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HW1sol10 - ECE2280 Homework#1 Spring 2010 1 Given Vg=10mV...

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Unformatted text preview: ECE2280 Homework #1 Spring 2010 1. Given Vg=10mV, find V0. Find the Thevenin equivalent between terminals a-b. (Note: v1¢ Vg) I 109 5!) First Jr‘md VH‘ZVG Va: + {2(202 = boil 20+(H01WO) ”0 ~5V‘ (20) , \oov 10+20 {Agg m*b Vo QQWVhWK \t (—55.30% m) M “5“ W Von-Vh ” 'QfimV 1. Given Vg=10mV, find V0. Find the Thevenin equivalent between terminals a—b. (Note: v1 ¢ Vg) IOQ SQ 1. Given Vg=10mv, find V0. Find the Thevenin equivalent between terminals a-b. (Note: v1 ¢ Vg) 10!) SQ 2. Sketch the following waveforms. Identify the dc component of the waveform and the ac component of the waveform. a. Vs=5cos(20t) V DC component = 0V; AC component = 5005(20t) b. Vs=7V +7cos(21rt) V DC component = 7V; AC componenF7cos(21ct) c. Vs=8V i 1V DC component=8V; AC componenF 1t 1V (a) 500890!) 1'=(2-pwzo=.3142 (b) 7+7cos(2~prt) l T=(2'Pi)l(2'PD=1 (0) 9 8.8 8.6 Z: N A I, V l 7.4 \ 7.2 3. Explain in your own words the procedural steps for plotting Bode Plots. (Note: I would prepare this question for use during an exam) ?rocedufal $41295 423/ Each Q’lo‘V‘S. ' ' ' o_._....._. i. Determine. ‘H'le FMS ana W Zeros é? Marmot flu. SIS/1 a Hafiz fifiugz “1+ 0 magic Mion‘flv- mifwl“ We}. 3 Dana: 19¢ 6‘ Hugh \o‘\'-’ I a8: ‘HUL Shr‘hh S‘lar‘i 44%me Pop. beswnw '9qu or $30“- oar-C) " (1+ and- Zero 091de320 dB klmée and a‘i- each \e Subi'moi' 2043 dcméc. (e {as-rt: ordef Pgi‘hzrmines hop: mama/(:dglucqh 3:9. «aw-“9* Or gdb+f chad CO‘Q'MUE Clodfi ' +le $lope mail reac Lung 4M anal o 4;; if???) 315 ‘1' Drum ‘Hu. phasz pb‘r. S-hzr’r VCX‘UUL‘; C30 '19 abnsi-qn'is >0 350‘, 5-? consihn+5<o ‘ ‘ mo" Qor W «zero whore“ ”‘15 53c: eggk PM or? Marla?“ cl. \e/z-c o c n'i‘r bytes 0» 45 (mercena- if?“ 501831.; 89mm T3: 2.932932% M EL: mlaE/ 35m finA‘Hu anti“; {.1EQ341n-flkr Habenrfifiszpra. Use the following equation for problem 4 and 5: 2 +. H (s) = 000(s 5) s(s +10)(s + 50) 4. (a) Plug in values of (0 from 0.1 to 105 rad/sec to obtain the magnitude and phase plots. (b) Sketch the Bode plots using a straight-line approximation (procedures described in class) Standard Form: 2000(0.5)[—s— + 1) 0.5 S S Magnitude when 03=0.1: H(s) = 2000(0.5)(%%’ +1) H<jw>=—————~_ 2900‘”? '5) or “WV—WT Jw(Jw+10)(Jw+50) (10)(50)]w(fi+1][§+1] 2 2000*0.5*[ fl +12] |H(01j)|_ 2000!Jo.12+0.52! or |H(011')|‘ 0-5 \/0.12+/0.12+102+/0.12+502 2 0,12 2 0,12 2 (10)(50) 0.1 — +1 — +1 10 50 |H(0.1j)| = 20.4 = 20 * log10(20.4) z 26dB Phase: (recall that £(jw) = 90°) 4H(jm)=M=m=i(o)+m=l(g_)_9oo —tan=‘[fl)—tan=1(m] 4w +410; +10) + 40a) + 50) 0.5 10 30 At (0=1: 4H(jw) = tan=l (0)+ tan=l i) + 90° — tan=1 (i) — tan=1 (i) = 0.5 10 50 %Matlab code: >>atand(0)+atand((l/0.5))—90—atand(1/10)-atand(1/50)=~33.4degrees Straight—line approximation: (See matlab graph) Magnitude: Plot a beginning value for the magnitude(See above) m=lo ale Phase Chane -_E_ 0.05< 00< 5 01:5; zero +45°/decade 1< m< 100 01:10; . 1e —45°/decade 5< w< 500 ((0:50; pole) Magnitude Plot (in dB) phase(degrees) 2 100039892 01 2039501754 -—-mzzm 2175965693 1739990883 448.01 199740506 0.199997 001999999 0.000 ——73 979 , 4 l Phase Plot (in degrees) I 1 o - ‘20} 0.01 0.01. 01 1 10 100 1000 10000 4 ~40 : l -60 5 g —100 i ’ -120 ‘~ r 2000‘(SQR‘HAZ"2+0.S"2))/(SQRT(A2A2+10*2)‘SQRT(A2"2+SO"2)J 1 "4° ' ' } EI—_- "6° i7 ‘ -——— 48° ‘ __- -200 i DEGREEsuATAN(O)+ATAN(A2/D.5)-ATAN(A2]10)-ATAN(A2/SO)))-90 , ‘ 5. Using the equation from problem 4 above. (a) Use Matlab to obtain the Bode Plots. Matlab code: >> s=tf('s'); >>H=(2000*(s+0.5))/(s*(s+10)*(s+50)); >> bode(I-I) , Bode Diagram 60 '7777777777 1 17777T7'7 7 777777 7 777 7T7 7‘ 7 ' ' 7' '7 ' 7 7 7 77 ' 1777777 '7‘7 7'77 7 7 ' ’7 1 ' '1' 1T" . 