ch12_solutions - 12-378 CHAPTER 12 12.1a( 29C2=C=0.005mF|...

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Unformatted text preview: 12-378 CHAPTER 12 12.1a( 29C2=C=0.005mF| C1=2C=0.01mF| R=12wOC=1240000p(290.005mF=1.13 kFor Q=12: Avs( 29=wo2s2+2wos+wo2| A( 29=1 | A jwo(29=12wH=wo! | fH=20 kHzb( 29For fo= 40kHz: C'C2: C1'=0.005 mF| C2'=0.0025 mF| R=1.13 k12.2(a) r1=1130; r2=1130; c1=1e-8; c2=5e-9; wo=1/sqrt(r1*r2*c1*c2) n=wo*wo; d=[1 2/(r1*c1) wo*wo]; bode(n,d) -40-30-20-10Frequency (rad/sec)Gain dBFrequency (rad/sec)Phase deg-200-150-100-50104105106104105106(b) *PROBLEM 12.94 - Low-pass Filter VS 1 0 AC 1 R1 1 2 1.125K R2 2 3 1.125K C1 2 6 0.01UF C2 3 0 0.005UF ISEN 3 6 DC 0 EC 4 0 3 6 1 RC 4 5 1K CC 5 0 15.915UF E1 6 0 5 0 100000 .AC DEC 40 1 1MEG .PRINT AC IM(VS) IP(VS) VDB(6) VP(6) 12-379 .PROBE I(VS) V(6) .END 1 Hz100 Hz10 kHz1 MHzFrequency40-20-80-100-200-300Phase (Degrees)Voltage Gain (dB)(c) The magnitude response is very similar to the ideal case. However, note the excess phase shift as one approaches the fTof the amplifier. In this case, it is not causing a problem, but for higher gain filters the situation would be different. 12.3Using Eq. 12.1 : V1s( 29=G1VSs( 29sC2+G2(29s2C1C2+sC2G1+G2(29+G1G2| IS=G1VS-V1(29IS=G1VS1-G1sC2+G2(29s2C1C2+sC2G1+G2(29+G1G2 =G1VSs2C1C2+sC2G2s2C1C2+sC2G1+G2(29+G1G2ZSs( 29=VSIS=R1s2C1C2+sC2G1+G2(29+G1G2s2C1C2+sC2G2=R1s2+swoQ+wo2s s+1R2C1wo2=1R1R2C1C2| woQ=1C11R1+1R212.4r1=2260; r2=2260; c1=2e-8; c2=1e-8; wsq=1/(r1*r2*c1*c2); n=r1*[1 2/(r1*c1) wsq]; d=[1 1/(r1*c1) 0]; bode(n,d) 12-380 50100150200Frequency (rad/sec)Gain dB10-110101102103104105106-200-150-100-50Frequency (rad/sec)Phase deg10-11010110210310410510612.5G1VS=sC1+G1+G2(29V1-s KC1+G2(29V2= -G2V1+sC2+G2(29V2VO=KV2| VOVS=Ks2R1R2C1C2+s R1C11-K(29+C2R1+R2(29+1wo=1R1R2C1C2| Q=wo31-KR2C2+1R1R2(29C1ForR1=R2=Rand C1=C2=C, wo=1RCQ=wo23-KSKQ=K3-K12.6wo=1R1R2C1C2| SR1wo=R1wowoR1=R1wo1R2C1C2-12R1( 29-32=R1wo-12woR1= -12SR1wo= -12| By symmetry, SC1wo= -1212.7For C1=C=C2: wo=1CR1R2| Q=12R2R1| wo=12R1CQSQwo=QwowoQ=Qwo-12R1CQ2=Qwo-woQ= -1 | SQwo= -112-381 12.8As noted in Example 12.1, the maximally flat response corresponds to Q=12.For Q=12: AVs( 29=wo2s2+2wos+wo2| A( 29=1 | A jwo(29=12wH=wo| fH=1 kHzR1=R=R2and C1=2C2=2Cyields Q=12| wo=12R2C2=12RCRC=12p21000(29=1.125 x10-4| For C= 0.001 mF, R=112.5 k. Thenearest 5% value is 110 k. The nearest 1% value is 113 k. Using 1% values,C1=0.002 mF| C2=0.001 mF| R1=R2=113 k12.9Using the R1=R2=Rand C1=C2=Ccase, AHPs( 29=Ks2s2+swoQ+ +wo2wo=1RC| Q=13-K| Q=1K=2 | AHPs( 29=2s2s2+swo+ +wo2We need to find the relationship between wLand wo.22=2wL2wo2-wL2(292+wLwo(292wo4-2wo2wL2+wL4+wo2wL2=2wL4| wL4+wo2wL2-wo4=wL2=-wo2wo4+4wo42wL2=wo25-12| wL=0.7862wo| 2p20kHz(29=0.7862wowo=1.599x105| RC=1wo=6.256x10-6| ForC= 270 pF, R= 23.17 kThe nearest 1% resistor value is 23.2 k....
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This note was uploaded on 01/17/2011 for the course ECE 3040 taught by Professor Hamblen during the Fall '07 term at Georgia Institute of Technology.

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ch12_solutions - 12-378 CHAPTER 12 12.1a( 29C2=C=0.005mF|...

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