Spectrum of PCM signal depends on Bit rate: Correlation of PCM data PCM waveform (pulse shape) Line encoding. For no aliasing:

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2 Spectrum of PCM signal depends on Bit rate: Correlation of PCM data PCM waveform (pulse shape) Line encoding For no aliasing: Bandwidth of PCM waveform:

3 Quantizing noise caused by the M-step quantizer Bit errors in the recovered PCM signal (channel noise + improper channel filtering ISI) Aliasing noise Intersymbol Interference Average Signal Power Average Noise Power = probability of bit error # of quantization levels 6-dB Law: Depends on: input waveshapes quantification characteristics

4 A-law Characteristics (Europe: A=87.6) m-law Characteristics (US, Canada, Japan: m=255)

5 m =255 Quantizer

6 Compandor (Compressor + Expandor) SNR for Different Quantizers Uniform quantizing: m-law companding: A-law companding: n: # of bits used in the PCM word V: the peak design level of the quantizer x rms : the rms value of the input analog signal V/x rms : loading factor

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8 Analog voice signal: Hz f s khz=6.8 khz & peak percentage error: 10

9 DARBE KOD MODÜLASYONU (PCM)

10 Düzgün Kuantalama Uniform Quantization Boş kanal gürültüsünü önleyici düzgün kuantalama eğrisi Giriş a adımından küçükse, daima 0 çıkışı elde edilir.

11 Düzgün Olmayan Kuantalama Nonuniform Quantization Sıkıştırma (compression) ve genleştirme (expansion) eğrileri. A/D çeviricide sıkıştırma yapılmışsa, D/A çeviricide genleştirme işlemi yapılmalıdır. Düzgün Kuantalayıcı W Bazı haberleşme sistemlerinde, sıkıştırma işlemi doğrudan analog ses işareti üzerinde yapılır.

12 Çok kanallı sistemlerde kullanılan işaret seviyesi değişimi (düzgün olmayan sıkıştırma + düzgün kuantalama) A-tipi sıkıştırma m-tipi sıkıştırma

13 A-tipi sıkıştırma eğrisinin parçalı gösterimi III II I Lokal kuantalama seviye (adım) sayısı: M IV V VI VIII VII Amaç; giriş genliğinin herhangi bir değeri için belirli sınırlar içinde kalan bir kuantalama hatası elde etmektir.

14 Örnek işaret biti 0 (+) 1 (-) parça numarasını belirleyen bitler: 2 nolu parça parça içinde işaretin kaçıncı dilime karşı geldiğinin belirlenmesi için kullanılan bitler: 3. dilim kodlanmış işaretin genliği: (0.25/16) =

15 1 Hertz can transmit a maximum of 2 pieces of information per second Noiseless channel of B Hz can transmit a signal of B Hz error-free Can reconstruct this signal with 2B samples Thus, channel of B Hz can transmit 2B pieces of information or 2 pieces of information/hertz Minimum theoretical channel bandwidth is: B T = n B hertz Information / Hz

16 Transmission Bandwidth Binary systems M (# of levels) = 2 n or n=log 2 M Signal m(t), bandlimited to B m Hz requires at least 2B m samples/sec (Nyquist). For reconstruction, we need 2nB m bits/sec or 2nB m pieces of information T 2T 3T 4T 5T 6T t Maximum Signal Rate: D Encoder Transmission System/Channel Bandwidth=B Decoder Nyquist Theorem (1920): For a system/channel bandwidth B, T min =1/2B maximum signal rate: D=2B pulses/sec (baud rate, Baud) = 2Blog2M bits/sec (bit rate, bps) To transmit data in bit rate D, the minimum bandwidth of a system/channel must be B D/2log 2 M (Hz)

17 Relationship between Transmission Speed and Noise t t Encoder Transmission System/Channel Bandwidth=B s(t) + Decoder Maximum Signal Rate Channel Capacity Noise n(t) Shannon Theorem (1948): For a system/channel bandwidth B and signal-to-noise ratio S/N, its channel capacity is, C = Blog2(1+S/N) bits/sec (bps, bit rate) C is the maximum number of bits that can be transmitted per second with a Pe=0. To transmit data in bit rate D, the channel capacity of a system/channel must be C D

18 Channel Capacity Shannon theorem C = Blog2(1+S/N) shows that the maximum rate or channel Capacity of a system/channel depends on bandwidth, signal energy and noise intensity. Thus, to increase the capacity, three possible ways are 1) increase bandwidth; 2) raise signal energy; 3) reduce noise. Shannon theorem tell us that we cannot send data faster than the channel capacity, but we can send data through a channel at the rate near its capacity. Examples 1. For an extremely noise channel S/N 0, C 0, cannot send any data regardless of bandwidth 2. If S/N=1 (signal and noise in a same level), C=B 3. The theoretical highest bit rate of a regular telephone line where B=3000Hz and S/N=35dB. 10log10(S/N)=35 log2(s/n)= 3.5x log210 C= Blog2(1+S/N) =~ Blog2(S/N) =3000x3.5x log210=34.86 Kbps If B is fixed, we have to increase signal-to-noise ratio for increasing transmission rate.