6.003: Signals and Systemsdspace.mit.edu/.../lecture-notes/MIT6_003S10_lec24.pdfCommunications Systems Signals are not always well matched to the media through which we wish to transmit

6.003: Signals and Systems

Modulation

May 6, 2010

Communications Systems

Signals are not always well matched to the media through which we

wish to transmit them.

signal applications

audio telephone, radio, phonograph, CD, cell phone, MP3

video television, cinema, HDTV, DVD

internet coax, twisted pair, cable TV, DSL, optical fiber, E/M

Amplitude Modulation

Amplitude modulation can be used to match audio frequencies to

radio frequencies. It allows parallel transmission of multiple channels.

x1(t)

z1(t)

cos 1t

z2(t) z(t)

z3(t)

cos 2t cos ct

cos 3t

LPFx2(t) y(t)

x3(t)

Superheterodyne Receiver

Edwin Howard Armstrong invented the superheterodyne receiver,

which made broadcast AM practical.

Edwin Howard Armstrong also invented and patented the “regenerative” (positive feedback) circuit for amplifying radio signals (while he was a junior at Columbia University). He also invented wide-band FM.

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Amplitude, Phase, and Frequency Modulation

There are many ways to embed a “message” in a carrier. Here are

three.

Amplitude Modulation (AM): y1(t) = x(t) cos(ωct)

Phase Modulation (PM): y2(t) = cos(ωct + kx(t)) � tFrequency Modulation (FM): y3(t) = cos ωct + k −∞ x(τ)dτ

� ��

Frequency Modulation

In FM, the signal modulates the instantaneous carrier frequency.

� � t � y3(t) = cos ωct + k x(τ)dτ

−∞ φ(t)

ωi(t) = ωc + dφ(t) = ωc + kx(t)dt


Compare AM to FM for x(t) = cos(ωmt).

AM: y1(t) = (cos(ωmt) + 1.1) cos(ωct)

t

FM: y3(t) = cos(ωct + m sin(ωmt))

t

Advantages of FM:

• constant power • no need to transmit carrier (unless DC important)

• bandwidth?

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� � � � � � �

� � �


Early investigators thought that narrowband FM could have arbitrar

ily narrow bandwidth, allowing more channels than AM. Wrong! t

y3(t) = cos ωct + k x(τ )dτ −∞ � � � � � �t t

= cos(ωct) × cos k x(τ )dτ − sin(ωct) × sin k x(τ )dτ −∞ −∞

If k → 0 then t

cos k x(τ)dτ → 1 −∞ t t

sin k x(τ)dτ → k x(τ )dτ −∞ −∞

t y3(t) ≈ cos(ωct) − sin(ωct) × k x(τ )dτ

−∞ Bandwidth of narrowband FM is the same as that of AM!

(integration does not change bandwidth)

Phase/Frequency Modulation

Find the Fourier transform of a PM signal.

x(t) = sin(ωmt)

y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))

= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))

2πx(t) is periodic in T = ωm , therefore cos(m sin(ωmt)) is periodic in T .

m sin(ωmt) m

0

−m

t

cos(m sin(ωmt))

t increasing m

1

0

−1



x(t) = sin(ωmt)




cos(m sin(ωmt))1

0 t

−1 m = 0

|ak|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




cos(m sin(ωmt))1

0 t

−1 m = 1

|ak|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




cos(m sin(ωmt))1

0 t

−1 m = 2

|ak|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




cos(m sin(ωmt))1

0 t

−1 m = 5

|ak|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




cos(m sin(ωmt))1

0 t

−1 m = 10

|ak|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




cos(m sin(ωmt))1

0 t

−1 m = 20

|ak|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




cos(m sin(ωmt))1

0 t

−1 m = 30

|ak|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




cos(m sin(ωmt))1

0 t

−1 m = 40

|ak|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




cos(m sin(ωmt))1

0 t

−1 m = 50

|ak|

k0 10 20 30 40 50 60

� ��


Fourier transform of first part.

x(t) = sin(ωmt)



ya(t)

|Ya(jω)| m = 50

ω ωc ωc

50ωm

increasing m



x(t) = sin(ωmt)



2πx(t) is periodic in T = ωm , therefore sin(m sin(ωmt)) is periodic in T .

m sin(ωmt) m

0

−m

t

sin(m sin(ωmt))1

0

−1

t

increasing m



x(t) = sin(ωmt)




sin(m sin(ωmt))1

0 t

−1 m = 0

|bk|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




sin(m sin(ωmt))1

0 t

−1 m = 1

|bk|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




sin(m sin(ωmt))1

0 t

−1 m = 2

|bk|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




sin(m sin(ωmt))1

0 t

−1 m = 5

|bk|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




sin(m sin(ωmt))1

0 t

−1 m = 10

|bk|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




0 t

sin(m sin(ωmt))1

−1 m = 20

|bk|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




sin(m sin(ωmt))1

0 t

−1 m = 30

|bk|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




sin(m sin(ωmt))1

0 t

−1 m = 40

|bk|

k0 10 20 30 40 50 60



x(t) = sin(ωmt)




sin(m sin(ωmt))1

0 t

−1 m = 50

|bk|

k0 10 20 30 40 50 60

� ��


Fourier transform of second part.

x(t) = sin(ωmt) y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))

= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt))) yb(t)

|Yb(jω)| m = 50

ω

ωc ωc

50ωm


Fourier transform.

x(t) = sin(ωmt) y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))

= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt))) � �� ya(t) yb(t)

|Y (jω)| m = 50

ω

ωc ωc

50ωm


Wideband FM is useful because it is robust to noise.

AM: y1(t) = (cos(ωmt) + 1.1) cos(ωct)

t

FM: y3(t) = cos(ωct + m sin(ωmt))

t

FM generates a very redundant signal, which is resilient to additive

noise.

Summary

Modulation is useful for matching signals to media.

Examples: commercial radio (AM and FM)

Close with unconventional application of modulation – in microscopy.

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6.003 Signals and Systems Spring 2010

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6.003: Signals and Systemsdspace.mit.edu/.../lecture-notes/MIT6_003S10_lec24.pdfCommunications Systems Signals are not always well matched to the media through which we wish to transmit