Pulse Code Modulation

Pulse Code Modulation EE 44 Spring Semester Lecture 9 Analog signal Pulse Amplitude Modulation Pulse Width Modulation Pulse Position Modulation Pulse Code Modulation (3-bit coding) 1

Advantages of Digital Over Analog For Communications 1. Digital is more robust than analog to noise and interference. Digital is more viable to using regenerative repeaters 3. Digital hardware more flexible by using microprocessors and VLSI 4. Can be coded to yield extremely low error rates with error correction 5. Easier to multiplex several digital signals than analog signals 6. Digital is more efficient in trading off SNR for bandwidth 7. Digital signals are easily encrypted for security purposes 8. Digital signal storage is easier, cheaper and more efficient 9. Reproduction of digital data is more reliable without deterioration 10. Cost is coming down in digital systems faster than in analog systems and DSP algorithms are growing in power and flexibility Analog signals vary continuously and their value is affected by all levels of noise.

amplitude amplitude amplitude Analog to Digital Conversion Process (ADC) Three Step Process time time time 0100101101011001 1110101010101000 0100011000100011 0101001111010101 1110110111010001 Analog Signal Analog signal is continuous in time & amplitude Sample Quantize Encode Sampling selects the data points we use to create the digital data Captured Sampled Data Values Discrete time values: few amplitudes from analog signal Quantizing chooses the amplitude values used to encode Quantized Sampled Data Now have discrete Values in both time & amplitude Encoding assigns binary numbers to those amplitude values Digital Signal Now have the digital data which is the final result Note: Discrete time corresponds to the timing of the sampling. 3

Next Topic Pulse Code Modulation Pulse-code modulation (PCM) is used to digitally represent sampled analog signals. It is the standard form of digital audio in computers, CDs, digital telephony and other digital audio applications. The amplitude of the analog signal is sampled at uniform intervals and each sample is quantized to its nearest value within a predetermined range of digital levels. Four-bit coding (16 discrete levels) 4

Allowed quantization levels Second Step Quantization I Quantization is the process of changing a continuous-amplitude signal into on with discrete amplitudes. m p m(t) Quantized samples of m q (t) t m p L = m p L = 16 levels 4 bits Maximum value = m p 5

Four-Bit Binary Pulse Code (Example) Table 5.1 on page 49 To communicate sampled values, we send a sequence of bits that represents the quantized value. For 16 quantization levels, 4 bits are required. PCM can use a binary representation of value. The PSTN uses PCM 6

Quantization II We start with a sampled signal {call it m(t)} and now we want to quantize it. The quantized amplitude is limited to a range, say from m p to +m p. (Note: the range of m(t) may extend beyond (-m p, m p ) in some cases.) Divide the range (-m p, m p ) into L uniformly spaced intervals. The number intervals is L and the separation between quantized levels is mp L The k th sample point of m(t) is designated as m(kt S ) and is assigned a value equal to the midpoint between two adjacent levels. Define: m(kt S ) = k th sample s value, and m q (kt S ) = k th quantized sample s value. Then the quantization error q(kt S ) is equal to m q (kt S ) - m(kt S ) 7

Error Generated by Quantization (Quantization Noise) Quantization noise q( t) m( t) m ( t) q Quantization fluctuation or noise 8

The quantized levels are separated by Quantization III The maximum error for any sample point s quantized value is at most ½. The time average mean-square quantization error is Let N q = q. Thus N q is proportional to the fluctuation of the error signal. This is usually called quantization noise. We know that m(t) = m q (t) + q(t) q m p m L The signal (or message) power S 0 is proportional to the square of m(t), thus 3L 1 p S m ( t), but if m( t) is sinusoidal, S 0 0 Note: denotes time average m p 9

Quantization IV We want a measure of the quality of received signal (that is, the ratio of the strength of the received signal S 0 relative to the strength of the error N q due to quantization). This is the Signal-to-Quantization Noise Ratio (SQNR) and is given by S m ( t) m ( t) 0 SQNR 3L N q m m p p 3L m t It is usually expressed in decibels, Note: () 1 m p SQNR db S 0 3L 10log10 10log 10 Nq Conclusion: To reduce the quantization error relative to the message signal level, use smaller quantization steps. 10

