1 Digital vs. Analog Transmission Digital vs. Analog Transmission Nyquist Nyquist and Shannon Laws and Shannon Laws CSE 3213, Fall 2010 Instructor: N. Vlajic Required reading: Garcia 3.1 to 3.5
1
Digital vs. Analog TransmissionDigital vs. Analog TransmissionNyquistNyquist and Shannon Lawsand Shannon Laws
CSE 3213, Fall 2010Instructor: N. Vlajic
Required reading:Garcia 3.1 to 3.5
2Transmission Impairments
Transmission / SignalTransmission / SignalImpairmentsImpairments
– caused by imperfections of transmissionmedia• for analog signals, impairments can
degrade signal quality
• for digital signals, impairments can cause bit errors
• main types of transmission impairments:
SentSent Received
Sent Received
3Transmission Impairments: Attenuation
AttenuationAttenuation – 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
receiver can detect signal, but should not be too strongto overload transmitter/receiver circuitry
(2) signal must maintain a level sufficiently higher thannoise 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
Less of a problem for
digital signal !!!
4
Attenuation (cont.)Attenuation (cont.) – def. loss in signal power as it is transferredacross a system (medium)• determined for each individual frequency
• apply sinwave of freq. f and power Pin to channelinput and observe signal power Pout at output
• aka ‘magnitude of frequency response’
[dB](f)P(f)Plog10L(f) (f)
out
in10⋅==nAttenuatio
22out
2in
out
in
A(f)1
(f)A(f)A
(f)P(f)PL(f) ===
[dB]A(f)1log20L(f)Atten.(f) 10⋅==
channel’s amplitude response function A(f)See Garcia pp. 125.
Transmission Impairments: Attenuation (cont.)
Pin Pout>
f
1
Amplitude Response
f
1
f
1
Amplitude Response
A(f)=Aout (f)Ain (f)
5
Which frequencies are better passed through
the medium?
6
1. Signal strength often falls off exponentially, so loss is easily expressedin terms of decibels – linear function in log-plot.
2. The net gain or loss in a cascaded transmission path can be calculatedwith 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 points2 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.
Why decibel (log function)?Why decibel (log function)?
3dB -7dB 3 dB
-1 dB
In this case, the attenuation can be calculated as: 3-7+3 = -1, which means that the signal has gained power.
Pin Pout
Transmission Impairments: Attenuation (cont.)
7
Example [ attenuation ]
Consider a series of transmission elements as shown in the figure below.
The input signal has the power of P1 = 4 mWW. The 1st 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 lossof 3 (x).
Calculate the output power P4.
loss = 5 gain = 7 loss = 3
P1 = 4 mW P4 = ???
0.4731
17
51
PP
PP
PP
PP
1
2
2
3
3
4
1
4 =⋅⋅=⋅⋅=
[mW] 1.88[mW] 40.47P4 =⋅=
Transmission Impairments: Attenuation (cont.)
8
G1 G3
P1 = 4 mW P4 = ???
321321
1
2
2
3
3
4
1
4
LLLGGG
PP
PP
PP
PP 111
⋅⋅=⋅⋅=⋅⋅=
G2
)log(G10)log(G10)log(G10)GGlog(G10[dB] PPlog10 321321
1
4 ⋅+⋅+⋅=⋅⋅⋅=⋅
[dB]G[dB]G[dB]G[dB] PPlog10 321
1
4 ++=⋅
Transmission Impairments: Attenuation (cont.)
9
Example [ attenuation ]
Consider a series of transmission elements as shown in the figure below.The input signal has the power of P1 = 4 mWW. The 1st element is aTransmission 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.
12 dB -35 dB 10 dB
-13 dB
P4 = ???P1 = 4 mW
[dB]13PPlog10
out
in −=⋅
[mW] 79.819.95410P10
PP 1.3in1.3-
inout =⋅=⋅==
Transmission Impairments: Attenuation (cont.)
G [dB] ↔ L [dB]
10Transmission Impairments: Delay Distortion
Delay DistortionDelay Distortion – change in signal’s form / shape• each signal component has its own propagation
speed through a medium, and therefore, its owndelay in arriving at the final destination
• critical for composite-analog & digital signals –some of the signal components of one bit positionwill spill over into other bit, causing ‘intersymbolinterference’
major limitation to achieving high bit rates
• in bandlimited channels, velocity tends to behighest near the center frequency and fall offtowards the edges of the band
11Transmission Impairments: Noise
Noise Noise – unwanted signals that get inserted / generated somewherebetween 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 receivercan correctly determine the information that was transmitted
• main categories of noise:(1) thermal noise(2) intermodulation noise(3) crosstalk(4) impulse noise
12
(1) Thermal Noise (1) Thermal Noise – result of random motion of electrons – appears inall electronic devices and transmission media –cannot be eliminated
• function of temperature
• uniformly distributed across frequency spectrum⇒ aka white noisewhite noise
• noise power density (No) = amount of thermal noiseto be found in a bandwidth of 1Hz
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]
Example Calculate N on 20C and 1GHz: N = k*(273+20)*109 = 3.8*10 -12 .
