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Baseband pulse transmission problem

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pnsakanjankumar

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for ISI to be zero we should chose Nyquist channel which has a data rate of Rb=1/Tb=2W where W is the channel Bandwidth.
Now when we use Raised cosine spectrum system Bandwidth is W(1+α) where α is the roll off factor
suppose α = 1
then Bandwidth is 2W but Rb the data rate should be the same because those sampled spectrum are space apart by Rb
given by
P(F)+p(F-2W)+P(F+2W)=1/2W -W<f<W
I want to know whether it is right or not?
Please Help me ?:sad:
 
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The data rate is still Rb. The functional "p" seems to be a spectral operator, do clarify on exactly what you mean by "p".
If you want to know the implication of the roll-off factor on the bandwidth, then that should not be a concern.
The roll off is there to create a more realizable filter. Ideally $\alpha = 0$.
ISI:
If $\alpha=1$ then you have a signal with an ISI assuming you have the same sampling rate of 2W, wherein the op= current_ip + past_ip.
Typically systems are causal so you consider the impact of the previous ip not the future ip.
 
I want to know whether it is right or not?
Right.
Although there are some upper/lower case issues in pnsakanjankumar's post, P(f) is the Fourier transform of the received pulse.
Note that not only alpha=0 is unrealizable; it is not desirable as a because:
- in time domain the pulse decays slowly (as 1/t), so truncation of the response causes more distortion than with a "decent" alpha
- the eye would be closed in practice: intollerant to synchronization misadjustment
Regards

Z
 
Now if we want to achieve Ideal bandwidth possible that is W in this case we use Duobinary Signalling
Explain how we get such a bandwidth using Duobinary Signalling ??
 

Duobinary signalling i.e. a partial response channel for transmission over W is viable but it must be noted that this is possible since we know the behaviour of the ISI e.g., '1-D' where D is the delay of one time sample. Since the ISI behaviour is known we are able to equalize it at the output.
 

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