Electrical Communication System: Block Diagram

MSIT 411 1

Electrical Communication System: Block Diagram

Transmitter

Channel

Receiver

Input Transducer

Output Transducer

Noise and signals from other sources

Information Source

Destination

Input electric signal

Output electric signal

Transmitted signal

Received signal

Electrical Communication System

•  Information Source – voice, audio, video, digital data, etc.

•  Input Transducer – translates information into electric signal, e.g., microphone

•  Output Transducer – reverts to the original information format, e.g., loudspeaker

MSIT 411 2

Electrical Communication System: Block Diagram (continued)

Transmitter

Channel

Receiver

Input Transducer

Output Transducer

Noise and signals from other sources

Information Source

Destination

Input electric signal

Output electric signal

Transmitted signal

Received signal

Electrical Communication System

•  Transmitter – modulation, filtering, encoding, transmitting (prepare the proper format for the channel)

•  Receiver – filtering, amplifying, demodulating, decoding, etc.

•  Channel – transmission medium such as twisted pair, coax, fiber, air, etc. Notes: 1. In a two-way communication system, both the transmitter and the receiver are integrated (transceiver)

2. The channel may be shared

MSIT 411 3

Waveform: Signal as a function of time

a(t)

t (second)

Analog Waveform examples: voice, audio, video

Waveform = “voltage fluctuation” in time: the signal is a function of time (information is represented by voltage variation)

d(t)

(Baseband) Digital Waveform examples: data, MPEG video, MP3 music

Examples: your weight, your portfolio value, temperature reading in Chicago, patient’s heartbeat as displayed on a monitor…

t (second)

(volts)

(volt)

MSIT 411 4

Sine Waves

T A

0 T t (second)

s(t) = A sin (2πf t) (volt)

A: Amplitude in “volts”

T: Period or Cycle in “seconds”

f = 1/T: Frequency in cycles/second or “Hertz” (Hz) s(t) = A sin (2πf t) = A sin (ωt), where ω = 2πf radians/sec = “angular velocity”

MSIT 411 5

Signal Power

T A

0 T t (second)

s(t) = A sin (2πf t) (volt)

Instantaneous energy in a signal = |s(t)|2 (measured in Joules)

Power is energy per unit time, measured in “Watts (Joules per second)”.

Average power of a sine wave: = A2/2

MSIT 411 6

Sine Wave Example: Household AC

1/60 156

0 1/60 t (second)

110-Volt AC source: s(t) = 156 sin (2π(60) t) (volt)

A = Amplitude = 156 Volts , f = Frequency = 60 Hz

Average power = 1562/2 = 12,168 watts

Same average power (or close) with constant 110 Volts ( DC equivalent)

MSIT 411 7

Sine Wave Animation

http://www.rkm.com.au/ANIMATIONS/animation-sine-wave.html

MSIT 411 8

Sine Waves with Different Amplitudes and Frequencies

High- and low-frequency sine waves

A –

B –

Freq. = fB

Freq. = fA

t

Noteworthy: In this example, the higher-frequency sine wave (in blue) has a smaller amplitude than the low-frequency sine wave (in green)

N O R T H W E S T E R NU N I V E R S I T Y

MSIT | Master of Science in Information Technology

Sinusoidal Signal

Amplitude A=1

Time delay = 12, Phase shift θ = 12/50 cycle = 86.4 degrees

Period= 50 sec, frequency f = 1/50 cycle/sec

Time t (seconds)

Electromagnetic wave s(t) = A sin (2 π f t + θ) s(

t)

MSIT 411 10

Sinusoids with Different Phases (same amplitude and same frequencies)

s1(t) = 100 sin (2π × 106 t) Phase = 0 s2(t) = 100 sin (2π × 106 t + 2π/3) Phase = 120° s3(t) = 100 sin (2π × 106 t - 2π/3) Phase = - 120°

Note: s1(t) + s2(t) + s3(t) = 0

•  “Phase” represents relative timing position between sinusoids. It becomes an important parameter when there are more than one sinusoid involved, which is typically the case in communication systems

t=6 µs t=0 t=1 µs t=2 µs t=3 µs t=4 µs t=5 µs

-150.00

-100.00

-50.00

0.00

50.00

100.00

150.00

100 sin(2pi 1000000 t) 100 sin(2pi 1000000 t +2pi/3) 100 sin(2pi 1000000 t - 2pi/3)



Two Signal Paths

s1(t)

s2(t)

Received signal r(t) = s1(t) + s2(t)

Suppose s1(t) = sin 2πf t. Then s2(t) = h s1(t - τ) = h sin 2πf (t - τ)

attenuation (e.g., h could be ½)

delay (e.g., τ could be 1 microsec.)



