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3-frequency_dependent_circuits

# 3-frequency_dependent_circuits - Frequency Dependent and...

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Frequency Dependent and Nonlinear Circuits Prof. Greg Kovacs Department of Electrical Engineering Stanford University

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EE122, Stanford University, Prof. Greg Kovacs 2 OH GREAT! MORE OP-AMP CIRCUITS! OBJECTIVES (Why am I sitting in this classroom?) • To gain insight into op-amp application circuits beyond those considered in the Intro to Op-Amps. • To understand the basics of analog filters. • To understand comparator and Schmitt-Trigger circuits. • To understand some linear and nonlinear oscillators.
EE122, Stanford University, Prof. Greg Kovacs 3 WHAT ARE FILTERS GOOD FOR? • FILTERS AMPLIFY OR ATTENUATE COMPONENTS OF SIGNALS ON THE BASIS OF THEIR FREQUENCY • EXAMPLES: AUDIO EQUALIZER VARIABLE GAIN OR ATTENUATION IN SEVERAL BANDS AUDIO BASS OR TREBLE CONTROL VARIABLE GAIN OR ATTENUATION WITH ONE BAND OSCILLOSCOPE TRIGGER CIRCUIT ATTENUATION OF HIGH OR LOW FREQUENCIES TELEPHONE VOICE-BAND FILTER LIMITS SIGNAL FREQUENCIES TO VOICE BAND ( 300 - 3KHz) TELEVISION INTERFERENCE FILTER LIMITS INTERFERENCE FROM CAR IGNITIONS, ETC. COFFEE FILTER KEEPS GROUNDS AND DIRT OUT OF YOU

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EE122, Stanford University, Prof. Greg Kovacs 4 THE BASIC TYPES OF FILTER • REAL FILTERS DON'T HAVE "BRICK-WALL" RESPONSES! • YOU CAN IMPLEMENT THEM AS PASSIVE (RLC) FILTERS, ACTIVE FILTERS, OR DIGITAL FILTERS.… • WE WILL CONSIDER ACTIVE FILTERS - USING OP-AMPS.
EE122, Stanford University, Prof. Greg Kovacs 5 REVIEW: FIRST ORDER LOW-PASS V V in out R C 0.707 V - 45° 1 KHz v out v in = a 0 S + ϖ 0 ϖ 0 = 2 π f c = 1 RC

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EE122, Stanford University, Prof. Greg Kovacs 6 REVIEW: FIRST-ORDER HIGH-PASS R C V V in out 0.707 V + 45° 1 KHz v out v in = a 1 S S + ϖ 0 "ZERO AT ZERO" POLE AT o ϖ 0 = 2 π f c = 1 RC
EE122, Stanford University, Prof. Greg Kovacs 7 POLES, ZEROS, AND CIRCUS TENTS? T S = A S B S = S - Z 1 S - Z 2 S - Z m S - P 1 S - P 2 S - P n LPF S = A S + 1+4j S + 1-4j + j - j

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EE122, Stanford University, Prof. Greg Kovacs 8 EXAMPLE: 2nd ORDER HIGH-PASS TWO ZEROS TWO POLES HPF S = S 2 S + 1+4j S + 1-4j
EE122, Stanford University, Prof. Greg Kovacs 9

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EE122, Stanford University, Prof. Greg Kovacs 10 THE BASIC 2nd ORDER RESPONSES LPF S = A S 2 + S ϖ o Q + ϖ o 2 HPF S = AS 2 S 2 + S ϖ o Q + ϖ o 2 BPF S = AS S 2 + S ϖ o Q + ϖ o 2 THESE ARE YOUR MAIN BUILDING BLOCKS!!!
EE122, Stanford University, Prof. Greg Kovacs 11 WHAT DOES "Q" DO? 6 7 8 9 1 2 2 3 4 5 6 7 8 9 1 2 3 4 Q > 0.707 Q < 0.707

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EE122, Stanford University, Prof. Greg Kovacs 12
EE122, Stanford University, Prof. Greg Kovacs 13 POLES OF 2nd ORDER SYSTEMS LPF S = A S 2 + S ϖ o Q + ϖ o 2 Q = 1 2 P 1,2 = - ϖ o 2Q ± j ϖ o 1 - 1 2Q 2 POLE DISTANCE FROM j ϖ AXIS = ϖ o 2Q FOR LEAST PEAKING ("MAXIMALLY-FLAT")

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EE122, Stanford University, Prof. Greg Kovacs 14 CASCADING FILTERS .... 1.0h 100h 10Kh 1.0Mh Frequency V(2) V(3) V(4) V(5) V(6) V(7) V(8) V(9) V(10) V(11) 1.0V 1.0mV 1.0uV 1.0h 100h 10Kh 1.0Mh Frequency V(2) V(4) V(6) V(8) V(10) V(12) V(14) V(16) V(18) V(20) 1.0V 1.0mV 1.0uV BUFFERING BETWEEN STAGES KEEPS THE POLES FROM INTERACTING ....
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