ESE_305_Notes_wk_3 - Let fl = 21121533 then.119 =-2 5...

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Unformatted text preview: Let fl? = 21121533.; then .119 = -2 + 5 cmfoZnfiJ9 - (II) - 2.4 cmfflonfiJ9 - 4)) or .119 = -2 + 5 msflma t- 4)) - 2.4 cmfymaf- (I9 ..F.ourier series: .119 =01: + a: [11$me 9 + a2 cmfzma 9 + . . . + in fl'flftflm + £12 fiflfE-ma9 + £13 sfflféma9 + . . . Here m? = -2 ; a2 = 4 ; 92 = 3 ; a9 = 4.4 ; 31 all others are D. Prob. 1.10: From prob. 1.9: XIHHI-H a“ MU : xiii—1H} Fig. 1.34 a]: Prob. 1.22: .119 = 2 + sfan.4t) - 4 castJ 1) Periodic: yes, since 1.4=2[D.T) & 2.1=3[D.T). .So..fundamental angular frequency.r is on = 0;?" and fundamental frequencyr is fi = e. M211) and fundamental period is To = (says. a Sines: 1.19 = 2 + sfan.4t) - 4 sinfzJ t + m2) = Zsfnfm) + 54111141) + 4 .5411er f- m?) Frequencies: of; 1.44211); 114211;); Amplitude s: 2; I; 4; at, Phases: -m‘°2; 0; so”? Cosines: .119 = Zoosffl) + cmfl.4f- W) + 4 fianJ t- it) Phases: 4:3; 3; m2 Complex Farm: Mt) = :1. firm (amt?) ~ .1..4_m(%,t) . amplitusiez Phaae; -.:-i Memorylsss systems: R1 up} _.|_ R2 wt)- o + 5mm Wt) 212 + fun 0 + wit} it fit) Dne input, one output. [linear]. Linear: 2 inputs, 3 outputs. anlinear Systems with memory: Linear system State: capacitor voltage; Hit} Zerosstate response: output yffl corresponding to input aft} where the initial capacitor voltage is zero. "Memoryless" component: operational amplifier [op-amp} Behavior: 1'. = fl II+ = fl ' ' yfl} =Afv+ffl - “(0) :‘L :L i where Areal and A is uerylarge. ..Ide.al.op-amp: A. —r .90 .and so. ”+139 = up), for...linear operation. |[Zip-amp examples: Since mfg = “(9, v1 = fl : and sum of currents into node [1] = D, so that (erg-eym + @(y-fl)flr = fl, and then y(9 = -(Hffifi1)fl(9 C + I _ ' + IIV 'ifljndflfl ifv'ti'flf'="¥f'I/'ffl Cflflffi ‘3?de = -(L’(HC3‘)flfl} ...
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