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Analogue to Digital Conversion (PCM and DM) 1. The advantages offered by digital modulation Performance Digital pulse modulation permits the use of regenerative.

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Presentation on theme: "Analogue to Digital Conversion (PCM and DM) 1. The advantages offered by digital modulation Performance Digital pulse modulation permits the use of regenerative."— Presentation transcript:

1 Analogue to Digital Conversion (PCM and DM) 1

2 The advantages offered by digital modulation Performance Digital pulse modulation permits the use of regenerative repeaters, when placed along the transmission path at short enough distances, can practically eliminate the degrading effects of channel noise and signal distortion. Ruggedness A digital communication system can be designed to withstand the effects of channel noise and signal distortion Reliability Can be made highly reliable by exploiting powerful error-control coding techniques. Security Can be made highly secure by exploiting powerful encryption algorithms Efficiency Inherently more efficient than analogue communication system in the trade-off between transmission bandwidth and signal-to-noise ratio System integration To integrate digitized analogue signals with digital computer data 2

3 3

4 Digitizing Analogue Data 4

5 Pulse Code Modulation (PCM)  The simplest technique for transforming analogue data into digital signals is pulse code modulation (PCM).  sampling theorem: “If a signal is sampled at regular intervals at a rate higher than twice the highest signal frequency, the samples contain all information in original signal” eg. 4000Hz voice message, requires 8000 samples per second  strictly have analog samples Pulse Amplitude Modulation (PAM)  assign each a digital value 5

6 Figure Components of PCM encoder 6

7 Figure Three different sampling methods for PCM 7 Sampling called pulse amplitude modulation (PAM), the result is still analogue signal with non-integral value.

8 Figure Nyquist sampling rate for low-pass and bandpass signals 8

9 For an intuitive example of the Nyquist theorem, let us sample a simple sine wave at three sampling rates: f s = 4f (2 times the Nyquist rate), f s = 2f (Nyquist rate), and f s = f (one-half the Nyquist rate). Figure 4.24 shows the sampling and the subsequent recovery of the signal. Example 9

10 PCM Block Diagram

11 Quantisation The results of sampling is a set of amplitude which can be infinite of non- integral values between two limits V min and V max. Divide the range into L steps, each of height Δ We approximate the sampled value to the quantised value (midpoint) (quantisation error) 11

12 Figure 4.26 Quantization and encoding of a sampled signal 12 Δ=[20V-(-20V)]/8 = 5V

13 13 Quantisation level L, depends on the range of the amplitudes and how accurately we need. L= 2 n, audio communication, L=256. less L more quantisation error. Quantisation error: Quantisation is a approximation process. –Δ/2 ≤ error ≤ Δ/2 The contribution of the Quantisation error to the SNR dB of the signal depends on the number of the quantisation level L, or the bits per sample n b. n b = log 2 L signal to quantisation error (or noise) ratio SNR db = 6.02n b + 1.76 dB Each additional bit used for quantizing increases SNR by about 6 dB, which is a factor of 4.

14 What is the SNR dB in the example of Figure 4.26? Solution We can use the formula to find the quantization SNR. We have eight levels and 3 bits per sample, so SNR dB = 6.02×(3) + 1.76 = 19.82 dB Increasing the number of levels increases the SNR. Example 14

15 15 Encoding Each quantised value can be coded to an n b bit code word. Bit rate = sampling rate × number of bit per sample = f s × n b = f s × log 2 L We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? The human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate and bit rate are calculated as follows:

16 PCM bandwidth 16 The minimum bandwidth of a PCM signal B min = n b × B analogue We have a low-pass analog signal of 4 kHz. If we send the analog signal, we need a channel with a minimum bandwidth of 4 kHz. If we digitize the signal and send 8 bits per sample, we need a channel with a minimum bandwidth of 8 × 4 kHz = 32 kHz. Maximum data rate of a channel R max = 2 × B × log 2 L Minimum required bandwidth

17 Delta Modulation (DM)  PCM finds value of amplitude of each sample; DM finds the change from the previous sample.  analog input is approximated by a staircase function can move up or down one level (Δ) at each sample interval  has binary behavior function only moves up or down at each sample interval hence can encode each sample as single bit 1 for up or 0 for down

18 Figure The process of delta modulation 18

19 Figure Delta modulation and demodulation components 19

20 PCM verses Delta Modulation  DM has simplicity compared to PCM but has worse SNR  issue of bandwidth used for good voice reproduction with PCM: want 128 levels (7 bit) & voice bandwidth 4khz need 8000 x 7 = 56kbps  data compression can improve on this  still growing demand for digital signals use of repeaters, TDM, efficient switching  PCM preferred to DM for analog signals 20


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