Sinusoid Addition (Constructive)

+ =

s1(t)

s2(t)

r(t)

Adding two sinusoids with the same frequency gives another sinusoid with the same frequency!



Sinusoid Addition (Destructive)

+ =

Signal is faded.

s1(t)

s2(t)

r(t)

MSIT 411 14

Sine Wave: General Expression

Mathematical representation of a sine wave:

s(t) = A sin (2π f0 t + φ) where –π ≤ φ ≤ π is the phase angle of the sine wave

In summary, a sine wave is characterized by three parameters:

A: amplitude – strength, power = A2/2 (Watt) f0: frequency – rate of voltage oscillation in time φ: phase - relative timing position (meaningful only when there are two or more sine waves involved)

Importance of sine waves in electrical communication systems:

•  They are the fundamental components (“genes”) of any signal, analog or digital: voice, audio, video, data,…

•  They are used as “carriers” in most communication channels for transmission of information-bearing signals

MSIT 411 15

The Frequency Domain

f 4000

f

amplitude spectrum phase spectrum

5 -90°

4000

-160°

•  The frequency concept originated from sine waves

•  “frequency-domain interpretation” of the time function s(t) = A sin (2πf0 t +φ):

It has an amplitude A (strength parameter) and a phase φ (timing parameter) at frequency f0.

•  These are illustrated by the “amplitude spectrum” and “phase spectrum, respectively

•  Example: s(t) = 5 sin (2π(4000)t - 160°):

4000

power spectrum

12.5

|S(f)|

Ps(f)

Arg[S(f)]

We shall deal mainly with amplitude spectrum

MSIT 411 16

Sine Waves and Their Spectra

High- and low-frequency sine waves

f

Line (Amplitude) Spectra

A –

B –

A

B–

Freq. = fB

Freq. = fA

fA fB

t

MSIT 411 17

Signal Spectrum: Frequency Profile of a Signal

Example: A 3-tone signal

x(t) = 20sin(2π1000t −90°)+10sin(2π2000t +0°)+5sin(2π4000t −160°)

1.  x(t) has an amplitude 20 and phase -90° at frequency f =1000 Hz

2.  x(t) has an amplitude 10 and phase 0° at frequency f =2000 Hz

3.  x(t) has an amplitude 5 and phase -160° at frequency f =4000 Hz

Hence, the above signal can be depicted in the frequency domain as a function of frequency X(f) called signal spectrum

f (Hz)

|X(f)|

1000 2000

f (Hz)

Arg[X(f)]

1000 2000

amplitude spectrum (magnitude) phase spectrum (angle)

20 •

-90°

4000

4000

-160°

In this example, the spectrum has three “spectral lines”

MSIT 411 18

a(t)

t

Analog Signal: voice, audio, video

Time domain: Waveform

t

d(t)

Pulse Rate = r (baud)

|A(f)|

f

Frequency domain: Spectrum

|D(f)|

Digital Signal: PCM, data, MPEG video

f

Signal Bandwidth (B Hz) Voice ~ 3 kHz HQ Audio ~ 15 kHz Video ~ 6 MHz

Signal Bandwidth B: proportional to r

Fourier Transform

Fourier Transform

Shape arbitrarily

drawn

Shape arbitrarily

drawn

B = k r, k being a positive constant

Signal Spectrum and Bandwidth Generally, any signal, whether it comes from voice, audio, video, or data, is made up of sine waves whose frequencies span a range. The size of the frequency range is called “signal bandwidth.” The collection of amplitudes for all of the sine waves constitute the “signal spectrum”.



