2071 Chaitra(Q).Explain E1 digital hierarchy as related to the telephony system. Evaluate the expression of SQNR in a uniform quantized PCM system :

E1 digital hierarchy as related to telephony system:

Vf band frequency: 4 kHz (300Hz-3400Hz)

Sampling frequency: 8kHz

Sampling rate = 8000 cycles per second

i.e 8000 frames per second

In each frame:

Time slot 0 or channel 0: – synchronous channel or frame alignment channels

Time slot 16 or channel 16:- Signaling channel for all 30 Vf channels.

Time slot or Vf channel {1-15 & 17-31} 30 Vf

or Telephony channels.

Each time slot or channel consists of an 8-bit digital signal.

->0 channel -> framing code

->16 channel -> signaling code

channel {1-15 and 17-31} information

-> Total no. of bits in a frame = 32*8=256 bits/frame

i.e 125 μ => 256 bits

Thus, bit rate =2.048 Mbps

Channel Speed = 8 bits/sample*8000 sample/sec

=64 Kb/Sec

SQNR in uniformly quantized PCM System:

We know that in a PCM system for linear quantization, the signal-to-quantization noise ratio is given as,

S/N= Normalized Signal Power/Normalized Noise Power

But, Normalised noise power has been calculated as

Therefore, S/N = Normalised Signal Power/(△² /12)……….(a)

We know that the number of bits ‘n’ and quantization levels are related as,

q= 2ⁿ ……………..(b)

let us assume that i/p X(nTs) to a linear quantizer has a continuous amplitude in the range -X_max to +X_max. Therefore, the total amplitude range

=X_max-(-X_max) = 2X_max

Now, the step size will be

△ = 2X_max/q……………………………..(c)

Here, substituting this value in eq(a), we get

S/N = Normalized Signal Power/((2X_max/2ⁿ)*1/12)

Let the Normalized Signal Power be denoted as ‘P’.

Then, S/N = P/((4X²_max/2^2q )*1/12) = 3p*2²ⁿ/X_max² ………….(d)

Hence, eq (d) is the required relation for the signal to quantization noise ratio for linear quantization in a PCM system.

This expression shows that signal to noise power ratio of the quantizer increases exponentially with increasing bits per second.