EEM 308 Lecture telecommunications introduction

EEM 308 Introduction to Communications
Lecture 7
Dr. Can Uysal
Department of Electrical and Electronics Engineering,
Eskisehir Technical University
April 15, 2025

Amplitude Modulation
Double sideband suppressed carrier (DSB-SC) AM
Conventional double-sideband (DSB) AM
Single-sideband (SSB) AM
Vestigial-sideband (VSB) AM
1/43

Single-Sideband (SSB) Amplitude Modulation
In our development of DSB (conv. AM or DSB-SC), we saw
that USB and LSB have even amplitude and odd phase
symmetry about the carrier frequency.
All the information is contained in either half – the other is
redundant.
Thus, transmission of both sidebands is not necessary, since
either sideband contains su!cient information to reconstruct
the message signalm(t).
Elimination of one of the sidebands prior to transmission
results in single-sideband (SSB), which reduces the bandwidth
of the modulator output from 2WtoW,whereWis the
bandwidth ofm(t).
However, this bandwidth savings is accompanied by a
considerable increase in complexity.
2/43

SSB Amplitude Modulation
Remember: Conventional AM & DSB-SC occupy twice the message
bandwidth (2→W).
Aside from suppressing the carrier, only the upper or lower sideband of
the message signal is transmitted.
3/43

SSB Amplitude Modulation
A single-sideband (SSB) AM signal is represented mathematically
as,
uSSB(t)=m(t)c(t)→ˆm(t)ˆc(t)
uSSB(t)=Acm(t) cos 2ωfct→Acˆm(t)sin2ωfct
where ˆm(t) is the Hilbert Transform ofm(t)
↑:↓upper sideband (USB)
+:↓lower sideband (LSB)
Recall that the Hilbert transform:
ˆx(t)=x(t)ε
1
ωt
F[ˆx(t)] =ˆX(f)=↑jsgn(f)X(f)
since
1
ωt
↔↗ ↑jsgn(f).
4/43

Single-Sideband (SSB) AM
Derivation of the Time-Domain Description for SSB Signals
M+(f)=M(f)u(f)=M(f)
1
2
[1 +sgn(f)] =
1
2
[M(f)+jˆM(f)]
M
→(f)=M(f)u(→f)=M(f)
1
2
[1→sgn(f)] =
1
2
[M(f)→j
ˆ
M(f)]
We can now express SSB signal in terms ofm(t)and ˆm(t). From
(d), it is clear that USB spectrumU
USB(f)is
U
USB(f)=M+(f→fc)+M
→(f+fc)
=
1
2
[M(f→fc)+M(f+fc)]→
1
2j
[
ˆ
M(f→fc)→
ˆ
M(f+fc)]
the inverse Fourier transform of this equation yields,
u
USB(t)=m(t)cos2ωfct→ˆm(t)sin2ωfct
Similarly,
u
LSB(t)=m(t)cos2ωfct+ˆm(t)sin2ωfct
Hence, a general SSB signal can be expressed as,
u
SSB(t)=m(t)cos2ωfct↑ˆm(t)sin2ωfct
5/43

nu/f)
v(-f)
1 -f -f
am(f)
-f
-
M
+
(f) am
=
(f)
>
f
>
f
E[m(f)
+
M(f)
syn(f)
--
j
-
M(f)
asgnlf)
1
-f
-I
-1
+
syn(f)
(f)
=
u(f)
2
-> 1

Generation of SSB
Method 1: Frequency Discrimination Method (Sideband Filtering)
Generate a DSB-SC AM signal
Employ a ﬁlter that selects either the upper sideband or the
lower sideband
Uses sharp cut-o”ﬁlters to eliminate the undesired sideband
Figure:Generation of an SSB-AM signal by ﬁltering one of the sidebands
of a DSB-SC AM signal.
6/43

Generation of SSB
Method 2: Phase Discrimination Method (Phasing Method) : (will
be implemented in the lab class)
Represents time-domain description of SSB signal
Use phase shifting network to achieve the same goal.
Has two parallel paths (each involving product modulator)
in-phase
quadrature
Figure:Generation of a lower/upper SSB-AM signal.
7/43

Practical Issues for Generation of SSB Signals
Frequency discrimination method:
Designing narrow, sharp cut-o”band-pass ﬁlters is very
di!cult, especially at high carrier frequencies.
Requires very selective ﬁlters to work
Message modulated using SSB should not have low frequencies
Phase discrimination method :
requires an exactly 90-degree phase shifter to completely
cancel one of the sidebands: It is very hard to realize. Not
suitable for the transmission of wideband signals
Amplitude and phase balance must be precise; any imbalance
results in incomplete sideband suppression.
8/43

Single-Sideband (SSB) AM
Example 1
An SSB-AM signal is generated by modulating an 800 kHz carrier
by the signalm(t) = cos 2000ωt+2sin2000ωt. The amplitude of
the carrier isAc= 100.
a.Determine the signal ˆm(t).
b.Determine the (time-domain) expression for the lower
sideband of the SSB-AM signal.
c.Determine the magnitude spectrum of the lower-sideband-SSB
signal.
9/43

Demodulation of SSB AM Signals
There are several methods that can be employed to demodulate SSB.
The simplest technique is to multiply the modulated signal by a
demodulation carrier and lowpass ﬁlter the result, which is similar to
DSB-SC demodulation
We assume a demodulation carrier having a phase errorωthat yields
uSSB(t)→c(t)=[Acm(t)cos2εfct↑Acˆm(t)sin2εfct]→2cos(2εfct+ω)
The preceding expression can be written as,
=[Acm(t)cos2εfct→2cos(2εfct+ω)]↑[Acˆm(t)sin2εfct→2cos(2εfct+ω)]
=Acm(t)cosω+Acm(t)cos(4εfct+ω)↑Acˆm(t)sinω±Acˆm(t)sin(4εfct+ω)
Lowpass ﬁltering yield
yε(t)=Acm(t)cosω↑Acˆm(t)sinω
Require aphase coherentorsynchronous demodulator!
10 / 43

Demodulation of SSB AM Signals
Example 2
FinduSSB(t) for a simple case of a tone modulation, that is, when
the modulating signal is a sinusoidm(t) = cos 2ωfmt.Also
demonstrate the coherent demodulation of this SSB signal.
11 / 43
e
->
C(t)
=
cosz!fct
->
USSB(t)
X CPF
yelt)
USSB(t)
=
m(t)c(t)
+
m(t)
!(t)
=
cos2Tfmt
.
casat!fct
I
s!nztfmt
-
s!n2tfct
cosaifct
=
E(ces2π(fc
+
fm)t
+
cos2π(fc
-
fm)t)
==(cos2t(fc-
fm)t
-
Cos25)(fc
+
fm)t)
->
After
CPF;
yu(t)
=cos
2πfrntUus(t)
=cos
2T(fa
+
fm)t
VCs!(t)
=
cos2T(fc
-
fm)
t
nm(f) -VDSB-sc
->
LOWERUPPER
-
f!n
for t
farf!refrufcflefonfatfor
For
coherent
demodulat!on
:
Usalt).
2 cos21fct
=
2c0s25(fcIfm)
t
.coszift
=cos
2!f!nt
+
10321(2fc
=
fru!t

About SSB
SSB relies on being able to ﬁlter out one sideband. For audio
this is possible because the voice spectrum drops o”below
300 Hz, allowing space for a transition band
This is not possible for other signals, like video, that have
strong components at low frequencies.
12 / 43

Vestigial Sideband Modulation
The solution isVestigial Sideband Modulation, VSB,wherea
small portion (a vestige) of the unneeded sideband.
13 / 43

Vestigial-Sideband (VSB) AM
SSB modulation works for a speech signal with an energy gap
centered around zero frequency. But, the spectrum of
wideband signals (like TV video signals) contains a large
power concentrated in the vicinity off=0(lowfrequencies),
impractical to use SSB modulation (The sideband ﬁlter with a
sharp cuto”is very di!cult to implement in practice.)
The generation of DSB-SC signals is much simpler
But, DSB-SC requires a transmission bandwidth equal to twice
the message bandwidth, which violates the bandwidth
conservation requirements.
To overcome these practical limitations, we need a
compromise method of modulation that lies between SSB &
DSB-SC in its spectral characteristics.
AcompromisebetweenSSBandDSB-SC!
VSB signals are relatively easy to generate, and their
bandwidth is only a little (typically 25%) greater than that of
SSB signals.
14 / 43

Vestigial-Sideband (VSB) AM
In VSB modulation, instead of completely removing a sideband, a small
portion (vestige)ofthatsidebandistransmitted;hencethename
vestigial sideband.
with VSB, the design of the sideband ﬁlter is simpliﬁed since the need for
a sharp cut-o!at the carrier frequency is eliminated.
This modulation type is appropriate for signals with a strong
low-frequency component, such as video signals.
15 / 43

Vestigial-Sideband (VSB) AM
In VSB modulation, one of the sidebands is partially suppressed
and a vestige of the other sideband is transmitted to compensate
for that suppression.
16 / 43

Review: Passing Through a Filter
h(t): impulse response of the ﬁlter
H(f) : Frequency response of the ﬁlter
Convolution in Time = Multiplication in Frequency
17 / 43

Vestigial-Sideband (VSB) AM
Generation of VSB
To generate a VSB-AM signal, we begin by generating a
DSB-SC AM signal and passing it through a sideband ﬁlter
with the frequency responseH(f).
In the time domain, the VSB signal may be expressed as,
u(t)=[Acm(t) cos 2ωfct]↘h(t)
whereh(t) is the impulse response of the VSB ﬁlter. In the
frequency domain,
U(f)=
Ac
2
[M(f↑fc)+M(f+fc)]H(f)
18 / 43

Vestigial-Sideband (VSB) AM
Generation of VSB
fais the frequency parameter that determines the frequency extent
of the vestige of the sideband
19 / 43

Vestigial-Sideband (VSB) AM
Demodulation of VSB
To determine the frequency-response characteristics of the ﬁlter, we will
consider the demodulation of the VSB signalu(t).
We multiplyu(t)bythecarriercomponentcos2εfctand pass the result
through an ideal lowpass ﬁlter,
v(t)=uVSB(t)cos(2εfct)
or equivalently,
V(f)=
1
2
[UVSB(f↓fc)+UVSB(f+fc)] (1)
As we know,
UVSB(f)=
Ac
2
[M(f↓fc)+M(f+fc)]H(f)
UVSB(f↓fc)=
Ac
2
[M(f↓2fc)+M(f)]H(f↓fc)
UVSB(f+fc)=
Ac
2
[M(f)+M(f+2fc)]H(f+fc)
20 / 43

Vestigial-Sideband (VSB) AM
Demodulation of VSB
Then we can rewrite Eq. (1) as,
V(f)=
Ac
4
[M(f→2fc)+M(f)]H(f→fc)
+
Ac
4
[M(f)+M(f+2fc)]H(f+fc)
V(f)=
Ac
4
M(f)[H(f→fc)+H(f+fc)] +
Ac
4
[M(f→2fc)H(f→fc)+M(f+2fc)H(f+fc)]
After LPF,
Vl(f)=
Ac
4
M(f)[H(f→fc)+H(f+fc)]
We require that the message signal at the output of the LPF be undistorted. Hence,
H(f→fc)+H(f+fc)=constant|f|↑W.
21 / 43

Vestigial-Sideband (VSB) AM
Demodulation of VSB: Freq. Domain
22 / 43

Vestigial-Sideband (VSB) AM
VSB Filter Characteristics: ChoosingH(f)
VSB ﬁlter, Sideband shaping ﬁlter
H(f) must satisfy the condition that the transmitted vestige compensates
for the spectral portion missing from the other sideband.
This requirement ensures that coherent detection of the VSB modulated
wave recovers a replica of the message signal, except for amplitude
scaling.
H(f) must satisfy the following condition,
H(f↓fc)+H(f+fc)=constant|f|↔W. (2)
Figure:VSB ﬁlter characteristics
23 / 43

