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DigitalDigital vs.vs. AnalogAnalog TransmissionTransmission NyquistNyquist andand ShannonShannon LawsLaws
Required reading: Garcia 3.1 to 3.5
CSE 3213, Fall 2010 Instructor: N. Vlajic Transmission Impairments 2
Transmission / Signal – caused by imperfections of transmission Impairments media • for analog signals, impairments can degrade signal quality
Sent Received
• for digital signals, impairments can cause bit errors
Sent Received
• main types of transmission impairments: Transmission Impairments: Attenuation 3
Attenuation – reduction / loss in signal power
• when a signal travels through a medium it loses some of its energy by overcoming the resistance of medium
• main challenges in combating attenuation: (1) received signal must have sufficient strength so that Less of a receiver can detect signal, but should not be too strong problem for to overload transmitter/receiver circuitry digital signal !!! (2) signal must maintain a level sufficiently higher than noise at all times, to be received without error
• to compensate for loss, analog amplifiers / digital repeaters are used to boost the signal at regular intervals Transmission Impairments: Attenuation (cont.) 4
Attenuation (cont.) –def. loss in signal power as it is transferred across a system (medium)
• determined for each individual frequency Pin > Pout • apply sinwave of freq. f and power Pin to channel input and observe signal power Pout at output
Pin(f) Attenuation(f) = L(f) = 10⋅log10 [dB] Pout (f)
Amplitude Response • aka ‘magnitude of frequency response’
Aout (f) 1 A(f)= 2 Ain (f) Pin(f) Ain (f) 1 L(f) = = 2 = 2 Pout (f) Aout (f) A(f)
1 Atten.(f) = L(f) = 20⋅log [dB] 10 A(f)
See Garcia pp. 125. f channel’s amplitude response function A(f) 5
Which frequencies are better passed through the medium? Transmission Impairments: Attenuation (cont.) 6
Why decibel (log function)?
1. Signal strength often falls off exponentially, so loss is easily expressed in terms of decibels – linear function in log-plot.
2. The net gain or loss in a cascaded transmission path can be calculated with simple addition and subtraction. In figure below, 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 attenuation just by adding the decibel measurements between each set of points.
-1 dB 3dB -7dB 3 dB
Pin Pout
In this case, the attenuation can be calculated as: 3-7+3 = -1, which means that the signal has gained power. Transmission Impairments: Attenuation (cont.) 7
Example [ attenuation ] Consider a series of transmission elements as shown in the figure below.
st The input signal has the power of P1 = 4 mW. The 1 element is a transmission line with a loss of 5 (x), the 2nd element is an amplifier with a gain of 7 (x), nd the 3rd element is a transmission line with a loss of 3 (x).
Calculate the output power P4.
loss = 5 gain = 7 loss = 3
P1 = 4 mW P4 = ???
P P P P 1 7 1 4 = 4 ⋅ 3 ⋅ 2 = ⋅ ⋅ = 0.47 P1 P3 P2 P1 5 1 3
P4 = 0.47 ⋅4 [mW] = 1.88 [mW] Transmission Impairments: Attenuation (cont.) 8
G1 G2 G3
P1 = 4 mW P4 = ???
P4 P4 P3 P2 1 1 1 = ⋅ ⋅ = G1 ⋅G2 ⋅G3 = ⋅ ⋅ P1 P3 P2 P1 L1 L2 L3
P4 10⋅log [dB] = 10⋅log(G1 ⋅G2 ⋅G3 ) = 10⋅log(G1) +10⋅log(G2 ) +10⋅log(G3 ) P1
P4 10⋅log [dB] = G1[dB] + G2[dB] + G3 [dB] P1 Transmission Impairments: Attenuation (cont.) 9
Example [ attenuation ] Consider a series of transmission elements as shown in the figure below. st The input signal has the power of P1 = 4 mW. The 1 element is a Transmission line with a 12 dB loss, the 2nd element is an amplifier with a 35 dB gain, and the 3rd element is a transmission line with a 10 dB loss.
Calculate the output power P4.
-13 dB 12 dB -35 dB 10 dB
P1 = 4 mW P4 = ???
