12- In the given code, statement 2 will executed at what time step?: initial fork 5x 1'b0; // statement 1 # 15 y= 1b'1; //statement 2 join a) 15 b) 20 c) 5 d) Current simulation time e) 1 0 None of the above 13- Assume an 8-bit word sized 2K byte of memory. The number of address lines required for transferis: a) 11 lines. b) 10 lines. c) 8 lines. d) 2048 line. e) none of the given options 14- In coincident decoding, the decoder size required to decode 16K bytes of memory should be: a) 5x 32. b) 4 x 16. c) 6x 64. d) 7x 128 e) none of the given options 15- The slowest type of ROM to be erased is: a) EEPROM b) EPROM c) PROM d) Mask fabricated ROM e) ROM cannot be erased. 16- What is the total memory size reserved in the following instruction? reg (7: 0] memram [0: 8192); a) 16 MByte b) 16 Kbyte c) 8 Mbyte d) 4 Kbyte e) None of the given Page 5 of 8

Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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This QUESTION FROM ADVANCED DIGITAL SYSTEMS DESIGN course.

12- In the given code, statement 2 will executed at what time step?:
initial
fork
#5x= 1'b0; // statement 1
# 15 y= 1b'1; //statement 2
join
a) 15
b) 20
c)
5.
d) Current simulation time
e) 1
O None of the above
13- Assume an 8-bit word sized 2K byte of memory. The number of address lines required for
transfer is:
a) 11 lines.
b) 10 lines.
c) 8 lines.
d) 2048 line.
e) none of the given options
14- In coincident decoding, the decoder size required to decode 16K bytes of memory should be:
a) 5x 32.
b) 4 x 16.
c) 6x 64.
d) 7x128
e) none of the given options
15- The slowest type of ROM to be erased is:
EEPROM
a)
b) EPROM
c) PROM
d) Mask fabricated ROM
e) ROM cannot be erased.
16- What is the total memory size reserved in the following instruction?
reg (7: 0] memram [0: 8192];
a) 16 MByte
b) 16 Kbyte
c) 8 Mbyte
d) 4 Kbyte
e) None of the given
Page 5 of 8
AAO-PI0-RO1 3 30/9/2018
Transcribed Image Text:12- In the given code, statement 2 will executed at what time step?: initial fork #5x= 1'b0; // statement 1 # 15 y= 1b'1; //statement 2 join a) 15 b) 20 c) 5. d) Current simulation time e) 1 O None of the above 13- Assume an 8-bit word sized 2K byte of memory. The number of address lines required for transfer is: a) 11 lines. b) 10 lines. c) 8 lines. d) 2048 line. e) none of the given options 14- In coincident decoding, the decoder size required to decode 16K bytes of memory should be: a) 5x 32. b) 4 x 16. c) 6x 64. d) 7x128 e) none of the given options 15- The slowest type of ROM to be erased is: EEPROM a) b) EPROM c) PROM d) Mask fabricated ROM e) ROM cannot be erased. 16- What is the total memory size reserved in the following instruction? reg (7: 0] memram [0: 8192]; a) 16 MByte b) 16 Kbyte c) 8 Mbyte d) 4 Kbyte e) None of the given Page 5 of 8 AAO-PI0-RO1 3 30/9/2018
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