BMEN_321_Fall_2008_example_exam

BMEN_321_Fall_2008_example_exam - BMEN 321 Dr Maitland...

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Unformatted text preview: BMEN 321, Dr. Maitland Practice Exam 1 ‘ September 27, 2008 This practice exam has three purposes: 1. expose you to my testing style; 2. communicate to you a NON- EXAUSTIVE set of topics that are important on the actual exam (I reserve the right and will test you on some topics not on this practice); 3. provide structure for the review day. The length of this problem set is not necessarily the same as the exam next Thursday. THIS WILL NOT BE TURNED IN OR GRADED 1. Short answer questions: a. For 0) = 411 [rad/s], f = [Hz] b. True or False (circle one): Chebyscheff filters are used for their flat response over the passband. c. True or False (circle one): A bandpass filter with high Q has a narrow bandwidth relative to the center frequency. d. True or False (circle one): Bode plots are a graphical representation of the magnitude and phase frequency-dependent response of circuits. e. Write the expression for the equivalent impedance, Zeq, for three parallel passive components: R1, R2, and C. : 2. Draw the following circuits, labeling all, elements, input and output functions and write out the transfer functions for each. Higher order circuits can use more than one op—amp. a. Non-inverting, 1St order LPF b. 2nd order inverting HPF 3. At time t=0, the switch in the circuit is open after having been in the current position for a long time. Derive and sketch the voltage across the capacitor, Vc(t),as a function of time. Make sure to include the time at which Vc(t) is 63% of its final value. 4. For the Bode diagram below, magnitude and phase - ..... KCu,%<>f‘f féequeficy 301cm 5 SloPe‘ ‘29§d§fl5l‘???§¢. a- -l-.........w.._._..1, . Gain ((13) Stopband Phase (degrees) 0.001 0.0 ‘l 0. l l 10 100 1000 Angular frequency (rad/s) a. Write the transfer function T(s) that is represented in the diagram b. Given that Vout = |T(s)| V0 sin (2 1r f t + (1)), where the input voltage is of the form Vin = Vosin (2 1: f t) and the magnitude and phase Bode plots provide 20 log |T(s)| and (1), respectively, find Vout for the following input signal (note the phase lag at fc is -45 degrees) Vin =10 sin (2 1t (0.01) t) + 10 sin (2 1t (1) t) + 10 sin (2 1: (100) t) Vout = 4. For the following transfer fianctions, sketch an asymptotic log-log Bode magnitude plot for the gain (G(f)=20 log |T(s)| = 2010g(Vo/Vin)). Make sure to identify key features like cut-off frequencies, the passband, gains at fc and maximum gain. Plot G(f), not G(w), where w=27cf S T 2100—— a- (5) (ea? .S 2 2 Tm: (2+1) b. (i 1 0,203+” 5. The infamous “Maitland” filter was first used in 2008 with no other known purpose than breaking students from wanting to calculate transfer function magnitudes from T(s) rather than simply taking the number off a plot, converting from dB to V/V and quickly calculating the effect of the filter on the magnitude of various input frequencies. For the filter function plotted below [07g {’42-} .*Tn(s). Move left to right across the figure and determine what Ti(s) will cause the break “III"IIIII IIIIIII.““II- "Hilllufl "u “In""IuHMI IIIIIIII'I IIIIII'I'I IIIIIIIII‘ II‘II'I'I‘ II‘I‘ 1‘1“! IIII‘IIII IIIHIIIIIII II :I.IIII IIIII‘II .- I‘“‘I‘IIII IIIIIIIIII III.IIIII.II IIII‘II‘II IIIIIII“ n ‘IIIII I... IIIIIIIIII IIIIIIIIII IIIII‘.IIII IIR'III“‘II IIIIII IIII IIIIIII III. 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IIIIIIIIIIIII'h‘II IIIIIIII'IIIII "IIIIIIIIII . IIIIIIIIIIIIIIII'.‘ IIIIIIIIIIIIII IIIIIIIIIII IIIIIIIIIIIII IIIII'III'I II....I'IIII IIIIIIII r I IIIIIIIIIIIIIIIIII nullmullillnll IIIIIIIIIIIA III-ll"-Ill-l-InIs-I-lll-Inl- I II II 1‘. I. IIII-IIIIIIIIIIIII .- IIIII-I-IIInn-J-nnu-I-I-III-I-u —- .I'll I I I ' n /O ‘A8 L{ 71C) a. (Yes, a little difficult) Write the transfer function that corresponds to the plot above. Remember that T(s)=T1(s)*T2(s)* . .. (turn) in the asymptotic Bode plot (HP and LP T(s) will get you there). Start with a higher order HP. b. F ill out the table below. You may assume asymptotic values for the magnitude, INCLUDING all cutoff frequencies. Now you see why they are called “comer” frequencies too — the log magnitude plot turns “comers” at these frequencies. All sin and cos functions are of the form A cos[27t(f)t] where A is amplitude in [V] and f is frequency in [Hz]. Neglect phase in all cases. (6 pts) f [Hz] 10 cos[21t(3)t] 1 C08[27‘C(20)t] 1 sin[21t(2000)t] 30000 100 cos[2n(30000)t] 100000 100 cos[21t(105)t] 6. The first figure below shows a carotid stenotic lesion. As blood passes across the stenosis, it accelerates and creates vortices, both of which result is higher than normal flow rates (frequencies). The second line of personal health, the family doctor (you are the first), listens for high frequency components as the first diagnosis of this disease. Imaging will confirm. You, an engineer, decide that the frequency spectrum of the flow, as measured by an acoustic microphone is the way to go for diagnosing if these high flows are present. Pieces of plaque can break free, travel to the ' brain, and block. blood ' vessels that supply blood to the brain (‘2‘ fifififiz‘; , if”; 5:. For the flow below taken as a voltage off the microphone, draw a block diagram of the circuit that you would build to reduce the high frequency components to a 0-5 V DC voltage that can be used quantitatively measure the magnitude of the high frequency element. The peak voltage on the plot below is 1 mV. The period between major peaks is 1 second. What frequencies is the higher component in the time-based signal below? 10-100 Hz. ...
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