Pulse Code Modulation Lesson 16, Exams of Signals and Systems

– 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 ...

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BME 333 Biomedical Signals and Systems
- J.Schesser
10
Pulse Code Modulation
Lesson 16
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BME 333 Biomedical Signals and Systems 10

Pulse Code Modulation

Lesson 16

BME 333 Biomedical Signals and Systems 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.

13

Shannon’s Sampling Theory

Channel Capacity

C = Channel Capacity B = Channel Bandwidth L = Quantization Levels C = 2 × B × log 2 ( L ) C = 2 × 3000 × log 2 ( 8 ) = 18 , 000 bps BME 333 Biomedical Signals and Systems

BME 333 Biomedical Signals and Systems 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. 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. Time Slot Bit Slot Actual 2.823212 3.58678 4.207355 4.660195 4. Code 3 4 4 5 5 Binary-Coded 011 100 100 101 101 Absolute Error -0.17679 -0.41322 0.207355 -0.3398 -0. % Error -6.26% -11.52% 4.93% -7.29% -1.48%

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 × log 2 ( 1 + SNR ) C = 3000 × log 2 ( 1 + 1000 ) = 29 , 901. 67 bps SNRdb = 10 × log 10 ( SNR ) SNR = 10 SNRdb 10 C = B × log 2 ( 1 + 10 SNRdb (^10) ) C = 1000 × log 2 ( 1 + 10 30 (^10) ) = 29 , 901. 67 bps BME 333 Biomedical Signals and Systems

BME 333 Biomedical Signals and Systems 17

A PCM System

Sampler Quantizer^ Encoder Quantizer* Decoder^ LP Filter Analog-to-digital converter Analog signal m(t) Analog samples PCM Digital encoded Signal transmitted over a communications channel

  • To remove the digital signal from any noise on the communications channel Digital Encoded Signal Samples of Quantized m(t) Reconstructed m(t) with possible quantization errors Digital-to-analog converter Some Bit Rates: Voice: 4kHz BL Spectrum =4x10^3 x 2 samples/sec x 8 bits/sample =64kbits/sec Video: 4MHz BL Spectrum=4x10^6 x2x8 = 64Mb/s

BME 333 Biomedical Signals and Systems 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 20

Angle Coding Modulation

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

y ( t ) =Acos [ ω

o

t+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 ( ω

o

± Ω) t where

+ Ω for a one

- Ω for a zero