P 10⋅log in = −13 [dB] Pout G [dB] ↔ L [dB] P P = in = P ⋅101.3 = 4⋅19.95 = 79.8 [mW] out 10-1.3 in Transmission Impairments: Delay Distortion 10
Delay Distortion – change in signal’s form / shape
• each signal component has its own propagation speed through a medium, and therefore, its own delay in arriving at the final destination
• critical for composite-analog & digital signals – some of the signal components of one bit position will spill over into other bit, causing ‘intersymbol interference’ major limitation to achieving high bit rates
• in bandlimited channels, velocity tends to be highest near the center frequency and fall off towards the edges of the band Transmission Impairments: Noise 11
Noise – unwanted signals that get inserted / generated somewhere between transmitter and receiver
• major limiting factor in communications system performance cannot be predicted – appears at random!
• presence of noise limits the reliability with which the receiver can correctly determine the information that was transmitted
• main categories of noise: (1) thermal noise (2) intermodulation noise (3) crosstalk (4) impulse noise Transmission Impairments: Noise (cont.) 12
(1) Thermal Noise – result of random motion of electrons – appears in all electronic devices and transmission media – cannot be eliminated
• function of temperature
• uniformly distributed across frequency spectrum ⇒ aka white noise
• noise power density (No) = amount of thermal noise to be found in a bandwidth of 1Hz
No = k ⋅T [W/Hz]
where k = Boltzmann’s constant = 1.3803*10-23 [J/K] T = temperature [K]
• thermal noise (N) in [W], in a bandwidth of B [Hz]
N = k ⋅T ⋅B [W]
Example Calculate N on 20C and 1GHz: N = k*(273+20)*109 = 3.8*10 -12 . Transmission Impairments: Noise (cont.) 13
(2) Intermodulation Noise – signals that are sum / difference of original frequencies sharing a medium
• result of nonlinearity in transmission medium – output signal is a complex function of the input
Vo(t) Vo(t)
Vi(t) Vi(t) linear channel non-linear channel
(3) Crosstalk – effect of one wire on the other – one wire acts as a sending antenna and the other as the receiving antenna
• can be reduced by careful shielding and using twisted pairs • of the same magnitude, or less, than thermal noise
(4) Impulse Noise – non-continuous, consisting of irregular pulses or noise spikes of short duration and of relatively high amplitude • induced by external electromagnetic disturbances, such as lightening, faults and flaws in communication system Transmission Impairments: Noise (cont.) 14
Signal to Noise Ratio – ratio of the power in the desired signal (SNR) to the power in the superimposed noise average signal power SNR = average noise power
SNR (dB) = 10 log10 SNR
• high SNR ⇒ high-quality signal & low number of required amplifiers / repeaters Analog Transmission 15
Analog Long-Distance • each repeater attempts to restore analog Communications signal to its original form • restoration (noise removal) is imperfect – Goals: noise gets amplified too ! 1) restore amplitude 2) remove delay distortion if signal only had components in certain freq. 3) remove noise band, repeater could remove noise components outside signal band – but, not those inside /
• signal quality decreases with # of repeaters ⇒ communications is distance-limited
• analogy: copy a song using cassette recorder
Attenuated and Recovered signal distorted signal + + residual noise noise
Amp Equalizer
Repeater 16 Digital Transmission 17
Digital Long-Distance • regenerator does not need to completely Communications recover the original shape of the transmitted signal – it only needs to determine whether the original pulse was positive or negative
• original signal can be completely recovered each time ⇒ communication over very long distance is possible
• analogy: copy an MP3 file
compensate for distortion introduced by the channel sample signal at midpoint of each pulse to determine its Decision circuit polarity Amplifier and signal equalizer regenerator
keep track of Timing intervals that recovery define each pulse Digital Transmission (cont.) 18
Example [ transmission impairments in digital transmission ]
Digital transmission can easily recover from various types of channel impairments.