The Dilemma of Strong Signals versus Weak Signals Strong Signal Weak Signal (a) Linear encoding Note different encoding levels on each side. (b) With non-linear encoding Companding 11

Use Compression and Expansion Companding Compression Restoration m(t) (m) m(t) m(t) http://www.slideshare.net/91pratham/unit-ipcmvsh 1

Output (y/y max ) Output (y/y max ) Companding Laws A-Law Companding (Europe) -Law Companding (North America) Input (m/m p ) A m m 1 y for 0 1 log e A m p mp A A Am 1 m y 1 log e for 1 1 log e A m p A mp y Input (m/m p ) 1 m m log e 1 for 0 1 log e(1 ) m p mp 13

Flattening of the S/N Ratio Using the -Law For optimal S 0 /N q ratio in North America = 55 is used. An approximately constant S 0 /N q ratio is the most desirable. S N 0 q (8 bits) Relative signal power S 0 (db) 14

Transmission Bandwidth In binary PCM, we have a group of n bits corresponding to L levels with n bits. Thus, L = n or n = log (L) Signal m(t) is band-limited to B Hz which requires B samples per second. For nb elements of information, we must transfer nb bits/second. Thus, the minimum bandwidth B T needed to transmit nb bits/second is B T = nb Hz Practically speaking, usually we choose the transmission bandwidth to be a little higher than the minimum bandwidth required. 15

Example Problem: A band-limited signal m(t) of 3 khz bandwidth is sampled at rate of 33⅓ % higher than the Nyquist rate. The maximum allowable error in the sample amplitude (i.e., the maximum quantization error) is 0.5% of the peak amplitude m p. Assume binary encoding. Find the minimum bandwidth of the channel to transmit the encoded binary signal. Solution: The Nyquist rate is R N = x 3000 Hz = 6000 Hz (samples/second), but the actual rate is 33⅓ % higher, so that is 6000 Hz + (⅓ x 6000) = 8000 Hz. The quantization step is and the maximum quantization error is plus/minus /. Hence, we can write mp 0.5 mp L 00 L 100 For binary coding, L, must be a power of two; therefore, knowing that L = 7 = 18 and 8 = 56, we must choose n = 8 to guarantee better than a 0.5% error. 16

Example Continued Solution (continued): Having chosen n = 8 to guarantee < 0.5% error, to find the bandwidth required we start with Total number of bits per second C = 8 bits 8000 Hz = 64,000 bits/second However, we know we can transmit bits/hz of bandwidth, so it requires a bandwidth B T of B T = C/ = 3,000 Hz = 3 khz If 4 such signals are multiplexed onto a single line (known as a T1 Line in the Bell telephone system), then C T1 = 4 x 64 kb/s = 1.536 Mb/s, and the bandwidth is 768 khz A maximum of B independent elements of information per second can be transmitted, error-free, over a noiseless channel of bandwidth B Hz. 17

Exponential Increase of the Output SNR (S/N Ratio) We start with the SNR (signal-to-noise ratio) equation from slide 10 above: S N () 3 S0 m t q mp N 0 q m () t 3 L m p The number of levels L can be expressed as L = n where n = log (L) and is the number of bits to generate L levels. The SNR can now be expressed as Using the expression for bandwidth, B T = nb, then we arrive at S N Taking the logarithm gives 0 q m () t 3 m p n B / B S 0 S m () t 0 10 log 10 10 log 10 3 10 nlog 10 6n db Nq Nq m p db T denotes time average 18

SNR Example Given a sinusoidal modulating signal m(t) of amplitude A m into a load resistance R = 1 ohm, find the signal-to-quantization noise ratio (sometimes called SNR): Setting m max = A m Am Pave and set mmax A S m () t 0 3Pave 3 3 () () () Nq m p mmax n n n m L n SNR S 0 10log10 1.76 6n db Nq 3 5 31.8 db 64 6 37.8 db 18 7 43.8 db 56 8 49.8 db 19

