page 7.1

CHAPTER 7

AMPLITUDE

Transmit information-b earing message or baseband signal voice-music

through a Communications Channel

Baseband = band of frequencies representing the original signal

for music 20 Hz - 20,000 Hz, for voice 300 - 3,400 Hz

write the baseband message signal mt $ M f 

Communications Channel

Typical frequencies 10 KHz ! 300 GHz

write ct= A cos2f ct

c

ct = Radio Frequency

A = Carrier Amplitude

c

fc = Carrier Frequency

Amplitude Mo dulation AM

! Amplitude of carrier wavevaries  a mean value in step with the baseband signal mt

st= A [1 + k mt] cos 2f t

c a c

Mean value A .

c 31

page 7.2

Recall a general signal

st= at cos[2f t + t]

c

For AM

at = A [1 + k mt]

c a

t = 0 or constant

k = Amplitude Sensitivity

a

Note 1 jk mtj < 1or [1 + k mt] > 0

a a

2 f  w = bandwidth of mt

c 32

page 7.3

AM Signal In Time and Frequency Domain

st = A [1 + k mt] cos 2f t

c a c

j 2f t j 2f t

c c

e + e

st = A [1 + k mt]

c a

2

A A

c c

j 2f t j 2f t

c c

e + e st =

2 2

A k

c a

j 2f t

c

+ mte

2

A k

c a

j 2f t

c

+ mte

2

To nd S f  use:

mt $ M f 

j 2f t

c

e $  f f 

c

j 2f t

c

e $  f + f 

c

expj 2f tmt $ M f f 

c c

expj 2f tmt $ M f + f 

c c

A

c

S f  = [f f +f +f ]

c c

2

A k

c a

+ [M f fc+Mf +f ]

c

2 33

page 7.4

st = A [1 + k mt] cos 2f t

c a c

A

c

= [1 + k mt][expj 2f t+ expj 2f t]

a c c

2

If k mt > 1, then

a

! Overmo dulation

! Envelop e

see Text p. 262 - Fig. 7.1

max value of k mt= Percent Mo dulation

a

Lo ok at AM signal in frequency domain

A

c

S f  = [ f f +f +f ]

c c

2

k A

a c

+ [M f f +Mf +f ]

c c

2

where M f =FT of mt 34

page 7.5

Example 1 p. 265 - Single Tone Mo dulation

mt = A cos 2f t f = 1 KHz message frequency

m m m

st = A 1 + k mt] cos 2f t

c a c

st = A 1 +  cos 2f t] cos 2f t

c m c

 = k A < 1

a m

= Mo dulation Factor Percentage

Draw st for <1

A 1 +  A

c max

=

A A 1 

min c

A A

max min

 =

A + A

max min

Rewrite st using trig identity

1 1

cos cos = cos  + cos +  with = f ; = f

c m

2 2



st = A [cos 2f t + cos 2 f + f t

c c c m

2



+ cos 2 f f t]

c m

2

A

c

S f  = [ f f +f +f ]

c c

2

A 

c

+ [ f f f +f +f +f ]

c m c m

4

A 

c

+  f f + f +f +f f ]

c m c m

4 35

page 7.6

2 2

[st] =power into 1 jSfj

2

2

A A

c

c

= Carrier Power 2

2 2

Positive and

NegativeFrequencies 

2

2 2

A  A 

c

c

Upp er Sideband 2 =

4 8

Power

2

2 2

A  A 

c

c

Lower Sideband 2 =

4 8

Power

2 2 2 2

A  A 

2

c c

+

USB + LSB Power 

8 8

= =

2 2 2 2 2

A   A A

c c c

Total Power 2+

+ +

2 8 8

Example

Mo dulation

Audio

Carrier  Power Total Audio Power

Power Mo dulation USB + LSB Power Total

2 2 2 2

A A A 

c c c

2 2

  100  2 +  

2

2 4 4 2+

100 100 50 150 0.33

100 50 12.5 112.5 0.11

100 20 4 104 0.04

100 0 0 100 0 36

page 7.8

Generation of AM Waves - P. 267

Square Law Mo dulator -

3 stages: Adder, Non-Linearity NL, Bandpass Filter BPF

2

For this example, we use non-linearity v t= a v t+a v t

2 1 1 2

1

Here v t = A cos 2f t + mt

1 c c

z }| {

 

