CommunicatingInformationPartOne.ppt

(1)

Communicating Information

(2)

Syllabus Objectives

30.1 Principles of Modulation 30.2 Sidebands and bandwidth

Candidates should be able to:

(a) understand the term modulation and be able to distinguish between amplitude modulation (AM) and frequency

modulation (FM).

(b) recall that a carrier wave, amplitude modulated by a single audio frequency, is equivalent to the carrier wave

frequency together with two sideband frequencies. (c) understand the term bandwidth.

(3)

Communications Systems

Essentials of a communication system

(4)

Audio Frequencies

Audible sound lies in the range 20Hz – 20kHz

• To transmit these frequencies as radio waves would require a VERY LONG aerial

i.e.  = 1.5 x 107m to 1.5 x 104m!

(5)

Radio Frequency Spectrum

Band Frequency Wavelength

LW Low frequency (LF) 30-300kHz 1-10km MW Medium freq (MF) 300-3000kHz 100-1000m

SW High freq (LF) 3-30MHz 10-100m

FM Very high freq (VHF) 30-300MHz 1-10m TV Ultra high freq (UHF) 300-3000MHz 10-100cm

Microwave /satellite

Super high freq (SHF) Extra high freq (EHF)

3-30GHz 30-300GHz

(6)

Modulation

• High frequency

carrier wave

• Transmitter requires only a short aerial

• Audio frequency is superimposed on this

signal by:

(7)

Amplitude Modulation

(8)

Amplitude Modulation

After some P2/3 mathematics…

x_AM = A_csin_ct + ½A_i[cos(_c– _i)t – cos(_c+ _i)t]

Information signal: x

_i

= A

_i

sin



_i

t

Carrier signal: x

_c

= A

_c

sin



_c

t

(9)

Amplitude Modulation

(10)

Amplitude Modulation

A

m

pl

it

ud

e

frequency

Carrier, f_c

High frequency carrier signal modulated by a single audio signal

Sideband

f_c – f_a

Sideband

f_c + f_a

(11)

Amplitude Modulation

Lower side band

frequency = f_c – f_i

amplitude = A_i/2

Upper side band

frequency = f_c + f_i

amplitude = A_i/2

Note

(12)

Amplitude Modulation

A

m

pl

it

ud

e

frequency

Carrier, f_c

High frequency carrier signal modulated by a voice spectrum audio signal

Sideband Sideband

(13)

AM Bandwidth

• AM radio station bandwidth = 9kHz • Maximum audio frequency = 4.5kHz

Signal lacks higher audible frequencies and so lacks quality

4.5kHz

(14)

AM Bandwidth

The bandwidth required for an AM

waveform is twice the maximum

frequency in the information signal

4.5kHz

(15)

AM in Excel

SAQ3.1 (

Page 20, Telecommunications

)

A 100kHz carrier of amplitude 20V is

amplitude modulated by a 10kHz

square

wave

of amplitude 5V.

(16)

AM in Excel

Amplitude modulation.xls

Amplitude Modulated wave

-30 -20 -10 0 10 20 30

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00

Time (10-4s)

A

m

p

lit

u

d

e

(m

ax

=

A

+

a

) A = 20  5V

1 x 10-4 _{sec (10kHz)}

20V, 100kHz modulated by

(17)

Frequency Modulated wave

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00

Time (sec)

A

m

p

lit

u

d

e

Time (10-4_s)

Frequency Modulation

3kHz sine wave frequency modulated by a 100Hz signal of modulation index = 20

1kHz 5kHz

(18)

FM Modulation Index

Carrier wave frequency = f_c

Audio frequency = f_a

Modulation Index = f = m_if_a

Signal frequency = f_c f Example

3kHz carrier modulated by 100Hz audio signal with m_i = 20. f = 20 x f_a = 2.0kHz

(19)

Frequency Modulation

(20)

FM Bandwidth

• Maximum audio frequency = 15kHz

• FM radio station bandwidth (VHF) = 200kHz

 FM signal has a higher quality than AM

15kHz

(21)

Bandwidth

Long Wave band (LW) = 30kHz – 300kHz

No. of AM stations = 270kHz/9kHz =

30 FM stations in same region = 270/200



1

(22)

FM Example

Worked Example (Page 21, Telecommunications)

A 500kHz sinusoidal carrier of amplitude 10V is frequency modulated by a 2kHz information signal of amplitude 2V. The carrier frequency deviation is 20kHz per Volt. Describe the resulting FM

(23)

FM Example

2V information signal

f_c = 2V x 20kHz/V = 40kHz

f_max = 500kHz + 40kHz = 540kHz

f_min = 460kHz

3V information signal

f_c = 3V x 20kHz/V = 60kHz

f_max = 500kHz + 60kHz = 560kHz

f_min = 440kHz

(24)

AM versus FM

• AM signals depend on amplitude variation

• FM signals depend on frequency variation

(25)

AM versus FM

• AM bands LW, MW and SW travel long

distances by reflection from the ionosphere

• FM range only



30km by line-of-sight

Many transmitters required

(26)

AM versus FM

• FM transmitters are more complex

(27)

Syllabus Objectives

30.3 Transmission of digital information

Candidates should be able to:

(e) recall the advantages of the transmission of data in digital form.