Bell System s T1 Carrier System (196) The T-carrier is a member of the group of carrier systems developed by AT&T Bell Laboratories for digital transmission of multiplexed telephone calls using Pulse Code Modulation and Time Division Multiplexing. The first version, the Transmission System 1 (T1), was introduced in 196 in the Bell System, and could transmit up to 4 telephone calls simultaneously over a single transmission line consisting of copper wire. 193 bit frame 1 sec/frame 1.544 Mbit/s data rates 0

T1 Carrier Time Division Multiplexing 1

Comparison of T-Carrier (North America) and E-Carrier (Europe) Carrier Level T-Carrier Data Rates E-Carrier Data Rates Zero-level 64 kbits/s (DS-0) 64 kbits/s First-level Second-level Third-level Fourth-level Fifth-level 1.544 Mbits/s (DS-1) T1 4 channels 6.31 Mbits/s (DS-) T 96 channels 44.736 Mbits/s (DS3) T3 67 channels 74.176 Mbits/s (DS4) T4 403 channels 400.35 Mbits/s (DS5) T5 5760 channels.048 Mbits/s (E1) 3 user channels 8.448 Mbits/s (E) 18 channels 34.368 Mbits/s (E3) 51 channels 139.64 Mbits/s (E4) 048 channels 565.148 Mbits/s (E5) 819 channels

Worked PCM Example We are given a signal m(t) = cos(50 t) as a single-tone signal input. (a) Find the SNR with 8-bit PCM. For 8-bit encoding, L = n where n = 8, therefore, the number of levels = 56. The amplitude A m of the sinusoidal waveform means that m p = volts. The total signal swing possible (- m p to + m p ) will be m p = 4 volts, therefore, the average signal power is P ave = [(A m ) /] = [ /] = watts. (See slide 19) The interval = [m p /L] = 4 volts/56 levels = 1.565 10 - volt. (See slide 19) Now we can find the SNR (signal-to-quantized noise ratio) (See side 18) Using for the quantization noise N q = [ /1], and taking P ave = W, the SNR is given by S P ave 4 1 98, 304 Nq Nq (1.56510 ) SNR 10log 98, 304 49.93 db db 10 3

Worked Example for PCM (continued) We are given a signal m(t) = cos(50 t) as the signal input. (b) If the minimum SNR is to be at least 36 db, how many bits n are needed to encode the signal (i.e., find n)? Other parameters such as signal power remain the same as in part (a) on previous slide. Note that 36 db is numerically equivalent to 3,981. Remembering that the interval is = [m p /L] and m p = 4 volts. m p 4 4 3, 981 ; 0.001005 and 0.0317 volt 3981 Therefore, we can determine the number of levels L and then find n. m p 4 L 31.5 0.0317 The lowest integer number of bits n that will give at least 31.5 levels is n = 5 because 5 = 3 levels. So the answer is 5 bits. 4

Differential Pulse Code Modulation (DPCM) PCM is not really efficient because it generates so many bits taking up a lot of bandwidth. Can we improve on this? YES. Suppose we have a slowly varying signal m(t), then we exploit this by using the difference between two adjacent samples. This will form the basis of differential pulse code modulation (DPCM). Let m[k] be the k th sample reading of signal m(t). Then we can express the difference between two adjacent samples as d[k] = m[k] m[k-1] Principle: Instead of transmitting m[k], we transmit d[k]. 5

Differential Pulse Code Modulation (continued) At the receiver knowing d[k] and the previous value of m[k-1] allows us to construct the value of m[k]. How do we benefit from doing this? The difference of successive samples almost always is much smaller than the full range of the sample values of m(t) (full range covers -m p to +m p ). We use this fact to improve upon the efficiency of PCM by requiring fewer bits. Furthermore, we can make use of the estimate of m[k], denoted by m est [k]. We use previous sample values of m(t) to make this estimate. Suppose m est [k] is the estimate of the k th sample, then the difference d[k] is defined by d[k] = m[k] m est [k] and it is the difference d[k] that is transmitted. 6

