What is the minimum bandwidth for transmitting data at a rate of 33.6 kbps without ISI?
Answer:
The minimum bandwidth is equal to the Nyquist bandwidth. Therefore, (BW)min = W = Rb/2 = 33.6/2 = 16.8 kHz
• Note: If a 100% roll-off characteristic is used, bandwidth = W(1+α) = 33.6 kHz
BT.67
Example
Bandwidth requirement of the T1 system
T1 system
– multiplex 24 voice inputs, based on an 8-bit PCM word.
– bandwidth of each voice input (B) = 3.1 kHz
For converting the voice signal into binary sequence,
• The minimum sampling rate = 2B = 6.2 kHz
• Sampling rate used in telephone system =8 kHz
BT.68
1
Example
With a sampling rate of 8 kHz, each frame of the multiplexed signal occupies a period of 125µs.
:
:
8 bit from
1st input
No. of bits = 8 ·24+1=193
8 bit from
2nd input
….
8 bit from
1 bit for
24th input Synchronization
125 µs
BT.69
Example
Correspondingly, the bit duration is 125 µs/193 = 0.647 µs.
For eliminating ISI, the minimum transmission bandwidth is
1 / 2Tb = 772kHz
BT.70
2
Eye diagrams
This is a simple way to give a measure of how severe the ISI
(as well as noise) is.
This pattern is generated by overlapping the incoming signal elements. Example: bipolar NRZ PAM
1
0
1
1
0
0
Tb
BT.71
Eye diagrams
Eye pattern is often used to monitoring the performance of baseband signal.
– The best time to sample the received waveform is when the eye opening is largest.
Effects of noise are ignored
1
BT.72
3
Eye diagrams
The maximum distortion and ISI are indicated by the vertical width of the two branches at sampling time.
2
BT.73
Eye diagrams
The noise margin or immunity to noise is proportional to the width of the eye opening.
3
BT.74
4
Eye diagrams
The sensitivity of the system to timing errors is determined by the rate of closure of the eye as the sampling time is varied. 4
BT.75
Equalization
In preceding sections, raised-cosine filters were used to eliminate ISI. In many systems, however, either the channel characteristics are not known or they vary.
Example
The characteristics of a telephone channel may vary as a function of a particular connection and line used.
It is advantageous in such systems to include a filter that can be adjusted to compensate for imperfect channel transmission characteristics, these filters are called equalizers. BT.76
5
Before equalization
After equalization
BT.77
Transversal filter (zero-forcing equalizer)
xk
T is the bit duration.
BT.78
6
Equalization
The problem of minimizing ISI is simplified by considering only those signals at correct sample times.
The sampled input to the transversal equalizer is x(kT ) = x k
x0
The output is
x2
y (kT ) = y k
For zero ISI, we require that
1 k = 0 yk =
0 k ≠ 0 …(*)
x1
BT.79
aN xk − N
The output can be expressed as yk =
N
∑ an xk −n
xk
n=− N
−N ≤k≤N
a0 xk a− N xk + N
There are 2N+1 independent equations in terms of an . This limits us to 2N+1 constraints, and therefore (*) must be modified to
1
k =0 yk =
0 k = ±1,±2,...,± N
BT.80
7
Equalization
The 2N+1 equations becomes
xo x−1 L x− N L x− 2 N −1
x0
L x− N +1 L x− 2 N
x1
M
M
xN −1 L x0 L x− N −1
xN
M
M
x2 N −1 x2 N − 2 L xN −1 L x− 2
x
x2 N −1 L xN x−1 L
2N
x− 2 N a − N 0
x− 2 N +1 a− N +1 0
M M
x− N a0 = 1
M M
x−1 a N −1 0 x0 a N 0
BT.81
Example
Determine the tap weights of a three-tap, zero-forcing equalizer for the input where x− 2 = 0.0, x−1 = 0.2, x0 = 1.0, x1 = −0.3, x2 = 0.1 , xk = 0 for k > 2
The three equations are
+ 0.2a0 a−1 − 0.3a−1
+ a0
+ 0.2a1
0.1a−1
− 0.3a0
+ a1
N=1
=0
=1
=0
Solving, we obtain a−1 = −0.1779, a1 = 0.2847, a0 =