Continue to Site

# bandwidth of digital signal

Status
Not open for further replies.

#### electronics_kumar

i want a proof tp prove that digital systems needs more bandwidth when compared to analog signals.. how to prove it( week point for your digital signal)

There can be no such proof as sometimes a digital signal uses less bandwidth than a comparative analog one. In the case of digital TV for example. Digital TV signals use less bandwidth than an analog TV signal does.

There can be no such proof as sometimes a digital signal uses less bandwidth than a comparative analog one. In the case of digital TV for example. Digital TV signals use less bandwidth than an analog TV signal does.
we cannot say like that ...two week points of digital signal/systems are the real world is analog..they will occupy more bandwidth

There can be no such proof as sometimes a digital signal uses less bandwidth than a comparative analog one. In the case of digital TV for example. Digital TV signals use less bandwidth than an analog TV signal does.

'Digital' is for me same as signal going inside databus, IDE-bus, RS232, SCSI, ethernet, optical fibers etc. with pure '1' and '0' with simple lower or higher than '0.5' detection in reciver side, and in this world 'digital' modulation waste very much bandwith (10 - 50 times) compare to more advanced analog modulation.

for exampe IDE-bus can transporting around 166 MB/s as max in digital way - but using same advanced modulation as 56K-modem a analog way you can recive 1.5 GB/s on IDE-bus using same bandwidh (43.5 MHz per cord)

(

166 *1024 *1024 = 174063616 byte/s

174063616 byte/s * 8 = 1392508928 bit/s

1392508928 / 16 = 87031808 bit/s per cord i IDE-bus

if thinking '1010101...' - give 43515904 Hz = 43.5 MHz

3400Hz - 300Hz give 3100 Hz BW for 56-kbit transfer in modem (best possible case)

43515904 / 3100 = 14037 'ekv. phone canal'

14037 * 56000 = 786093750 bit/s = 786.1 Mbit/s for one IDE-cord

786.1 Mbit/s * 16 = 12.6 GB/s = 1499.3 MByte/s or 9 bit per 'digital' ide- bit

)

In practical using IDE-bus need also lowest third over tone (130.5 MHz) in signal for make little bit more square-wave than sinus - digital circurit not want pure sinus to decision if '0' or '1' - so using this bandwidth give around 4.5 GB/s! via advanced modulation as 56K-modem - and digital IDE in near 27 time lower BW effectivity compare modulation as 56K-modem

In many optical fiber system using so bad and sluggish modulation is need step over every other laser 'color' in DWDM (dence wave division multiplex) matris
depend each laser have so much sideband from badly 'digital' modulation is make nearst neighbour 10 nm (~ 1.25 THz between channel)) laser frequence unusable for transporting data - here talk we more than >50 wasting BW compare shannon teroretical value for actual data volume...

'digital_TV' is not digital - transport layer using OFDM and is analog modulation (modulated from digital source), and if not using advanced compress on TV and waste most of data in picture and still transporting in raw form (around 132 Mbit/s, 12 bit/pixel + sound) give arounde ~ 24 bit per Hz squeezed in 5.5 MHz BW, compare to 56K moden have 18 bit/Hz) - you need more bandwidth compare to 'old' analog TV also with modern OFDM... advantage her is depend of lossy compression of data (TV), not ODFM- transport layer itself.

modern 56K phone modem using analog transport layer and using near 90% of teoretic capacity as shannon law on phone line.

/xxargs

electronics_kumar said:
i want a proof tp prove that digital systems needs more bandwidth when compared to analog signals.. how to prove it( week point for your digital signal)

In my opinion, the answer is in the Shannon's theorem.
If you have an analog signal with a limited bandwidth you can digitize it sampling the signal in question with a sampling rate at least twice as high as the highest frequency component (Fourier) in that signal (i.e. you have to get at least two samples per the "shortest" period within the signal).
In such a case the original analog signal can be restored from the digitized signal without "losses".

Regards,
Eric

Status
Not open for further replies.