Vestigial-Sideband (VSB) AM
VSB Filter Characteristics: ChoosingH(f)
The constraint given in Equation (2) implies thatH(f) should be
symmetric forfc↑fa≃f≃fc+fa
Figure:Symmetry condition forH(f).
24 / 43

Vestigial-Sideband (VSB) AM
Example 1
Suppose that the message signal is given as
m(t) = 2 cos(2ω10t) + 3 cos(2ω30t)
We wish to transmit this signal using VSB with varrier
c(t) = 5 cos(2ω500t). The ﬁlter response is shown below. Find
uVSB(t)aswellasitsbandwidth.
25 / 43

VSB-AM
Example 2
Show that we can demodulateuVSB(t) of the previous example to
obtainGm(t), whereGis a constant.
26 / 43

Use of VSB in Broadcast Television
Vestigial sideband modulation plays a key role in commercial (analog)
television.
The video signal exhibits a large bandwidth and signiﬁcant low-frequency
content, which suggest the use of vestigial sideband modulation.
In the transmission of television signals in practice, a controlled amount
of carrier is added to the VSB modulated signal. This is done to permit
the use of and envelope detector for demodulation. The design of the
receiver is thereby considerably simpliﬁed.
The baseband video signal of television occupies an enormous bandwidth
of 4.5 MHz, and a DSB signal needs a bandwidth of 9 MHz. The
resulting VSB spectrum bandwidth is 6 MHz
27 / 43

Implementation of Amplitude Modulators and
Demodulators
Many modulator circuits have been developed to generate AM-modulated
signals
We will describe some of the more common and widely used
modulators/demodulators in practice.
The circuits we will see show individual components, but keep in mind,
today most circuits are in integrated circuit form. Furthermore,
modulation and demodulation functions are commonly implemented in
software in digital signal processing circuits.
28 / 43

Amplitude Modulators
Power-Law Modulation
Switching Modulator
Balanced Modulator
Ring Modulator
29 / 43

Implementation of Amplitude Modulators and
Demodulators
Recall : Conventional AM
u(t)=Ac[1 +m(t)] cos(2ωfct)
Figure:Amplitude modulation. (a) Message signal. (b) Modulator output
fora<1. (c) Modulator.
30 / 43

Amplitude Modulators
Power-Law Modulation
Awaytogeneratetheproductofthecarrierandmodulatingsignalisto
apply both signals to a nonlinear component or circuit
Figure:Linear and square-law response curves. (a) A linear voltage-current relationship. (b) A
nonlinear or square-law response.
Alinearcomponentorcircuitisoneinwhichthecurrentisalinear
function of the voltage [see Fig. (a)]. A resistor or linearly biased
transistor is an example of a linear device. The current in the device
increases in direct proportion to increases in voltage.
Anonlinearcircuitisoneinwhichthecurrentisnotdirectlyproportional
to the voltage. A common nonlinear component is a diode that has the
nonlinear parabolic response shown in Fig. (b)
31 / 43

Amplitude Modulators
Power-Law Modulation
For such a diode, the current variation is apower-lawor
square-law function
Suppose that the voltage input to such a device is the sum of
the message signalm(t)andthecarrierAccos 2ωfct
Suppose that the nonlinear device has an input-output
(square-law) characteristic of the form
vo(t)=a1vi(t)+a2v
2
i(t)
vi(t) is input signal,vo(t) is output signal, anda1,a2are
constants.
32 / 43

Amplitude Modulators
Power-Law Modulation
vo(t)=a1vi(t)+a2v
2
i
(t)
If the input to the nonlinear device is
vi(t)=m(t)+Accos 2ωfct
The output is,
vo(t)=a1[m(t)+Accos 2ωfct]
+a2[m(t)+Accos 2ωfct]
2
=a1m(t)+a2m
2
(t)+a2A
2
c
cos
2
2ωfct
+Aca1
!
1+
2a2
a1
m(t)
"
cos 2ωfct.
The output of the bandpass ﬁlter with a bandwidth 2Wcentered atf=fcyields
u(t)=Aca1
!
1+
2a2
a1
m(t)
"
cos 2ωfct,
where
2a2
a1
|m(t)|>→1bydesign. Thus,thesignalgeneratedbythismethodisa
conventional AM signal.
33 / 43

Amplitude Modulators
Switching Modulator
Another method for generating an AM-modulated signal is by
means of a switching modulator.
34 / 43

Amplitude Modulators
Switching Modulator
This switching operation may be viewed as a multiplication ofvi(t)withthefollowing
switching functions(t)
vo(t)=vi(t)s(t)=[m(t)+Accos (2ωfct)]s(t)
s(t)isaperiodicfunction,anditsFourierSeriesrepresentation
s(t)=
1
2
+
2
ω
↓
#
n=1
(→1)
n→1
2n→1
cos [2ωfct(2n→1)]
s(t)canalsobewrittenas
s(t)=
1
2
+
2
ω
(cos 2ωfct→
1
3
cos 6ωfct+
1
5
cos 10ωfct...)
Hence,vo(t)=
$
m(t)
2
ω
+
1
2
Ac
%
cos (2ωfct)+ other terms
Passingvo(t)throughBPF(fcand the bandwidth 2W):
u(t)=
Ac
2
!
1+
4
ωAc
m(t)
"
cos (2ωfct)
35 / 43

Amplitude Modulators
Balanced Modulator
Abalanced modulatoris a circuit that generates a DSB (conv
AM) signal, suppressing the carrier and leaving only the sum
and di”erence frequencies at the output.
To generate a DSB-SC AM signal, use two conventional AM
modulators shown in ﬁgure
36 / 43

Amplitude Modulators
Ring (Lattice) Modulator
Another type of modulator for generating a DSB-SC AM
signal is the ring modulator
It consists of an input transformer T1, an output transformer
T2, and four diodes connected in a bridge circuit.
Figure:Ring modulator for
generating a DSB-SC AM signal
https://wiki.analog.com/university/courses/electronics/electronicslabdioderingmodulator
37 / 43

Example
The system shown in the ﬁgure generates a conventional AM
signal. The modulating signalm(t)haszeromeanandits
maximum (absolute) value isAm=max|m(t)|. The nonlinear
device has the input–output characteristic
y(t)=ax(t)+bx
2
(t).
1Expressy(t) in terms of the modulating signalm(t)andthe
carrierc(t) = cos 2ωf0t.
2Specify the ﬁlter characteristics that yield an AM signal at its
output.
3What is the modulation index?
38 / 43

Amplitude Demodulators
Envelope Detector
Demodulators,ordetectors,arecircuitsthataccept
modulated signals and recover the original modulating
information. The demodulator circuit is the key circuit in any
radio receiver. In fact, demodulator circuits can be used alone
as simple radio receivers.
The simplest and most widely used amplitude demodulator is
the envelope detector
Conventional DSB AM signals are easily demodulated by
means of an envelope detector (diode detector)
During the positive half-cycle of the input signal, the diode
conducts and the capacitor charges up to the peak value of
the input signal.
When the input falls below the voltage on the capacitor, the
diode becomes reverse-biased and the input disconnects from
the output. During this period, the capacitor discharges slowly
through the load resistor R.
As a result, the voltage across R is a series of positive pulses
whose amplitude varies with the modulating signal. A
capacitor C is connected across resistor R, e”ectively ﬁltering
out the carrier and thus recovering the original modulating
signal.
39 / 43

Amplitude Demodulators
Envelope Detector
When the input falls below the voltage on the capacitor, the
diode becomes reverse-biased and the input disconnects from
the output. During this period, the capacitor discharges slowly
through the load resistor R.
On the next cycle of the carrier, the diode again conducts
when the input signal exceeds the voltage across the
capacitor. The capacitor again charges up to the peak value
of the input signal and the process is repeated.
40 / 43

Recall: Time Constant of RC Circuits
The RC circuit’s time constant is deﬁned as the product of the
resistance and capacitance values (RC), representing the time it
takes for the capacitor to charge or discharge to 63.2% of its
maximum voltage.
ϑ=RC[seconds]
The time constantϑis related to the cuto”frequency fcut→o!,an
alternative parameter of the RC circuit, by
ϑ=RC=
1
2ωfcut→o!
or, equivalently,
fcut→o!=
1
2ωRC
=
1
2ωϑ
41 / 43

Amplitude Demodulators
Envelope Detector: Selecting R and C
The time constant RC must be selected to follow the variations in the
envelope of the carrier-modulated signal. If RC is too small, then the
output of the ﬁlter falls very rapidly after each peak and will not follow
the envelope of the modulated signal closely. This corresponds to the case
where the bandwidth of the lowpass ﬁlter is too large. If RC is too large,
then the discharge of the capacitor is too slow and again the output will
not follow the envelope of the modulated signal. This corresponds to the
case where the bandwidth of the lowpass ﬁlter is too small. In e!ect, for
good performance of the envelope detector, we should have
1
fc
↗RC↗
1
W
42 / 43

Next Week
Angle Modulation
43 / 43

EEM 308 Introduction to Communications
Lecture 8
Dr. Can Uysal
Department of Electrical and Electronics Engineering,
Eskisehir Technical University
April 22, 2025

Angle Modulation
So far we have modulated the
amplitude of the carrier to transmit
the message signal.
Another class of modulation methods
includes frequency modulation (FM)
and phase modulation (PM),
In FM systems, the frequency of the
carrier is changed by the message
signal; in PM systems, the phase of
the carrier is changed according to
the variations in the message signal.
Both FM and PM are non-linear and
are often calledangle modulation
methods.
Both techniques are forms of
angle modulation because they
involve varying the phase
angle of the carrier signal to
convey information.
1/35

Angle Modulation
Angle modulation, due to its inherent nonlinearity, is more
complex to implement and much more di!cult to analyze
Poor spectral e!ciency
Requires much wider bandwidth compared to amplitude
modulation schemes (DSB, SSB, etc.)
The major beneﬁt of these systems is their high degree of
noise immunity.
these systems sacriﬁce bandwidth for high-noise immunity
FM systems are widely used in high-ﬁdelity music
broadcasting and point-to-point communication systems,
where the transmitter power is quite limited.
Can use non-linear ampliﬁers
2/35

Concept of Instantaneous Frequency
Instantaneous frequencyis a concept commonly used in the
analysis of non-stationary signals. In simpler terms, it refers to
the frequency of a signal at a particular point in time.
For a stationary signal, like a sine wave, the frequency remains
constant over time. However, many real-world signals, are
non-stationary, meaning their frequency content changes over
time. Instantaneous frequency helps describe how the
frequency of such signals varies.
While AM signals carry a message with their varying
amplitude, FM signals can vary theinstantaneous frequencyin
proportion to the message signalm(t).
This means that the carrier frequency is changing
continuously every instant.
Let us consider a generalized sinusoidal signals(t) given by
s(t)=Accos[ω(t)]
whereω(t) is the generalized angle and is a function oft.
3/35

Concept of Instantaneous Frequency
Figure shows a hypothetical case ofω(t). The generalized angle for
a conventional sinusoidAccos (2εfct+ω0) is a straight line
2εfct+ω0, as shown in Figure. The hypothetical case general
angle ofω(t) is tangential to the angle (2εfct+ω0) at some
instantt. The crucial point is that, aroundt, over a small interval
”t→0, the signals(t)=Accosω(t) and the sinusoid
Accos (2εfct+ω0)areidentical;thatis,
s(t)=Accos (2εfct+ω0)t1<t<t2
4/35

