EE456 – Digital Communications · 2016-09-20 · Chapter 3: Frequency Modulation (FM) Double-Sideband Suppressed-Carrier (DSB-SC) Amplitude Modulation Let m(t) be the message signal,

Chapter 3: Frequency Modulation (FM)

EE456 – Digital CommunicationsProfessor Ha Nguyen

September 2016

EE456 – Digital Communications 1


ANALOG MODULATIONAmplitude Modulation (AM) & Frequency Modulation (FM)



The Concept of Modulation

Modulation refers to a process that puts the message signal into a specificfrequency band in order to match the transmission characteristics of the physicalchannel (e.g. telephone channel, wireless LAN channel, etc.)

Modulation can be classified into baseband and passband.

The term “baseband” refers to the frequency band of the original message signal,which is usually near the zero frequency. For example, the band of audio (voice)signals is between 0 to 3.5kHz, the video baseband occupies 0 to 4.3 MHz.

In baseband modulation, the message signals are directly transmitted over thechannels (e.g. twisted pairs of copper wires, coaxial cables).

Passband modulation is also known as carrier modulation, in which the spectrumof the message signal is shifted to a higher-frequency band by means of asinusoidal carrier.

Two important advantages of carrier modulation are: (i) To ease radio-frequency(RF) transmission, and (ii) to allow frequency division multiplexing.

Carrier modulation can be analog or digital. Traditional communications such asAM/FM radios are based on analog modulation, while many current and newcommunications systems are all digital (cellular phone systems, HDTV, etc.)



Double-Sideband Suppressed-Carrier (DSB-SC) Amplitude Modulation

Let m(t) be the message signal, whose Fourier transform is denoted as M(f).

To move the frequency band of m(t) to a new frequency band centered at fc, themessage signal m(t) is simply multiplied by a sinusoid of frequency fc (i.e., thecarrier):

sDSB−SC(t) = m(t) cos(2πfct).

The above signal is seen as a sinusoid of frequency fc whose amplitude is variedaccording to the message signal, hence the name amplitude modulation (AM).

The spectrum of the AM signal s(t) is obtained as a convolution between thespectrum of m(t) and the spectrum of cos(2πfct).

Since the spectrum of cos(2πfct) is 12δ(f + fc) +

12δ(f − fc), one has

SDSB−SC(f) = M(f) ∗

[1

2δ(f + fc) +

1

2δ(f − fc)

]

=1

2M(f − fc) +

1

2M(f + fc).



Graphical Illustration of DSB-SC Amplitude Modulation

( )M f

fBB− 0

( )m t

( ) cos(2 )c

c t f tπ=

DSB-SC( ) ( ) cos(2 )

cs t m t f tπ= ×

( )m t

( )C f

f

12

cf

12

cf−

f

2K

cf

2K

cf−

DSB-SC( )S f

cf B+

cf B−0 0

2B

t

t

( )cos(2 )c

m t f tπ

( )m t

( )m t−

The process of amplitude modulation shifts the spectrum of the modulatingsignal to the left and to the right by fc.

If the bandwidth of m(t) is B Hz, then the modulated AM signal has bandwidthof 2B Hz.

There are upper sideband (USB) and lower sideband (LSB) portions of the AMspectrum.

The modulated signal does not contain a discrete component of the carrierfrequency fc. For this reason, the modulated signal in this scheme isdouble-sideband, suppressed-carrier (DSB-SC) modulation.



Coherent (Synchronous) Demodulation of DSB-SC AM Signals

( ) cos(2 )c

c t f tπ=

( )

2

m t

2

( )m t

DSB-SC( ) ( )r t s t=

( )e t

fc

f

2K

cf−

( )E f

2c

f

4K

2c

f−

4K

0f

BB− 0

2K

BB−

e(t) = m(t) cos2(2πfct) =1

2m(t) +

1

2m(t) cos(2π(2fc)t)

E(f) =1

2M(f) +

1

4[M(f + 2fc) +M(f − 2fc)]

This method of AM demodulation is called synchronous detection, or coherentdetection, where the receiver requires to have a carrier of exactly the samefrequency (and phase) as the carrier used for modulation.