2a

2

! v t = a A 1+ mt cos 2f t

2 1 c c

a

1

2 2 2

+a mt+ a m t+a A cos 2f t

1 2 2 c

c

The rst bracketed term is desired AM wave at carrier f with

c

2a

2

k =

a

a

1

Remaining terms are at baseband or 2f and are ltered out by BPF. BPF has center

c

2

frequency f bandwidth 2w . The cos term has comp onents at baseband and at 2f b ecause

c c

1

2

= cos 1 + 2 cos 2  with =2f t

c

2 37

page 7.10

Given v t nd V f  Curve

2 2

v t = a A cos 2f t a

2 1 c c

2a A mt cos 2f t b

2 c c

+a mt c

1

2

+a m t d

2

1

2

+a A 1 + cos 2  2f t e

2 c

c

2

| {z }

2

cos 2f t

c

A

c

V f  = a [ f f +f +f ]

2 1 c c

2

+2a A [M f f +Mf +f ]

2 c c c

+a M f 

1

2

+a FT[m t]

2

2

a A

2

c

[ f + f 2f +f +2f ] +

c c

2

Assignment

Text: p. 384, p. 7.1, p. 7.3, p. 269 Ex. 2 38

page 7.11

Generation of AM Waves Continued-P. 269-270

Switching Mo dulator

Same as square law except non linearityisnow: !



v t v t>0

1 1

v t=

2

0 v t<0

1

v t = A cos 2f t + mt

1 c c

= ct+ mt

Assume jmtj A

c



v t ct> 0

1

Thus v t=

2

0 ct< 0

Thus switch dio de on and o at rate f

c

dio de on only when ct > 0

v t=[A cos2f t + mt]g t

2 c c p

where g t is a unip olar square waveat f .

p c 39

page 7.12

Recall Fourier Series ELEC 260 p. 106

f t with p erio d T

1

X

2nt 2nt

f t = a + a cos + b sin

o n n

T T

n=1

Z

T=2

1

f tdt a =

o

T

T=2

Z

T=2

2 2nt

a = dt f t cos

n

T T

T=2

Z

T=2

2 2nt

b = f t sin dt

n

T T

T=2

Consider g t with p erio d T and frequency f =1=T

p 0 c 0

Z

T =4

o

1 1 1 T

o

= a = 1  dt =

o

T T 2 2

T =4

o

o o

T

o

Z

T =4

o

4

2 2nt 2 T 2nt

o

a = cos dt = sin

n

T

o

T T T 2n T

T =4

o

o o o o

4

"  !  !

2nT =4 1 2nT =4

o o

= sin sin

n T T

o o

8 9

2

> >

n =1; 5; 9 :::

n1

< =

n

1 n 2 1

= 2 sin =  =

> >

n 2 2n 1 

: ;

2

n =3; 7; :::

n

n =2m1; m=1;2;3;::: 40

page 7.13

Relab el m ! n

 "

1

n1

X

1 2 1 2 2n 1f t

c

with f =1=T g t= + cos

c o p

2  2n 1 T

o

n=1

write g tasFourier cosine series see ELEC 260

p

1

n1

X

1 2 1

g t = + cos[2f t2n 1]

p c

2  2n 1

n=1

v t = [A cos 2f t + mt]g t

2 c c p

 

1 2

= [A cos 2f t + mt] + cos 2f t + :::

c c c

2 

A 2 mt

c

= cos 2f t + mt cos 2f t + + :::

c c

2  2

 

4 A

c

1+ + ::: = mt A cos 2f t

c c

|{z}

2 A

c

{z } |

4

= AM wave with k = unwanted comp onents

a

A

c

at frequencies removed from f

c 41

page 7.14

Demo dulation of AM Waves

Square Law Demo dulator

No go o d unless mt  A i.e.  mo dulation very low see P. 271 - Skip this

c

Envelop e Detector -P. 272 42

page 7.15

7.2 P. 274 - Double Sideband Suppressed Carrier Mo dulation DSBSC

Carrier ct= A cos 2f t do es not carry information ! waste of p ower

c c

Power in mt 1

Recall = max at 100 mo dulation

Power in ct 3

! Suppress the Carrier

AM: s t = [1 + mt]ct

AM

= ct+mt ct k =1

a

DSB: st = mtct

= A cos 2f tmt

c c

1

Sf = A [Mf f +Mf f ]

c c c

2

Phase reversal when mt < 0. 43

page 7.16

DSBSC

st = ctmt

= A cos 2f t mt

c c

A

c

= [expj 2f t + expj 2f t]mt

c c

2

Recall shifting prop erty

mt $ M f 

expj 2f tmt $ M f f 

c c

A

c

[Mf +f +Mf f ] S f =

c c

2

Single tone mo dulation

mt = A cos 2f t

m m

A

m

M f  = f f +f +f ]