Concept of Instantaneous Frequency
Over this small interval” t, the angular frequency ofs(t)is
ϑc=2εfc.
Because (2εfct+ω0) is tangential toω(t), the angular
frequency ofs(t) is the slope of its angleω(t) over this small
interval.
We can generalize this concept at every instant and deﬁne
that the instantaneous frequencyfiat any instanttis the
slope ofω(t)att. Thus, the instantaneous frequency and the
generalized angle are related,
fi(t)=
1
2ε
dω(t)
dt
ω(t)=2ε
!
t
→↑
fi(ϖ)dϖ
5/35

Angle Modulation
As a generalized angle-modulation, the angleω(t)is
ω(t)=2εfct+ϱ(t)
We will use two methods to modulate message signal with the
angle,
1.Phase Modulation (PM): Ifm(t) is the message signal, the
angle is proportional to the message as,
ω(t)=2εfct+kpm(t)
wherekpis a phase deviation constant (phase sensitivity). So,
the phase-modulated signal is described as,
uPM(t)=Accos(2εfct+kpm(t))
The instantaneous frequencyfi(t) in this case is given by
fi(t)=
1
2ε
dω(t)
dt
=fc+
kp
2ε
dm(t)
dt
6/35

Angle Modulation
2.Frequency Modulation (FM):instantaneous frequencyis
proportional to the message signal,
fi(t)=fc+kfm(t)
wherekfis the frequency deviation constant (frequency
sensitivity). The angleω(t)isnow
ω(t)=
!
t
→↑
[2εfc+2εkfm(ϖ)]dϖ
=2εfct+2εkf
!
t
→↑
m(ϖ)dϖ
Here we have assumed the constant term inω(t)tobezero
without loss of generality. The FM wave is
uFM(t)=Accos
"
2εfct+2εkf
!
t
→↑
m(ϖ)dϖ
#
7/35

FM - PM Comparison
8/35

Angle Modulation
Representation of FM and PM Signals
From the preceding relationships, we have
ϱ(t)=
$
kpm(t), PM
2εkf
%
t
→↑
m(ϖ)dϖ,FM
d
dt
ϱ(t)=
&
kp
d
dt
m(t),PM
2εkfm(t),FM
,
The maximum phase deviation in a PM system is given by
”ϱmax=kpmax[|m(t)|],
and the maximum frequency deviation in an FM system is given by
”fmax=kfmax[|m(t)|].
9/35

Frequency Deviation
In FM, the carrier frequency (fc) changes
based on the instantaneous level of the
modulating signal.
The current distance (in Hz) of the carrier
from the nominal center frequency (f0)isthe
frequency deviation (” f)
Can be negative or positive
A deviation of +15 kHz means that the
carrier is currently 15 kHz abovef0
Di#erence between the carrier frequency (for
the frequency at maximum amplitude of the
message signal) and nominal carrier
frequency is termed as maximum frequency
deviation,” fmax
10 / 35

Example
Message signal is given in the ﬁgure. This signal will be modulated
using FM. Carrier frequency is 3000 Hz. Frequency deviation
constant is 40 Hz/V. Plot instantaneous frequency graphic with
respect to time, ﬁnd maximum frequency deviation.
11 / 35

Angle Modulation
Relation Between FM and PM
From PM and FM equations, it is apparent that PM and FM
not only are very similar but are inseparable.
Replacingm(t)inPMeq.with
%
m(ς)dςchanges PM into
FM.
Thus, a signal that is an FM wave corresponding to m(t) is
also the PM wave corresponding to
%
m(ς)dς.
Similarly, a PM wave corresponding to m(t) is the FM wave
corresponding to
d
dt
m(t).
Therefore, by looking only at an angle-modulated signal, there
is no way of telling whether it is FM or PM.
12 / 35

Angle Modulation
Relation Between FM and PM
13 / 35

Example 0
Sketch FM and PM waves for the modulating signalm(t)shownin
Figure. The constantskfandkpare 10
5
and
ω
2
,respectively,and
fc= 100 MHz.
14 / 35

Answer to Example 0
15 / 35

Properties of Angle-Modulated Signals
1.Constancy of Transmit Power
16 / 35

Properties of Angle-Modulated Signals
2.Nonlinearity of Modulation Process
17 / 35

Properties of Angle-Modulated Signals
3.Visualization Di!culty of the message waveform
18 / 35

Figure:Angle modulation for sinusoidal message signal. (a) Message
signal. (b) Unmodulated carrier. (c) Output of phase modulator with
m(t). (d) Output of frequency modulator withm(t).
19 / 35

Example 1
The message signalm(t)=acos(2εfmt)isusedtoeither
frequency modulate or phase modulate the carrierAccos 2εfct.
Find the modulated signal in each case.
20 / 35
->
Inphase
modulat!on
(PM)
,wehave:(t)
=
kp
m(t)
=Kp
.acos255
f!t
->
In
Fre
,wehave:(t)
=
25kffm(t)
&E=2
K7
.
s!nzt!f!nt
=
Ka
s!nzlf
e
->modulated
s!gnals
:
Ac
-
cos(25Tfct
+kp
.a
.
cos25fmt)
->PM
u(t)
=
E
Ac.
cosl24fct
+
19
.
S!ne!fnt)
-ee
->
by
def!n!ngBp
=kp.aand
Bf
=If.
fr
Ac
.
cos(25fct
+
Bpcos2!fmt)
;inPM
v(t)
=
S
Ac-
cos(2T!fct
+
Bes!n2Tfmt)
;inFreBy
,
Bf
=modulationindex

Angle Modulation
We can extend the deﬁnition of the modulation index for a
general nonsinusoidal signalm(t)as
φp=kpmax[|m(t)|]
φf=
kfmax[|m(t)|]
W
whereWdenotes the bandwidth of the message signalm(t).
In terms of the maximum phase and frequency deviation” ϱ
and” f,wehave
φp=”ϱmax
φf=
”fmax
W
Depending on the value of modulation index,φ,wemay
distinguish two cases of frequency modulation:
Narrowband Angle Mod. :φ<1
Wideband Angle Mod. :φ↑1
21 / 35

Example 2
An FM transmitter has a frequency deviation constant of 15 Hz/V.
Assuming a message signal ofm(t) = 9 cos(40εt), write the
expression for modulated FM signal and determine the maximum
frequency deviation and modulation index.
22 / 35

Narrowband Angle Modulation
Recall the angle-modulated signal in the form as,
u(t)=Accos(2εfct+ϱ(t))
=Accos(2εfct) cos(ϱ(t))↓Acsin(2εfct)sin(ϱ(t))
If the deviation constantskpandkfand the message signal
m(t)aresuchthatforallt
ϱ(t)↔1
Then, we can use an approximation
u(t)↗Accos(2εfct)↓Acϱ(t)sin(2εfct)
where we have used the approximations cosϱ(t)↗1and
sinϱ(t)↗ϱ(t)forϱ(t)↔1.
23 / 35

Narrowband FM Analysis
in FM, for single tone modulation, we found that
ϱ(t)=2εkf
%
t
→↑
m(ϖ)dϖ=
kfa
fm
sin(2εfmt)=φsin(2εfmt)
Then, we can write the approximation
u(t)↗Accos(2εfct)↓Acϱ(t)sin(2εfct)as,
uFM(t)↗Ac[cos(2εfct)↓φsin(2εfct)sin(2εfmt)]
or
uNBFM(t)↗Accos(2εfct)+
φAc
2
[cos(2ε(fc+fm)t)↓cos(2ε(fc↓fm)t)]
24 / 35

Conv. AM vs. Narrowband FM
The narrowband FM signal is very similar to a conventional AM signal.
The only di!erence is that the sign of the lower side frequency in the
narrowband FM is reversed.
The bandwidth of this signal is similar to the bandwidth of a conventional AM
signal, which is twice the bandwidth of the message signal.
BW=2W
Seldom used in practice for communication purposes
Used as an intermediate stage for generation of wideband angle-modulated
signals
25 / 35

Conv. AM vs. Narrowband FM
Figure:Comparison of AM and narrowband angle modulation. (a) Phasor
diagrams. (b) amplitude spectrum. (c) phase spectrum.
26 / 35

Spectral Characteristics of Angle-Modulated Signals
Angle Modulation by a Sinusoidal Signal
Consider the case where the message signal is a sinusoidal signal (to be more
precise, sine in PM and cosine in FM)
u(t)=Accos(2ωfct+εsin 2ωfmt)
whereεis the mod index that can be eitherεporεf,andinPMsin2ωfmtis
substituted by cos 2ωfmt.
Using Euler’s relation, the modulated signal can be written as
u(t)=Re(Ace
j2ωfct
e
jεsin 2ωfmt
)
Since sin 2ωfmtis periodic with periodTm=
1
fm
,thesameistrueforthe
complex exponential signale
jεsin 2ωfmt
.SoitcanbeexpandedinaFourier
Series representation. The FS coe”cients:
xn=fm
!1
fm
0
e
jεsin 2ωfmt
e
→jn2ωfmt
dt;u=2ωfmt
=
1
2ω
!
2ω
0
e
j(εsinu→nu)
du
This integral:Bessel function of the ﬁrst kind of order nand denoted byJn(ε).
We have the Fourier series for the complex exponential
e
jεsin 2ωfmt
=
↑
"
n=→↑
Jn(ε)e
j2ωnfmt
.
27 / 35

Spectral Characteristics of Angle-Modulated Signals
By substituting inu(t)
u(t)=Re
#
$Ac
↑
"
n=→↑
Jn(ε)e
j2ωnfmt
e
j2ωfct
%
&
=
↑
"
n=→↑
AcJn(ε) cos [2ω(fc+nfm)t]
The spectrum ofu(t)isobtainedbytakingtheFouriertransformsofbothsides
U(f)=
Ac
2
↑
"
n=→↑
Jn(ε)[ϑ(f→fc→nfm)+ϑ(f+fc+nfm)]
Even in this simple case (a sinusoid modulating signal), the angle modulated signal
contains all frequencies of the formfc+nfmforn=0,±1,±2,...
The actual bandwidth of the modulated signal is inﬁnite!
However, the amplitude of the sinusoidal components of frequenciesfc±nfmfor large
nis very small. Hence, we can deﬁne a ﬁnite e!ective bandwidth for the modulated
signal.
Bc=2(ε+1)fm
Contains 98% of the signal power, in general.
28 / 35

Spectral Characteristics of Angle-Modulated Signals
Figure:Bessel functions for various values of n.
29 / 35

Properties of Bessel Function
30 / 35

Spectrum of Angle-modulated Signal
Figure:Spectrum of an angle-modulated signal. (a) amplitude spectrum.
(b) phase spectrum. They are single-sided spectrums. We don not show
the negative frequencies.
31 / 35

Example 3
Let the carrier and the message signal be given by
c(t) = 10 cos 2εfct,m(t) = cos 20εt
Assume that the message is used to frequency modulate the carrier
withkf= 50.
a.Find the expression for the modulated signal
b.Determine how many harmonics should be selected to contain
99% of the modulated signal power
32 / 35

Table
Values of Selected Bessel Functions
33 / 35

Summary
Angle modulation conveys information by changing the angle
(frequency or phase) of the carrier
High quality audio and relatively immune to amplitude
variations (fading, noise, interference, etc.)
Requires wider bandwidths than other modulation types.
Frequency deviation is the excursion of the carrier from the
nominal center frequency
Modulation index (deviation ratio) is max frequency deviation
divided by max modulating frequency
Narrowband and wideband based on modulation index
34 / 35

