × = π Finally, the output with for all three components is: ) 9 . 88 10 2 cos( 100 . ) 2 . 68 40 cos( 86 . 1 ) 71 . 5 40 cos( 95 . 9 ) ( 4 out o o o − × + − + − = π π π E6.6 dB 98 . 33 ) 50 log( 20 ) ( log 20 ) ( dB = = = E6.7 (a) dB 15 ) ( log 20 ) ( dB = = 0.75 /20 15 ) ( log = = 623 . 5 10 ) ( 0.75 = = 3 (b) dB 30 ) ( log 20 ) ( dB = = f H 1.5 /20 30 ) ( log = = 62 . 31 10 ) ( 1.5 = = E6.8 (a) Hz 4000 2 1000 2 = × is two octaves higher than 1000 Hz....
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 Spring '07
 Haris
 Frequency, Lowpass filter, fb, Vout, Hz

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