Digital Electronics Test 6

Concept:

Flag Register is an 8-bit flag register out of which only 5 bits are used as shown:

D₂ (Parity Flag) is set when the result contains an even number of 1’s.

Application:

For the given flag register, we have S = 1 and P = 0 (Odd Parity)

Since MSB bit (D₇) of the accumulator is ‘1’, the result has to have odd parity, i.e. odd number of 1’s.

We can now analyse the options as shown:

7E : (0 1 1 1 1 1 1 0)₂ : Even number of 1’s

6C: (0 1 1 0 1 1 0 0)₂: Even number of 1’s

DB: (1 1 0 1 1 0 1 1)₂ : Even number of 1’s

B6: (1 0 1 1 0 1 1 0)₂ : Odd number of 1‘s

Memory size = 2¹¹× 8

= 2 × 2¹⁰ × 8

= 2 × 1024 × 8

= 2048 Bytes

i.e., 0^th byte to 2047^th byte.

Concept:

CALL Addr performs the following operations:

1. Save (store) the contents of PC into the stack.

Note: In PUSH operation, the stack pointer is decremented and then data is stored at the address location of SP

2. The address is stored in PC

Application:

Now, the sequential operation involved in the execution of “CALL Addr” instruction are:

(SP) decremented

((SP)) ← (PC)

The operation for the given instruction set is as explained:

The number 96, 8-bit format is stored at 2501. The instructions executed are

LDA 2501 H;   loads the data 96 H in A from 2501 H

CMA;               Complements the contents of A

ADI 01 H;        Adds 01 H to A to find 2’s complement

STA 2502H;    stores the 2’s complement at 2502H

HLT;                 Halts the execution of program

∴ The contents of 2502 H is 2’s complement 8-bit number representing 96, i.e. 10100000.

96H=1001 0110 binary..A's

1s complement is complement of 96h ie= 69H 69H= 0110 1001 in binary.

adding one to it becomes 6A H=0110 1010 in binary

OR Simply 2s complement of 96h is 0110 1010.