Name: SOLUTION Section: 8:30_Chang 11:30_Meckl ME 365 FINAL EXAM Monday, April 29, 2013 3:30 pm-5:30 pm LILY 1105 Problem Score Problem Score Problem Score Problem Score Problem Score 1 5 9 13 17 2 6 10 14 18 3 7 11 15 19 4 8 12 16 Sub. Sub. Sub. Sub. Sub. Total Score: /300 Problems 1 and 3 are worth 40 pts each, 2 and 4 are worth 35 pts each. All others are 10 pts each. Don t forget your name and please circle your section instructor. Exam is 15 pages including this cover page: make sure you re not missing any pages. If you use extra pages, indicate on the problem page that you re continuing onto an extra page. Pay attention to units. Explain your reasoning. Correct answer with wrong explanation = no credit.
2 Problem 1 An op-amp circuit is shown below: (a) What is the frequency response function (V o /V in ) of this device? Sol: (V in -0)/R 2 = (0-V o )/(Z f ) Z f = Z C1 //R 1 = (b) From the FRF derived in (a), what function does it perform (circle the correct answer)? It is a first / second order low pass / high pass / band pass / band stop filter.
3 (c) Sketch the magnitude Bode Plot of this device, given that R 1 = 800Ω, C 1 = 2µF, and R 2 = 80Ω. Sol: (r/s). (db) (d) What is the input impedance of this device? Sol: R 2 (or 80 Ω)
4 Problem 2 A bridge circuit shown in the figure is to be used together with strain gages. (a) If both Z 1 and Z 3 are wired to strain gages with a rest resistance of 120 Ω, while Z 4 is a resistor with resistance of 100 Ω, what is the impedance of Z 2 in order to balance this bridge circuit? Sol: Z 1 Z 3 -Z 2 Z 4 = (120) 2 Z 2 *100 = 0 Z 2 = 14400/100 = 144 Ω (b) If the strain gages wired to Z 1 and Z 3 are to be glued to a cantilever beam for force measurement (the configuration is shown in the figure), where would you glue the gages in order to obtain maximum sensitivity? (circle all that apply) (i) A and B (ii) C and D (iii) A and C (iv) both to A (v) both to D (c) In the configuration of (a) and (b), will temperature-induced strains have considerable impact on the output voltage? Why (briefly explain)? Sol: YES Z 1 Z 3 -Z 2 Z 4 R T From the numerator of the bridge circuit equation, the changes of resistance due to temperature effect are on the same side of the minus sign ( - ). This would result in maximum difference between Z 1 Z 3 and Z 2 Z 4 and hence the temperature-induced strains would have considerable impact on the output voltage.
5 (d) To calibrate a one-arm-active bridge circuit whose gage factor is 2, a resistor (R c = 1KΩ) is connected in parallel to Z 3 (as shown in the figure). What is the equivalent strain resulting from R c (when the switch is closed) if the rest resistance (R 0 ) is 120Ω? ((d) is independent from (a)-(c)) Sol: R c = R 0 R 0 //R c = 120 (120*1000)/(120+1000) = 120 107.14 = 12.86 (Ω) One arm active case: Gε = R c /R 0 ε = (1/2)(12.86/120) = 0.054 (in/in)
6 Problem 3: Given the following periodic signal: x(t) - volts 15 10 5 0-5 -10-15 -5 0 5 10 Time -Seconds (a) Determine the period and fundamental frequency of this signal. Sol: Period is 5 sec. Fundamental frequency ω 1 = 2π/5 rad/sec = 0.4π rad/sec (b) Calculate the dc component A 0. Sol: Since the average value of the signal is zero, A 0 = 0. (c) Is this signal symmetric? If so, is it even or odd symmetry? Explain. Sol: YES. It has odd symmetry, since the signal takes on values of opposite sign on either side of time zero. (d) Calculate the Fourier series coefficients (A k and/or B k ). Sol: Since the signal has odd symmetry, A k = 0. 2.5 5 ( πkt) ( πkt) 2 2.5 2 5 4 cos 0.4 4 cos 0.4 Bk = 10sin 0.4πktdt 10sin 0.4πktdt = = 5 5 0.4πk 0.4πk B k 0 2.5 0 2.5 10 10 20 20 = + + = = πk πk πk πk 20 ( 1 ( 1) k = ) πk k ( cosπk 1) ( cos 2πk cosπk) ( 1 cosπk) ( 1 ( 1) )
