What is the minimum bandwidth for transmitting data at a rate of 33.6 kbps without ISI?
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
Bandwidth requirement of the 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
With a sampling rate of 8 kHz, each frame of the multiplexed signal occupies a period of 125µs.
8 bit from
No. of bits = 8 ·24+1=193
8 bit from
8 bit from
1 bit for
24th input Synchronization
Correspondingly, the bit duration is 125 µs/193 = 0.647 µs.
For eliminating ISI, the minimum transmission bandwidth is
1 / 2Tb = 772kHz
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
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
The maximum distortion and ISI are indicated by the vertical width of the two branches at sampling time.
The noise margin or immunity to noise is proportional to the width of the eye opening.
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
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.
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
Transversal filter (zero-forcing equalizer)
T is the bit duration.
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
The output is
y (kT ) = y k
For zero ISI, we require that
1 k = 0 yk =
0 k ≠ 0 …(*)
aN xk − N
The output can be expressed as yk =
∑ an xk −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
k =0 yk =
0 k = ±1,±2,...,± N
The 2N+1 equations becomes
xo x−1 L x− N L x− 2 N −1
L x− N +1 L x− 2 N
xN −1 L x0 L x− N −1
x2 N −1 x2 N − 2 L xN −1 L x− 2
x2 N −1 L xN x−1 L
x− 2 N a − N 0
x− 2 N +1 a− N +1 0
x− N a0 = 1
x−1 a N −1 0 x0 a N 0
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
Solving, we obtain a−1 = −0.1779, a1 = 0.2847, a0 =