GATE 2023 CS – Question 42
A 4 kilobyte (KB) byte-addressable memory is realized using four 1 KB memory blocks. Two input address lines (IA4 and IA3) are connected to the chip select (CS) port of these memory blocks through a decoder as shown in the figure. The remaining ten input address lines from IA11–IA0 are connected to the address port of these blocks. The chip select (CS) is active high. [Figure: 2-to-4 decoder with inputs IA4 (MSB) and IA3 (LSB), outputs Q0–Q3 driving CS of blocks X1–X4; IA11–IA5 and IA2–IA0 go to the 10-bit Addr port.] The input memory addresses (IA11–IA0), in decimal, for the starting locations (Addr=0) of each block (indicated as X1, X2, X3, X4 in the figure) are among the options given below. Which one of the following options is CORRECT?

Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: (C) (0, 8, 16, 24)
Explanation
Block start needs Addr=0, so all other address bits are 0. Only IA4IA3 = 00, 01, 10, 11 select the blocks, giving addresses 0, 2³=8, 2⁴=16 and 8+16=24.