Datornätverk A – lektion 3 Kapitel 3: Fysiska signaler. Kapitel 4: Digital transmission.

Physical Layer PART II

Chapter 3 Signals

Figure 3.1 Comparison of analog and digital signals

Signals can be analog or digital. Analog signals can have an infinite number of values in a range; digital signals can have only a limited number of values. Note:

Sinusvågor Periodtid T = t 2 - t 1. Enhet: s. Frekvens f = 1/T. Enhet: 1/s=Hz. T=1/f. Amplitud eller toppvärde Û. Enhet: Volt. Fasläge: θ = 0 i ovanstående exempel. Enhet: Grader eller radianer. Momentan spänning: u(t)= Ûsin(2πft+θ)

Figure 3.4 Period and frequency

Tabell 3.1 Enheter för periodtid och frekvens EnhetEkvivalentEnhetEkvivalent Sekunder (s)1 sHertz (Hz)1 Hz Millisekunder (ms)10 –3 sKilohertz (kHz)10 3 Hz Mikrosekunder (μs)10 –6 sMegahertz (MHz)10 6 Hz Nanosekunder (ns)10 –9 sGigahertz (GHz)10 9 Hz Pikosekunder (ps)10 –12 sTerahertz (THz)10 12 Hz Exempel: En sinusvåg med periodtid 1 ns har frekvens 1 GHz.

Exempel 1 Vilken frekvens i kHz har en sinusvåg med periodtid 100 ms? Solution Alternativ 1: Gör om till grundenheten. 100 ms = 0.1 s f = 1/0.1 Hz = 10 Hz = 10/1000 kHz = 0.01 kHz Alternativ 2: Utnyttja att 1 ms motsvarar 1 kHz. f = 1/100ms = 0.01 kHz.

Figure 3.5 Relationships between different phases

Example 2 A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians? Solution We know that one complete cycle is 360 degrees. Therefore, 1/6 cycle is (1/6) 360 = 60 degrees = 60 x 2  /360 rad = rad

Figure 3.6 Sine wave examples

Figure 3.6 Sine wave examples (continued)

Figure 3.7 Time and frequency domains (continued)

Figure 3.7 Time and frequency domains

Figure 3.8 Square wave

Figure 3.9 Three harmonics

Figure 3.10 Adding first three harmonics

Figure 3.11 Frequency spectrum comparison

Figure 3.12 Signal corruption

Figure 3.13 Bandwidth

Example 3 If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V. Solution B = f h  f l = 900  100 = 800 Hz The spectrum has only five spikes, at 100, 300, 500, 700, and 900 (see Figure 13.4 )

Figure 3.14 Example 3

Example 5 A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium? Solution The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies between 3000 and 4000 Hz; the signal is totally lost.

Figure 3.16 A digital signal

Example 6 A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval) Solution The bit interval is the inverse of the bit rate. Bit interval = 1/ 2000 s = s = x 10 6  s = 500  s

Figure 3.17 Bit rate and bit interval

Figure 3.18 Digital versus analog

A digital signal is a composite signal with an infinite bandwidth. Note:

Table 3.12 Bandwidth Requirement Bit Rate Harmonic 1 Harmonics 1, 3 Harmonics 1, 3, 5 Harmonics 1, 3, 5, 7 1 Kbps500 Hz2 KHz4.5 KHz8 KHz 10 Kbps5 KHz20 KHz45 KHz80 KHz 100 Kbps50 KHz200 KHz450 KHz800 KHz

The bit rate and the bandwidth are proportional to each other. Note:

Figure 3.19 Low-pass and band-pass

The analog bandwidth of a medium is expressed in hertz; the digital bandwidth, in bits per second. Note:

Digital transmission needs a low-pass channel. Note:

Analog transmission can use a band- pass channel. Note:

Figure 3.20 Impairment types

Figure 3.21 Attenuation

Förstärkning mätt i decibel (dB) 1 gång effektförstärkning = 0 dB. 2 ggr effektförstärkning = 3 dB. 10 ggr effektförstärkning = 10 dB. 100 ggr effektförstärkning = 20 dB ggr effektförstärkning = 30 dB. Osv.