Next Week
Angle modulation (continued) and 4th Quiz
35 / 35

EEM 308 Introduction to Communications
Lecture 9
Dr. Can Uysal
Department of Electrical and Electronics Engineering,
Eskisehir Technical University
April 29, 2025

Last Week
Last week, We deﬁned an angle-modulated signal,
u(t)=Accosω(t)=Accos(2εfct+ϑ(t))
To analyze the spectral properties of angle-modulated signal
with a sinusoidal signal, we assume that
ϑ(t)=ϖsin(2εfmt)
whereϖis themodulation index. The angle-modulated signal
is,
u(t)=Accos(2εfct+ϖsin 2εfmt)
1/26

Last Week
The angle-modulated signal can be written in terms ofBessel
functionsas
u(t)=Ac
→
!
n=↑→
Jn(ϖ) cos [2ε(fc+nfm)t]
The spectrum has components at the carrier frequency and
has an inﬁnite number of sidebands separated from the carrier
frequency by integer multiples of the modulation frequencyfm.
The amplitude of each spectral component can be determined
from a table of values of the Bessel function.
2/26

Spectral Characteristics of Angle-Modulated Signals cont’d
the e!ect of the amplitude and frequency of the sinusoidal message signal on
the bandwidth and the number of harmonics in the modulated signal
message signalm(t)=acos(2ωfmt)andrecallthee!ectivebw Bc=2(ε+1)fm
Bc=2(ε+1)fm=
!
2(kpa+1)fm,PM
2
"
k
fa
fm
+1
#
fm,FM
or
Bc=
!
2(kpa+1)fm,PM
2(kfa+fm),FM
This relation shows that increasinga,theamplitudeofthemodulatingsignal,in
PM and FM has almost the same e!ect on increasing the bandwidth Bc.
Increasingfm,hasagreatere!ectinincreasingthebandwidthofaPMsignalas
compared to an FM signal
In both PM and FM, theBcbandwidth increases asfmincreases; however, this
increase in PM is a proportional increase, and in FM it is only an additional
increase and is generally not signiﬁcant (for largeε).
3/26

Spectral Characteristics of Angle-Modulated Signals cont’d
Now if we look at the number of harmonics in the bandwidth (including
the carrier) and denote it byMc,wehave
Mc=2(→ω↑+1)+1=2 →ω↑+3=
!
2→kpa↑+3,PM
2
"
k
fa
fm
#
+3,FM
In both cases, increasing the amplitudeaincreases the number of
harmonics in the bandwidth of the modulated signal.
However, increasingfmhas no e!ect on the number of harmonics in the
bandwidth of the PM signal, and it almost linearly decreases the number
of harmonics in the FM signal.
This explains the relative insensitivity of the FM-signal bandwidth to the
message frequency.
First, increasingfmdecreases the number of harmonics in the bandwidth,
and at the same time, it increases the spacing between the harmonics.
The net e!ect is a slight increase in the bandwidth.
In PM, however, the number of harmonics remains constant and only the
spacing between them increases. Therefore, the net e!ect is a linear
increase in bandwidth. Figure in the next slide shows the e!ect of
increasing the frequency of the message in both FM and PM.
4/26

Spectral Characteristics of Angle-Modulated Signals cont’d
5/26

Angle Modulation
Angle Modulation by an Arbitrary Message Signal
The spectral characteristics of an angle-modulated signal for a
general message signal m(t) is quite complicated due to the
nonlinear nature of the modulation process.
However, there exists an approximate relation for the e!ective
bandwidth of the modulated signal. This is known asCarson’s
ruleand is given by
Bc= 2(ϖ+ 1)W
where the modulation index
ϖ=
"
kpmax[|m(t)|],PM
kfmax[|m(t)|]
W
,FM
andWis the bandwidth of the message signalm(t).
In wideband FM, the value ofϖis usually around 5 or more.
Thus, the bandwidth of an angle-modulated signal is much
greater than the BW of various amplitude-modulation
schemes.
6/26

Angle Modulation
Example 1
Assuming thatm(t) = 10sinc(10
4
t), determine the transmission
bandwidth of an FM-modulated signal withkf= 4000 Hz/V.
7/26

Review: FM Spectrum
Narrowband approximation
of FM wave has essentially
the same bandwidth as an
AM wave
lower side-frequencies of
narrowband FM and AM
have di!erent signs
Wideband FM has an
inﬁnite bandwidth since
Bessel functions extend from
→↑to +↑
Majority of an FM wave’s
spectral components are
contained within 2” f+2fm
8/26

Review: Transmission BW of FM
FM wave is e!ectively limited to a ﬁnite number of signiﬁcant
side-frequencies compatible with a speciﬁed amount of
distortion. Thus, we can deﬁne an e!ective bandwidth
required for transmission of FM wave
Carson’s Rule:
Bc= 2(ϖ+ 1)fm= 2”f+2fm
Observations:
For largeϖ(wideband FM), bandwidth approaches, and is
slightly greater than 2” f
For smallϖ(narrowband FM), the spectrum is e!ectively
limited by the carrier and its two immediate side frequencies.
Thus, BW = 2fm
9/26

Example 2
An angle-modulated signal with carrier frequencyfc= 100 kHz is
described by the equation
u(t) = 10 cos(2εfct+ 5 sin 3000t+ 10 sin 2000εt)
a.Find the power of the modulated signal
b.Find the maximum frequency deviation” fmax
c.Find modulation indexϖ
d.Find the bandwidth ofu(t)
10 / 26

Example 3
The message signalm(t) = 10sinc(400t)frequencymodulatesthe
carrierc(t) = 100 cos(2εfct). The modulation index is 6.
a.Write an expression for the modulated signalu(t).
b.What is the maximum frequency deviation of the modulated
signal?
c.What is the power content of the modulated signal?
d.Find the bandwidth of the modulated signal.
11 / 26

Implementation of Angle Modulators and Demodulators
Generation of FM Waves
Basically, there are two ways of generating FM waves:indirectanddirect
We ﬁrst describe the narrowband FM (NBFM) generator that is utilized in the
indirect FM generationof wideband angle modulation signals.
NBFM Generation:
Due to the similarity of conventional AM signals, the generation of narrowband
angle-modulated signals is straightforward.
Figure:Generation of a narrowband angle-modulated signal
Recall
u(t)↓Accos(2ωfct)↑Acϑ(t)sin(2ωfct)
Next, we use the narrowband angle-modulated signal to generate a wideband
angle-modulated signal
12 / 26

Indirect Method
Frequency Multiplier
Frequency multipliersare used in the narrowband-wideband
conversion.
The ﬁrst stage of narrowband-wideband conversion is to
create a narrowband (NB) angle modulator
The NB angle-modulated signal enters a frequency multiplier
that multiplies the instantaneous frequency of the input by
some constantn.
This is usually done by applying the input signal to a
nonlinear element and then passing its output through a
bandpass ﬁlter tuned to the desired central frequency
13 / 26

Indirect Method
Frequency Multiplier
If the narrowband modulated signal is represented by
un(t)=Accos(2εfct+ϑ(t))
the output of the frequency multiplier is given by
y(t)=Accos(2εnfct+nϑ(t))
In general, this is a wideband angle-modulated signal.
However, there is no guarantee that the carrier frequency of
this signal,nfc,willbethedesiredcarrierfrequency.
In the last stage, the modulator performs an up/down
conversion to shift the modulated signal to the desired center
frequency. This stage consists of a mixer and a bandpass ﬁlter.
14 / 26

Indirect Method
Frequency Multiplier
Assuming that the output of the local oscillator is
eLO(t) = 2 cos (2εfLOt)
results in
e(t)=Accos [2ε(nfc+fLO)t+nϑ(t)]
+Accos [2ε(nfc→fLO)t+nϑ(t)]
for the multiplier output. This signal is then ﬁltered, using a
bandpass ﬁlter having center frequencyfc, given by
fc,wb=fLO+nfcorfc,wb=fLO→nfc
This yields the output
uWBFM(t)=Accos [2εfc,wbt+nϑ(t)]
15 / 26

Example 4
A narrowband-to-wideband converter is implemented as shown in
Figure. The output of the narrowband frequency modulator is with
fc= 100,000Hz. The max frequency deviation (” fmax) ofϑ(t)is
50 Hz and the bandwidth ofϑ(t) is 500 Hz. The wideband output
u(t) is to have a carrier frequency of 85 MHz and a modulation
index (deviation ratio) of 5. In this example we determine the
frequency multiplier factor,n, two possible local oscillator
frequencies, and the center frequency and the bandwidth of the
bandpass ﬁlter.
16 / 26

Generation of FM Waves
Direct Method
One method for directly generating an FM signal is to design
an oscillator whose frequency changes with the input voltage.
The frequency of the oscillator generating the carrier signal
varies directly depending on the instantaneous amplitude of
the information signal.
This oscillator is called asvoltage-controlled oscillator, VCO
avaractor diodewhose capacitance changes with the applied
voltage.
areactance tubewhose inductance varies with the applied
voltage. The analysis is very similar to the analysis for the
varactor diode.
Note: Due to the close relation between FM and PM signals,
basically, the same methods can be applied to the generation
of PM signals.
17 / 26

Demodulation of FM Signals
Frequency demodulation is the process by means of which the
original message signal is recovered from an incoming FM
wave.
In other words, frequency demodulation is theinverse of
frequency modulation.
Since a frequency modulator is a device that produces an
output signal whose instantaneous frequency (from carrier
frequency) varies linearly with the amplitude of the input
message signal, for frequency demodulation we need a device
whose output amplitude is linearly sensitive to changes in the
instantaneous frequency of the input FM wave.
For frequency demodulation, we will describe a device called a
frequency discriminator, based on slope detection followed by
envelope detection.
18 / 26

Demodulation of FM Signals
Frequency Discriminator
Recall that an FM signal has the following expression
uFM(t)=Accos
#
2εfct+2εkf
$
t
↑→
m(ϱ)dϱ
%
The question to be addressed is:how do we recover the
message signalm(t) from the modulated signalu(t)?
If we take the derivative of the FM signal, we obtain
duFM(t)
dt
=→2εAc[fc+kfm(t)] sin
&
2εfct+2εkf
$
t
0
m(ϱ)dϱ
'
the derivative is a band-pass signal with amplitude modulation
deﬁned by the multiplying term [fc+kfm(t)]
19 / 26

Demodulation of FM Signals
Frequency Discriminator
iffcis large enough such that the carrier is not overmodulated, then we can
recover the message signalm(t)withanenvelopedetectorlikethatdescribed
for AM signals.
This idea motivates the frequency discriminator, which is basically a
demodulator that consists of a di!erentiator followed by an envelope detector.
However, there are practical issues related to the implementation of the
discriminator: the di!erentiator.
we showed that di!erentiation corresponds to a linear transfer function in the
frequency domain; that is,
d
dt
↭j2ωf
In practice, it is di”cult to construct a circuit that has such a transfer function
for all frequencies.
20 / 26

Demodulation of FM Signals
Frequency Discriminator
The role of di!erentiator can be replaced any linear system
whose frequency response is approximately a straight line in
the frequency band of the FM signal.
The frequency response of such a system is given by
|H(f)|=V0+k(f→fc)for|f→fc|<
Bc
2
H(f) is the frequency response of the system.
V0is the amplitude at the carrier frequencyfc.
kis the slope of the frequency response.
Bcis the bandwidth aroundfcwhere the response is deﬁned.
21 / 26

