Pulse Code Modulation Lesson 16, Exams of Signals and Systems

Download Pulse Code Modulation Lesson 16 and more Exams Signals and Systems in PDF only on Docsity! BME 333 Biomedical Signals and Systems - J.Schesser 10 Pulse Code Modulation Lesson 16 BME 333 Biomedical Signals and Systems - J.Schesser 11 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 - J.Schesser 14 Quantization Error 0 1 2 3 4 5 6 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 Time Slot 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% 15 Signal-to-Noise Ratio ( ) ( ) noise s dB noise signal dB dB noise s noise signal L noise noise L S signal X XSNR P P SNR SNRSNR X X P P SNR R XAP R XAP log20 log10 )log(10 )( )( 2 2 2 1 2 1 = = = == = = RL Σ Xo Xs A1 + + Xnoise BME 333 Biomedical Signals and Systems - J.Schesser 16 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) SNR =10 SNRdb 10 C = B × log2 (1+10 SNRdb 10 ) C =1000 × log2 (1+10 30 10 ) = 29, 901.67bps BME 333 Biomedical Signals and Systems - J.Schesser BME 333 Biomedical Signals and Systems - J.Schesser 19 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 - J.Schesser 20 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 - J.Schesser 21 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?