Lec 30 - 5.74. The input signal to the integrating...

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Unformatted text preview: 5.74. The input signal to the integrating amplifier cir- cuit in Fig. 5.4? is the triangular voltage shown in Fig. P534. The integrating-summer parameters. are: HI = 15 k3}: CI = {Jfilénfi and HI = 15 V. Assume in. is zero when r is zero. [a] Derive the espressien fer up when U 1: t E 25 ms. {1]} Derive the expressien fer nfi when 25 ms 5 r :2 F5 ms. {e} Derive the expression fer n“ when T5 ms 5 r 5 100 ms. (Ll) Plot 11, vs. t for {J 5' r E MD Ins. nhrw] .l I EQIE. sen [3-3415 25 —EUD Figure P534 r5 .7 % luxmg‘afiék‘iyk R5 :T'SE’LQ“ 2': 833.35 C5; =- . “an ESQ? » 7 1°" 0;). C: 1:: «F... as“? Data = 2% v L. Va: 7: "833.535 (l,z_2_i.\)()c£}x —-Cc-Z-5 ‘ .025 = ' X573: ~'-' “me x t +Za,ocoxz‘\ #5,.2‘3 “.025 E xzpz-S = {0,0mt’1— zthaW fifty/ms UDCSOY‘MS} ’= ~425V Cw.wa \fde V1§$Q$JJC0> We: (79ws) .«_- _{D.ELS V m = #83333 (—2-H+24x)cfix $59.25 .015 - —-lb oth+-‘ZOQO‘E -- looU J 75§:té_— wows I —. so 15 “’0 tfiw<.\ was MQMM M.o%.o<a L. -=_ 16H Q = (.(peeemuF bit) = -u eq‘mfi zoe’qm v,t7/o_ 3‘ == -1oo hath/5 $L= "fCOMA/s S1+$L = -Zu. =-5oo ‘4 =750 “mg/‘5 first = aha—sz =3oo ' (535’: 2002 = J— 0;: eat at he. ($1+§—+'L3_ “RC. LC» -0 = L 0L 2R; new '9qu Co-"a I‘M/5 Low \ TESMM+ W-Cfltw (0 l7 ‘SLJL MA/s z Z. (s +2u¢§ +w¢ 7:3 SI=_K+sz-m:h1 {WW I mtg/.5 $7.: -°'L" Noel—w} Cat's 5‘1: sit ULC— . 5500': UM / dawd 2. - +' ._ Md SOOOW/s 310m”, A WqMA Ovacmpecfi 5&7 Lug?- Ufl&ef‘c‘awq>ec0 Lu: 7 «‘2' Qowflex Cami (5.50.12, Vcd'kto [DA -=- \Smoz __ 0L2. Com VoHaqes, qumwk M ‘HVS RN” IWH‘ \cJ ComA§+\éM$: (-6 = O+> FNM Qwun-r 17mm 63mit'ov‘ U‘Co+\ Bl 0.5g; (0+3 -- (£5, + («9432 $®\\)€. £9? $‘ )B2 3 \Am: Ms T; .—. ESL 3‘: .— My» éLl Cw: VG: 'OV 'I°= EOMA .. _J_ ._., ‘ - 62.552. @ 0" "igc = Zocg 2(‘amoXlo‘7 _. __ .19.. =—.24A " *— e25 $20% = = —--.'Z""(~IOQ C. 5|? L, " L. .ot ~83: Ag A \oao = ——(%eooX.o%\ +Ltaooo\B4 ‘3” = .Z735A ‘ ~3°°°t( «38cm amt +£12.39“ moot) A, t 7/0 ...
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Lec 30 - 5.74. The input signal to the integrating...

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