University of Michigan EECS 461_EECS 461 Problem Set 1: SOLUTIONS 1. Questions & Answers

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EECS 461 Problem Set 1: SOLUTIONS1 1. Consider a thermocouple that gives an output voltage of 0:5 mV/◦F. Suppose we wish to measure temperatures that range from −20◦F to 120◦F with a resoluti ... on of 0:5◦F. (a) If we pass the output voltage through an n-bit A/D converter, what word length n is required in order to achieve this resolution? SOLUTION: The range of temperatures that needs to be measured is 140◦ × 0:5 mV/◦F = 70mV, with a resolution of 0:5◦ × 0:5 mV/◦F = 0:25mV. Hence n should be chosen so that 70 2n ≤ 0:25 ) n ≥ 8:13: Since n must be an integer, a 9-bit wordlength is required. (b) Does you answer to (1a) change if the maximum binary value 1 : : : 1 is used to represent Vmax uniquely? Explain. SOLUTION: The voltage range is now divided into one less interval than previously, so the resolution is Vmax 2n − 1 : However, the answer is unchanged, as a 9-bit wordlength is still required: 70 2n − 1 ≤ 0:25 ) n ≥ 8:13: (c) Return to the scheme of (1a), and choose the minimum value of n that will achieve the desired resolution. The binary representation of −20◦F is equal to zero. What is the lowest temperature that has a nonzero binary representation, assuming unipolar coding without centering? SOLUTION: With 9 bits, the least significant bit represents 70mV 29 = 0:137mV: The LSB corresponds to 0:137mV 0:5mV=◦F = 0:274◦F: Hence the lowest temperature that has a nonzero binary representation is −20◦+0:274◦ = −19:726◦F. (d) How does the answer to (1c) change if centering is used? SOLUTION: With centering, the smallest voltage that yields a nonzero digital value corresponds to 1/2 LSB, or 1 2 70mV 29 = 0:068mV; which in turn represents 0:068mV 0:5mV=◦F = 0:137◦F: Hence the lowest temperature that has a nonzero binary representation is −20◦+0:137◦ = −19:863◦F. 2. Consider a rotating wheel with one mark painted on it, and suppose that the wheel rotates with a constant rate R revolutions/second in a counterclockwise (CCW) direction. Suppose also that we make a movie (as with a camcorder), which consists of still photographs taken at regularly spaced intervals of T seconds. We saw in class that if T = 1=R, then the wheel will appear to be stationary. Furthermore, in order that the movie depicts the correct direction of rotation of the wheel, it is necessary that T < 1=2R. Recall the movie from [2] in which the wheel had four arrows, 90◦ apart, painted on it, and suppose also that this wheel rotates at a constant rate CCW. 1Revised November 25, 2016. 1(a). What is the smallest value of T that will now make the wheel appear stationary in the movie? (b). What is the largest value T can take in order that the direction of rotation appear correctly? (c). Suppose that the camcorder shoots pictures at a rate of 30 frames/second. What is the maximum value of rotation rate R for the wheel with one painted arrow that will allow the direction of rotation to appear correctly? (d). Suppose again that the camcorder shoots pictures at a rate of 30 frames/second. What is the maximum value of rotation rate R for the wheel with four painted arrows that will allow the direction of rotation to appear correctly? (e). How do the answers to these questions change if the wheel has N equally spaced arrows? [Show More]

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