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yousef_ob
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Posts posted by yousef_ob
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hello,
fom the maths point of view this is forbidden because m(t) as shown is a function of time not a constant
bye ;D -
thanx i solved it
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i think that you didnt noticed that the cos is multiplied by another time variable function so you cant integrate that simple. anyway thanx abt ur try
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thanx a lot. nice try!!
i need a mathemetical solution. -
Hello,
I thought that i give u the time to think in my quwstion. but to this moment i didnt get a solution to this problem.
i hope someone will help me.
bye -
hello,
in the attached equation m(t) is a power signal - not an energy signal. we know that DSB modulation is defined by the equation m(t) cos (2*pi * fc*t). so that the power of this signal is the integration of the sequared equation over the period - i.e power = integration over period of (m(t) cos(2*pi* fc*t)^2. we know that cos^2(wc*t)= .5(1+ cos(2*wc*t))
then multiply this by m^2(t) and integrate over the period, then the first term in 0.5 * the power of the mudulating signal (m(t)). and the second term is zero i am asking why it is zero. i gonna attatch the equation again
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C'mon i have asked a question it wasnt a joke or a story. plz is there anyone could solve that for me??!!
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sure m(t) is the modualting power signal
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hello, plz could anyone solve this for me:
in a DSB modulation if we assume that the modulating signal is a power signal, why this integration is zero, plz i want the mathemetical prove for the attached equation
thanx very much
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hello
about using dBs:
we use dB because when we use dBs multiplication turns into addition as well as division turns into subtraction and this is very useful characteristic. and we can draw Bode plot which is a very helpful chart. -
HELLO,
I AM SO SORRY, BUT I DIDNT GET IT YET!!
CAN YOU EXPLAIN THE ATTACHED CICUITS TO ME?? PLEASE
"I AM A BIGINNER.. FORGIVE ME"
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HELLO,
YOU CAN USE THIS METHOD IN FINDING THE TOTAL RESISTANCE SEEN BY A VOLTAGE SOURCE.
1. AT THE TERMINALS OF THE VOLTAGE SOURCE NAME (a,b), THEN WHEN U FIND ANY RESISTANCE NAME ©, THEN GO ON d,e AND SO ON.
2. REDRAW THE CCT IN A NEAT FORM BY PUTTING THE LETTERS (a,b,c,d...) THEN SEE WHERE IS THE SERIES AND WHERE IS THE PARALLEL CONFIGURATIONS AND SIMPLFY THE CCT.
3. DO ANY DELTA-STAR OR STAR-DELTA TRANSFORMATIONS.
SAY THANX TO ME -
HELLO,
ABOUT YOUR COMMENT, YOU MAY BE WRONG BECAUSE IF YOU USE ONE +VE POWER SUPPLY THEN THE OP-AMP WILL NOT WORK PROPERLY BUT IT WILL SATURATE IN THE -VE REGION AND WE CANT GET A -VE O/P FROM THIS OP-AMP. WE WILL GET THE SAME THING IF WE USE A -VE POWER SUPPLY.
THANX VERY MUCH -
HELLO VISHO,
ABOUT UR AMPLIFIER CCT U DONT TAKE IN UR ACCOUNTS THE LOADING EFFECTS FROM THE FIRST STAGE TO THE SECOND STAGE.
U WOULD BETTER USE TRANSISORS OTHER THAN OPAMPS, BECAUSE HERE U NEED ONE POWER SUPPLY BUT IN OP-AMP CCT U NEED TWO POWER SUPPLIES ONE -VE AND THE OTHER +VE.
BUT THE ADAVANTAGE OF THE OP-AMP ON THE TRANSISTOR THAT THE OP-AMP HAS AN INFINITE INPUT RESISTANCE WHICH IS NOT PRESENT IN THE TRANSISOR SO THE LOADING EFFECTS ARE NEGLECTED.
IF I WERE U VISHO I WOULD SAY THANX YOUSEF
BYE
Study about GSM
in Electronics chit chat
Posted
plz can anybody help me in making a study about GSM?? links wanted