7 (e) What would happen if the time axis were shifted to the right so that time zero would now occur at the point corresponding to time 1.25 sec in the current plot? What is the symmetry now? What are A k and/or B k? Sol: The signal has even symmetry, thus B k = 0. ( πkt) ( πkt) k ( sin 0.5πk) ( sinπk sin 0.5πk) ( sin 0.5πk) ( 1 ( 1) ) 1.25 2.5 2 1.25 2 2.5 8 sin 0.4 8 sin 0.4 Ak = 2 10cos 0.4πktdt 2 10cos 0.4πktdt = = 5 5 0.4πk 0.4πk 0 1.25 0 1.25 20 20 40 20 = = = πk πk πk πk 20 ( 1 ( 1) k Ak = ) πk Note that this is the same coefficient value as in part (d). (f) What if the original signal shown on the previous page were shifted upward by 5 volts? What would the new Fourier series expression look like? Sol: The new signal has an average value of 5 volts. Thus, the new Fourier series is: k k= 1 k= 1 k ( ) A 0 20 xt ( ) = + B sin 0.4kπt= 5+ 1 ( 1) sin 0.4kπt 2 πk
8 Problem 4: A sensor measurement is given by x(t) = 1 + 2 sin 50t. This signal is used to modulate a carrier signal given by V s (t) = 5cos 5000t. (a) Sketch the amplitude spectrum ( M k ) of the modulated signal given by Sol: m(t) = 5(1 + 2 sin 50t) cos 5000t volts. Label the values of components in the spectrum and don t forget to label axes and indicate units. 1 1 mt ( ) 5cos 5000t 10 sin 50t cos 5000t 5cos 5000t 10 = + = + sin 5050t sin 4950t 2 2 M k (volts) 5 (b) Sol: If this modulated signal is demodulated by multiplying by the carrier signal, sketch the amplitude spectrum ( M k ) of the demodulated signal. Label the values of components in the spectrum and don t forget to label axes and indicate units. 2 1 1 d( t) = 25( 1+ 2 sin 50t) cos 5000t= 25( 1+ 2sin 50t) cos10000t = 2 2 = 12.5 1+ 2 sin 50t 12.5 1+ 2sin 50t cos10000t= ( ) ( ) 4950 5000 5050 frequency (rad/sec) = 12.5+ 25sin 50t 12.5 cos10000t 12.5sin10050t+ 12.5sin 9950t M k (volts) 25 12.5 (c) 0 50 9950 10000 10050 frequency (rad/sec) Explain what you need to do to recover the original sensor signal from this demodulated signal. Sol: The demodulated signal must be low-pass filtered with a cut-off frequency between 50 and 9950 rad/sec to remove the high-frequency components around 10,000 rad/sec.
9 Problem 5. A potentiometer is used to measure the displacement of a moving object. The static input-output relationship is V o = 0.04 x + 0.001 volts, where x is displacement in mm. The output voltage is read by a voltmeter whose display is shown in the figure. What is the resolution of the overall system? Sol: Ires = Ores/K = 0.1 (mv)/ 0.04 (V/mm) = 0.0001/0.04 = 0.0025 mm =2.5 µm Problems 6-7 A measurement system has a frequency response function: 4 17 4 34 Problem 6. Find the differential equation that models this system. Sol: Problem 7. If this system receives an input 2cos 3, what is the steady state output y(t)? Sol: G(j3) = 5/10 = 0.5, Arg[G(j3)] = tan -1 (3/4) tan -1 (6/8) = 0.
10 Problems 8-10 A 10-bit ADC with a nominal input range of ± 5V has a sampling rate of 1 khz and an aperture time (t a ) of 5 µsec. Problem 8. What are the ranges of the incoming voltage that could have resulted in the following codes? (a) 128 (b) 999 Sol: Q = 10/2 10 = 0.009765625 Volts Range of V in = (Q*code + V min ) ± Q/2 (a) [(Q*128-5) - Q/2, (Q*128-5) + Q/2] = [-3.7549, -3.7451] Volts (b) [(Q*999-5) - Q/2, (Q*999-5) + Q/2] = [4.751, 4.761] Volts Problem 9. If the highest frequency of the input signal is 120 Hz, how many bits are significant? Sol: n = 9.0512 (bits) Problem 10. Suppose you observe a 400 Hz signal at the output of this ADC. What does this tell you about the actual frequency of the incoming signal? If aliasing occurs, what are the possible frequencies of the actual incoming signal? It is not mentioned whether there s a low pass filter (anti-aliasing filter) used before ADC or not. If there is no filter, then we will not be able to find out the true frequency of the incoming signal. It could be 400 Hz, or any frequency = k*1000 ± 400 Hz, where k= 0, 1, 2... If an anti-aliasing filter is used and the cutoff frequency is smaller than f s /2 = 500 Hz, Then the frequency of the input signal is 400 Hz.