In practice, such a requirement is typically fulfilled with a phase-locked loop(PLL) circuit.



Example 1

Consider a single tone baseband signal m(t) = cos(ωmt) = cos(2πfmt). Find theDSB-SC signal and sketch its spectrum. Identify the USB and LSB. Also verify thatthe coherent demodulation works and recovers m(t).

(Partial) Solution:



Effect of Using a Non-Coherent Carrier at the Receiver

( ) cos(2 )c

c t f tπ=

DSB-SC( ) ( ) cos(2 )

cs t m t f tπ= ×

( )m t

cos(2 ( ) )c

f f tπ + ∆

0

1( )cos(2 )

2dm t t ftπ θ− ∆ +

0 0( ) ( ) cos(2 ( ))

cr t m t t f t tπ= − × −

( )e t

To see the effect on the demodulated signal when the local oscillator at thereceiver is not synchronized in frequency and phase with the incoming carrier,consider the case that the received signal at the receiver is a delayed version ofthe transmitted AM signal (due to propagation time):

r(t) = m(t − t0) cos[2πfc(t − t0)] = m(t − t0) cos(2πfct− θd),

where θd = 2πfct0 is the equivalent phase shift.

Furthermore, assume that, due to the lack of a good PLL circuit, the oscillator inthe receiver generates cos(2π(fc +∆f)t), i.e., there is a frequency offset of ∆fcompared to the carrier generated by the oscillator at the transmitter.



Then the signal e(t) is

e(t) = m(t − t0) cos(2πfct− θd) cos(2π(fc +∆f)t)

=1

2m(t − t0) cos(2π∆ft + θd) +

1

2m(t − t0) cos(2π(2fc +∆f)t − θd)

The first component of e(t) is the message signal modulated with a very small“carrier” frequency of ∆f , while the second component is the message signalmodulated with a high carrier frequency of 2fc +∆f . Thus, the output of theLPF will be the first component only, i.e., v(t) = 1

2m(t − t0) cos(2π∆ft + θd).

In the frequency domain, it can be shown that the spectrum of v(t) is

V (f) =1

4ejθdM(f −∆f)e−j2π(f−∆f)t0 +

1

4e−jθdM(f +∆f)e−j2π(f+∆f)t0

In essence, the spectrum at the output of the LPF is the spectrum of the originalmessage moved to ±∆f .

In Lab #1 you will hear the effect of this frequency translation for different valuesof ∆f when m(t) is an audio signal.



Amplitude Modulation with Carrier

( )M f

fBB− 0

( )m t

( ) cos(2 )c

c t f tπ=

DSB( ) [ ( ) ] cos(2 )

cs t t tAm fπ= + ×

( )m t

( )C f

f

12

cf

12

cf−

f

2K

cf

2K

cf−

DSB( )S f

cf B+

cf B−0 0

2B

∑

(( )) AM ff δ+

fBB− 0

( )m t A+

2A

2A

tt

t

t

0

C R

( )m t

t



To enable a simpler receiver than the coherent receiver, the transmitter can senda carrier along with the modulated signal:

sAM(t) = A cos(2πfct)+m(t) cos(2πfct) = [A+m(t)] cos(2πfct), where A > 0.

The spectrum of sAM(t) is basically the same as that ofsDSB−SC(t) = m(t) cos(2πfct), except for two additional impulses at ±fc:

SAM(f) =1

2M(f − fc) +

1

2M(f + fc) +

A

2δ(f − fc) +

A

2δ(f + fc).

If the carrier component is large enough, the message signal m(t) can berecovered with a very simple envelope detector (see the next slides).

The option of AM with carrier is very desirable in broadcasting systems as itoffers a trade-off to have one expensive high-power transmitter and (many)inexpensive receivers.



