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In "duo-binary signalling" how exactly is the transmission capacity increased?

Because , when we encode the binary bits, the bit duration remains the same, but the amplitude will change. Then how can the transmission capacity be better?

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  • $\begingroup$ You say duobinary modulation doubled the capacity. Compared to what modulation scheme? Do you have a reference for this claim? $\endgroup$ – Deve Apr 29 '15 at 7:32
  • $\begingroup$ correlative coding is generally used to increase the bit rate, given the same bandwidth.I am not comparing it with any specific modulation scheme.but not understanding how capacity will be increased. $\endgroup$ – spectre Apr 29 '15 at 7:46
  • $\begingroup$ Capacity has a particular meaning in communications circles which is different from what is being considered here. What is happening is that the use of duobinary signaling allows for an increase in the transmission rate (# of bits per second) or spectral efficiency (bits per second per Hertz). $\endgroup$ – Dilip Sarwate Apr 29 '15 at 13:15
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If you want to transmit a symbol sequence $A_k$ using baseband pulse amplitude modulation (PAM), the transmitted signal is

$$s(t)=\sum_kA_kp(t-kT)$$

where $p(t)$ is the transmit pulse, and $T$ is the symbol interval. If you want to avoid intersymbol interference (ISI), the pulse function $p(t)$ must satisfy the Nyquist criterion, which says that its value must be zero at multiples of the symbol interval:

$$p(0)=1,\quad p(kT)=0,\quad k\neq 0\tag{1}$$

The minimum bandwidth required to satisfy (1) is $1/2T$, but the corresponding pulse $p(t)$ is an ideal low-pass filter, i.e. a sinc function, which is of course impractical. So in practice, Nyquist pulses with a higher bandwidth (excess bandwidth) are used. This means that for a given bandwidth, the possible symbol rate is reduced as compared to the theoretically maximum possible value.

With duobinary signalling practical systems can be built with zero excess bandwidth, because the data symbols are filtered such that a spectral zero is introduced at half the baud rate. Note that this is a very simple filter unlike the ideal low-pass filter required for the minimum bandwidth Nyquist pulse. The price paid is the deliberate introduction of ISI and the introduction of a ternary (instead of binary) signal, which requires a higher SNR to achieve a given error probability.

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  • $\begingroup$ I think the data are still binary but at the receiver, instead of needing to discriminate between two levels (say, $+A$ and $-A$), it is necessary to discriminate between three levels ( $+B$ , $0$, and $-B$) to make a decision as to whether a $0$ or a $1$ was transmitted, and so, for a given noise level, $B$ must be larger than $A$ to achieve the same error probability. $\endgroup$ – Dilip Sarwate Apr 29 '15 at 13:20
  • $\begingroup$ @DilipSarwate: Of course you're right, I shouldn't have referred to ternary data, but to a ternary signal (at the output of the filter). Edited. $\endgroup$ – Matt L. Apr 29 '15 at 14:55
  • $\begingroup$ thanks a lot for reply. But i didnt get 2 points,1]what is the meaning of "for a given bandwidth, the possible symbol rate is reduced as compared to the theoretically maximum possible value." 2]"the data symbols are filtered such that a spectral zero is introduced at half the baud rate" $\endgroup$ – spectre Apr 30 '15 at 1:19
  • $\begingroup$ @spectre: 1) If you use excess bandwidth in order to be able to implement a Nyquist pulse, you use more bandwidth than theoretically necessary, i.e. for a given bandwidth you need to lower the data rate as compared to the theoretical maximum value. 2) The frequency response of the filter becomes zero at the frequency $1/2T$, where $T$ is the symbol period. $\endgroup$ – Matt L. Apr 30 '15 at 16:46
  • $\begingroup$ as far as i know, bandwidth=(Rb)(1+a)/2, where, a=roll off factor,Rb=rate of transmission. So , for a given Bandwidth , data rate Rb is fixed right?@MattL. $\endgroup$ – spectre Apr 30 '15 at 17:15

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