Pulse Width vs. Bandwidth

frequency

Power bandwidth = 1/T

frequency

Power bandwidth = 1/T

time

signal pulse

time

signal pulse

T

T

Narrowband

Wideband

19

MSIT 411 20

Linear Filtering of a Signal

Channel or Filter Frequency Response: |H(f)| |X(f)| |Y(f)| = |H(f)| × |X(f)|

Input signal spectrum (A collection of sine waves within a “band” of frequencies)

Output signal spectrum (A collection of sine waves within a “band” of frequencies)

To avoid/limit distortion: Channel Bandwidth ≥ Signal Bandwidth (Size and location)

|H(f)|

f

Ideal

Channel bandwidth

Practical

Passband Stopband Transition band

Transition band Stopband

•  Passband: frequency range within which the frequency response is at or near its maximum (width = channel bandwidth)

•  Stopband: frequency range within which the frequency response is at or near zero

•  Transition band: frequency range between passband and stopband (the smaller, the better)

Frequency Response of a channel or filter: |H(f)|

MSIT 411 21

Linear Filtering of a Signal

Note: “the function multiplier” |H(f)| creates the “filtering” effect

|H(f)|

f

Ideal

Channel bandwidth

Practical


Transition band Stopband

•  Output |Y(f)| ≈ 0 if |H(f)| ≈ 0 whatever the value of |X(f)|

•  |Y(f)| ≈ |X(f)| if |H(f)| ≈ 1

•  Ideally, |H(f)| = constant, for f in the passband and |H(f)| = 0 for all other f

|H(f)| |X(f)| |Y(f)| = |H(f)| × |X(f)|

|H(f)| Ideal

Channel bandwidth

Practical


Lowpass channel or filter Bandpass channel or filter

f

MSIT 411 22

Power Loss (Attenuation) and Signal-to-Noise Ratio

Transmitted Signal

Power Loss ≡ L ≡ PT / PR > 1 Power Gain ≡ G ≡ PR / PT < 1

Power PT

Channel Received

Signal

Power PR

Input Signal

Power Gain ≡ G ≡ Po / Pi > 1 Power Loss ≡ L ≡ Pi / Po < 1

Power Pi

Repeater/Amplifier

Output Signal

Power Po

Decibel Convention Gdb ≡ 10 log G dB Ldb ≡ 10 log L dB

Remarks: •  Attenuation and delay are unavoidable when a

signal travels over any communication channel, wired or wireless.

•  Attenuation and delay increase with distance •  Attenuation and delay are not distortions as long

as they are uniform across all the frequencies within the signal spectrum

MSIT 411 23

Power Examples

Hop A Repeater Hop B Receiver Amplifier

Source

PT=10W LA= 30 dB

Example 1. Consider the following two-hop communication channel with a repeater in between. Find the received signal power following the receiver amplifier.

G 1 = 20 dB LB= 30 dB G 2 = 20 dB

Destination

PR = ?

L = 30 dB – 20 dB + 30 dB – 20 dB = 20 dB

L = 20 dB L = 100 ⇒ PR = PT/100 = 0.1 W

Example 2. A transmitting node on a coax link uses PT = 0.5 W. The coax attenuation is rated at 1 dB per 10m. If the receiver node has a receiver sensitivity (i.e., minimum received power required) of 5 µW, what is the maximum length of the link between the two nodes?

Solution

At the maximum length, the received power is 5 µW which translates to a power loss of

L = 0.5W/5µW = 105 (largest loss that can be tolerated)

⇒ L = 10 log10 L = 50 dB

⇒ Range x (1 dB/10m) = 50 dB ⇒ Range = 500 meters

Solution



Power Gain/Loss: Wired Channels

Pr (dB)

Distance d

slope = G dB per meter

PT

0

Transmitted Signal

Power PT

Channel Received Signal

Power PR

Twisted pair, cable



Power Gain/Loss: Free Space

reference distance d0=1

distance d

Reference power at reference distance d0 Path loss exponent=2

In dB: Pr = P0 (dB) – 20 log (d)

Pr (dB)

log (d)

slope = -20 dB per decade

P0

0

P0 = Gt Gr (λ/4π)2

antenna gains wavelength



Wavelength

•  Wavelength >> size of object è signal penetrates object.

•  Wavelength << size of object è signal is absorbed and/or reflected by object.