Demodulation of FM Signals
Frequency Discriminator
1Instantaneous Frequency of the Input FM Signal:The
instantaneous frequencyfi(t) of the FM signal is:
fi(t)=fc+kfm(t)
The frequency deviation is” f(t)=kfm(t).
2Amplitude Scaling by the Frequency Response:The
amplitude of the output signal at any timetdepends on the
instantaneous frequencyfi(t):
|H(fi(t))|=V0+k(fi(t)→fc)=V0+k·kfm(t)
Thus, the amplitude is scaled byV0+k·kfm(t).
3Output Signal:The output signaly(t) retains the phase of
the input FM signal but with amplitude modulation:
y(t)=|H(fi(t))|·Accos
&
2εfct+2εkf
$
t
↑→
m(ϱ)dϱ
'
22 / 26

Demodulation of FM Signals
Frequency Discriminator
Substituting|H(fi(t))|:
y(t)=(V0+k·kfm(t))Accos
&
2εfct+2εkf
$
t
↑→
m(ϱ)dϱ
'
Simplifying:
y(t)=AcV0
&
1+
kkf
V0
m(t)
'
cos
&
2εfct+2εkf
$
t
↑→
m(ϱ)dϱ
'
This represents anamplitude-modulated FM signal.Nextstep
is applying envelope detector to obtainAc(V0+kkfm(t)) from
which the messagem(t) can be recovered.
23 / 26

Demodulation of FM Signals
Phase-Locked Loop (PLL)
A Phase-Locked Loop (PLL) is a closed-loop system that
locks the frequency and phase of a VCO (Voltage Controlled
Oscillator) to an input signal.
Main components:
Phase Comparator
Loop Filter
Voltage Controlled Oscillator (VCO)
Basic Operation:
1The Phase Comparator measures the phase di!erence between
the input signal and VCO output.
2The Loop Filter smooths the phase di!erence signal.
3The VCO adjusts its frequency based on the ﬁlter output to
minimize the phase error.
24 / 26

Demodulation of FM Signals
Phase-Locked Loop (PLL)
FM Demodulation with PLL:
The FM signal changes the instantaneous frequency of the
carrier according to the information signal.
The PLL locks to the instantaneous frequency of the FM
signal.
The VCO control voltage becomes directly proportional to the
message signalm(t).
Summary:
PLL enables simple and e#cient FM demodulation.
The loop automatically adjusts to track frequency deviations.
The VCO control voltage represents the demodulated signal.
Proper design of loop parameters (bandwidth, gain) is crucial
for performance.
25 / 26

Next Week
RANDOM PROCESSES
Probability and Random Variables
(Study Chapter 5.1 in the textbook
or
Review your IST 244 Engineering Probability notes)
26 / 26

EEM 308 Introduction to Communications
Review of Probability Theory and Random Variables
Department of Electrical and Electronics Engineering,
Eskisehir Technical University

Review: Basic Concepts of Probability Theory
Sample Space (!): The set of all possible outcomes.
Event: A subset of the sample space (or collection of
outcomes)
Probability Axioms:
1P(A)→0
2P(!) = 1
3IfA↑B=↓,thenP(A↔B)=P(A)+P(B)
Conditional Probability:P(A|B)=
P(A→B)
P(B)
Bayes’ Theorem:
P(A|B)=
P(B|A)P(A)
P(B)
1/7

Review: Random Variables
Arandom variableis a numerical outcome of a random
experiment.
It assigns a real number to each possible outcome of a sample
space.
Random variables help usquantify uncertaintyand perform
mathematical analysis on random phenomena.
There are two main types of random variables:
Discrete: Takes on countable values (e.g., 0, 1, 2, ...)
Continuous: Takes on values in a continuous range
Examples:
Discrete: Number of heads in 10 coin tosses
Continuous: Voltage level measured by an oscilloscope.
Notation: Often denoted by uppercase letters (e.g.,X,Y),
and values by lowercase (e.g.,x,y)
2/7

Review: Probability Density Function vs. Cumulative
Distribution Function
Probability Density Function (PDF)
Denoted asfX(x)
Describes the likelihood of a
random variable taking a speciﬁc
value (continuous case)
P(a<X<b)=
!
b
a
fX(x)dx
Properties:
fX(x)→0forallx
!
→
↑→
fX(x)dx=1
Cumulative Distribution Function
(CDF)
Denoted asFX(x)
Gives the probability thatXis less
than or equal to a value:
FX(x)=P(X↑x)=
"
x
→↑
fX(t)dt
Properties:
Non-decreasing
limx↓↑→FX(x)=0
limx↓→FX(x)=1
3/7

Uniform and Gaussian Random Variables
Uniform Random Variable
A continuous random variable taking values betweenaandbwith
equal probabilities for equal-length intervals.
fX(x)=
"
1
b↓a
,a<x<b
0,otherwise
Example:Random phase of a sinusoid between 0 and 2ω.
Gaussian (Normal) Random Variable
A continuous random variable described by the density function:
fX(x)=
1
↗
2ωε
exp
#
↘
(x↘µ)
2
2ε
2
$
Characterized by meanµand standard deviationε.
4/7

Review: Expectation and Moments
Expected Value (Mean):
E[X]=µX=
%
↔
↓↔
xfX(x)dx
The expected value ofY=g(X)is
E(g(X)) =
%
↔
↓↔
g(x)fX(x)dx
Variance:
Var(X)=ε
2
X
=E[(X↘E[X])
2
]=E[X
2
]↘E[X]
2
Higher Moments:
n-th moment:E[X
n
]
Central moment:E[(X↘µX)
n
]
5/7

Review: Expectation and Moments
6/7

Review: Joint Distributions and Independence
Joint PMF / PDF:
P(X=x,Y=y)orfX,Y(x,y)
Marginal Distributions:
fX(x)=
"
fX,Y(x,y)dy
Independence:
fX,Y(x,y)=fX(x)·fY(y)
Covariance:
Cov(X,Y)=E[XY]↓E[X]E[Y]
E[XY]iscalledthecorrelationofXandY. Note that ifXandYare
independent, then
E[XY]=E[X]E[Y].
If COV(X,Y)=0,i.e.,ifE(XY)=E(X)E(Y), thenXandYare called
uncorrelated.
Correlation coe!cient :
ωXY=
Cov(X,Y)
εXεY
7/7

EEM 308 Introduction to Communications
Lecture 10
Dr. Can Uysal
Department of Electrical and Electronics Engineering,
Eskisehir Technical University
May 6, 2025

Probability and Random Variables
Study Chapter 5.1 in the textbook
or
Review your IST 244 Engineering Probability notes
(A review document for probability theory and
random variables is published)
1/39

Deterministic vs. Random Signals
Deterministic Signal
Predictable at any time.
Described by a known function.
No randomness involved.
Example:
x(t)=Acos(2ωft+ε)
Used in idealized models.
Random Signal (Stochastic Process)
Unpredictable; varies from one
realization to another.
Described statistically.
Uses probability distributions.
Example:
X(t)=Acos(2ωft+!),!→U(0,2ω)
Common in real-world data (e.g.,
noise, fading, ECG).
2/39

Introduction – Why Study Random Processes?
In real-world communication systems, signals are a!ected by
noise, fading, and interference.
These signals are not deterministic — they are random in
nature.
Random processeshelp us model, analyze, and predict signal
behavior in such systems.
3/39

Random Processes
An extension of the concept of a random variable
In the description of arandom variable, for each outcome
wi,thereexistsarealnumber.
For arandom process, for each outcomewi,thereexistsa
signalx(t;ωi).
Thus, a random process can be viewed as a collection of time
functions, or signals, corresponding to various outcomes of a
random experiment.
x(t;ωi)iscalledasample function,orarealizationof the
random process.
The collection of all sample functions is called anensemble.
4/39

Random Processes
Noisy sine waves (Each curve (Realization 1, 2, 3) represents
one possible outcome of the random process)
In communication systems, this could model a noisy carrier
wave, showing how the received signal might vary over
di!erent trials or due to random disturbances.
5/39

Random Processes
One such example is thereﬂection of
radio waves from di!erent layers of
the ionosphere
-makes long-range broadcasting of
short-wave radio possible.
Due to the randomness of this reﬂection,
the received signal can again be modeled
as a random signal.
Another example is the case ofthermal
noise in electronic circuits.
-due to the random movement of
electrons
Therefore, the resulting current and
voltage can only be described statistically.
6/39

Random Processes
Example:
Assume that we have a signal generator that can generate one
of the six possible sinusoids.
amplitudes→>1, phases→>0. But the frequencies can be
100, 200, . . . , 600 Hz.
We throw a die, and depending on its outcome, which we
denote byF, we generate a sinusoid whose frequency is 100
times what the die shows (100F).
This random process can be deﬁned as
X(t) = cos(2ε↑100Ft).
7/39

Description of Random Processes
Example: A random process is deﬁned by
X(t)=Acos(2εf0t+ϑ)whereϑis a random variable
uniformly distributed on [0,2ε).
Figure:Samples of the random process
8/39

Description of Random Processes
Example: The processX(t)isdeﬁnedbyX(t)=X,whereX
is a random variable uniformly distributed on [-1, 1].
9/39

Random Processes
Corresponding to each outcome
ωi,thereexistsasignalx(t;ωi).
for eachωi,wehavex(t;ωi),
which is called asample function,
or arealizationof the random
process.
For the di!erent ωi’s at a ﬁxed
timet0,thenumbersx(t0;ωi),
constitute a random variable
denoted byX(t0).
After all, a random variable is an
assignment of real numbers to the
outcomes of a random
experiment.
In other words,at any time
instant the value of a random
process constitutes a random
variable.
10 / 39

Statistical Averages
Mean
Although it is not possible to predict the exact value of the random signal or
process, it is possible to describe the signal in terms of the statistical parameters
Themean,orexpectationof the random processX(t)isadeterministic
function of time denoted bymX(t)thatateachtimeinstantt0equals the mean
of the random variableX(t0). That is,mX(t)=E[X(t)] for allt
Since at anyt0the random variableX(t0)iswelldeﬁnedwithaprobability
density functionf
X(t0)(x), we have
E[X(t0)] =mX(t0)=
!
→
↑→
xf
X(t0)(x)dx
Pictorial description of this deﬁnition:
11 / 39

Statistical Averages
Mean
Example 1: Find the mean of the random process deﬁned by
X(t)=Acos(2εf0t+ϑ)
Its pdf given as,
fϑ(ϑ)=
!
1
2ω
0↓ϑ<2ε
0 otherwise
12 / 39

Statistical Averages
Autocorrelation
Another statistical average that plays a very important role in
our study of random processes is theautocorrelation function.
Theautocorrelation of a random processdescribes how
the values of the process at di!erent times are statistically
relatedto each other.
The autocorrelation function is especially important because it
completely describes thepower spectral densityand the
power contentof a large class of random processes.
The autocorrelation function of the random processX(t),
denoted byRX(t1,t2), is deﬁned by
RX(t1,t2)=E[X(t1)X(t2)]
RX(t1,t2)=
"
↑
↓↑
"
↑
↓↑
x1x2f
X(t1),X(t2)(x1,x2)dx1dx2
13 / 39

Statistical Averages
Autocorrelation
Example 2: FindRX(t1,t2) for the random process given in
previous example.
Example 2.1: Letv(t)=X+3twhereXis an RV withµx=0
andE[X
2
] = 5. What are the values ofµvandRv(t1,t2)?
14 / 39

Stationary Processes
an RP observed at any given time is just a random variable
(RV), and the properties of this RV depend on the time at
which the RP is observed.
Some properties of this random variable can be independent
of time.
Depending on what properties are independent of time,
di!erent notions of stationaritycan be deﬁned.
1.Strictly Stationary
AprocessX(t) is strictly stationary if for allnand all
(t1,t2,...,tn), and all”(time-shift)
f
X(t1),X(t2),...,X(tn)(x1,x2,...,xn)=f
X(t1+”),X(t2+”),...,X(tn+”)(x1,x2,...,xn)
A shift in time does not change the statistical properties of
the process.
For many applications, we do not require the process to be
strictly stationary.
15 / 39