11 Problem 11. Given that p(x) is a probability density function. If p(x) = 1/32 when 3 x A, and p(x) = 0 otherwise. Find the value of A. Sol: A+3 = 32 A = 29 Problem 12. The heights of adults in the US follow a Gaussian distribution and the mean is 66.65 inches while the standard deviation is 2.8 inches. What is the probability for an adult living in the US to be at least 6 feet (72 inches) tall? Sol: 6 ft = 72 in (72-66.65)/2.8 = 1.91 = Z Q When Z Q = 1.91, probability = 94.39% (from the table in Appendix A) Taller than 6ft (100% - 94.39%) / 2 = 2.8 % Problem 13. The 95% confidence interval of the mean mass of a pollen grain after 81 measurements is [244, 256] nanogram (ng). How many more measurements need to be made in order to shrink the 95% confidence interval to [248, 252] ng? Sol: mean mass = (244+256)/2 = 250 (ng) Z Q S x /[81^(1/2)] = 256-250 =250-244 = 6 (ng) Z Q S x = 54 Z Q S x / [N^(1/2)] = 252-250 = 2 54/ [N^(1/2)] = 2 [N^(1/2)] = 27 N=729 How many more measurements are needed? 729-81 = 648
12 Problem 14. For a second-order system, give the range of damping ratio (ζ) so that the system has an overshoot in the output response when given a step input but does not have a resonant peak in the magnitude Bode Plot. Sol: 0.707 < ζ < 1 Problem 15: Shown below is the frequency response of a first order system. The system input is in psi, the system output is in Volts. (i) Calculate the sensitivity of the system (K). Sol: 20 log 10 K = 20 => K = 10 Volts/psi (ii) Calculate the cut-off frequency and time constant of the system. Sol: f c = 200 Hz = 400π rad/sec = 1256.64 rad/sec τ = 1/1256.64 sec = 0.796 msec
13 Problem 16: G 1 (jω) G 2 (jω) G 3 (jω) x(t) v 1 (t) v 2 (t) Filter Amp. Tape Recorder Z in = 10 5 Ω Z in = 10 6 Ω Z in = 10 4 Ω Z out = 500Ω Z out = 1000Ω (i) What is the frequency response of the system relating v 2 (t) to x(t) taking into account the effect that connecting the tape recorder has on v 2? V G= 2 = G 1 G 2 L 12 L 23 = G 1 G 2 X G = 0.909 G 1 G 2 6 4 10 10 = G 6 4 1 G 2 (0.99995) (0.9091) 10 + 500 10 + 1000 (ii) What would you change to ensure that loading effects were negligible? Sol: Increase the input impedance of the tape recorder or decrease the output impedance of the amp.
14 Problem 17: Johnson noise is a problem in a measurement because the signal being measured is very small. The noise, considered to be Johnson noise, is passed through a filter with gain as shown below: G FILTER 1.5 Boltzmann s constant = 1.38 10-23 J/K Temp. = 127 o C frequency Resistance = 25 10 6 Ω 20,000 (Hz) Calculate the noise power. Sol: Φ noise = G 2 Φ J = G 2 4k R T Noise Power = (20,000) (1.5) 2 4(1.38 10 23 )(400)(25 10 6 ) Noise Power = 2.48 10-8 volts 2 or watts Problem 18: When taking measurements near power lines 60 Hz noise can be a problem due to coupling between the measurement, circuit and the power circuit. Provide two ways of reducing this interference, one involving filtering, and the other involving a reduction at the source: (i) Notch filter the signal with a narrow notch filter centered at 60 Hz. (ii) Reduce interference with twisted wire pairs, shielding, and/or avoiding ground loops.
15 Problem 19: The noise power in a measurement x(t) = s(t) + n(t) is 2 3 2 n ( t) = 1.3 10 volts. The noise-free signal is st ( ) = 0.1+ 0.2 sin 8π t volts (i) Calculate the signal power Sol: (0.1) 2 + (0.2) 2 /2 = 0.01 + 0.02 = 0.03 volts 2 (ii) Calculate the signal to noise ratio in db. 0.03 Sol: SNR db = 10 log10 = 13.63 db 3 1.3 10 (iii) If the bandwidth of the noise signal is 0 to 10,000 Hz, how can the signal to noise ratio be improved? Sol: Low pass filter the signal, with a filter whose cut-off frequency is above the signal frequency.