ch14_ism - 14 Introduction to Frequency-Selective Circuits...

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14 Introduction to Frequency-Selective Circuits Assessment Problems AP 14.1 f c =8 kHz c =2 πf c =16 π krad/s ω c = 1 RC ; R =10 k Ω; · .. C = 1 ω c R = 1 (16 π × 10 3 )(10 4 ) =1 . 99 nF AP 14.2 [a] ω c c π (2000) = 4 π krad/s L = R ω c = 5000 4000 π =0 . 40 H [b] H ( )= ω c ω c + = 4000 π 4000 π + When ω π (50 , 000) = 100 , 000 π rad/s H ( j 100 , 000 π 4000 π 4000 π + j 100 , 000 π = 1 1+ j 25 . 04/ 87 . 71 · .. | H ( j 100 , 000 π ) | . 04 [c] · .. θ (100 , 000 π 87 . 71 AP 14.3 ω c = R L = 5000 3 . 5 × 10 3 . 43 Mrad/s 14–1
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14–2 CHAPTER 14. Introduction to Frequency-Selective Circuits AP 14.4 [a] ω c = 1 RC = 10 6 R = 10 6 100 =10 krad/s [b] ω c = 10 6 5000 = 200 rad/s [c] ω c = 10 6 3 × 10 4 =33 . 33 rad/s AP 14.5 Let Z represent the parallel combination of (1 /sC ) and R L . Then Z = R L ( R L Cs +1) Thus H ( s )= Z R + Z = R L R ( R L +1)+ R L = (1 /RC ) s + R + R L R L 1 RC = (1 /RC ) s + 1 K 1 RC where K = R L R + R L AP 14.6 ω 2 o = 1 LC so L = 1 ω 2 o C = 1 (24 π × 10 3 ) 2 (0 . 1 × 10 6 ) =1 . 76 mH Q = ω o β = ω o R/L so R = ω o L Q = (24 π × 10 3 )(1 . 76 × 10 3 ) 6 =22 . 10 Ω AP 14.7 ω o =2 π (2000) = 4000 π rad/s ; β π (500) = 1000 π rad/s ; R = 250 Ω β = 1 RC so C = 1 βR = 1 (1000 π )(250) . 27 µ F ω 2 o = 1 LC so L = 1 ω 2 o C = 10 6 (4000 π ) 2 (1 . 27) =4 . 97 mH AP 14.8 ω 2 o = 1 LC so L = 1 ω 2 o C = 1 (10 4 π ) 2 (0 . 2 × 10 6 ) =5 . 07 mH β = 1 RC so R = 1 βC = 1 400 π (0 . 2 × 10 6 ) =3 . 98 k
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Problems 14–3 AP 14.9 ω 2 o = 1 LC so L = 1 ω 2 o C = 1 (4000 π ) 2 (0 . 2 × 10 6 ) =31 . 66 mH Q = f o β = 5 × 10 3 200 =25= ω o RC · .. R = Q ω o C = 25 (4000 π )(0 . 2 × 10 6 ) =9 . 95 k AP 14.10 ω o = 8000 π rad/s C = 500 nF ω 2 o = 1 LC so L = 1 ω 2 o C =3 . 17 mH Q = ω o β = ω o L R = 1 ω o CR · = 1 ω o CQ = 1 (8000 π )(500 × 10 9 )(5) =15 . 92 Ω AP 14.11 ω o =2 πf o π (20 , 000) = 40 π krad/s ; R = 100 Ω; Q =5 Q = ω o β = ω o L RC so L = RQ ω o = 100 40 π × 10 3 . 98 mH ω 2 o = 1 LC so C = 1 ω 2 o L = 1 (40 π × 10 3 ) 2 (3 . 98 × 10 3 ) . 92 nF
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14–4 CHAPTER 14. Introduction to Frequency-Selective Circuits Problems P 14.1 [a] ω c = R L = 127 10 × 10 3 =12 . 7 krad/s · .. f c = ω c 2 π = 12 , 700 2 π = 2021 . 27 Hz [b] H ( s )= ω c s + ω c = 12 , 700 s +12 , 700 H ( 12 , 700 12 , 700 + H ( c 12 , 700 12 , 700 + j 12 , 700 =0 . 7071/ 45 H ( j 0 . 2 ω c 12 , 700 12 , 700 + j 2540 . 981/ 11 . 31 H ( j 5 ω c 12 , 700 12 , 700 + j 63 , 500 . 196/ 78 . 69 [c] v o ( t ) | ω c =7 . 07 cos(12 , 700 t 45 ) V v o ( t ) | 0 . 2 ω c =9 . 81 cos(2540 t 11 . 31 ) V v o ( t ) | 5 ω c =1 . 96 cos(63 , 500 t 78 . 69 ) V P 14.2 [a] ω o = R L = 2000 π rad/s R = o =(0 . 005)(2000 π )=31 . 42 Ω [b] R e =31 . 42 k 270=28 . 14 Ω ω loaded = R e L = 5628 rad/s · loaded = ω loaded 2 π = 895 . 77 Hz P 14.3 Note: add the resistor to the cirucit in Fig. 14.4(a). [a] H ( s V o V i = R sL + R + R l = ( R/L ) s +( R + R l ) /L
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Problems 14–5 [b] H ( )= ( R/L ) R + R l L + | H ( ) | = ( R/L ) r R + R l L 2 + ω 2 | H ( ) | max occurs when ω =0 [c] | H ( ) | max = R R + R l [d] | H ( c ) | = R 2( R + R l ) = R/L r R + R l L 2 + ω 2 c · .. ω 2 c = ± R + R l L ² 2 ; · c =( R + R l ) /L [e] Note – add 75 Ω resistor in series with the 10 mH inductor. ω c = 127+75 0 . 01 =20 , 200 rad/s H ( 12 , 700 20 , 200 + H ( j 0)=0 . 6287 H ( j 20 , 200) = 0 . 6287 2 / 45 . 4446/ 45 H ( j 6060) = 12 , 700 20 , 200 + j 6060 . 6022/ 16 . 70 H ( j 60 , 600) = 12 , 700 20 , 200 + j 60 , 600 . 1988/ 71 . 57 P 14.4 [a] ω c = 1 RC = 1 (10 3 )(100 × 10 9 ) =10 krad/s f c = ω c 2 π = 1591 . 55 Hz [b] H ( ω c s + ω c = 10 , 000 s +10 , 000 H ( 10 , 000 10 , 000 + H ( c 10 , 000 10 , 000 + j 10 , 000 . 7071/ 45
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14–6 CHAPTER 14. Introduction to Frequency-Selective Circuits H ( j 0 . 1 ω c )= 10 , 000 10 , 000 + j 1000 =0 . 9950/ 5 . 71 H ( j 10 ω c 10 , 000 10 , 000 + j 100 , 000 . 0995/ 84 . 29 [c] v o ( t ) | ω c = 200(0 . 7071) cos(10 , 000 t 45 ) = 141 . 42 cos(10 , 000 t 45 ) mV v o ( t ) | 0 . 1 ω c
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ch14_ism - 14 Introduction to Frequency-Selective Circuits...

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