0.5 1
0
So, is digital transmission the ultimate winner?! Analog vs. Digital Transmission 19
Low-pass Channel – bandwidth = [0, f1) • entire medium/bandwidth dedicated to 2 devices • devices alternate in transmission
Band-pass Channel –bandwidth = [f1, f2) • medium is shared among multiple users • each pair of users gets a portion of overall bandwidth Analog vs. Digital Transmission (cont.) 20
Digital Transmission • signal can be transmitted over long-distance Advantages without loosing any quality • can operate with lower signal levels ⇒ lower system cost
• easier to apply encryption
• easier integration of voice, video and data
Digital Transmission • digital signal theoretically needs a bandwidth Disadvantages [0, ∞) – upper limit can be relaxed if we decide to work with a limited number of harmonics ⇒ digital transmission needs a low-pass channel
• analog transmission can use a band-pass channel
Both analog and digital data may be transmitted on suitable transmission media using either digital coding or analog modulation. Analog vs. Digital Transmission (cont.) 21
digital or digital or analog low-pass channel analog data data (digital signal)
digital or digital or analog analog band-pass channel data data (analog signal) Analog vs. Digital Transmission (cont.) 22
Example [ digital transmission of digital and analog data ]
Digital Data → Digital Signal: Line Coding
Analog Data → Digital Signal: PCM (Pulse Code Modul.) or Delta Modulation Analog vs. Digital Transmission (cont.) 23
Example [ analog transmission of digital and analog data ]
Digital data → Analog Signal: Digital Modulation
Analog data → Analog Signal: Analog Modulation Last Note about Signals … 24
Throughput – measurement of how fast data can pass through an entity in the network (computer, router, channel, etc.)
• if we consider this entity as a wall through which bits pass, throughput is the number of bits that can pass this wall in one second
e.g. R=56 kbps
Example [ throughput ] If the throughput at the connection between a device and the transmission medium is 56 kbps, how long does it take to send 100,000 bits out of this device? N [bits] 100000 [bit] t = = = 1,786 [sec] R[bps] 56000 [bps] Last Note about Signals … (cont.) 25
Propagation Time – measures the time required for a signal (or a bit) to travel from one point of transmission medium to another d p = [sec] c
• d – length of physical link [m] • c – signal propagation speed in medium ∼ 2*108 [m/s]
Example [ propagation time ] The light of the Sun takes approximately 8 minutes to reach the Earth? What is the distance between the Sun and the Earth? m m d = p [sec]⋅c [ ] = 8 * 60 [sec] ⋅ 3⋅108 [ ] = 144⋅109 [m] = 144⋅106 [km] sec sec Last Note about Signals … (cont.) 26
Overall Delay
• L [bits] number of bits in message • R [bps] speed of digital transmission system • d [m] distance in meters • c [m/s] speed of light (3x108 m/s in vacuum)
Time to deliver a block of L bits:
Delay = tpropagation + ttransmission = d/c + L/R seconds
Use data compression to reduce L. How can theUse time higher to speeddownload modem/cable a file to be increase minimized??? R. Place server closer to reduce d.
http://media.pearsoncmg.com/aw/aw_kurose_network_2/applets/transmission/delay.html Data Rate Limits in Digital Transmission 27
Max Data Rate [bps] – depends on three factors: Over a Channel? • bandwidth available • # of levels in digital signal • quality of channel – level of noise
Nyquist Theorem – defines theoretical max bit rate in noiseless channel [1924] • even perfect (noiseless) channels have limited capacity
Shannon Theorem – Nyquist Theorem extended [1949]– defines theoretical max bit rate in noisy channel • if random noise is present, situation deteriorates rapidly!