m m

2

A

m

M f f  = [ f f f +f +f f ]

c m c m c

2

A

m

M f + f  = [ f f + f +f +f +f ]

c m c m c

2 44

page 7.18

3.5 Generation of DSBSC Waves -P. 276

DSBst = ctmt

= A cos 2f tmt

c c

Balanced mo dulator

s t = A [1 + k mt] cos 2f t

1 c a c

s t = A [1 k mt] cos 2f t

2 c a c

st = s t s t

1 2

= 2k A cos 2f tmt

a c c

= 2kctmt

a 45

page 7.19 P. 277 To pro duce DSB signal st= mtct

Ring mo dulator mixer multiplies mt with ct.

For p ositive half-cycles of ct

D1 and D3 conduct, D2 and D4 op en

p oint a connected to p oint b

p oint c connected to p oint d

current through T2 primary prop ortional to mt.

For negative half-cycles of ct

D2 and D4 conduct, D1 and D3 op en

p oint a connected to p oint d

p oint b connected to p oint c

current through T2 primary prop ortional to mt.

Use square wave cttoavoid problems with dio de o sets. 46

page 7.20

Ring Mo dulator

st = ctmt

1

n1

X

4 1

= cos[2f t2n 1] mt

c

{z } |

 2n 1

n=1

| {z }

mo d

Square Wave Carrier

Compare ct here with g t on page 7.12.

p

Here ct range 1 to +1, no DC term

g t range 0 to 1, with DC term.

p

Z

1

S f =Cf Mf= C M f d

1

Cf

z }| {

1

n1

X

1 4 [ f f 2n 1 +  f + f 2n 1]

c c

S f = M f 

 2n 1 2

n=1

For n =1

4

Sf = [f f +f +f ] M f 

c c

2

Z

1

4

=   f ++f ]M f d

c c

2

1

4

= [M f f +Mf +f ]

c c

2

For n =2:::

n =3::: 47

page 7.20A

Sup erhetero dyne Receiver

Since f varies from one signal to the next, the receiver is designed to convert all f to a

c c

xed frequency f . This waywe can use the same lter and detector for any signal at any

IF

f .

c

RF 540-1600 KHz AM Broadcast

LO 995-2055 KHz

IF=LO-RF 455 KHz

Example Consider CFAX at 1070 KHz

Here LO = 1525 KHz SUM = 2595

IF = LO = RF = 455 KHz DIF = 455

Mixer outputs sum and di erence frequencies

thus another station at 1980 KHz image frequency will also b e received since

LO = 1525 KHz

RF = 1980 KHz

RF-LO = 455 KHz

Image is ltered out by RF AMP/ lter

Homo dyne receiver uses zero frequency IF, i.e. LO=RF, IF=0 48

page 7.20X

General I-Q receiver - demo dulator

applies for any signal, AM, DSB and others.

Input signal

st = atcos[2f t + t]

IF

= xt cos 2f t yt sin 2f t

IF IF

Output signals I = xt and Q = y t.

Exercise - show this using algebra 49

page 7.23

DSBSC Demo dulation

Costas Receiver

same as general I-Q receiver with extra feedback signal

Two coherent detectors Output

1

One LO cos2f t +  A cos  mt

c c

2

1

A sin mt Other LO sin2f t + 

c c

2

Here f = f = f .

c IF

Phase discriminator output et is a DC control signal

1 1

et = A cos mt  Asin  mt

c c

2 2

2 2

A A

c c

2 2

= m t cos  sin  = m t sin 2

4 8

The DC control signal is used as a feedback signal to reduce 

TP11 = cos 2f t cos2f t + 

c c

TP19 = cos 2f t sin2f t + 

c c

TP13 = 50

page 7.24

Costas receiver in more detail

Costas receiver is same as general I-Q receiver with extra feedback signal et=IQ used

to adjust oscillator frequency f to b e close to f . In general f 6= f .

IF IF

IF input signal s t=mt cos2f t +  .

