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Exam 1 Form B Solutions

# Exam 1 Form B Solutions - EE319K Spring 2011 Exam 1B...

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EE319K Spring 2011 Exam 1B Solution Page 1 Jonathan W. Valvano February 25, 2011 2:00pm-2:50pm (4) Question 1. The basis elements are 1000=27, 0100=9, 0010=3, and 0001=1* 27+0*9+2*3+1=34 (3) Question 2. Answer true/false for each of the following three statements Part a) True, the stack pointer (SP) points to the data on top of the stack. Part b) False, the order in which I add the numbers does not affect the final value of RegA. Part c) False, dropout error cannot occur on a logical left shift (e.g., lsla ). Overflow can occur. (4) Question 3. Consider ldab #-6 subb #251 Convert to signed, 251 = 251-256 = -5. Subtract two signed -6 - -5 is -1. This fits so V=0. Convert to unsigned -6 = -6+256 = 250. Subtract unsigned 250-251 is -1. Does not fit, C=1. (4) Question 4. What is the binary representation of 8-bit signed number -11? Method 1) +11 is 8+2+1 or 00001011. Negative is 2’s complement. Complement 1111,0100, then add 1. 11110101 Method 2) Look at basis elements, need -128,64,32,16,4,1, so 11110101 Method 3) -11 is the same binary as -11+256 = 245. 245/16=15 remainder 5. So hex is \$F5 (20) Question 5. The current through LED resistor 25mA = (5-2-0.5)/R. Solve for R= 2.5V/25mA = 100

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Exam 1 Form B Solutions - EE319K Spring 2011 Exam 1B...

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