•  The antenna size should a fraction of a wavelength (say ¼ to ½).

λ (meters) = c (speed of light) / frequency



Indoor Propagation Measurements

Hypothetical large indoor environment

Ceiling

Normalized received power vs. distance





Ceiling


Large-scale variation (average over many wavelengths)





Ceiling


Small-scale variations (over fractions of a wavelength)



Power Attenuation: Urban Environment

In dB: Pr = P0 (dB) – 10 n log (d)

reference distance d0=1

distance d

Reference power at reference distance d0 Path loss exponent

Pr (dB)

log (d)

slope (n=2) = -20 dB per decade P0

0

slope = -40 (n=4)



Large-Scale Path Loss (Scatter Plot)

Aver

age

Rec

eive

d Po

wer

(dB

m)

Distance (meters, log scale)



Attenuation: Wireless vs. Wired

•  Path loss ~ 13 dB / 100 m or 130 dB / 1 km –  Increases linearly with

distance

•  Requires repeaters for long distances

•  Path loss ~ 30 dB for the first meter + 20 dB / decade –  70 dB / 100 meters

(2 decades) –  90 dB / 1 km

(3 decades) –  130 dB / 100 km! –  Increases as log (distance)

•  Repeaters are infeasible for satellites

Short distance à Wired has less path loss. Large distance à Wireless has less path loss.

Unshielded Twisted Pair 1 GHz Radio (free space)

MSIT 411 33

Noise and Signal-to-Noise Ratio

•  Noise is present throughout any communication system – transmitter, transmission medium, receiver •  Dominant noise: receiver noise, because it does not attenuate with the signal •  The bandwidth of white noise covers the entire frequency range occupied by the communication signal

•  Statistical noise model: Additive White Gaussian Noise (AWGN)

White Noise Spectrum + Signal Spectrum

Noise (power) spectrum Signal (power) spectrum Area = PS (Watts)

Signal band (B Hz)

•  A receiver filter typically removes “out-of-band” noise and interference. The bandwidth of this filter should be at least the bandwidth of the transmitted signal. The noise within the signal band are thus retained.

•  Reception quality is determined by the Signal-to-Noise Ratio (SNR) after filtering:

S/N ≡ Signal Power PS / Noise Power PN (after filtering, usually in dB)

where PN ≡ N0 × B, N0 ≡ Noise Power Spectral Density in Watt/Hz (readily measurable) B ≡ Filter Bandwidth in Hz (usually assumed to be the Signal bandwidth)

•  Can an ampliflier improve the SNR?

N0 Watt/Hz

Power spectrum: shows power (instead of amplitude) as a function of frequency

f (Hz)

No, it would boost both the signal and the noise.

MSIT 411 34

Physical versus Allocated Channel Bandwidth •  Physical Channel Bandwidth: passband of a linear channel

frequency range of a physical channel (e.g., a fiber link, coax, twisted pair, wireless link) with acceptable power attenuation

•  Allocated Channel Bandwidth: a portion of the physical channel bandwidth designated for a particular communication

•  Plain Old Telephone Service (POTS for voice) uses 0 to 4 kHz, DSL uses bandwidth above 4 kHz.

POTS DSL Upstream DSL Downstream

Up to 384 kbps Up to 7 Mbps

4k 30k 138k 1.1 M f (Hz) ISDN

Example: twisted-pair telephone line

excellent mediocre

Mediocre or poor (uneven and varies)

DSL

•  The DSL band is sub-partitioned into 4-kHz slices. Slices with acceptable attenuation carry data; the data rate is proportional to the spectral quality. The overall DSL data rate is the sum rate over all slices.

