chapter 13 - Chapter 13: Frequency Response Exercises Ex....

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Chapter 13: Frequency Response Exercises Ex. 13.3-1 o s 2 1 () 1 1 1 1 ( ) = tan jCR gain CR phase shift C R ω ωω == + = + V H V 46 o 1 When 10 , = 100, and 10 , then = = 0.707 and = 45 2 R C gain phase shift Ex. 13.3-2 o s 22 R RjL R gain RL + = + V H V 2 2 30 30 .6 30 0.6 20 rad s 2 30 (2 )    =⇒ = = + Ex. 13.3-3 s 1 1 1 tan R jL gain L phase shift R + = + =− I H V When 30 , 2 H, and 20 rad/s, then =Ω= = 1 1A 4 0 0.02 and tan 53.1 V3 30 40 gain phase shift = = ° + 0 13-1
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Ex. 13.3-4 o s 2 1 () 1 () 1 1 1 ( ) tan j CR gain CR phase shift C R ω ωω == + = + =− V H V 16 3 6 tan (45 ) 45 tan (20 10 ) 50 10 2010 RR ° −− −° = ⇒ = = ⋅ Ex. 13.3-5 o s 2 1 1 1 ( ) gain + = + V H V , , and are all positive, or at least nonnegative, so gain 1. These specifications cannot be met. Ex. 13.4-1 (a) dB = 20 log (.5) = 6.02 dB (b) dB = 20 log 2 = 6.02 dB Ex. 13.4-2 -2 2 2 212 1 1 21 1 20log 20log 20 log ( ) 40log 20 log ( ) 20log ( ) 40log 40log -40 log dB let 10 to consider 1 decade, then 40log10 40 decade slope slope  == =   =−= + = H HH = 13-2
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Ex. 13.4-3 10 10 10 10 When , ( ) ( ) ( ) in dB 20log ( ) 20log ( ) ( does not depend on so 0 When , ( ) ( ) in dB = 20 log ( ) = 20 log jA A CB jC C A d C b slope jA A j BB ω ωω >> =  ==   = << = H HH H H ± ± 10 10 +20 log dB ( ) The slope is the coefficient of 20 log , that is, = 20 decade ( ) The break frequency is the frequency at which , that is, A B c slope B aC C B = = Ex. 13.4-4 1 oc 2 1 s 2 o 1 s2 () 1 1 1( 1 1 1 1 R R R Rj C R R ) R jCR =+ + + + VV V V H V 1 2 When 0.1 and 3, 4 then ( ) 10 R RC R j = + H 13-3
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Ex. 13.4-5 a) o2 2 oo 1 s1 o 12 2 1 2 2 1 1 1 1 R + 1 1 where = 16.7 rad/s 1 and = 5.56 rad/s ( ) R jC j R R Rj RC RR C ω =+ + + == = ++ + = = + Z VZ () ss o s t 10cos20 or 10 0 20 1 16.7 20 1 5.56 1 1.20 0.417 24.3 1 3.60 vt j j j j + ° = + + + V V V ° b) So o o 4.17 24.3 ( ) 4.17cos(20 24.3 ) V t =∠ ° =− V ° 13-4
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Ex. 13.4-6 Ex. 13.5-1 a) 7 o 3 2.5 10 8000 20 40 10 C QR CR L ω × == = × b) 37 11 500 rad s 20 (40 10 ) (2.5 10 ) o BW Q QL C −− = = ×× Ex. 13.5-2 7 0 5 o 27 2 1 2 o 10 = = = 50 2 10 111 Now = 1 mH (10 ) (10 10 ) Q BW L LC C × ⇒= = = × Ex. 13.5-3 4 o 1 35 2 4 o 43 o 1 1 = 10 rad s (10 )(10 ) 10 = = 100 2 (15.9) (10 )(10 ) = = = 0.1 100 LC Q BW L R Q π   = 13-5
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Ex. 13.5-4 a) o 62 3 o 26 2 1 0 o 11 100 pF (10 ) (0.01) 10 1 = = = 10 (10 ) (10 ) o C LC BW QR BW RC C ω =⇒ = = ⇒= = b) 6 o 3 66 o 64 o 10 1000 10 1.05 10 10 1 1 1000 10 1.05 10 1 1 97.6 Q BW jQ j j ωω === ==  × +−  ×  = + H H 13-6
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