Differential Pulse Code Modulation (continued) Receiver Concept: At the receiver we determine the estimate m est [k] from previous sample values, and then generate m[k] by adding the received d[k] values to the estimate m est [k]. Thus the reconstruction of the samples is done iteratively. How do we carry out such an estimation? 7

Digression on Signal Prediction Starting with a Taylor series (with time step T s ), 3 3 d( m( t)) TS d ( m( t)) TS d ( m( t)) m[ t TS] m( t) TS... 3 dt! dt 3! dt dm() t m[ t TS ] m( t) TS for small TS dt We denote the k th sample of m(t) by m[k], that is, m[kt S ] = m[k], and m[kt S T S ] = m[k 1], and so on. This is a first-order predictor. In handling the derivatives, we write Thus, So we get an approximation of the (k+1) th sample, m[k+1], from the two prior samples, namely m[k] and m[k-1]. d dt m[ k] m[ k 1] m[ k 1] m[ k] TS TS m[ k 1] m[ k] m[ k 1] m( kts ) m( kts TS ) m ( kt S ) T S 8

Signal Prediction (continued) But we can do even better than this. In general, m[ k] a m[ k 1] a m[ k ]... a m[ k N] m [ k] 1 The set of {a i } are the predictor coefficients. This is the predicted value of m[k]. It is an N th order predictor. Note that the input consists of the weighted previous samples m[k-1], m[k-], etc. We say that input m[k] gives output m est [k]. For a first-order prediction, m est [k] = m[k-1]. The next slide shows how to implement this prediction of m[k]. N q 9

Linear Predictor Implemented With Transversal Filter m [ k] a m[ k 1] a m[ k ]... a m[ k N] est 1 N Input m[k] Delay T S Delay T S Delay T S Delay Delay... T T S S a 1 a a 3 a N Output m est [k] Transversal filter is a tapped delay line (with required weights {a i } ) 30

DPCM Transmitter Input m[k] + d[k] Quantizer Output d q [k] m est [k] + + Predictor m q [k] d[ k] m[ k] m [ k] and is quantized to yield, est d [ k] d[ k] q[ k] where q[ k] is the quantization error q The predictor output m est [k] is fed back to the input so the predictor input m q [k] is given by m [ k] m [ k] d [ k] m[ k] d[ k] d [ k] m[ k] q[ k] q q q q This shows that m q [k] is the quantized version of m[k]. 31

DPCM Receiver Input d q [k] + Output m q [k] + m est [k] Predictor The receiver s output (which is the predictor s input) is also the same, m q [k] = m[k] + q[k]. Hence, we are able to receive the desired signal m[k] plus the quantization noise, q[k]. It is important to note that from the difference signal d[k] is much smaller that the noise associated with m[k]. 3

DPCM SNR Improvement How much better is DPCM with regard to SNR? To determine this, define m p and d p as the peak amplitudes of m(t) and d(t), respectively. Assuming the same number of steps L for both, then the quantization step in DPCM is reduced in magnitude by d p /m p. The quantization noise is proportional to () the quantization noise power is reduced by a factor (d p /m p ) and the SNR is therefore increased by (m p /d p ). Maintaining the same SNR, the number of bits can be reduced. Example: The AT&T telephone system sometimes operates at 3 kbits/s (or even 4 kbits/s) when using DPCM. [The telephone system was initially designed to use a 64 kbits/second data rate.] 33

Adaptive Differential PCM Adaptive differential PCM (ADPCM) can further improve upon DPCM by Incorporating an adaptive quantizer (variable ) at encoding. The quantized prediction error d q [k] is a good measure of the predicted error size it can be used to change which minimizes d q [k]. When d q [k] fluctuates around large positive or negative values, the prediction error is large and needs to increase, but when d q [k] fluctuates around zero (small values), then needs to decrease. m[k] + Adaptive Quantizer n th order Predictor d q [k] To Channel Example: An 8-bit PCM sequence can be encoded into a 4-bit ADPCM sequence at the same sampling rate. This reduces the channel bandwidth by one-half with no loss in quality. 34

Adaptive Differential PCM Output Example time 35

Next Topic is Delta Modulation 36