[W/Hz] TkNo ⋅=
Transmission Impairments: Noise (cont.)
[W] B TkN ⋅⋅=
13Transmission Impairments: Noise (cont.)
(2) (2) IntermodulationIntermodulation Noise Noise – signals that are sum / difference of originalfrequencies sharing a medium• result of nonlinearity in transmission medium –
output signal is a complex function of the input
(3) Crosstalk (3) Crosstalk – effect of one wire on the other – one wire acts as a sendingantenna 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 (4) Impulse Noise – non-continuous, consisting of irregular pulses ornoise spikes of short duration and of relatively highamplitude• induced by external electromagnetic disturbances, such
as lightening, faults and flaws in communication system
Vi(t)
Vo(t)
Vi(t)
Vo(t)
linear channel non-linear channel
14
– ratio of the power in the desired signalto the power in the superimposed noise
• high SNR ⇒ high-quality signal & low numberof required amplifiers / repeaters
Transmission Impairments: Noise (cont.)
Signal to Noise RatioSignal to Noise Ratio(SNR)(SNR)
SNR (dB) = 10 log10 SNR
power noise averagepower signal averageSNR =
15Analog Transmission
Analog LongAnalog Long--DistanceDistanceCommunicationsCommunications
• each repeater attempts to restore analogsignal to its original form
• restoration (noise removal) is imperfect –noise gets amplified too !
if signal only had components in certain freq.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 anddistorted signal
+ noise
Equalizer
Recovered signal+
residual noise
Repeater
Amp
Attenuated anddistorted signal
+ noise
Equalizer
Recovered signal+
residual noise
Repeater
Amp
Goals:1) restore amplitude2) remove delay distortion3) remove noise
16
17Digital Transmission
Digital LongDigital Long--DistanceDistanceCommunicationsCommunications
• regenerator does not need to completelyrecover 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 longdistance is possible
• analogy: copy an MP3 file
compensate for distortion
introduced by the channel
keep track of intervals that
define each pulse
sample signalat midpoint
of each pulse to determine its
polarity Amplifierequalizer
Timingrecovery
Decision circuitand signal
regenerator
Amplifierequalizer
Timingrecovery
Decision circuitand signal
regenerator
18
Example [ transmission impairments in digital transmission ]
Digital Transmission (cont.)
Digital transmission can easily recover from various types of channelimpairments.
So, is digital transmission the ultimate winner?!So, is digital transmission the ultimate winner?!
0
10.5
19Analog vs. Digital Transmission
LowLow--pass Channelpass Channel – bandwidth = [0, f1)• entire medium/bandwidth dedicated to 2 devices• devices alternate in transmission
BandBand--pass Channelpass Channel – bandwidth = [f1, f2)• medium is shared among multiple users• each pair of users gets a portion of overall
bandwidth
20Analog vs. Digital Transmission (cont.)
Digital TransmissionDigital TransmissionAdvantagesAdvantages
• signal can be transmitted over long-distancewithout loosing any quality
• can operate with lower signal levels ⇒ lowersystem cost
• easier to apply encryption
• easier integration of voice, video and data
Digital TransmissionDigital TransmissionDisadvantagesDisadvantages
• digital signal theoretically needs a bandwidth[0, ∞) – upper limit can be relaxed if we decideto 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 transmissionmedia using either digital coding or analog modulation.
21Analog vs. Digital Transmission (cont.)
low-pass channel(digital signal)
band-pass channel(analog signal)
digital or analog data
digital or analog data
digital or analog data
digital or analog data
22Analog vs. Digital Transmission (cont.)
Example [ digital transmission of digital and analog data ]
Digital Data → Digital Signal: Line Coding
Analog Data → Digital Signal: PCM (Pulse Code Modul.) orDelta Modulation
23Analog vs. Digital Transmission (cont.)
Example [ analog transmission of digital and analog data ]
Digital data → Analog Signal: Digital Modulation
Analog data → Analog Signal: Analog Modulation
24Last Note about Signals …
Throughput Throughput – measurement of how fast data can pass through anentity 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 thiswall in one second
e.g. R=56 kbps
Example [ throughput ]
If the throughput at the connection between a device and the transmissionmedium is 56 kbps, how long does it take to send 100,000 bits out of thisdevice?
[sec] 1,786[bps] 56000[bit] 100000t ===
][][
bpsRbitsN
25
Propagation TimePropagation Time – measures the time required for a signal (or a bit) to travel from one point of transmission medium to another
• d – length of physical link [m]• c – signal propagation speed in medium ∼ 2*108 [m/s]
Last Note about Signals … (cont.)