Dämpning mätt i decibel Dämpning 100 ggr = Dämpning 20 dB = förstärkning 0.01 ggr = förstärkning med – 20 dB. Dämpning 1000 ggr = 30 dB dämpning = -30dB förstärkning. En halvering av signalen = dämpning med 3dB = förstärkning med -3dB.

Signal-brus-förhållande Ett signal-brus-förhållande på 100 dB innebär att den starkaste signalen är 100 dB starkare än bruset. Ljud som är svagare än bruset hörs inte utan dränks i bruset. Ljudets dynamik skillnaden mellan den starkaste ljudet och det svagaste ljudet som man kan höra, och är vanligen ungefär detsamma som signal-brus-förhållandet.

Example 12 Imagine a signal travels through a transmission medium and its power is reduced to half. This means that P2 = 1/2 P1. In this case, the attenuation (loss of power) can be calculated as Solution 10 log 10 (P2/P1) = 10 log 10 (0.5P1/P1) = 10 log 10 (0.5) = 10(–0.3) = –3 dB 10 log 10 (P2/P1) = 10 log 10 (0.5P1/P1) = 10 log 10 (0.5) = 10(–0.3) = –3 dB

Example 13 Imagine a signal travels through an amplifier and its power is increased ten times. This means that P2 = 10∙P1. In this case, the amplification (gain of power) can be calculated as 10 log 10 (P2/P1) = 10 log 10 (10P1/P1) 10 log 10 (P2/P1) = 10 log 10 (10P1/P1) = 10 log 10 (10) = 10 (1) = 10 dB = 10 log 10 (10) = 10 (1) = 10 dB

Example 14 One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are talking about several points instead of just two (cascading). In Figure 3.22 a signal travels a long distance from point 1 to point 4. The signal is attenuated by the time it reaches point 2. Between points 2 and 3, the signal is amplified. Again, between points 3 and 4, the signal is attenuated. We can find the resultant decibel for the signal just by adding the decibel measurements between each set of points.

Figure 3.22 Example 14 dB = –3 + 7 – 3 = +1

Figure 3.23 Distortion

Figure 3.24 Noise

Throughput (Genomströmningshastighet)

Figure 3.26 Propagation time (Utbredningstid)

Chapter 4 Digital Transmission

Figure 4.2 Signal level versus data level

Figure 4.3 DC component

Example 1 A signal has two data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows: Pulse Rate = 1/ = 1000 pulses/s Bit Rate = Pulse Rate x log 2 L = 1000 x log 2 2 = 1000 bps

Example 2 A signal has four data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows: Pulse Rate = = 1000 pulses/s Bit Rate = PulseRate x log 2 L = 1000 x log 2 4 = 2000 bps

Figure 4.4 Lack of synchronization

Example 3 In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Kbps? How many if the data rate is 1 Mbps? Solution At 1 Kbps: 1000 bits sent  1001 bits received  1 extra bps At 1 Mbps: 1,000,000 bits sent  1,001,000 bits received  1000 extra bps

Figure 4.5 Line coding schemes

Unipolar encoding uses only one voltage level. Note:

Figure 4.6 Unipolar encoding

Polar encoding uses two voltage levels (positive and negative). Note:

Figure 4.7 Types of polar encoding

In NRZ-L the level of the signal is dependent upon the state of the bit. Note:

In NRZ-I the signal is inverted if a 1 is encountered. Note:

Figure 4.8 NRZ-L and NRZ-I encoding

Figure 4.9 RZ encoding

A good encoded digital signal must contain a provision for synchronization. Note:

Figure 4.10 Manchester encoding

In Manchester encoding, the transition at the middle of the bit is used for both synchronization and bit representation. Note:

Figure 4.11 Differential Manchester encoding

In differential Manchester encoding, the transition at the middle of the bit is used only for synchronization. The bit representation is defined by the inversion or noninversion at the beginning of the bit. Note:

In bipolar encoding, we use three levels: positive, zero, and negative. Note:

Figure 4.12 Bipolar AMI encoding

4.3 Sampling Pulse Amplitude Modulation Pulse Code Modulation Sampling Rate: Nyquist Theorem How Many Bits per Sample? Bit Rate

