HW#8_Sol - (f[ACCD = A97A = 1010100101111010 after shift...

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EE 367 Introduction to Microprocessors (Fall 2011) Solution Set, Homework 8 7.9, The LDX #8F00 instruction produces [X] = 8F00 and the STX $FE instruction stores [X] in location 00FE and 00FF (i.e [00FE] = 8F and [00FF] = 00). The LDAA $FE instruction loads ACCA from address 00FE so that [ACCA]= 8F and LDAB $FF loads ACCB from address 00FF so that [ACCB] = 00. ACCD is the concatenation of A and B. Hence [ACCD] = 8F00. 7.15, (a) 8B 20 (c) 0A (l) CD AC 45 7.21, You can chnage first instruction to LDS #0286. Then interchange PSHA and PSHB so that ACCB gets pushed onto stack first. 7.43, (a) [0670] = 73 = 01110011: after shift – [0670] = 11100110 = E6, and C = 0. (b) [ACCB] = 7A = 01111010: after shift – [ACCB] = 00111101 = 3D, and C = 0. (c) [ACCB] = 7A = 01111010: after shift – [ACCB] = 00111101 = 3D, and C = 0. (d) Operand address = 0600 + 70 = 0670 [0670] = 73 = 01110011: after rotate – [0670] = 11100110 = E6, and C = 0. (e) [ACCA] = A9 = 10101001: after rotate – [ACCA] = 01010100 = 54, and C = 1.
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Unformatted text preview: (f) [ACCD] = A97A = 1010100101111010: after shift – [ACCD] = 0101010010111101, and C = 0. 7.93, (a) To achieve a 5ms delay, we want 5ms/0.5μs = 10,000 cycles Delay. 10,000 = 2+(Count * TL) Count * TL = 9,980 If we choose a COUNT = FA = 250 then TL = 9,980 cycles/250 = 40 Thus, we can add 17 NOPs (34 clock cycles) to the loop to achieve TL = 40. The exact delay will be: 2 + (250 * 41) = 5.126ms. (b) Refer to the second Program of Example 2.53. Using the X register: To achieve a 4ms delay, we want 4ms/0.5μs = 8,000 cycles delay. 8,000 = TE + (COUNT * TL) 8,000 = 3 + (COUNT * 8); COUNT = 999.6 The nearest integer value is COUNT 1000 = 03E8 (in Hex), the actual total delay will be 3 + (1000 * 8) = 8003 cycles = 4.002 ms. Using the Y register: 8,000 = TE + (COUNT * TL) 8,000 = 4 + (COUNT * 9); COUNT = 888.4 The nearest integer value is COUNT 888 = 0378 (in Hex), the actual total delay will be 4 + (888 * 9) = 7996 cycles = 3.998 ms....
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HW#8_Sol - (f[ACCD = A97A = 1010100101111010 after shift...

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