Pulse Code Modulation Lesson 16

Home , Analog signal, Digital signal, Telephony

Lesson 16

BME 333 Biomedical Signals and Systems 10 - J.Schesser How do we send the samples of f(t)

• Do we send the actual values of each sample? – This method is susceptible to noise • Do we convert the actual values of each sample into a binary digit and then transmit the samples as a binary stream? – Yes but? – How accurate, how precise, and how many bits (decimal places of the sample) do we use? • We use predefined levels and choose the level closest to the signal. This is called Pulse Code Modulation – Example: 8 levels needs 3 bits; 16 levels needs 4 bits, etc. – However we now have a new kind of distortion: Quantization Error.

BME 333 Biomedical Signals and Systems 11 - J.Schesser Pulse Code Modulation

0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6

Bit Slot Actual 2.823212 3.58678 4.207355 4.660195 4.927249 Code 3 4 4 5 5 Binary-Coded 011 100 100 101 101

Time Slot

BME 333 Biomedical Signals and Systems 12 - J.Schesser Shannon’s Sampling Theory Channel Capacity

C = Channel Capacity B = Channel Bandwidth L = Quantization Levels

C = 2 × B × log2 (L)

C = 2 × 3000 × log2 (8) = 18, 000bps

BME 333 Biomedical Signals and Systems 13 - J.Schesser Quantization Error

0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6

Bit Slot Actual 2.823212 3.58678 4.207355 4.660195 4.927249 Code 3 4 4 5 5 Binary-Coded 011 100 100 101 101 Absolute Error -0.17679 -0.41322 0.207355 -0.3398 -0.07275 % Error -6.26% -11.52% 4.93% -7.29% -1.48% Time Slot

BME 333 Biomedical Signals and Systems 14 - J.Schesser Signal-to-Noise Ratio 2 (A1 X S ) Psignal = RL (A X )2 P 1 noise noise = Xo R Σ A R L X + 1 L P 2 s + signal (X s ) SNR = = P X 2 noise ( noise ) Xnoise

SNRdB =10log(SNR)

Psignal SNRdB =10log Pnoise

X s SNRdB = 20log X noise BME 333 Biomedical Signals and Systems 15 - J.Schesser Shannon’s Sampling Theory Channel Capacity in Noise Shannon’s Limit C = Channel Capacity B = Channel Bandwidth SNR = Signal to Noise Ratio

C = B × log2 (1+ SNR)

C = 3000 × log2 (1+1000) = 29, 901.67bps

SNRdb = 10 × log10 (SNR)

SNRdb SNR = 10 10

SNRdb 10 C = B × log2 (1+10 )

30 10 C = 1000 × log2 (1+10 ) = 29, 901.67bps BME 333 Biomedical Signals and Systems 16 - J.Schesser A PCM System

Analog-to-digital converter Quantizer Sampler Encoder Samples of Quantized m(t) Digital Encoded Signal

Analog signal m(t) Quantizer* Decoder LP Filter Analog samples Digital-to-analog converter PCM

Digital encoded Signal Reconstructed m(t) transmitted over a with possible communications quantization errors channel Some Bit Rates: * To remove the digital signal from any noise on the communications Voice: 4kHz BL Spectrum channel =4x103 x 2 samples/sec x 8 bits/sample =64kbits/sec Video: 4MHz BL Spectrum=4x106x2x8 = 64Mb/s

BME 333 Biomedical Signals and Systems 17 - J.Schesser Examples of PCM • Digital Recording – Audio • CDs - 2 channel Stereo • DVDs - 6/7 channel– surround sound • MP3 – Video • DVDs • Telecommunications – Modern optical telecommunications • SONET/SDH – Basic SDH signal/frame – 9 rows X 270 columns = 2430 bytes – 9 rows X 9 columns = 81 bytes for overhead (line and equipment health, administration, timing, etc.) – 9 rows X 261 columns = 2349 bytes for payload – 2430 bytes x 8000 samples/second (based on telephony of 4kHz BL signal) x 8 bits/byte = line rate of 155.520 Mb/s – STS-3/STM-1 - 2349 bytes for payload x 8000 s/s x 8 bits/byte = 150.336 Mb/s • ATM Asynchronous Transfer Method • Etc. BME 333 Biomedical Signals and Systems 18 - J.Schesser Examples of PCM • Capacities – Requirements • Voice: 4k Hz x 2 = 8000 samples/s x 8 bits (256 levels) = 64 kbits/s • Video: 4M Hz x 2 = 8,000,000 samples/s x 8 bits = 64 Mb/s – SONET/SDH payload capacities: 2016 64 kbit channels in a payload of an STM-1 (301 channels can also be used for overhead) = 129.024 M bits • Cable TV supports 300 video channels = 19.2 Gigabits/s • STM-64 = 9.9 Gigabits • Need to use compression schemes to reduce the size of a video channel.

BME 333 Biomedical Signals and Systems 19 - J.Schesser Angle Coding Modulation

• Phase-shift Keying (PSK) send out a signal

y(t)=Acos[ωot+fϕ(t)] where f(t) = π for a one f(t) = -π for a zero • Frequency-shift Keying (FSK) send out a

signal y(t)=Acos(ωo± Ω)t where + Ω for a one - Ω for a zero

BME 333 Biomedical Signals and Systems 20 - J.Schesser Homework • Problem (1) – A BL signal with maximum frequency 1000 Hz is sampled at the Nyquist rate of 2000 samples per second. It is then quantized to 8 levels. What is the bit rate of the coded signal? Repeat for 16 levels and 256 levels. • Problem (2) – 20 BL signals (1000 Hz) are sampled at the Nyquist Rate and then quantized to 16 levels. Calculate the pulse width of each bit to support the multiplex of these 20 signals. Calculate the bit rate. Repeat for 200 signals. • Problem (3) – Consider N signals, each BL (1 Hz) and is quantized to 16 levels. If a transmission system can handle 40 bits per second, how many messages can be sent? Repeat for 10Mbps. • Problem (4) – Consider a BL signal with maximum frequency 1000 Hz is sampled at the Nyquist rate. The signal is quantized into 8 levels. The SNRdb for this channel is 10db. Will this channel work? What would you have to do to make this channel work?

BME 333 Biomedical Signals and Systems 21 - J.Schesser