Arun Segaran

• Nov 2nd, 2007

Not?many?synthesis?tools?are?available?to?synthesize?a?asynchronous?design?at?RTL?

swatias

 Profile Answers by swatias Questions by swatias

• Jan 11th, 2007

To avoid race effects and deadlock effect, synchronous circuits are used

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sumanyaram

 Profile Answers by sumanyaram Questions by sumanyaram

• Dec 5th, 2006

Praveen is right,1101 is not a valid BCD number........

Danny John

• Aug 8th, 2006

Praveen is right. we are supposed to do exactly what he did. For verification convert them to decimal that is 9 + 4 =13 that is in BCD it is 0001 0011.

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vijay

• Jun 30th, 2017

The size of the pointer will always be 2 bytes because the pointer variable is only used to store the address regardless of the data type. but the size will different from different compiler.

rvg

• Aug 8th, 2016

Any pointer irrespective of what it is referencing (pointing to), it stores address where that content is stored.
Any address is in hex, which is of type integer. So it takes up 2 bytes required to store int data type.

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Monika

• Sep 25th, 2017

ABC are the input of AND gate and A+B+C are the inputs of OR gate

nikhil4nigam

 Profile Answers by nikhil4nigam Questions by nikhil4nigam

• Jul 31st, 2010

Since NAND and NOR are the universal gates, hence the logics can be implemented simply by using NAND or NOR gate.

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vivek

• Dec 8th, 2014

Highest priority interrupt of RST series , call location 003CH can be enabled by two instructions EI(enable interrrupt ) and SIM(set interrupt mask)

Neeshu Agarwal

• Dec 24th, 2012

RST 7.5 - Maskable - Vector - 003C (H) - Edge triggered RST 6.5 - Maskable - Vector - 0034 (H) - Level triggered RST 5.5 - Maskable - Vector - 002C (...

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How much time does a serial link of 64 Kbps take to transmit a picture with 540 pixels.

japan_dave88

 Profile Answers by japan_dave88 Questions by japan_dave88

• Aug 20th, 2009

I would say that it also depends on length of the link as suppose Two devices named A and B are located 1000 kilometer far and we have link of 64kbps. Than when A will start sending first bit of ...

shreekanthss

 Profile Answers by shreekanthss Questions by shreekanthss

• Feb 2nd, 2009

Suppose a pixel is represented by 8 bits. Then total size=540*8=4320bitsIt takes 4320/64000=0.0675 seconds

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raval vikram

• Sep 16th, 2011

16 bit

karam_sodhi@yahoo.com

• Sep 11th, 2011

-128 to +127

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jaspreet

• Jul 11th, 2017

Firstly convert it into binary form and then into grey code

AJAY

• Jan 6th, 2016

(101011000010)B=(111110100011)GRAY

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MAMTA KHANNA

• Jul 28th, 2016

There will be less dissipation in cmos because of lower current gain from the power supply.

Aishi Sur

• Sep 18th, 2015

The advantages of CMOS technology over NMOS are:
Very low static power consumption
Reduce the complexity of the circuit
High density of logic functions on a chip
Low static power consumption
High noise immunity

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vijay

• Jun 30th, 2017

Parity bit is an additional bit which combines with the original message bit so as to detect and correct the error at receiver.

Satyam kumar dubey

• Nov 29th, 2016

Parity is the way to detection of error in original signal. It detect single bit error. To detect this error an extra bit is attached to original signal or original coad word. This extra bit is called parity bit

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adi

• Sep 30th, 2012

Bt practically i required 1.07 mb to record 1.31 min speech,
any1 kno its calculation?

adityajain

 Profile Answers by adityajain Questions by adityajain

• Nov 21st, 2009

Voice signal is sampled at 8000 samples per second. (Human voice is in the range of 300Hz-3400Hz, and according to nyquist sampling rate we take it more than double the rate).For a 4  min. voice signal, we have Memory = 4*60*8000 Bytes = 1875KB = 1.83 MB.

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Prakash

• Oct 21st, 2011

For gray 1 pixel = 1 byte.
for color 1 pixel = 3 byte;
64kbps =64 kilo bit per sec;
64kbps= 64/8 kilo byte per sec= 8000 byte per sec;
so 540 gray pixels, time=540/8000=0.067sec;
for color 540 pixels, time= 540*3/8000=0.202 sec;

vengat86

 Profile Answers by vengat86 Questions by vengat86

• Aug 18th, 2009

In transmission its considered as download and upload
so, we have 64kbps 
Divide by 8 to get the download speed so its 8kbps ie 8192 bits per sec
Now consider the image its 540Bytes convert into bits its 540*1024 = 552960
So divide both 552960 / 8192 = 67.5 sec
i.e., 1 min 7 secs

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k.v.bhavani

• Jul 11th, 2015

LIFO. Last in first out. Stack has two operations in one place that is push and pop.
Push means it inserts the value or data one bye one.
Pop means it removes the data one by one

rashs

• Jun 5th, 2015

Stack uses LIFO policy.

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virat ganesh

• Apr 14th, 2018

It is an outrput signal used to give information about AD0-AD7 contents

yogi

• Sep 26th, 2017

ALE is an active high pin. when we supplied active high pulse to this pin this pin is activated. When this pin is activated it de-multiplex lower address bus (A0-A7) from lower adress data bus(AD0-AD7).

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MANISH MANCHANDA

• Jun 16th, 2016

NOR gate

SUBASH

• Sep 11th, 2012

NOR gate

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Arvind Kumar

• May 12th, 2023

A.B+A(B+C)+B(B+C) A.B+A.B+A.C+B.B+B.C A.B+A.C+B+B.C [USING THIS RULE B.B=B] [ALSO USE THIS RULE A.B+A.B=A.B] A.B+A.C+B(1+C) [USE THIS RULE ...

Walta Napuo

• Jun 21st, 2020

AB + A( B + C ) + B ( B + C ) = = AB+AB+AC+BB+BC = AB+AC+B+BC = AB+AC+B(1+C) = AB+AC+BC

Code
c

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