0
Go4Expert Member
period 3 A1
period 4 A2
period 5 A2
period 6 A1
period 7 A2
period 8 A1

but how do tell if the signals are 1/(4f)... 2/(4f).... 0 etc ??
0
~ Б0ЯИ Τ0 С0δЭ ~
Check the phase difference between the first point of a period and the last point of the previous period

Another hint : phase difference of period 3 is 1/(2f) = 2/(4f).
0
Go4Expert Member
is there a phase where the difference between first point of a period and the last point of the previous period is = 3?

ok i give it a shot... so for

period 4 is 011 cuz amplitude is A2 and p.d is 1/4f
period 5 is 000 cuz amplitude is A1 and phase difference is 0
period 6 is 100 cuz amplitude is A1 AND P.D IS 1/2f (=2/4f)
period 7 is 101 cuz amplitude is A2 and p.d is 2/4f
period 8 is ?
0
~ Б0ЯИ Τ0 С0δЭ ~

I think you don't get it properly, try to follow my answer :
Period 1 is 001 [ Amplitude = A2 and PD = 0 ]
Period 2 is 010 [ Amplitude = A1 and PD = 1/(4f) ]
Period 3 is 100 [ Amplitude = A1 and PD = 2/(4f) ]
Period 4 is 011 [ Amplitude = A2 and PD = 1/(4f) ]
Period 5 is 101 [ Amplitude = A2 and PD = 2/(4f) ]
Period 6 is 000 [ Amplitude = A1 and PD = 0 ]
Period 7 is 011 [ Amplitude = A2 and PD = 1/(4f) ]
Period 8 is 110 [ Amplitude = A1 and PD = 3/(4f) ]

So, the bit sequence becomes : 001010100011101000011110

(I'm leaving, so I'll be able to read your reply after at least 2 hours)

Last edited by SaswatPadhi; 17May2009 at 12:46..
0
Go4Expert Member
mabbe u can explain why is period 3 1/2f and period 5 = 2/4f?
with your experience how can i tell?
0
Go4Expert Member
haha i think i got it! thanks for all your help! ... u have been a great help already...

i was trying out this question and not sure if i did it right

1. A data source produces a 7-bit ASCII characters and send them over a 1kbps transmission line. Calculate the effective rate of the transmission for the following cases.

a) Ascynchronous transmission with a start bit, 1 stop bit and a parity bit

b) Synchronous transmission with a frame consisting of 48 control bits and 1024 information bits. The information firld contains a numbers of ASCII characters. Each character consists of 7 information bits and a parity bit.

(effective data rate = useful data bits excluding overhead/total time)

so for a... this is what i did

Total transmit time = 7/10(to the power of -3) = 0.007 seconds

Useful bits = 4
therefore Effective data rate = 4 / 0.007 = 571 bps

not sure abt b though shd i be using the same approach?
0
~ Б0ЯИ Τ0 С0δЭ ~
Equation for the waveforms of : (here p stands for pi)

(1) period 2 is x = A1 cos (2pft)
(2) period 3 is x = - A1 cos (2pft)
(3) period 4 is x = A2 sin (2pft)
(4) period 5 is x = - A2 sin (2pft)

So, phase difference between periods 2,3 and periods 4,5 is obviously half time-period = 2/(4f) !
And phase difference between periods 3,4 is quarter time-period = 1/(4f).

PS : You posted the last post while I was writing this reply.

Last edited by SaswatPadhi; 17May2009 at 14:39..
0
Go4Expert Member
Dude i really want to thank you for all your help... its really kind of you to render so much help to me, a complete stranger over the net...

i hope i didn't irritate you by askin so many questions...thanks once again!
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~ Б0ЯИ Τ0 С0δЭ ~
Not at all, why do you think that way. You are not irritating me, you help me revise all these stuff.

PART A:
Total data to be transmitted = 7 + 1 + 1 + 1 = 10 bits
Total transfer time = 10/1024 s
Effective transfer rate = 7*1024/10 = 716.8 bps

PART B: (not sure abt this one)
Total data to be transmitted = 48 + 7 + 1 = 56 bits
Total transfer time = 56/1024 s
Effective transfer rate = 7*1024/56 = 1024/8 = 128 bps

But synchronous transfer should be faster than asynchronous, right ?
0
Go4Expert Member
yes synchronous shd be faster

http://www.computeruser.com/resource...ml?lookup=4916

effective data rate = useful data bits excluding overhead/total time

shdn't it be 7/(10/1024)??