Assignment 14 with Resolution | Analog Electronics | ECE 3050, Assignments of Electrical and Electronics Engineering

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ECE 3050 ~ Fall 2002 Page 1 ignment No. 14 18.30 c, Co lar l LaF 10 ka / roo ke ima IMO} 10kQ ls 1 rad 1 rad a, > = 0,909 Wg = = 58.5 110 (100kR + 1MQ) 3 | % 10°6(10k0 + 25kNILOkAfIMoa) 8 Separate widely spaced poles > ff =f, = 58.5 =9.32 Hz P L Qn 4.15 kQ 25kO[ }LMQ| | 1Oks| | 10k 2 of = 1 TaCy Ct (oo; opr +spr{+2ms(s.15%0) +8258) melt Hn (90.9kQ)(38. OpF) Vgu = 14(100KO}IMQ) = (90.9kQ)1, |v, =-(2x10)v,,(25kQ Oka okapMa) = 46.1 kHz As oa = +(2mS)(4.15k0)(90, 9kQ) = -7.55x10°Q | y%, =-10%s 1+ AB = 1+ (-7.55x10°Q)(~10°$s) = 1.76 {l= Ve =5.29Hz fy = 46.1kH2(1.76) = 81.0 kiz 18.32 2x10!4 q? {2x x10°)(2x x105) 5x108 fa} Als) * s $ = s 5 1 1+ 1 1 ( * on xio® \ on sar) ( * Sa x10? t * ox sa) A(s} represents a low - pass amplifier with two widely - spaced poles Open -loop: A, =5x105=114dB | f,=0 | fy +f, =1000 Hz (5)A common mistake would be the following: Closed - loop: fy = 1000Hz[1 + 5x10°(0.01)] = MHz Oopst - This exceeds f, = 100 kHz! This is a two - pole amplifier,