T Data Rate Limits: Nyquist Theorem 28
Intersymbol Interference – the inevitable filtering effect of any practical channel will cause spreading of individual data symbols that pass through the channel
• this spreading causes part of symbol energy to overlap with neighbouring symbols causing intersymbol interference (ISI)
• ISI can significantly degrade the ability of the data detector to differentiate a current symbol from the diffused energy of adjacent symbols
impulse response: narrow pulse delayed pulse with ringing
Bandwidth: B[Hz] Ts = 1/2B
As the channel bandwidth B increases, the width of the impulse response decreases ⇒ pulses can be input in the system more closely spaced, i.e. at a higher rate. Data Rate Limits: Nyquist Theorem 29
Impulse Response – response of a low-pass channel (of bandwidth B) to a narrow pulse h(t), aka Nyquist pulse: sin(2πBt) s(t) = 2πBt 1 • zeros: where sin(2πBt)=0 ⇒ t = 2B
1.2
1.21 0.81 0.80.6 0.60.4 0.40.2 0.20 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 -0.20 t -7T -6 T -5 T -4T -3T -2 T T -1 -0.2 -0.4 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 -0.41 T = S 2B 2 3 2B 2B What is the minimum pulse/bit duration time to avoid significant ISI?! Data Rate Limits: Nyquist Theorem 30
Example [ system response to binary input 110 ]
1 2
1
T 0 S 0 -2 -1 0 1 2 3 4 -2 -1 0 1 2 3 4 -1 TS TS
-1 -2
three separate pulses combined signal
Assume: channel bandwidth = max analog frequency passed = B [Hz].
New pulse is sent every TS sec ⇒ data rate = 1/TS [bps] = 2B [bps] The combined signal has the correct values at t = 0, 1, 2.
1 pulse ⎡ pulses ⎤ rmax = = 2W = 2B ⎢ ⎥ TS second ⎣second⎦
Maximum signaling rate that is achievable through an ideal low-pass channel. Data Rate Limits: Nyquist Theorem 31
Nyquist Law – max rate at which digital data can be transmitted over a comm. channel of bandwidth B [Hz] is
Cnoiseless = 2⋅B⋅log2M [bps]
• M – number of discrete levels in digital signal
• M ↑⇒C ↑, however this places increased burden on receiver – instead of distinguishing one of two possible signals, now it must distinguish between M possible signals especially complex in the presence of noise
max amplitude if spacing between levels becomes too small, noise signal can cause receiver to make wrong decision Typical noise min amplitude Four signal levels Eight signal levels Data Rate Limits: Nyquist Theorem 32
Example [ multilevel digital transmission ]
2-level encoding: C=2B [bps] one pulse – one bit
4-level encoding: C=2*2=4B [bps] 100110100011010010 ⇒ one pulse – two bits
8-level encoding: C=2*3=6B [bps] 100110100011010010 ⇒ one pulse – three bits Data Rate Limits: Shannon Theorem 33
Shannon Law – max transmission rate over a channel with bandwidth B, with Gaussian distributed noise, and with signal-to- noise ratio SNR=S/N, is
Cnoisy = B⋅log2(1+ SNR) [bps]
• theoretical limit – there are numerous impairments in every real channel besides those taken into account in Shannon's Law (e.g. attenuation, delay distortion, or impulse noise)
• no indication of levels – no matter how many levels we use, we cannot achieve a data rate higher than the capacity of the channel
• in practice we need to use both methods (Nyquist & Shannon) to find what data rate and signal levels are appropriate for each particular channel:
The Shannon capacity gives us the upper limit! The Nyquist formula tells us how many levels we need! Data Rate Limits 34
Example [ data rate over telephone line ]
What is the theoretical highest bit rate of a regular telephone line? A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually 35 dB (3162) on up-link channel (user-to-network).
Solution: We can calculate the theoretical highest bit rate of a regular telephone line as:
C = B log2 (1 + SNR) =
= 3000 log2 (1 + 3162) =
= 3000 log2 (3163)
C = 3000 × 11.62 = 34,860 bps Data Rate Limits 35
Example [ data rate / number of levels ]
We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and number of signal level?
Solution: First use Shannon formula to find the upper limit on the channel’s data-rate
6 6 C = B log2 (1 + SNR) = 10 log2 (1 + 63) = 10 log2 (64) = 6 Mbps
Although the Shannon formula gives us 6 Mbps, this is the upper limit. For better performance choose something lower, e.g. 4 Mbps. Then use the Nyquist formula to find the number of signal levels.
C = 2⋅B⋅log2M [bps]
4 Mbps = 2 × 1 MHz × log2 L Î L = 4