GATE EE 2008 | GATE EE Year Wise Previous Years Questions

GATE EE 2008

MCQ (Single Correct Answer)

-0.6

A $$3$$ line to $$8$$ line decoder, with active low outputs, is used to implement a $$3$$- variable Boolean function as shown in the figure: The simplified form of Boolean function $$F(X,Y,Z)$$ implemented in 'Product of Sum' form will be GATE EE 2008 Digital Electronics - Combinational Circuits Question 2 English

GATE EE 2008 Digital Electronics - Combinational Circuits Question 2 English

$$\left( {x + z} \right).\left( {\overline x + \overline y + \overline z } \right).\left( {y + z} \right)$$

$$\left( {\overline x + \overline z } \right).\left( {x + y + z} \right).\left( {\overline y + \overline z } \right)$$

$$\left( {\overline x + \overline y + z} \right).\left( {\overline x + y + z} \right).\left( {x + \overline y + z} \right).\left( {x + y + \overline z } \right)$$

$$\left( {\overline x + \overline y + \overline z } \right).\left( {\overline x + y + \overline z } \right).\left( {x + y + z} \right).\left( {x + \overline y + \overline z } \right)$$

GATE EE 2008

MCQ (Single Correct Answer)

-0.3

An input device is interfaced with Intel $$8085$$ $$A$$ microprocessor as memory mapped $${\rm I}/O,$$ the address of the device is $$2500H.$$ In order to input data from the device to accumulator, the sequence of instructions will be

$$\eqalign{ & LX{\rm I}\,\,\,H,2500H \cr & \,\,\,\,MOV\,\,A,M \cr} $$

$$\eqalign{ & LX{\rm I}\,\,\,H,2500H \cr & \,\,\,\,MOV\,\,M,A \cr} $$

$$\eqalign{ & LHLD\,\,\,2500H \cr & \,\,\,\,MOV\,\,A,M \cr} $$

$$\eqalign{ & LHLD\,\,\,2500H \cr & \,\,\,\,MOV\,\,M,A \cr} $$

GATE EE 2008

MCQ (Single Correct Answer)

-0.6

The contents (in Hexadecimal) of some of the memory locations in an $$8085A$$ based system are given below GATE EE 2008 Digital Electronics - Microprocessor Question 9 English

GATE EE 2008 Digital Electronics - Microprocessor Question 9 English

The contents of stack pointer $$(SP),$$ program counter $$(PC)$$ and $$(HL)$$ are $$270H,$$ $$2100H$$ and $$0000H$$ respectively. When the following sequence of instructions are executed,
$$2100H:$$ $$DAD$$ $$SP$$
$$2101H;$$ $$PCHL$$

The contents of $$(SP)$$ and $$(PC)$$ at the end of execution will be

$$\left( {PC} \right) = 2102H,\,\,\left( {SP} \right) = 2700H$$

$$\left( {PC} \right) = 2700H,\,\,\left( {SP} \right) = 2700H$$

$$\left( {PC} \right) = 2800H,\,\,\left( {SP} \right) = 26FEH$$

$$\left( {PC} \right) =2A02H,\,\,\left( {SP} \right) = 2702H$$

GATE EE 2008

MCQ (Single Correct Answer)

-0.6

The time constant for the given circuit will be GATE EE 2008 Electric Circuits - Transient Response Question 11 English

$$1/9$$ $$s$$

$$1/4$$ $$s$$

$$4$$ $$s$$

$$9$$ $$s$$

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