Typically high attenuation

MSIT 411 35

Cable and Fiber

f (MHz) 5 45 84 120

Upstream Cable

Data

Cable Channels

2-6

FM Cable

Channels

14-22 174 216 750

Cable Channels

7-13

Cable Channels 23 and up

Channel N: (78+6N) MHz to (84 + 6N) MHz

one or more of these can be used for Downstream Cable Data and/or digital TV

US Cable Spectrum

Fiber

850 nm ~ 0.8 dB/km (Cheaper HW - GaAs)

1300 nm ~ 0.2 – 0.4 dB/km

1550 nm ~ 0.2 – 0.3 dB/km

Each offers 25-30 THz spectrum

2.1



36



37

Useful for wireless telecommunications



38



39

The higher the frequency, the more bandwidth is available, but the worse the channel characteristics

104 4×109

VLF

LF

MF

HF, VHF, UHF, Satellite (including cellular, PCS)

ISM1: 902MHz – 928 MHz

ISM2: 2.4 -2.4835 GHz f (Hz)

108 1010

Satellite

ISM3: 5.735 GHz – 5.86 GHz

Penetration Loss, multi-path fading, rain effect ↑ as frequency ↑



40



41

“beachfront property”: Between ~ 700 and 900 MHz



Cellular/PCS Spectrum Allocation (1993)

Wavelengths too long; propagates too far

802.11b/g (WiFi), 2.4 GHz

802.11a 5.2 GHz

42



dB and dBm •  dB is a ratio of two powers:

We say that power P1 is x dB stronger than power P2 if x = 10 log (P1/P2), where log is base 10. –  Example: P1 is 3 dB more than P2 if P1/P2 ≈ 2.

•  dBm is power relative to a milliwatt (1 mW = 0.001 W): P in dBm = 10 log (P/0.001) –  Example: 1 mW = 10 log 1 = 0 dBm



Cellular Signal Strength Measurements

drive test plots



Link Budget How much transmit power is required to achieve a

target received power?

•  dBs add: Target received power (dBm)

+ path loss (dB) + other losses (components) (dB) - antenna gains (dB) Total power needed at transmitter (dBm)



Example

•  Recall that dBm measures the signal power relative to 1 mW (milliwatt) = 0.001 Watt. To convert from S Watts to dBm, use S (dBm) = 10 log (S / 0.001)

•  Transmitted power (dBm) = -30 + 40 = 10 dBm = 10 mW

•  What if the received signal-to-noise ratio must be 5 dB, and the noise power is -45 dBm?

Transmitter Receiver wireless channel

40 dB attenuation Received power must be > -30 dBm

What is the required Transmit power?

MSIT 411 47

Distortion and Countermeasures Linear Distortion

•  Uneven attenuation within signal bandwidth •  Unequal delays within signal bandwidth

Channel w/ Linear Distortion

Equalizer |X(f)|

|Hc(f)|

|Y(f)| = |Hc(f)| × |Heq(f)| ×|X(f)|

|Heq(f)| constant attenuation within signal bandwidth

|H(f)|

12 k 8 k 4 k

0.03 0.02

0.01

0 f

|Heq(f)|

12 k 8 k 4 k

1 1.5

3

0 f

|Hc(f)| × |Heq(f)| = constant

8 k 4 k

0.03 0.03 0.03

0 f

12 k

Example: Assume that the frequency range of |X(f)| is from 0 to 12 kHz

MSIT 411 48

Distortion and Countermeasures Non-linear Distortion

•  Uneven gain or attenuation with respect to input signal strength (power) (Practical connection: audio amplifier distortion when volume is tuned too high)

•  Nonlinear distortion cannot be “equalized” and thus must be avoided

•  Nonlinear distortion is one of the main reasons why signal power has to be constrained in communication systems and networks.

PT or Pi

PR or Po

Linear Range

Saturation

PT or Pi

PR or Po

Ideal amplifier (constant gain)

Ideal channel (constant loss)

Practical amplifier/channel

Slope = power gain

MSIT 411 49

Companding

•  The compressor ensures that signal entering the channel within the linear operating range of the channel

•  The expander is the “inverse function” of the compressor and thus restores the original signal strength contrast

µ-law Compressor (North America)

y = log (1+ µx)/log(1+µ), where µ is “degree of compression”, is used in North America telephone systems. Soft

voices actually get amplified while strong voices are suppressed. A-law Compressor (Europe)

y = Ax/[1 + ln A] for 0 ≤ x ≤ 1/A = (1 + ln Ax)/(1+ ln A) for 1/A ≤ x ≤ 1

A technique used to deal with nonlinear distortion in telephone systems

Compress Non-linear Channel Expand Si Sx Sy So

a

a Si

Sx

Sx

Sy

Sy

So

Electrical Communication System: Block Diagram

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