Envelope Detection of AM Signals

The time constant of the RC circuit should be such that the capacitor dischargesvery little between positive peaks and in a similar rate of the AM envelopevariation:

1/ωc ≪ RC < 1/(2πB), or 2πB <1

RC≪ ωc,

where B is the bandwidth of m(t).

The envelope detector output is A+m(t) with a ripple frequency ωc. The DCterm can be blocked by a simple highpass filter, while the ripple may be furtherreduced by another lowpass filter.



Envelope of an AM Signal

By definition, the envelope of the AM signal is |A+m(t)|.

If A is large enough to ensure that A+m(t) ≥ 0 for all t, then the envelope hasthe same shape as the message m(t). This means that we can detect the desiredsignal m(t) by detecting the envelope of the AM signal!

If A+m(t) < 0 for some time t, then the envelope |A+m(t)| does not have thesame shape as the message m(t) and envelope detection does not work correctly.



Modulation Index of an AM Signal

Typically, a message signal m(t) has no DC (i.e., zero offset) since a DC does notcarry any useful information.

Let ±mp be the maximum and the minimum values of m(t). This means thatm(t) ≥ −mp for all t. Then, the condition of envelope detection is

A+m(t) ≥ 0, for all t ⇔ A+min∀t

{m(t)} ≥ 0

⇔ A+ (−mp) ≥ 0 ⇔ A ≥ mp .

The modulation index is defined as

µ =mp

A

For envelope detection to work correctly, the condition is A ≥ mp and themodulation index satisfies

0 ≤ µ ≤ 1, or 0 ≤ µ ≤ 100%

When A < mp, µ > 100% (overmodulation) and envelope detection no longerworks correctly.



Modulation Index for Single Tone AM

µ =mp

A=

(Amax −Amin)/2

(Amax +Amin)/2=

Amax − Amin

Amax + Amin.

s(t)A+m(t)

Amax

Amin

A

t



Example 2

Consider the case of tone modulation in which the message signal is a pure sinusoid(e.g. a test tone) m(t) = mp cos(ωmt). Sketch the AM signals for modulation indicesof µ = 50% and µ = 100%

Solution:

sAM(t) = [A+m(t)] cos(ωct) = A[

1 +mp

Acos(ωmt)

]

cos(ωct) = A[1+µ cos(ωmt)] cos(ωct)



Power Efficiency of an AM Signal

Recall that the AM signal has two components: the carrier and sidebands

sAM(t) = A cos(ωct)︸︷︷︸

carrier

+m(t) cos(ωct)︸︷︷︸

sidebands

The carrier does not carry any information, hence its power is wasteful from thispoint of view. The carrier power is Pc = A2/2.

The sideband power is the power of m(t) cos(ωct). If Pm is the power of messagem(t), then the sideband power is well approximated by Pm/2.

The power efficiency of an AM signal is defined as:

η =useful power

total power=

sideband power

sideband power + carrier power≈

Pm/2

Pm/2 + A2/2=

Pm

A2 + Pm

Example: For the special case of tone modulation, m(t) = mp cos(ωmt). Thus

Pm = m2p/2 = (µA)2/2. It then follows that η =

µ2

2 + µ2100%

Observations: Under the condition that 0 ≤ µ ≤ 1, η increases monotonicallywith µ and reaches a maximum value of ηmax = 33% when µ = 100%. Thus, fortone modulation, under the best condition of µ = 100%, only one-third of thetransmitted power is used for carrying the message!

Quiz: Determine η for tone modulation when (a) µ = 0.5, and (b) µ = 0.25.



FCC AM Standards



AM Stations in Saskatoon



Application of AM: Frequency-Division Multiplexing



Example 3

The signals m1(t) and m2(t), both band-limited to 5 kHz, are to be transmittedsimultaneously over a channel by a multiplexing scheme shown below. The signal at

point b is the multiplexed signal, which then modulates a carrier of frequency 20kHz. The signal at point c is transmitted over a channel.

A

1( )M f

f5 kHz5 kHz− 0

A

2 ( )M f

f5 kHz5 kHz− 0

1( )m t

∑

2( )m t

2cos(20,000 )tπ

2cos(40,000 )tπa

bc

(a) Sketch the signal spectra at points a , b , and c .

