BASEBAND DIGITAL DATA
Common form of base band signals to represent digital information bits are shown below.
Voltage levels for uni-polar and bi-polar signals are also shown.
BIT-RATE, BAUD-RATE & BANDWIDTH
1


B denotes the duration of the 1 bit
1
Hence Bit rate =

B
bits per second
All the forms of the base band signalling shown transfer data at the same bit rate.

E denotes the duration of the shortest signalling element.
Baud rate is defined as the reciprocal of the duration of the shortest signalling element.
1 i.e. Baud Rate =

E
baud
In general
Baud Rate ≠ Bit Rate
For NRZ : Baud Rate = Bit Rate
RZ : Baud Rate = 2 x Bit Rate
Bi-Phase: Baud Rate = 2 x Bit Rate
(

B =

E )
(

E =

B
2
)
(

E =

B
2
)
AMI: Baud Rate = Bit Rate
A major consideration for base band signals is the minimum bandwidth and the upper and lower frequencies f
H
and f respectively, in the signal.
L
Considering bi-polar signal (i.e. –V or +V)
NRZ
The highest frequency occurs when the data is 1010101010……. i.e.
2

This sequence produces a square wave with periodic time
 
2

E
. From the Fourier series for a square wave, this waveform contains 1 st
, 3 rd
, 5 th
… (Odd) harmonics.
If we pass this signal through a LPF then the maximum bandwidth would be 1/T Hz, i.e. to just allow the fundamental (1 st
harmonic) to pass. As illustrated in the diagram below, the data sequence 1010…… could then be completely covered.
Hence the minimum channel bandwidth
3

B min

1
T

2

1
E

Baud
2
Rate
Since

1
E

Baud Rate
This corresponds to the ‘upper frequency’ f . The lower frequency limit is DC, for example
U if continuous 1’s were transmitted, the voltage is +V volts.
Example:

B
= 1 msec, Bit Rate = 1000 bits/sec
1
Baud Rate =

E
= 1000 Bauds
B min
 f
U

Baud
2
Rate

500 Hz f
L

' DC '
Return to Zero
Considering RZ signals, the max frequency occurs when continuous 1’s are transmitted.
4

This produces a square wave with periodic time
 
2

E
. Following the same reasoning as for
NRZ
B min
 f
U

Baud
2
Rate
If the sequence was continuous 0’s, the signal would be –V continuously, hence f
L

' DC ' .
Example
Again

B
= 1 msec, Bit Rate = 1000 bits/sec

E
= 0.5 msec
Therefore,
1
Baud Rate =

E
= 2000 Bauds
B min
 f
U

Baud
2
Rate

1000 Hz f
L

' DC ' i.e. twice the bandwidth of NRZ
Bi-Phase
Maximum frequency occurs when continuous 1’s or 0’s transmitted.
This is similar to RZ with
5

1
Baud Rate =

E
= 2 x Bit rate
B min
 f
U

Baud Rate
2
The minimum frequency occurs when the sequence is 10101010……. e.g.
In this case

B
= 
E
Baud Rate = Bit rate
B min
 f
L

Baud
2
Rate
Example:

B
= 1 msec, Bit Rate = 1000 bits/sec f

1000 Hz
U f
L

500 Hz
Comments:
6

RZ and NRZ require channel BW down to DC, i.e. DC coupled. This causes problems with drift and line capacitance. It is better to AC coupled lines.
RZ is not preferred because it has twice the bandwidth of NRZ.
Bi-phase may be AC coupled and is used for this reason and also because of other useful properties for e.g.
There is transition every bit ↑ or ↓ which eases clock generation at the receiver - describes as self clocking.
Bi-phase is actually a phase modulated signal; phase shift keying.
The main disadvantage with bi-phase is that although the BW I same as NRZ, it extends up to twice the upper frequency of NRZ.
Base band signal are not usually transmitted directly although bi-phase signals are used.
Usually NRZ signals are used to modulate carrier to produce
ASK – Amplitude Shift Keying
FSK – Frequency Shift Keying
PSK – Phase Shift Keying
These are discussed in later sections.
7

8