EXAM_1_Practice_Solution

EXAM_1_Practice_Solution - ME 370 Practice Exam 1 Fall 2010...

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Unformatted text preview: ME 370 Practice Exam 1 Fall 2010 ME 370: Engineering Measurements and Instrumentation Fall 2010 Practice Exam 1 Name: 52 15*" in Section: fl /‘ NOTE: T his is a practice exam, it, by no means, should be identical to the real exam. [2‘ is solely used for practice purpose, provides some general idea on the real exam, PLEASE READ THE FOLLOWING INSTRUCTIONS For the multiple choice and true/false problems, please WRITE IN your answer clearly and legibly in the indicated space for each problem. Please write down justification or calculations to support your answers. For problems involving calculations, you may not be awarded points if you do not show your work. This is a close-book, close note exam ~ you are NOT allowed to use the assigned course textbook. You are allowed a single 8.5 " x I I " sheet (both sides) of notes. The note sheet has to be handwritten and to be made by yourself Photocopies or printed version is not allowed. You must attach your note sheet with your exam and turn it in. By writing your name on this sheet, you agree to abide by the honor system and uphold academic integiity. WARNING: Cheating on the exam constitutes academic fraud and will result in an automatic failing grade for the course. ME437O Practice Exam 1 Fall 2010 Problem 1 (25 points): A multimeter is used to measure the voltage across resistances shown in circuit diagram below. We know that the driving voltage Ei= 5 V, R, = 10 M9, R2 = 15 M9, R3 = 25 M52. The measurement voltage between A and B is 2 V. (A) What is the loading impedance of the voltage meter used? (10 points) (B) If we use the meter to measure voltage across B and C, what is the measurement voltage? (5 points), and what is error percentage over the true value? (5 points) (C) What is the minimum loading impedance of the multimeter if we want to measure the voltage across B and C with a maximum error of 1%. 122Jll7—L- : HMS”? M0.“ 3 l. 0§QY1Gm :3 (at; (034514“. ME 370 ‘ Practice Exam 1 Fall 2010 Problem 3 (25 points): Consider the signal y(t) = 5 +10 COS(30t) + 15 cos(90t). Determine (A) The frequencies (in Hz) contained in the signal (5 points) (B) The minimum sampling rate (in Hz) to avoid aliasing (5 points) (C) The frequency resolution of the frequency spectmm if the signal is sampled at that rate for 2 seconds. (5 points) Finally, sketch ‘ (D) the amplitude~fiequency spectmm of y(t) (5 points) and (E) the amplitude—frequency spectrum if the signal is sampled at 20 samples/s. (5 points) rm WW . "o r 9° 4-!” Sabin“ 3 'Cl'L HQ: 3 I»: ’5‘? 2 T‘TTHE— d3) 4-572?me 3 27% HQr t6) «Magma, mam“ ‘“ re=‘% @ “We 93% = 7;» ‘ w NI ‘° “Va. “7% (E) BQCMM. ~95 :LBYl'E— (241mm appew avaiqu .Po-v:_p‘ ~ 417‘“?! =9 Ra,me c. a.“ ‘ s ‘5 t“ e = A t w the MW gm?!” ’ +1“ 935"" WP’M’ 4" 5‘ :fiét n; ’35 3 5H5 Outfit) {- W30 22er t) i m IS‘ (0 I“ > b «3711 bet 9 2 ME 370 Practice Exam 1 Fall 2010 WWW Problem 3 (50 points): If the output signal from a transducer is Ea = 0.1 sin(500t) V. This signal is pass to a signal conditioning device, which is shown in figure below. Finally, the output signal is measured by a 12—bit data acquisition card, with a sampling rate of 10 kHz, the range of the data acquisition card is —12 V to +12 V. Here R = 100 kg, Rgain = 50 k9 (A) Determine the quantization error if the device is to sample the original signal Ei(10 points) (B) Determine the output signal E0 (15 points) ' (C) Determine the quantization error if the device is to sample the output signal Eo(10 points) (D) Assume the input signal is E, = 5 Sin(27€fi) V, and the Op-Amps with a slew rate of 0.01 V/ms, are driven with a power supply of E V. Determine the maximum frequency that the system can handle properly (15 points). ': V'L gbtufr‘bblll a ‘ Q 0 -~ 2,, n. M) Qua—imam We 1617B; 1’ 2 i i 3* ~ _ .321 a“, 2' v a H, MUM-fie : I in? *507. z gxiob'zail‘tJ-Z I 0 902-3 \/ Us) Tim‘s :i a Low Wm 10M” balm ii lml‘rwwfi‘he Mr ‘ew Fu-r La? am; was“ Tet/m . l M Of) > I f 3‘ M. M“. m 3 k! H‘ eta" ulfitg‘of a) we) m ‘PG': fiat 22> ode-=74; ELé 7) [We—9): b :ofiéneg t‘ H (Per l$lo3>< (MK/0’6)" . ~ --3 _:E‘_ - - ($665,. a.“ Leg) j, " w'fi-g: -- _vtw (gel/3) L—OJN‘ PM Em: \ - a E t H 3—35 «I Iv. > wo‘o‘HQK 3w} cswt’ a“) : b 9-3“) ’9 ) 5 M q \ I «0&3? 1+ 9» L eut- om) ‘ but __3 .01 N‘bo a Wm M Wm Lu v 4 W79 04.?3wX2, mumccf) gdig’m‘t 'l‘d’fl'i} —.: 9°39“ gnu-eat ‘ 0&6") a» Eon-Me?)- 4w— A. méwfl 67)] . ‘ l ,r— Mes-3.: “(L-:1 GM 3 l'f Egg-M; 3; A, N H Pi)" ’ =9 130», .w Mw- saiwr’v WW] ale. 9L9“ rm '5 “Egg”: — M’Ml—h-le. oDSZIgft- >9 ’Mmm-wgf» < \D c) ‘F < omé'arHZ' de shah: Chg-4.615 $151M! ...
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EXAM_1_Practice_Solution - ME 370 Practice Exam 1 Fall 2010...

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