1) 4800 bits/sec 2) Shift the pulse so as to get 1 after f = 17kHz, so at intervals of f = 27kHz they add to 1. Therefore Rb = 27kHz. 3) f2-f1 = 2fx and 1 / (1 + f1/fx) = r. Therefore r = 0.3
Sir could I ask you a question? At 8:05 , does p(t) satisfy Nyquist criterion? I know a square wave that is penetrated by y axis on its middle satisfy criterion but not sure about p(t) which is a shifted form of it
Gaussian pulse requires infinite time. Even if we use truncated time pulse, it would be an option but not the one that maximize the rate with minimum bandwidth.
@@alimuqaibel7619 I'm kind of lost with part (3)... Everyone is saying 0.3 is the roll-off factor. but applying this to BW=(1+r)Rb/2 will give us a bandwidth of 1.3 instead of 2.6 Shouldn't the Bandwidth be the 2*f2 since there's also a negative side?
@@ryugalaw Basically , this BW formula with Rb/2 is not correct, it would be Rs/2. Rs is sampling per sec also u can say transmission rate , whereas Rb is bit per sec. And here Rs is 2M symbol per sec. Rb is how much data bit are sent in a single symbol multiplied with Rs.
Answers of the last exercises: 1- 4800 bits/s 2- shift the signal to n*(27 kHz) which is (n*Rb) where n=....,-3,-2,-1,0,1,2,3.... to get their summation equal to a constant. 3- Rb=2 MHz, fx=0.3 MHz, r=0.3
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Many thanks Mr. Saad Mann for your kind comment
1) 4800 bits/sec
2) Shift the pulse so as to get 1 after f = 17kHz, so at intervals of f = 27kHz they add to 1. Therefore Rb = 27kHz.
3) f2-f1 = 2fx and 1 / (1 + f1/fx) = r. Therefore r = 0.3
Good job
So nyquist criteria are, 2fm for sampling and fm/2 for symbol rate? Am I getting it correctly? And thanks for sharing the lecture
Sir could I ask you a question?
At 8:05 , does p(t) satisfy Nyquist criterion?
I know a square wave that is penetrated by y axis on its middle satisfy criterion but not sure about p(t) which is a shifted form of it
No it does not. Any squre wave requires invite bandwidth.
@@alimuqaibel7619 thank you sir
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اهلا وسهلا بك
Thanks Sir.It is very clear.
You are welocme.
Super
Thanks for the comment. Please spread.
Whats makes a gaussian pulse fail to meet the nyquist criterion?
Gaussian pulse requires infinite time. Even if we use truncated time pulse, it would be an option but not the one that maximize the rate with minimum bandwidth.
1 - Rb=2w/(1+r) = 4.8kbps
2 - Rb+fx = 17 >> Rb-fx = 10 >> 2 equations 2 unknowns >> Rb =27 kbps (I'm not sure)
3 - 0.7 + 2fx = 1.3 >> fx = 0.3
I have a final exam in two days, wish me luck
1 - 4.8 kbits/s
2- f = 27 kbits/s
3- max rate (Rb) 2 Mbits/s ... roll-off factor = 0.3
You are awesome...for part (2) since he is asking about the rate use kbits/s or kbps .. Also in Engineering we refer to kilo with small 'k'
@@alimuqaibel7619 I'm kind of lost with part (3)...
Everyone is saying 0.3 is the roll-off factor. but applying this to BW=(1+r)Rb/2 will give us a bandwidth of 1.3 instead of 2.6
Shouldn't the Bandwidth be the 2*f2 since there's also a negative side?
@@ryugalaw Basically , this BW formula with Rb/2 is not correct, it would be Rs/2. Rs is sampling per sec also u can say transmission rate , whereas Rb is bit per sec. And here Rs is 2M symbol per sec. Rb is how much data bit are sent in a single symbol multiplied with Rs.
1- 4800 bits/s
2- shift the signal to n*(17 kHz)
3- Rb=2 MHz, fx=0.3 MHz, r=0.3
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Answers of the last exercises:
1- 4800 bits/s
2- shift the signal to n*(27 kHz) which is (n*Rb) where n=....,-3,-2,-1,0,1,2,3.... to get their summation equal to a constant.
3- Rb=2 MHz, fx=0.3 MHz, r=0.3
Good job. The answer to part 2 is not explicit.