IF IF in

From table 3 page 626,

2 cos sin = sin  + sin + 

2 cos cos = cos   + cos + 

TP11 = mt cos2f t +   cos 2f t +  

IF in `

1

mt[cos2 f f t +    + cos2 f + f t +  +  ] =

IF in ` IF in `

2

TP19 = mt cos2f t +   sin2f t +  

IF in `

1

= mt[sin2 f f t +    + sin2 f + f t +  +  ]

IF in ` IF ` in

2

1

TP13 = mt cos[2 f f t +   

IF in `

2

1

TP12 = mt sin2 f f t +   

IF in `

2 51

page 7.25

Costas receiver continued

De ne 4f = f f

IF

=  

` in

et = TP12  TP13

1

2

= m t cos2 4ft  sin2 4ft 

4

1

2

= m t sin4 4ft 2 

8

1

Using sin cos = sin 2

2

et= 0 if 4f = 0 and =0

If et 6= 0, then VCO will b e adjusted so that et= 0

After LPF 5Hz

1

2

e t = sin4 4ft 2 

LP

| {z }

8

Average Value

= DC Control Signal

For small 4 4ft 2 =x; sin x  x.

Thus e t ' 4 4ft 2

LP 52

page 7.26

P. 283 - Quadrature Carrier Multiplexing

 Combine two indep endent DSBSC mo dulated waves on the same frequency f

c

with two mo dulations m t m t

1 2

o

 Toachieve this we use two carrier waves at f but 90 out of phase

c

st= A m t cos 2f t + A m t sin 2f t

c 1 c c 2 c

To receive the two indep endent mo dulations m t m t use the general I-Q receiver. The

1 2

o

I-Q receiver contains two separate DSBSC demo dulators with lo cal oscillators carriers 90

out of phase, cos 2f t and sin 2f t.

c c

For this to work, we need to have the lo cal oscillators to have frequency f exactly the same

c

as the frequency of st.

Note that st is in the general format with xt= m t and y t=m t.

1 2 53

page 7.28

Quadrature Carrier Multiplexing - continued

We will show that there is no mixing of the two mo dulations, even though the carrier

frequency is the same

v t = st cos 2f t

c

= A m t cos 2f t cos 2f t

c 1 c c

+ A m t sin 2f t cos 2f t

c 2 c c

1

2

1 cos 2  cos =

2

1

cos sin = sin 2

2

1

v t = A m t[1 cos 4f t]

c 1 c

2

1

+ A m t sin 4f t

c 2 c

2

After LPF

1

v t= A m t

o c 1

2 54

page 7.29

P. 284 - Single Sideband Mo dulation

In DSBSC, the two sidebands carry redundant information thus we can eliminate one side-

band to get SSB-SC usually referred to as SSB.

SSB with carrier is called SSB-AM

We b egin with frequency domain description of SSB

1. Single tone mo dulation mt = cos 2f t

m

2. General mo dulation mt

Lo oking only at p ositive frequencies

mt

AM

DSB-SC

SSB 55

page 7.31

3.9 Generation of SSB Waves

1 - Filtering Frequency Discrimination

2 - Phase Discrimination general I-Q transmitter

1 Filter Metho d

Pro duce Mo dulator + Filter

Need sharp lter to pass USB and reject LSB

e.g 8-p ole crystal lter at 10.7 MHz not practical

Up convert to desired carrier frequency

Assignment4P. 388 P. 7.4  16, 19a, 20a 56

page 7.31A

2 General I-Q transmitter

st= xt cos 2f t yt sin 2f t

c c 57

page 7.32

text P. 288

Time Domain Description of SSB

A

c

st= [mt cos 2f t  m^ t sin 2f t]

c c

2

For upp er sideband

+For lower sideband

m^ t = Hilb ert transform of mt see notes p. 3.3-3.4

o

90 Phase shift for all p ositive frequencies

o

+90 Phase shift for all negative frequencies

H f  sp ecial kind of \ lter"

Recall H f  = j sgn f p.3.3-3.4

ht = 1= t

Z

1

1 m 

m^ t = mt ht= d

 t 

1

^

M f  = M f H f 

= j sgn fMf 58

page 7.33

Single tone mo dulation for SSB

Example 3 - P. 291

If mt = A cos 2f t

m m

m^ t = A sin 2f t

m m

A

c

st = [mt cos 2f t  m^ t sin 2f t]

c c

2

A A

c m

= [cos 2f t cos 2f t + sin 2f t sin 2f t]

m c m c

2

Recall cos   = cos cos + sin sin ,thus

A A

c m

st= cos[2 f  f t]

c m

2

st is a single tone at frequency f  f . See also text gure 7.3 1 p. 266

c m 59

page 7.34

SSB - Time $ Frequency Domain

A

c

st = [mt cos 2f t  m^ t sin 2f t]