[sec] cdp =
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?
[km]10144[m]10144]secm[103 [sec] 60*8 ]
secm[c [sec] pd 698 ⋅=⋅=⋅⋅=⋅=
26Last Note about Signals … (cont.)
Overall DelayOverall Delay
Use data compression to reduce L.Use higher speed modem/cable to increase R.
Place server closer to reduce d.
• 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
http://media.pearsoncmg.com/aw/aw_kurose_network_2/applets/transmission/delay.html
How can the time to download a file be minimized???
27Data Rate Limits in Digital Transmission
– depends on three factors:• bandwidth available• # of levels in digital signal• quality of channel – level of noise
Max Data Rate [bps]Max Data Rate [bps]Over a Channel?Over a Channel?
T
NyquistNyquist TheoremTheorem – defines theoretical max bit rate in noiseless channel [1924]• even perfect (noiseless) channels have limited
capacity
Shannon TheoremShannon Theorem – Nyquist Theorem extended [1949]– definestheoretical max bit rate in noisy channel• if random noise is present, situation deteriorates
rapidly!
28Data Rate Limits: Nyquist Theorem
IntersymbolIntersymbol InterferenceInterference – the inevitable filtering effect of any practicalchannel will cause spreading of individual data symbols that pass through the channel• this spreading causes part of symbol energy
to overlap with neighbouring symbols causingintersymbol interference (ISI)
• ISI can significantly degrade the ability of thedata detector to differentiate a current symbolfrom the diffused energy of adjacent symbols
Ts = 1/2B
narrow pulseimpulse response:
delayed pulse with ringing
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.
Bandwidth: B[Hz]
29
IImpulsempulse ResponseResponse – response of a low-pass channel (of bandwidth B)to a narrow pulse h(t), aka Nyquist pulse:
• zeros: where sin(2πBt)=0 ⇒
Data Rate Limits: Nyquist Theorem
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7t
T T T T T T T T T T T T T T -0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 tT T T T T T T T T T T T T T
Bt2Bt)sin(2s(t)
ππ
=
2Bt 1=
2B1TS =
2B2
2B3
What is the minimum pulse/bit duration time to avoid significant ISI?!
30Data Rate Limits: Nyquist Theorem
-1
0
1
-2 -1 0 1 2 3 4
-2
-1
0
1
2
-2 -1 0 1 2 3 4
Example [ system response to binary input 110 ]
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.
TS
TS
2B2W T 1r
second
pulses
second
pulse
Smax ⎥⎦
⎤⎢⎣⎡===
Maximum signaling rate that is achievable through an ideal low-pass channel.
TS
31
NyquistNyquist LawLaw – max rate at which digital data can be transmittedover a comm. channel of bandwidth B [Hz] is
• M – number of discrete levels in digital signal
• M ↑ ⇒ C ↑, however this places increased burden on receiver – instead of distinguishing one of twopossible signals, now it must distinguish between M possible signals
especially complex in the presence of noise
Data Rate Limits: Nyquist Theorem
[bps] MlogB2C 2noiseless ⋅⋅=
Four signal levels Eight signal levels
Typical noise
Four signal levels Eight signal levels
Typical noise
if spacing between levels becomes too small, noise signal
can cause receiver to make wrong decision
max amplitude
min amplitude
32
Example [ multilevel digital transmission ]
2-level encoding: C=2B [bps]one pulse – one bit
4-level encoding: C=2*2=4B [bps]one pulse – two bits
Data Rate Limits: Nyquist Theorem
8-level encoding: C=2*3=6B [bps]one pulse – three bits
100110100011010010 ⇒
100110100011010010 ⇒
33Data Rate Limits: Shannon Theorem
Shannon LawShannon Law – max transmission rate over a channel with bandwidthB, with Gaussian distributed noise, and with signal-to-noise ratio SNR=S/N, is
• theoretical limit – there are numerous impairments in everyreal channel besides those taken into account in Shannon'sLaw (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 ofthe channel
• in practice we need to use both methods (Nyquist & Shannon) to find what data rate and signal levels are appropriate foreach particular channel:
The Shannon capacity gives us the upper limit!The Nyquist formula tells us how many levels we need!
[bps] SNR)(1logBC 2noisy +⋅=
34Data Rate Limits
Example [ data rate over telephone line ]
Solution:
We can calculate the theoretical highest bit rate of a regular telephone line as:
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).
C = B log2 (1 + SNR) = = 3000 log2 (1 + 3162) == 3000 log2 (3163)
C = 3000 × 11.62 = 34,860 bps
35Data Rate Limits
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
C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 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.
4 Mbps = 2 × 1 MHz × log2 L L = 4
[bps] MlogB2C 2⋅⋅=