Pulse Code Modulation PCM, Slides of Digital Communication Systems

Download Pulse Code Modulation PCM and more Slides Digital Communication Systems in PDF only on Docsity! Communications Dr. Ahmed Gomaa Electronics and Communications Engineering Department El-Madina Higher Institute for Engineering and Technology Spring 2021 2 Pulse Code Modulation PCM  Unipolar uniform quantization: this method is used in case of unipolar Signal. Where is divides the level of signal into an equal values M. The value of M is changed depending on the number of bits N, that spicificated to codificate each level . Actually there are some parameters we have to know how to calculated. These parameters are as the following: 1- Normalized Step Size denoted by Δxu and calculated by 2- Actual Step Size that denoted by Δvu and calculated by Where is Vfs is denoted with Full-Scale Voltage L Vp L Vfs  5 Pulse Code Modulation PCM - To calculate signal to noise ratio SNR is used the following function - To calculate signal to noise ratio SNR in dB is used the following function 22 222 2 2 2 , 4 * 2 LMSNRhence eq MVpp Pswhere M eq Ps Pn Ps SNR             )log(10)log(10 2 2 M eq Ps SNRdB  Power is often expressed in dBm (dBm is dB from 1mW) Example: if the power, P = 10 mW then P in dBm is P=10mW=10log(10) if the power is in dBm then Example: if the power, PdBm = 27 dBm then find P in mW P =27 dBm = mWmWdBP 501)10(127 10/27  Pulse Code Modulation PCM 7 Pulse Code Modulation PCM  Example: for the following voice signals: Is required converting it into a digital signals using 4 bits for a quantization, so calculate the following parameters: 1- signal power Ps 2- sampling frequency that required to obtain 2500 samples per second. 3- quantization step size 4- Quantization error 5- it’s Signal to nose ratio SNR in dB 7- number of bits that is required to obtain a SNR = 40dB 8- minimum storage capacity Solution tctf tctf   1000sin10)( 500sin3)( 2 10 bitsNthen NthenMwhere levelsofpowerbemustM M M M eq Ps SNR N dB 7 )128(2log2 128)2(100 1000010 )log(1040 )log(10)log(10 42 2 2 2       7- number of bits that is required to obtain a SNR = 40dB Pulse Code Modulation PCM 8- minimum storage capacity. Capacity=n. of samples * n of bits per sample 7*2500=17500bits Then capacity= 17.5Kbits 11 Pulse Code Modulation PCM 1- for the analog signal that shown in the figure do the following: a- draw the quantization curve, corresponded that the sampling frequency fs=50KHZ, and N=4 bits. b- Make a codification of the quantized signal. c- total number of bits (capacity). Pulse Code Modulation PCM 10v Vin 8.75v 7.50v 6.25v 5.00v 3.75v 2.50v 1.25v 0.00v resistor a] Digital Code _— | | | | | | ! a4 t—— Output Octal to - Binary Encoder Comparator Example- If Vin = 6.00v, then the first 4 comparators from the bottom will return a logic high signal while the top three will return a low signal. Octal 4 = Binary 100 15 Delta Modulation Types of Digital Modulation: 1- Pulse code Modulation. 2- Differential Pulse Code Modulation. 3- Delta Modulation. 4- Adaptive Delta Modulation. 16  Differential Pulse Code Modulation: differential Pulse Code Modulation based on codificate the difference between the value the pulse at a moment X(K) and the value of the pulse at the previous moment X(K-1). By this method a little number of bits is necessary to codificate the difference. This system can be used with digital signals and analog signals but analog signal must be firstly inter to sampler and then to DPCM Encoder. Delta Modulation 17 Delta Modulation  Differential Pulse Code Modulation: the following figure shows the circuit diagram for Differential Pulse Code Modulation (DPCM). 20 Delta Modulation  Delta Modulation: Delta Modulation system corresponded as a simple model of Differential Pulse Code Modulation, where is compared each sample with the previous sample and then is specificated if mayor or minor. In this system is utilized only one bit ‘1’ or ‘0’ and this is don by, if the sample value is mayor than the previous sample the result will be ‘1’ and will be ‘0’when the sample value is smaller than the previous.  Dela Modulation block diagram shown in the following fig 21  Example: for the analog signal shown in the following fig. it is required converting it to a digital signal by using Delta Modulation system. 1- Draw the output quantized signal after modulation 2- codificate the quantized signal. Solution Delta Modulation 22 Delta Modulation 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 Example: for the analog signal shown in the following fig. is required converting it to a digital signal by using Delta Modulation system. 1-Draw the output quantized signal after modulation corresponding that sampling frequency fs= 1MHz 2- codificate the quantized signal.