(f) understand that the digital transmission of speech or music involves analogue-to digital conversion (ADC) on transmission and digital-to-analogue conversion (DAC) on reception.

(g) show an understanding of the effect of the sampling rate and the number of bits in each sample on the reproduction of an input signal.

(h) appreciate that information may be carried by a number of different channels, including wire-pairs, coaxial cables, radio and microwave links, and optic fibres.

(28)

Analogue Voice Signal

microphone

amplifier

transmitter V/ Ｖ

ｔ / ｓ

Analogue audio signal contains frequencies 20–20kHz

(29)

Repeater Stations

• Signal attenuation requires amplification

• This reduces signal-to-noise ratio (SNR)

(i.e. increased noise level)

V

t _repeater

amplifie r

V

t

V

t

attenuated & noisy

regenerated & noisy

(30)

Signal Noise

(31)

Digital Regeneration

• Signal attenuation necessitates amplification • Regenerator amplifier recreates original signal

V

t _regenerator

amplifier V

t

V

t

(32)

Digital Regeneration

Original signal can be reproduced perfectly by binary sampling

V

t

V

(33)

Decimal to binary (4-bit)

0 0000 8 1000

1 0001 9 1001

2 0010 10 1010

3 0011 11 1011

4 0100 12 1100

5 0101 13 1101

6 0110 14 1110

7 0111 15 1111

1 x 2 x 2 = 4

1 x 2 = 2

1 x 2 x 2 x 2= 8

(34)

Analogue-to-Digital

Conversion (ADC)

4-bit sampling (2 x 2 x 2 x 2 = 16 levels)

t/s Signal

V/V 16 14 12 10 8 6 4 2 0

0 125 250 375 500 625 750 875

Analogue signal 8V

13V 15V

2V _1V

8V

13V 15V

Digitally sampled

signal

1000 1101 1111 0010 0001 1000 1101 1111

(35)

_out

= – V

₁

– 2V

₂

– 4V

₃

V₁ V₂ V₃ –V_out

1 0 0 1

0 1 0 2

1 1 0 3

0 0 1 4

1 0 1 5

0 1 1 6

1 1 1 7

R_f

R_f/2

R_f/4

R_f

–

+

V₁

V₂

(36)

Modulation process for transmitting the data sequence "110100" at 10,000 bits/sec using sinc-shaped pulses. The

thick black waveform is the only signal transmitted.

Binary Transmission

1

(37)

Sampling

Rate

2.5Hz signal sampled at 10Hz

Good quality recovered signal

at the correct frequency

Sampling.xls

Modelling Simple Harmonic Motion

Rate

Poor quality and lower frequency

Sampling rate must be at least

twice the signal frequency

• A good signal can be recovered at just over twice the actual signal frequency

• At lower sampling rates the signal appears to possess a lower frequency

(46)

Sampling Rates

Audible range 20-20kHz

Speech sampling

(Telephone)

11.025 KHz (8KHz)

Low Grade Audio

(WWW Audio, AM Radio) 22.05 KHz

(47)

Creating a square wave

Infinite addition of harmonics

Single sine wave at 1Hz, amplitude, A = 1

Signal

-1.500 -1.000 -0.500 0.000 0.500 1.000 1.500

0.000 0.200 0.400 0.600 0.800 1.000 1.200 1.400 1.600 1.800 2.000

Time,t (sec)

A

m

p

lit

u

d

e

, A

(

V

)

(48)

Creating a

square wave

1

st

harmonic (A) + 3

rd

harmonic (A/3)

(49)

Creating a

square wave

1

st

(A) + 3

rd

(A/3) + 5

th

(A/5)

Sig nal -1 .50 0 -1 .00 0 -0 .50 0 0.0 00 0.5 00 1.0 00 1.5 00 0.0 00 0.2 00 0.4 00 0.6 00 0.8 00 1.0 00 1.2 00 1.4 00 1.6 00 1.8 00 2.0 00 Tim e, t (s ec ) A m p lit u d e , A (V ) Frequency Spectrum 0 0.2 0.4 0.6 0.8 1 1.2

1 2 3 4 5 6 7 8 9 10 11

Angular frequency,  (2Hz)

(50)

Creating a

square wave

1

st

(A) + 3

rd

(A/3) + 5

th

(A/5) + 7

th

(A/7)

(51)

Creating a

square wave

1

st

to 17

th

harmonics

(52)

Creating a

square wave

1 to 31 harmonics

Signal -1.500 -1.000 -0.500 0.000 0.500 1.000 1.500

0.000 0.500 1.000 1.500 2.000

Time,t (sec)

(53)

Fourier Addition Applet

(54)

Information Channels

The medium through which data is

transmitted is referred to as the

channel

e.g.

• Wire-pairs

• Co-axial cables

(55)

• Require regular amplification due to attenuation • Energy lost as heat due to resistance

• Wires act as aerials leading to cross-talk interference

• Easy to ‘tap’/ Low security • Low bandwidth (500kHz)

(56)

Co-axial Cables

• _{Co-axial conductors:}

Copper wire & braid with polythene insulator • _{Outer mesh ‘shields’ inner wire}

• Outer mesh earthed

• Less prone to interference – more secure • _{Higher bandwidth (}50MHz)

Cu wire

Polythene insulator

Cu braid Outer wire