Stationary Processes
WSS
2.Wide-Sense Stationary, WSS
One of the most useful concepts of stationarity is the
wide-sense stationary (WSS) random processes.
AprocessX(t) is WSS if the following conditions are
satisﬁed:
a.The mean is constant over time (is independent oft)
E[X(t)] =mX(t)=mx
b.The autocorrelation depends only on the time di!erence
ϖ=t2→t1
RX(t1,t2)=RX(ϖ)
16 / 39

Examples
Example 3: For the random process in the previous example
determine whether the process is WSS or not.
Example 4: Let the random process Y(t) be similar to the random
process X(t) deﬁned in previous example but assume thatϑis
uniformly distributed between 0 andε.Determinewhetherthe
process is WSS or not.
17 / 39
②°#x
-
->
Theprocessg!ve
in
Ext
isluss ->
Mx
isconstant
->
(t)
=
Acos(25fot
+
3)
->
1x1t
,
t2)dependsonly
T=t
,
-tz
s
:
(
,O</
3
0
,ow
,My
=
E(y(t))=
Sales (2
+
fot
+
3).
O=i
"Ife(e)
/
-s!n (2
+
fot
+
0)#
0=0
+>
=#(s!n
/255f0t
+
5)
-sin
/25fo
+
)
I

Properties of Autocorrelation of WSS Process
Property 1: Power of a WSS Process
18 / 39

Properties of Autocorrelation of WSS Process
Property 2: Symmetry
19 / 39

Properties of Autocorrelation of WSS Process
Property 3: Maximum Value
20 / 39

Stationary Processes
3.Cyclostationary
In cyclostationary processes, the statistical properties are
periodic with time.
A random processX(t)withmeanmX(t) and autocorrelation
functionRX(t+ϖ,t) is called cyclostationary, if both the
mean and the autocorrelation are periodic intwith some
periodT0;i.e.,if
mX(t+T0)=mX(t)
RX(t+ϖ+T0,t+T0)=RX(t+ϖ,t)
for alltandϖ.
HW : LetX(t)=Acos(2εf0t)whereAis a random variable
uniformly distributed on [0, 1]. Determine whether the
process is cyclostationary or not.
21 / 39

Ergodicity
Time and Ensemble Average
22 / 39

What is Ergodicity?
A random process is said to beergodicif itstime averages
are equal to itsensemble averages.
This means a single, su#ciently long observation of the
process contains all the statistical properties of the entire
population.
Why is this important?
In practice, we observe only one realization (e.g., one signal),
not the whole ensemble.
If the process is ergodic, we can compute statistics (mean,
variance, etc.) from that single realization.
Implication:Time averaging over one realization is enough
to describe the whole process.
23 / 39

Example: Ergodic Process
Suppose we observe a signalX(t) over a long period of time.
If the time average ofX(t):
X=lim
T↔↑
1
2T
"
T
↓T
X(t)dt
equals the expected valueE[X(t)] computed across many
realizations, thenX(t)isergodic in the mean.
Similar deﬁnitions exist for ergodicity in variance,
autocorrelation, etc.
Conclusion:Ergodicity allows us to analyze random processes
with just a single long-term measurement.
24 / 39

Multiple Random Processes
If an RP is passed through an LTI system, we have a second RP which is
the output of the systemY(t)=X(t)→h(t)
When dealing with two random processes, we naturally question the
dependence between the random processes.
Two random processesX(t)andY(t)areindependentif for allt1,t2,the
random variablesX(t1)andY(t2)areindependent.
Similarly,X(t)andY(t)areuncorrelatedifX(t1)andY(t2)are
uncorrelated for allt1,t2.
Thecross-correlation functionbetween two random processesX(t)and
Y(t)isdeﬁnedas
RXY(t1,t2)=E[X(t1)Y(t2)]
andRXY(t1,t2)=RYX(t2,t1)
The random processesX(t)andY(t)arejointlywide-sensestationaryif
bothX(t)andY(t)areindividuallyWSSandthecross-correlation
functionRXY(t1,t2)dependsonlyonε=t1↑t2.ForjointlyWSS
random processesRXY(ε)=RYX(↑ε)
25 / 39

Multiple Random Processes
Example 5
LetX(t)andY(t) be two jointly WSS processes. Determine the
autocorrelation function of the processZ(t)=X(t)+Y(t).
26 / 39

Recall: LTI System Response
27 / 39

Random Processes and Linear Systems
In the section on multiple random processes, we saw that when a random
process passes through an LTI system, the output is also a random
process deﬁned on the original probability space.
We are assuming that a stationary processX(t)istheinputtoanLTI
system with the impulse responseh(t)andtheoutputprocessisdenoted
byY(t)
If a stationary processX(t)withmeanmXand auto-correlation function
RX(ε)ispassedthroughanLTIsystemwithimpulseresponseh(t), the
input and output processesX(t)andY(t)willbejointlystationarywith
mY=mX
!
→
↑→
h(t)dt
RXY(ε)=RX(ε)→h(↑ε)
RY(ε)=RX(ε)→h(↑ε)→h(ε)
28 / 39

Power Spectral Density of Stationary Processes
A random process is a collection of signals,
spectral characteristics of these signals determine the spectral
characteristics of the random process
A useful function that determines the distribution of the power
of the random process at di!erent frequencies is the power
spectral densityorpower spectrumof the random process.
The power spectral density of a random processX(t)is
denoted bySX(f), and denotes the strength of the power in
the random process as a function of frequency. The unit for
power spectral density isW/Hz.
29 / 39

Power Spectral Density of Stationary Processes
For stationary random processes, we use a very useful theorem
that relates thepower spectrum of the random process to
its autocorrelation function.Thistheoremisknownasthe
Wiener–Khinchin theorem
The power spectral density of a stationary random process
X(t) is equal to the Fourier transform of the autocorrelation
function as,
SX(f)=F[RX(ϖ)]
SX(f)=
"
↑
↓↑
RX(ϖ)exp(→j2εfϖ)dϖ
and
RX(ϖ)=
"
↑
↓↑
SX(f)exp(j2εfϖ)df
30 / 39

PSD of Stationary Processes
Example 6
A random process is deﬁned byX(t)=Acos(2εf0t+ϑ)whereϑis
a random variable uniformly distributed on [0,2ε). Find the
autocorrelation function and power spectral density of this
stationary random process.
31 / 39

PSD of Stationary Processes
The power content, or simply the power, of a random process is the
sum of the powers at all frequencies in that random process. In
order to ﬁnd the total power, we have to integrate the power
spectral density over all frequencies. This means that the power in
the random processX(t), denoted byPX, is obtained using the
relation
PX=
"
→
↑→
SX(f)df
SinceSX(f) is the Fourier transform ofRX(ϖ), thenRX(ϖ)willbe
the inverse Fourier transform ofSX(f). Therefore, we can write
RX(ϖ)=
"
→
↑→
SX(f)e
j2ωfε
dϖ.
Substitutingϖ= 0 into this relation yields
RX(0) =
"
→
↑→
SX(f)df.
Comparing this with the ﬁrst equation, we conclude that
PX=RX(0).
32 / 39

Example 7
Find the power in the process given in the previous example.
33 / 39

Transmission over LTI Systems
We have seen that
mY=mX
!
→
↑→
h(t)dt
RXY(ε)=RX(ε)→h(↑ε)
RY(ε)=RX(ε)→h(ε)→h(↑ε)
Translation of these relations into the frequency domain is straightforward.
Noting thatF{h(↑ε)}=H
↓
(f)and
"
→
↑→
h(t)dt=H(0), we have
mY=mXH(0)
SXY(f)=SX(f)H
↓
(f)
SY(f)=SX(f)|H(f)|
2
SinceRXY(ε)=RYX(↑ε), we have
SYX(f)=S
↓
XY(f)=SX(f)H(f)
34 / 39

Transmission over LTI Systems
Figure:Input–output relations for the power spectral density and the
cross-spectral density
35 / 39

Example 8
A random process is deﬁned byX(t)=Acos(2εf0t+ϑ)whereϑis
a random variable uniformly distributed on [0,2ε).X(t)passes
through a di!erentiator, ﬁnd SY(f)andSXY(f).
36 / 39

PSD of a Sum Process
In practice, we often encounter the sum of two random processes
For example, in the case of communication over a channel with additive noise,
the noise process is added to the signal process.
Let assume thatZ(t)=X(t)+Y(t)whereX(t)andY(t)arejointlystationary
random processes.Z(t)isastationaryprocesswith
RZ(ϖ)=RX(ϖ)+RY(ϖ)+RXY(ϖ)+RYX(ϖ)
Taking the FT of b oth sides
SZ(f)=SX(f)+SY(f)+SXY(f)+SYX(f)
"#$%
S
→
XY
(f)
=SX(f)+SY(f)+2Re[SXY(f)]
The preceding relation shows that the power spectral density of the sum process
is the sum of the power spectra of the individual processes plus a third term,
which depends on the cross-correlation between the two processes.
If the two processesX(t)andY(t)areuncorrelated,then
RXY(ϖ)=mXmY
Now if at least one of the processes is zero mean, we will haveRXY(ϖ)=0and
SZ(f)=SX(f)+SY(f)
37 / 39

PSD of a Sum Process
Example 9
LetX(t) represent the process in the previous example and let
Z(t)=X(t)+
d
dt
X(t). Then ﬁndSZ(t)
38 / 39

Next Week
Gaussian Processes, Bandpass Process, E!ect of
Noise on Analog Comm. Systems
39 / 39

EEM 308 Introduction to Communications
Lecture 11
Dr. Can Uysal
Department of Electrical and Electronics Engineering,
Eskisehir Technical University
May 12, 2025

Last Week: PSD of Stationary Processes
A random process is a collection of signals, and the spectral
characteristics of these signals determine the spectral
characteristics of the random process
A useful function that determines the distribution of the power
of the random process at di!erent frequencies is the power
spectral densityorpower spectrumof the random process.
The power spectral density of a random processX(t)is
denoted bySX(f), and denotes the strength of the power in
the random process as a function of frequency. The unit for
power spectral density isW/Hz.
1/34

Last Week: PSD of Stationary Processes
For stationary random processes, we use a very useful theorem
that relates the power spectrum of the random process to its
autocorrelation function. This theorem is known as the
Wiener–Khinchin theorem
According to this theorem, the power spectral density of a
stationary random processX(t) is equal to the Fourier
transform of the autocorrelation function as,
SX(f)=F[RX(ω)]
SX(f)=
!
→
↑→
RX(ω)exp(→j2εfω)dω
and
RX(ω)=
!
→
↑→
SX(f)exp(j2εfω)df
2/34

Last Week: PSD of Stationary Processes
The power content, or simply the power, of a random process is the
sum of the powers at all frequencies in that random process. In
order to ﬁnd the total power, we have to integrate the power
spectral density over all frequencies. This means that the power in
the random processX(t), denoted byPX, is obtained using the
relation
PX=
!
→
↑→
SX(f)df
SinceSX(f) is the Fourier transform ofRX(ω), thenRX(ω)willbe
the inverse Fourier transform ofSX(f). Therefore, we can write
RX(ω)=
!
→
↑→
SX(f)e
j2ωfε
dω.
Substitutingω= 0 into this relation yields
RX(0) =
!
→
↑→
SX(f)df.
Comparing this with the ﬁrst equation, we conclude that
PX=RX(0).
3/34