(b) What must be the bandwidth of an ideal channel to pass all the frequencycontent of the signal at point c undistorted?

(c) Design a receiver to recover signals m1(t) and m2(t) from the modulated signalat point c .



Solution to Part (a):

A

5 10 15 (kHz)f

5−10−15− 0

aPoint

A

5 10 15 (kHz)f

5−10−15− 0

bPoint

A

5 (kHz)f

5−20− 0

cPoint

20



Solution to Part (b): The required channel bandwidth is 30 kHz.

Solution to Part (c):

LPF

0-15 kHz

c

LPF

0-5 kHz

HPFLPF

0-5 kHz

cos(20,000 )tπ

1( )m t

2 ( )m t

cos(40,000 )tπ



Building AM Transmitter and Receiver in Lab # 1

Transmitter:

[ ]m n +

A/D DSP D/ARF

Filter

NCO

[ ]

2

m n

12

12

ˆ (cycles/sample)cf

ˆcos( )cnω

1 [ ]

2ˆcos( )m n

cnω+1 [ ]

2

m n+

full-scale

input

+

microphone

∑

ˆˆ 2

(radians/sample)

c cfω π=

You will be able to set the carrier frequency and the DC level.



Receiver: This is known as a “square-law” receiver. It does require (i) a localoscillator, but the local carrier does not need to synchronize (lock) to the in comingcarrier; (ii) Modulation index less than 100%.

message

A/D DSP D/A

NCO

RF

Filter

ˆ ˆcos[( ) ]c

nω ω θ+ ∆ +

ˆ ˆsin[( ) ]c

nω ω θ+ ∆ +

LPF

ˆ ˆcf f+ ∆

LPF

1 [ ]

2ˆcos( )m n

cnω+ [ ]cx n

[ ]sx n

[ ]cy n

[ ]sy n

2[ ]cy n

2[ ]sy n

1 [ ]

4

m n+

this does not need to synchronize

to the incoming carrier

∑

You will be able to set the value of the carrier frequency offset.



Analysis of the AM Demodulator

xc[n] =1 +m[n]

2cos(ω̂cn) cos[(ω̂c +∆ω̂)n+ θ]

=1 +m[n]

2

cos[(ω̂c + ω̂c +∆ω̂)n+ θ]

2︸︷︷︸

this term is removed by the LPF

+1 +m[n]

2

cos(∆ω̂n+ θ)

2

yc[n] =1

4(1 +m[n]) cos(∆ωn+ θ)

xs[n] =1 +m[n]

2cos(ω̂cn) sin[(ω̂c +∆ω̂)n+ θ]

=1 +m[n]

2

sin[(ω̂c + ω̂c +∆ω̂)n+ θ]

2︸︷︷︸

this term is removed by the LPF

+1 +m[n]

2

sin((−∆ω̂)n− θ)

2

ys[n] = −1

4(1 +m[n]) sin(∆ω̂n+ θ)



output =√

y2c [n] + y2s [n]

=

√(1

4(1 +m[n])

)2

cos2(∆ω̂n+ θ) +

(

−1

4(1 +m[n])

)2

sin2(∆ω̂n+ θ)

=

√(1

4(1 +m[n])

)2

=1

4(1 +m[n])

Observations:

The receiver only works correctly if (1 +m[n]) > 0 for all n. This is the samecondition required by the envelope detector implemented in analog approach.

The receiver is noncoherent since it does not require the NCO to lock infrequency and phase to the incoming carrier.

The discrete-time output has a DC component. A DC blocking filter should be

appended. The output of the blocking filter would then bem[n]4

.

A gain of 4 would follow the blocking filter. Then the signal sent to the D/A ism[n], which is a full-scale signal.


EE456 – Digital Communications · 2016-09-20 · Chapter 3: Frequency Modulation (FM) Double-Sideband Suppressed-Carrier (DSB-SC) Amplitude Modulation Let m(t) be the message signal,

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