c c

2

^

M f  = M f H f =j sgn fMf p. 7.32

1 A

c

[M f  f f f +f +f g S f  =

c c

2 2

1

j sgn fMf ffff+fg]

c c

2j

A

c

[Mf ff =

c

4

+Mf f+f

c

sgn fMf ff

c

+ sgn fMf f+f]

c

A

c

= [M f f +Mf +f 

c c

4

sgn f f  M f f  + sgn f + f  M f + f ]

c c c c

A

c

= M f f ;ff >0;f > f

c c c

2

similarly

A

c

S f  = M f + f ;f+f >0;f < f

c < c

2 60

page 7.36

P. 293 Demo dulation of SSB

c t = cos 2f t

LO c

A

c

st = [mt cos 2f t +^mt sin 2f t]

c c

2

vt = st cos 2f t

c

A

c

mt cos 2f t cos 2f t =

c c

2

A

c

+ m^ t cos 2f t sin 2f t

c c

2

1

2

cos = 1 + cos 2 

2

1

cos sin = sin 2

2

A A A

c c c

v t = mt + mt cos 4f t + m^ t sin 4f t

c c

4 4 4

{z } {z } | |

Message Signal Unwanted Terms

This assumes no phase or frequency error in lo cal oscillator C t

LO 61

page 7.37

Demo dulation of SSB with Frequency Error

C t = cos 2 f + 4f t

LO c

A

c

[mt cos 2f t +^mt sin 2f t] st =

c c

2

vt = st cos 2 f + 4f t

c

A

c

= mt cos 2 f + 4f t cos 2f t

c c

2

A

c

+ m^ t cos 2 f + 4f t sin 2f t

c c

2

1

cos cos = [cos  + cos + ]

2

1

cos sin = [sin +  sin ]

2

= 2 f + 4f t =2f t =24ft

c c

1 A

c

mt cos 2 4ft vt =

2 2

A m^ t

c

+ sin 2 4ft + Double Frequency Terms

2 2

After LPF, v t is an SSB signal on a carrier 4f .

0 62

page 7.38

Carrier Plus SSB Wave

st = A cos 2f t + mt cos 2f t m^ t sin 2f t

c c c c

| {z }

{z } |

Carrier

USB

= A + mt] cos 2f t m^ t sin 2f t

c c c

= at cos2f t + 

c

q

2 2

where at= [A +mt] +[^mt]

c

at = output of envelop e detector

q

2 2 2

= A +2A mt+ m t+m ^ t

c

c

If A jmtj and A jm^tj

c c

q

1

1=2

2

at = A +2A mt recall 1 + x ' 1+ x

c

c

2

s

2

1+ = A mt

c

A

c

 

1

= A 1+ mt = A +mt

c c

A

c

A jmtj

c

Thus at=mt+ DC bias term as long as

A jm^tj

c 63

page 7.39 Sup erhetero oyne Receiver for USB

To tune in 2 change LO to 10.559 MHz

SIG -LO = 10.559 - 10.104 = 455 Khz

Tunable lter to eliminate image frequency

could also cho ose LO = 9.645 MHz for Station 1.

Question: Can we demo dulate LSB with this receiver? 64

page 7.40

In practice it is not cost e ective to lter at frequencies much greater than 10.7 MHz. This

is b ecause the lter skirts cannot b e made steep enough .

Typical Crystal Filter Sp ec.

Typical Voice Audio signal 300 Hz - 2600 Hz

If 2300 Hz passband centered on audio

then -60 dB attenuation will b e at 400 Hz in opp osite sideband. 65

page 7.50

Summary of mo dulation typ es, assuming the message jmtj < 1

AM

st = [1 + k mt] cos2f t + 

a c

DSB

st = mt cos2f t + 

c

USB

st = mt cos2f t +  m^ t sin2f t + 

c c

FM

Z

t

st = cos[2f t +2k m d ]

c f

0

= cos[2f t + t] = cos[ t]

c i

PM

st = cos[2f t + k mt]

c p

For mt = A cos 2f t with A =1

m m m

AM mt = [1 + k cos 2f t] cos2f t + 

a m c

DSB st = cos 2f t cos 2f t + 

m c

USB st = cos 2f t cos 2f t + 

m c

sin 2f t sin2f t + 

m c

k

f

FM st = cos2f t + sin 2f t]

c m

f

m

PM st = cos[2f t + k cos 2f t

c p m 66