Sampling och DA- omvandling Microfon- membranets läge Tid Viloläge 3 mm bakom 2 mm framför 2 mm bakom T = 0,4ms 0.1ms 0.2ms 0.3ms0.4ms 0.5ms 0.6ms T s = 0,1ms Nästan sinusformat ljud med periodtidT =0.4ms och frekvens f = 1/T = 2500 Hz = 2,5kHz. Samplingsperiod T s = 0.1ms,dvssamplingsfrekvensf s = 1/T =. Kvantisering (=avrundning) till 8 värden. Digitalisering ger 3 bit per värde: sampels/sek = 10kHz

PCM = Pulse Code Modulation = Digitalisering av analoga signaler och seriell överföring Sampler AD-omvand- lare med seriell utsignal DA- omvandlare Antiviknings- filter Interpola- tionsfilter Sifferexempel från PSTN = publika telefonnätet: Hz filter 8000 sampels per sek 8 bit per sampel dvs bps per tfnsamtal 2 8 = 256 spänningsnivåer 0 1 Mikrofon Högtalare

Exempel En 6 sekunder lång ljudinspelning digitaliseras. Hur stor är inspelningens informationsmängd? a) sampels/sekund, 256 kvantiseringsnivåer. b) sampels/sekund, 16 kvantiseringsnivåer. c) 5500 sampels/sekund, 256 kvantiseringsnivåer sampels * 6 s * 8 bit = bit sampels * 6 s * 4 bit = bit. 5500sampels * 6 s * 8 bit = bit.

Samplingsteoremet f < f s /2 Den högsta frekvens som kan samplas är halva samplingsfrekvensen. Om man samplar högre frekvens än f s /2 så byter signalen frekvens, dvs det uppstår vikningsdistorsion (aliasing). För att undvika vikningsdistorsion så har man ett anti-vikningsfilter innan samplingen, som tar bort frekvenser över halva samplingsfrekvensen. Interpolationsfiltret används vid rekonstruktion av den digitala signalen för att ”gissa” värden mellan samplen. Ett ideal interpolationsfilter skulle kunna återskapa den samplade signalen perfekt om den uppfyller samplingsteoremet. I verkligheten finns inga ideala filter. Följdregel: Nyqvist’s sats säger att max datahastighet = 2B 2 log M, där M är antal nivåer, och B är signalens bandbredd, oftast lika med signalens övre gränsfrekvens.

Figure 4.18 PAM

Pulse amplitude modulation has some applications, but it is not used by itself in data communication. However, it is the first step in another very popular conversion method called pulse code modulation. Note:

Figure 4.19 Quantized PAM signal

Figure 4.20 Quantizing by using sign and magnitude

According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency. Note:

Example 4 What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)? Solution The sampling rate must be twice the highest frequency in the signal: Sampling rate = 2 x (11,000) = 22,000 samples/s

Example 5 A signal is sampled. Each sample requires at least 12 levels of precision (+0 to +5 and -0 to -5). How many bits should be sent for each sample? Solution We need 4 bits; 1 bit for the sign and 3 bits for the value. A 3-bit value can represent 2 3 = 8 levels (000 to 111), which is more than what we need. A 2-bit value is not enough since 2 2 = 4. A 4-bit value is too much because 2 4 = 16.

Example 6 We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? Solution The human voice normally contains frequencies from 0 to 4000 Hz. Sampling rate = 4000 x 2 = 8000 samples/s Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps

Distorsion till följd av digitalisering Vikningsdistorsion ○Inträffar om man inte filtrerar bort frekvenser som är högre än halva samplingsfrekvensen. Kvantiseringsdistorsion (kvantiseringsbrus) ○Avrundningsfelet låter ofta som ett brus. ○Varje extra bit upplösning ger dubbelt så många spänningsnivåer, vilket ger en minskning av kvantiseringsdistorsionen med 6 dB. 16 bit upplösning ger ett signal-brus-förhållande på ca 16*6 = 96 dB (beroende på hur man mäter detta förhållande.) ○Svaga ljud avrundas bort, eller dränks i kvantiseringsbruset.