1 ‘ gavel .1 3.9. l 1 1 11 “$055K ‘f’w ‘ OAQAtC‘" ‘ 1253599 1 1 1 ‘3» 4° \‘\\‘%33 " ' ' ‘3 j' ""fifiaé " ' ‘ ‘1'” " ' " / “I, Magnitude (dB) 0 j / l ‘1 [I I \ l ‘ v 1 3 .40‘ ‘ ‘ H ‘ ,7 ‘ H 77\»1 l 1 ‘ ‘ ‘ ‘ ‘ I 0 ‘ “ ‘ ‘ ' ‘ ‘ \ -50 71141914,,LL%, , ,4, ,7 [yr f7) , [ 111,4 ,l_l .1 , , H, ; _ g A, r 0 77777 777 ‘77: \bgt°§g’ ‘7 7“ M7 (20167777; 1 77777 7717770," 7777 7 777777; ‘ ‘ 3 Huey? : 37m, «1 ,qoo/Accodmv :; .45177 7, t j &6\// . H Y "Q\\\ . . : V .‘i ’7 ‘ ‘ ‘ ‘ ‘ ’ ‘ ‘ ‘ ‘ ‘3 1 ‘17‘\ f ‘ ‘ ‘ ‘ 1'8, 1 I 1% 0 ‘ I ‘ I : \K ‘ 0 h ‘ ”55.0.6?" .1, -90.—.5;_—:::_4r7 1 g ‘1 H 0 r 0 33: §\; ‘ ’U“) ‘ 43577 1 \{\ 1 l 1 j 3 \“\\§ \ -180L1, 1,1,17,§_,7, , 11 707.174; ,, :17 71 lg :71. ,, z .7 a 1. 171717;»,9iwrwhhw.w’ _ __ __ ,,. 10‘2 .05 10‘1 .6 10° 5 1 50 102 500 103 __———l—/-———— \M dFreque cy (radg‘ec) 10 7‘75“" W‘ l we" Shawl {(0. l , 1 L190 Slave I decade (b) Compare the 4(a), 4(b) and 5(a). What drfferences do you see? The shape of all the plots look the same, the exact values differ dependent on the frequency. 6. Sketch the Bode plots for the equation below using a straight-line approximation {procedures described in class) and than use Matlab to obtain the Bode Plot Compare the two. '3 Hm I . .0,000[s +10) s‘(s+100k)(s+tk) Straight-line approximation: 20,000 * 10 * (1 +1) 1. Put equation into standard form. H (s) = 10 S 100k*1k*s2( S +1)(—+1) 100k 1k 2. Start by finding the locations of the poles and zeros. {10, 100k, 1 k} 3. Pick a value below the first location that creates a change. {c0=1} 4. Find the magnitude from the original equation for this value. 2 20,000 * 10 * iii +12 |H(1j)[ = — = 0.002 2010g10(0.002)=—54dB / 1 2 12 100k*1k*\/12 ~— +12 — +12 100k 1k 5. Create tables for the magnitude and the phase changes 6. (a) Plot the magnitude by starting at the value found in (3) above. If there is a pole/zero at the origin, place a slope through the start value(+20dB/decade for each zero and -20dB/decade for each pole). At every magnitude change, add or subtract the change from the current slope value. (b) For the phase plot, determine if there is a pole or zero starting phase. This is determined by the number of poles/zeroes at the origin. For each one, there is a +900 shift for each zero or a —90 0 shift for each pole. The phase change occurs one decade before the pole/zero to one decade alter the pole/zero. Note that this is NOT continuous and only has the phase change happen for this range of frequencies. M y 'tude Cha _e Location Range Phase Chane Sloe is 2 * —20dB/decade s at the oriin ole 2* -90°) 03=l(from above) -54dB l< c0< 100 ((0:10; zero) +45°/decade +20dB/decade 100< c0< 10k ((9:1k; .ole -45°/decade w=1k aole —20dB/decade 10k< co< lMe; co=100k; ole -45°/decade co=100k gpole) -20dB/decade See sketches created over the matlab figures below for the straight line approximation Matlab code: >> s=tf(‘s’) >> hl=(20000*(s+10))/(s"2*(s+100e3)*(s+1e3)); >> bode(h1) Magnitude (dB) Phase (deg) System: h1 Frequency (rad/sec): 1.13 ‘ Magnitude (dB): -56.1 OK £00K I requency (rad/sec) 1'0 Mogaw F W «- NS°S\ maecodl V0(s) 7. Use PSPICE to draw the circuit below. Derive the transfer function H (s) = V1( ) s 8. Use PSPICE to obtain the Bode Plots for the circuit below 9. Sketch the straight-line approximation over the bode plot of problem 7 and 8. 2 yd. v\\ Bqud @} Q, 88 $65?er 43 m2? fin: zocwswwnm $25 :ifiztanmezmu p finmaooa 283 $8 Nmsoom Nmeom 2%, SEE 5256 2:5; 10. Analyze the following circuit to find the transfer function Vo/V 1. Solve the circuit symbolically first (with R1, R2, and C1) and then plug in their values. Create a rough sketch of the transfer function using a straight—line approximation procedure. \/0 Whats) \oK(\+\\L(\h)s3 Jew ‘ ’ mc. + IOKOQS '(“K ”Cilia a? , II||||IIIII||||III|||||I|I|II||||||III||||I|I|II|||I|I| . 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