Transmission over LTI Systems
We have seen that
mY=mX
!
→
↑→
h(t)dt
RXY(ω)=RX(ω)→h(↑ω)
RY(ω)=RX(ω)→h(ω)→h(↑ω)
Translation of these relations into the frequency domain is straightforward.
Noting thatF{h(↑ω)}=H
↓
(f)and
"
→
↑→
h(t)dt=H(0), we have
mY=mXH(0)
SXY(f)=SX(f)H
↓
(f)
SY(f)=SX(f)|H(f)|
2
We can also deﬁne a frequency-domain relation for the cross-correlation
function. Let us deﬁne thecross-spectral densitySXY(f)as
SXY(f)=F[RXY(ω)]
SinceRXY(ω)=RYX(↑ω), we have
SYX(f)=S
↓
XY(f)=SX(f)H(f)
4/34

Transmission over LTI Systems
Figure:Input–output relations for the power spectral density and the
cross-spectral density
5/34

Example 0
A random process is deﬁned byX(t)=Acos(2εf0t+ϑ)whereϑis
a random variable uniformly distributed on [0,2ε).X(t)passes
through a di!erentiator, ﬁnd SY(f)andSXY(f).
6/34

PSD of a Sum Process
In practice, we often encounter the sum of two random processes
For example, in the case of communication over a channel with additive noise,
the noise process is added to the signal process.
Let assume thatZ(t)=X(t)+Y(t)whereX(t)andY(t)arejointlyWSS
random processes.Z(t)isastationaryprocesswith
RZ(ω)=RX(ω)+RY(ω)+RXY(ω)+RYX(ω)
Taking the FT of b oth sides
SZ(f)=SX(f)+SY(f)+SXY(f)+SYX(f)
!"#$
S
→
XY
(f)
=SX(f)+SY(f)+2Re[SXY(f)]
The preceding relation shows that the power spectral density of the sum process
is the sum of the power spectra of the individual processes plus a third term,
which depends on the cross-correlation between the two processes.
If the two processesX(t)andY(t)areuncorrelated,then
RXY(ω)=E[X(t)]E[Y(t+ω)] =mXmY
Now if at least one of the processes is zero mean, we will haveRXY(ω)=0and
SZ(f)=SX(f)+SY(f)
7/34

PSD of a Sum Process
Example 1
LetX(t) represent the process in the previous example and let
Z(t)=X(t)+
d
dt
X(t). Then ﬁndSZ(t)
8/34

Gaussian Processes
What is Gaussian Distribution?
Gaussian Distribution
TheGaussian distribution,alsoknownasthenormal
distribution, is a continuous probability distribution that is
symmetric about its mean. It describes how the values of a
variable are distributed.
9/34

Gaussian Processes
Gaussian Distribution
Key Characteristics
Bell-shaped curve
Symmetrical around the meanµ
Deﬁned by:
Meanµ: the center of the
distribution
Standard Deviationϖ:the
spread of the data
Mathematical Formula
fx(x)=
1
↑
2εϖ
2
e
↑
(x↑µ)
2
2ω
2
This function deﬁnes the probability
density of the Gaussian distribution.
Figure:Gaussian (Normal)
Distribution Curve
10 / 34

Gaussian Processes
Gaussian processes play an important role in
communication systems
provides a mathematically tractable and realistic
framework for modeling noise and probabilistic signal
processing
Gaussian processes can e!ectively model
real-world communication noise
many types of noise encountered in communication
systems — such asthermal noise—are
well-approximated by Gaussian distributions due to
the Central Limit Theorem.
For Gaussian processes, knowledge of the mean
and autocorrelation; i.e,mX(t)andRX(t1,t2)
gives a complete statistical description of the
process.
11 / 34

Why Do Thermal Noise Has Gaussian Distribution?
Thermal noise follows aGaussian (normal) distributiondue
to theCentral Limit Theorem.
Inside a resistor, there aremillions of free electrons.
These electrons moveindependently and randomly.
Each individual electron’s contribution isvery small,butthe
total e!ect is measurable .
According to theCentral Limit Theorem, the sum of a large
number of independent and random e!ects tends to follow a
normal distribution.
12 / 34

Gaussian Processes
Deﬁnition:A random processX(t)isaGaussianprocessifforalln
and all (t1,t2,...,tn), the random variables{X(ti)}
n
i=1
have a
jointly Gaussian density function.
From this deﬁnition, we know that at any time instantto,the
random variableX(to) is Gaussian; at any two pointst1,t2,
random variables (X(t1),X(t2)) are distributed according to a
two-dimensional jointly Gaussian distribution.
13 / 34

Gaussian Processes
Example 2
LetX(t) be a zero-mean WSS Gaussian random process with the
power spectral densitySX(f) = 5”
"
f
1000
#
.Determinethe
probability density function of the random variableX(3).
14 / 34

Gaussian Processes
Deﬁnition:If the random processesX(t)andY(t) are jointly
Gaussian, then each of them is individually Gaussian; but the
converse is not always true. That is, two individual Gaussian
random processes are not always jointly Gaussian.
Gaussian and jointly Gaussian random processes have some
important properties that are not shared by other families of
random processes. Two of the most important properties are
given here:
15 / 34

Gaussian Processes
Property 1:IftheGaussianprocessX(t) is passed through
an LTI system, then the output processY(t)willalsobethe
Gaussian process.
For a non-Gaussian process, knowledge of-the statistical
properties of the input process does not easily lead to the
statistical properties of the output process. However, for a
Gaussian process, we know that the output process of an LTI
system will also be Gaussian.
Property 2: For jointly Gaussian processes, uncorrelatedness
and independence are equivalent.
16 / 34

Gaussian Processes
Example 3
The random processX(t) in Example 2 is passed through a
di!erentiator and the output process is denoted byY(t), i.e.,
Y(t)=
d
dt
X(t), determine the probability density functionY(3)
17 / 34
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O

White Processes
Deﬁnition
Awhite processis a random process whose values at di!erent
time instants are uncorrelated and have constant power over all
frequencies.
Key Characteristics
Zero mean:E[X(t)] = 0
Constant power spectral density:SX(f)=N0/2
Autocorrelation:RX(ω)=
N0
2
ϱ(ω)
Samples at di!erent times are uncorrelated:
E[X(t1)X(t2)] = 0 fort1↓=t2
18 / 34

White Processes
The termwhite processis used to denote the processes in whichall
frequency components appear with equal power;i.e.,the
power-spectral density is a constant for all frequencies. This parallels the
notion of “white light” in which all colors exist.
AprocessX(t)iscalledawhiteprocessifithasaﬂatspectraldensity,
i.e., ifSX(f) is constant for allf.Powerspectraldensityofawhite
process:
SX(f)=
N0
2
19 / 34

White Processes
In practice, the importance of white processes stems from the
fact thatthermal noise can be closely modeled as a white
process over a wide range of frequencies.
Also, a wide range of processes used to describe a variety of
information sources can be modeled as the output of LTI
systems driven by a white process.
If we ﬁnd the power content of a white process using
SX(f)=C, a constant, we will have,
PX=
!
→
↑→
SX(f)df=
!
→
↑→
Cdf=↔
20 / 34

White Processes
Thermal Noise
Figure:spectrum of thermal noise
Thermal noise, though not precisely white, for all practical
purposes can be modeled as a white process with a power
spectrum equal to
kT
2
k: Boltzmann’s constant, 1.38↗10
↑23
Joules/Kelvin
T: temperature in degrees Kelvin
The valuekTis usually denoted byN0;therefore,thepower
spectral density of thermal noise is usually given as
Sn(f)=
N0
2
.
21 / 34

White Processes
The autocorrelation function for a white process:
Rx(ω)=F
↑1
$
N0
2
%
=
N0
2
ϱ(ω)
This shows that for allω↓=0wehaveRx(ω)=0
If we sample a white process at two pointst1andt2(t1↓=t2),
the resulting random variables will be uncorrelated.
If in addition to being white, the random process is also
Gaussian, the sampled random variables will also be
independent (recall that for jointly Gaussian R.Vs,
uncorrelatedness and independence are equivalent.)
22 / 34

White Processes
Thermal Noise
Properties of the Thermal Noise.The thermal noise is
assumed to have the following properties:
1.Thermal noise is a WSS process.
2.Thermal noise is a zero-mean process.
3.Thermal noise is a Gaussian process.
4.Thermal noise is a white process with a power spectral density
Sn(f)=
kT
2
.
It is clear that the power spectral density of thermal noise
increases with increasing the ambient temperature; therefore,
keeping electric circuits cool makes their noise level low.
.
23 / 34

White Processes
Example 4
LetX(t) be a white noise and
Y(t)=
&
2
T
!
T
0
X(t) cos (2εfct)dt
Find the power of the modulated processY(t)where
1
T
↘fc.
24 / 34

Bandpass Processes
Filtered Noise Processes
In many cases, the white noise generated in one stage of the
system is ﬁltered by the next stage; therefore, in the following
stage, we encounterﬁltered noise that is a bandpass process,
i.e., its power spectral density is located away from the zero
frequency and is mainly located around some frequencyfc,
which is far from zero and larger than the bandwidth of the
process.
Bandpass random processes are the equivalents of bandpass
deterministic signals.
In a bandpass process, sample functions are bandpass signals;
like the bandpass signals discussed in previous lectures, they
can be expressed in terms of thein-phase and quadrature
components.
25 / 34

Bandpass Processes
Filtered Noise Processes
Let us assume that the processX(t)istheoutputofanidealbandpassﬁlterof
bandwidthWwhich is located at frequencies aroundfc.Forexample,onesuchﬁlter
can have a transfer function of the form
H1(f)=
%
1|f↑fc|↓W
0otherwise
Since thermal noise is white and Gaussian, the ﬁltered thermal noise will be Gaussian
but not white. The power spectral density of the ﬁltered noise will be
SX(f)=
N0
2
|H(f)|
2
=
N0
2
H(f),
for ideal ﬁlters|H(f)|
2
=H(f). For the ﬁltered noise process corresponding toH1(f),
we have the following power spectral density:
SX1
(f)=
%
N0
2
|f↑fc|↓W
0otherwise
26 / 34

Bandpass Processes
Filtered Noise Processes
Another example of a such ﬁlter would be
H2(f)=
'
1fc≃|f|≃fc+W
0 otherwise
.
SX2
(f)=
(
N0
2
fc≃|f|≃fc+W
0 otherwise
27 / 34

Bandpass Processes
Filtered Noise Processes
All bandpass ﬁltered noise signals have anin-phase and
quadrature component that are lowpass signals.
This means that the bandpass random process X(t) can be
expressed
X(t)=Xc(t) cos (2εfct)→Xs(t)sin(2εfct),
whereXc(t)andXs(t)-the in-phase and quadrature
components-are lowpass processes. Parallel to the
deterministic case, deﬁne
Xc(t)=X(t) cos (2εf0t)+ˆX(t)sin(2εf0t)
Xs(t)=ˆX(t) cos (2εf0t)→X(t)sin(2εf0t)
Note that, we can also represent the ﬁltered noise in terms of
its envelope and phase as
X(t)=A(t) cos (2εfct+ϑ(t))
whereA(t)andϑ(t) are lowpass random processes.
28 / 34

Bandpass Processes
Properties of In-Phase and Quadrature Processes
For ﬁltered white Gaussian noise, the following properties forXc(t)andXs(t)
are obtained,
1.Xc(t)andXs(t)arezero-mean,lowpass,jointlyWSS,andjointly
Gaussian random processes.
2.If the power in processX(t)isPX,thenthepowerineachofthe
processesXc(t)andXs(t)isalsoPX. In other words,
PX=PXc=PXs=
!
→
↑→
SX(f)df
3.ProcessesXc(t)andXs(t) have a common power spectral density (after
low-pass ﬁltering)
SXc(f)=SXs(f)=
#
SX(f↑fc)+SX(f+fc)=N0,|f|↓W
0, otherwise
29 / 34

