# Exercise 4.1: PCM System 30/32

Binary display with dual code

For many years,  the  PCM system 30/32  was used in Germany,  which has the following specifications:

• It allows digital transmission of 30 voice channels in time division multiplex together with one channel each of synchronization and dial character   ⇒   the total number of channels is  $Z = 32$.
• Each individual voice channel is bandlimited to the frequency range of  $300 \ \rm Hz$  to  $3400 \ \rm Hz$.
• Each individual sample is represented by  $N = 8$  bits,  assuming the so-called  "dual code".
• The total bit rate is  $R_{\rm B} = 2.048\ \rm Mbit/s$.

The graph shows the binary representation of two arbitrarily selected samples.

Hints:

• The exercise belongs to the chapter  Pulse Code Modulation.
• Reference is made in particular to the page  PCM encoding and decoding.
• For the solution of subtask  (2)  it is to be assumed:  All speech signals are normalized and limited to the amplitude range  $±1$.

### Questions

1

What is the quantization step number  $M$?

 $M \ = \$

2

How is the sample value  "$-0.182$"  represented? With

 the bit sequence 1, the bit sequence 2, neither of them.

3

What is the bit duration  $T_{\rm B}$?

 $T_{\rm B} \ = \$ $\ \rm µ s$

4

At what distance  $T_{\rm A}$  are the speech signals sampled?

 $T_{\rm A} \ = \$ $\ \rm µ s$

5

What is the sampling rate $f_{\rm A}$?

 $f_{\rm A} \ = \$ $\ \rm kHz$

6

Which of the following statements is correct?

 The sampling theorem is not satisfied. The sampling theorem is just fulfilled. The sampling frequency is greater than the smallest possible value.

### Solution

#### Solution

(1)  With  $N = 8$  bits a total of  $2^8$  quantization intervals can be represented   ⇒   $\underline{M = 256}$.

(2)  Numbering the quantization intervals from  $0$  to  $255$,  the  "bit sequence 1"  represents

$$\mu_1 = 2^7 + 2^5 +2^4 +2^2 +2^1 +2^0 = 255 -2^6 -2^3 = 183\hspace{0.05cm},$$

and the  "bit sequence 2"  represents

$$\mu_2 = 2^6 + 2^5 +2^3 = 104\hspace{0.05cm}.$$
• With the value range  $±1$  each quantization interval has width  ${\it Δ} = 1/128$.
• The index  $μ = 183$  thus represents the interval from  $183/128 - 1 = 0.4297$  to  $184/128 - 1 = 0.4375$.
• $μ = 104$  denotes the interval from  "$-0.1875$"  to  $-0.1797$.
• The sample $-0.182$ is thus represented by  bit sequence 2.

(3)  The bit duration  $T_{\rm B}$  is the reciprocal of the bit rate  $R_{\rm B}$:

$$T_{\rm B} = \frac{1}{R_{\rm B} }= \frac{1}{2.048 \cdot 10^6\,{\rm 1/s} } \hspace{0.15cm}\underline {= 0.488\,{\rm µ s}} \hspace{0.05cm}.$$

(4)  During duration  $T_{\rm A}$   ⇒   $Z \cdot N$  binary symbols are transmitted:

$$T_{\rm A} = Z \cdot N \cdot {T_{\rm B} } = 32 \cdot 8 \cdot 0.488\,{\rm µ s} \hspace{0.15cm}\underline {= 125\,{\rm µ s}} \hspace{0.05cm}.$$

(5)  The reciprocal of  $T_{\rm A}$  is called the sampling rate:

$$f_{\rm A} = \frac{1}{T_{\rm A} } \hspace{0.15cm}\underline {= 8\,{\rm kHz}} \hspace{0.05cm}.$$

(6)  The sampling theorem would already be given by  $f_{\rm A} ≥ 2 \cdot f_\text{N, max} = 6.8 \ \rm kHz$.  Thus the last proposed solution is correct.