Example 6
Bandpass Processes
A noise process has a power spectral density given by
Sn(f)=
(
10
↑8
)
1→
|f|
10
8
*
,|f|<10
8
0, |f|>10
8
This noise is passed through an ideal bandpass ﬁlter with a
bandwidth of 2MHz, centered at 50MHz.
1.Find the power content of the output process
2.Write the output process in terms of the in-phase and
quadrature components and ﬁnd the power in each
component. Assumef0= 50MHz
3.Find the power spectral density of the in-phase and
quadrature components
30 / 34

Noise-Equivalent Bandwidth
Suppose that we have a source of white noise of zero mean
and power spectral densityN0/2 connected to the input of an
arbitrary lowpass ﬁlterof transfer functionH(f). The
resulting output noise power is therefore
Py=
!
→
↑→
SY(f)df=
N0
2
!
→
↑→
|H(f)|
2
df
Consider next the same source of white noise connected to
the input of anideal lowpass ﬁlterof zero frequency
responseH(0) (orHmax)andbandwidthB.Inthiscase,the
output noise power is
Py=N0BH
2
max
Therefore, equating these two output noise powers to each
other, we may deﬁne thenoise equivalent bandwidthas
B=Bneq=
+
→
↑→
|H(f)|
2
df
2H
2
max
Hmax:themaximumof|H(f)|in the passband of the ﬁlter.
31 / 34

Noise-Equivalent Bandwidth
Thus, the procedure for calculating the noise equivalent
bandwidth consists of replacing the arbitrary low-pass ﬁlter of
transfer functionH(f) by an equivalent ideal low-pass ﬁlter of
zero-frequency responseH(0) and bandwidthB.
In a similar way, we may deﬁne a noise equivalent bandwidth
for band-pass ﬁlters.
32 / 34

Noise-Equivalent Bandwidth
Example 5
33 / 34

Next Week
E!ect of Noise on Analog Comm. Systems
34 / 34

EEM 308 Introduction to Communications
Lecture 12
Dr. Can Uysal
Department of Electrical and Electronics Engineering,
Eskisehir Technical University
May 20, 2025

E!ect of Noise on Analog Communication Systems
Up to now, we studied the important characteristics of analog
communication systems.
time domain and frequency domain representations of the
modulated signal
bandwidth requirements
power content of the modulated signal
the modulator and demodulator implementation of various
analog communication systems.
In this section, the e!ect of noise on various analog
communication systems will be analyzed.
1/35

Signal-to Noise Ratio (SNR)
Signal-to-noise ratio (SNR or S/N) is a measure used in science and engineering
thatcompares the level of a desired signal to the level of background noise.
SNR is the ratio of signal power to noise power, often expressed in decibels. A
ratio higher than 1 (greater than 0 dB) indicates more signal than noise.
SNR is the measure of quality for analog systems; this statistical parameter is
also important in digital systems.
The received signal in many communication systems can be modeled as the sum
of the desired signal,s(t), and a noise signal,n(t), as shown by
x(t)=s(t)+n(t)
Both of the terms on the right-hand side are random.
For zero-mean processes, a simple measure of signal quality is theratio of the
variances of the desired and undesired signals.Onthisbasis,the
signal-to-noise ratio is formally deﬁned by
SNR=
average signal power
average noise power
=
Psignal
Pnoise
=
E
!
s
2
(t)
"
E[n
2
(t)]
A higher SNR!clearer signal, less noise, better quality (e.g.,
audio/video clarity, data reliability)
A lower SNR!more noise, degraded quality, higher error rate
For a communication signal, a squared signal level is usually proportional to
power. Consequently, the signal-to-noise ratio is often considered to be aratio
of the average signal power to the average noise power.
2/35

Signal-to Noise Ratio (SNR)
3/35

SNR - Decibels
Because many signals have a very wide dynamic range, signals
are often expressed using the logarithmic decibel scale. Based
upon the deﬁnition of decibel, signal and noise may be
expressed in decibels (dB) as
Psignal,dB= 10 log
10(Psignal)
and
Pnoise,dB= 10 log
10(Pnoise)
Similarly, SNR may be expressed in decibels as
SNRdB= 10 log
10(SNR) = 10 log
10(
Psignal
Pnoise
)
SNRdB=Psignal,dB→Pnoise,dB
4/35

E!ect of Noise on Amplitude Modulation Systems
E!ect of Noise on a Baseband System
Sincebaseband systems serve as a basis for comparison of various modulation
systems,webeginwithanoiseanalysisofabasebandsystem
Baseband transmission involves sending the message signal withoutmodulation
(ordemodulationin receiver).The receiver consists only of anideal lowpass
ﬁlter with the bandwidthW.
The noise power at the output of the receiver (for a white noise input)
Pno
=
#
+W
→W
N0
2
df
=N0W.
If we denote the received power byPR, the baseband Signal-to Noise Ratio
(SNR)isgivenby
$
S
N
%
b
=
PR
N0W
5/35
-m(t)
,
modulasyonolmadandogrudan!let!l!r
.
=
W ofm(t)
-
->
Ew
,
w]
=2w
T!mfrekanslardad!eg!nspektruma
-
>
ps!)
sah!polan
g!rlt
tr!
.
W/Hz
WA
Noise↑
SNR

Channel Attenuation
Attenuation is the reduction in signal strength as it
propagates through a communication channel.
Resistance in cables (wireline)
Free-space path loss (wireless)
Absorption, reﬂection, scattering
The amount of signal attenuation generally depends on the
physical channel, the frequency of operation, and the distance
between the transmitter and the receiver.
We deﬁne the loss (attenuation in the channel)Lin signal
transmission as theratio of the transmitted power to the
received power of the channel,i.e.,
L=
PT
PR
,
or, in decibels,
LdB↑10 log
10L= 10 log
10
PT
PR
= 10 log
10PT→10 log
10PR
6/35

Example 1
Find the SNR in a baseband system with a bandwidth of 5 kHz
and withN0= 10
→14
W/Hz.Thetransmitterpoweris1kilowatt
and the channel attenuation is 10
12
.
7/35

E!ect of Noise on AM Systems
E!ect of Noise on DSB-SC AM
DSB-SC signal:
u(t)=Acm(t)cos(2ωfct)
The received signal (modulated signal plus ﬁltered noise(can be written in
terms of in-phase and quadrature components))
r(t)=u(t)+n(t)
=Acm(t)cos(2ωfct)+nc(t)cos(2ωfct)→ns(t)sin(2ωfct)
Demodulate the received signal:
1. Multiplyr(t)withcos(2ωfct+ε)
r(t)cos(2ωfct+ε)=
1
2
Acm(t)cos(ε)+
1
2
Acm(t)cos(4ωfct+ε)
+
1
2
[nc(t)cosε+ns(t)sinε]
+
1
2
[nc(t)cos(4ωfct+ε)→ns(t)sin(4ωfct+ε)]
2. Pass the product signal through an ideal LPF having a bandwidthW
y(t)=
1
2
Acm(t)cos(ε)+
1
2
[nc(t)cosε+ns(t)sinε]
8/35

E!ect of Noise on AM Systems
E!ect of Noise on DSB-SC AM
Assume that a coherent demodulator is employed, without
loss of generality, we assume thatω=0
y(t)=
1
2
[Acm(t)+nc(t)]
Therefore, at the receiver output, the message signal and the
noise components are additive and we are able to deﬁne a
meaningful SNR. The message signal power (after
demodulation) is given by
Po=
A
2
c
4
Pm
The noise power
Pno=
1
4
Pnc=
1
4
Pn
where we have used the fact that the power contents ofnc(t)
andn(t)areequal
9/35
-
>
quadrate
hask!lan!r

E!ect of Noise on AM Systems
E!ect of Noise on DSB-SC AM
The PSD ofn(t)isgivenby
Sn(f)=
!
N0
2
,|f→fc|<W
0,otherwise
The noise power
Pn=
"
↑
→↑
Sn(f)df=
N0
2
↑4W=2WN0
Thus, the output SNR
#
S
N
$
o
=
Po
Pno
=
A
2
c
4
Pm
1
4
Pn
=
A
2
cPm
2WN0
The received signal power (DSB-SC Signal)PR=
A
2
c
2
Pm
#
S
N
$
oDSB→SC
=
PR
N0W
which is identical to
%
S
N
&
b
.
So, DSB-SC does not provide any SNR improvement over a simple baseband
communication system!
10 / 35

E!ect of Noise on AM Systems
E!ect of Noise on SSB-AM
In SSB, we have
u(t)=Acm(t) cos (2εfct)±Acˆm(t)sin(2εfct)
The received signal
r(t)=u(t)+n(t)
=Acm(t) cos (2εfct)±Acˆm(t)sin(2εfct)+n(t)
=[Acm(t)+nc(t)] cos (2εfct)+[±Acˆm(t)→ns(t)] sin (2εfct)
Demodulate the received signal:
1. Multiplyr(t) with cos (2εfct)
r(t) cos (2εfct)=
1
2
[Acm(t)+nc(t)] +
1
2
[Acm(t)+nc(t)] cos (4εfct)
+
1
2
[±Acˆm(t)→ns(t)] sin (4εfct)
2. AfterLPF↓y(t)=
1
2
[Acm(t)+nc(t)]
11 / 35

E!ect of Noise on AM Systems
E!ect of Noise on SSB-AM
Parallel to the discussion of DSB-SC:
Po=
A
2
c
4
PmandPno=
1
4
Pnc=
1
4
Pn
In the USSB (similar is valid for LSSB) case,Sn(f)=
!
N0
2
,fc<|f|<fc+W
0,otherwise
Pn=
"
↑
→↑
Sn(f)df=
N0
2
↑2W=WN0
Thus, the output SNR
#
S
N
$
oSSB
=
Po
Pno
=
A
2
c
4
Pm
1
4
Pn
=
A
2
cPm
WN0
In this case,PR=PU=A
2
cPm
#
S
N
$
oSSB
=
PR
N0W
=
#
S
N
$
b
Therefore, the signal-to-noise ratio in a single-sideband system is equivalent to
that of a DSB-SC system.
12 / 35

E!ect of Noise on AM Systems
E!ect of Noise on DSB (Conv. AM)
InDSB,wehave
u(t)=Ac[1 +amn(t)] cos (2εfct)
The received signal:
r(t)=u(t)+n(t)
={Ac[1 +amn(t)] +nc(t)}cos (2εfct)→ns(t)sin(2εfct)
The received signal is demodulated (when a synchronous
demodulator is employed, similar to DSB-SC and SSB), after
lowpass ﬁltering we have
yl(t)=
1
2
{Ac[1 +amn(t)] +nc(t)}
However, the desired signal ism(t), not 1 +amn(t). The DC
component is removed by a DC blocking device:
y(t)=
1
2
Acamn(t)+
1
2
nc(t)
13 / 35

E!ect of Noise on AM Systems
E!ect of Noise on DSB (Conv. AM)
The received signal power for a DSB signal (assuming the message
signal is zero mean):
PR=
A
2
c
2
!
1+a
2
Pmn
"
The output SNR,
#
S
N
$
oDSB
=
1
4
A
2
ca
2
Pmn
1
4
Pnc
=
A
2
ca
2
Pmn
2N0W
!
1+a
2
Pmn
"
[1 +a
2
Pmn]
=
a
2
Pmn
[1 +a
2
Pmn]
PR
N0W
=
a
2
Pmn
[1 +a
2
Pmn]
#
S
N
$
b
=ϑ
#
S
N
$
b
ϑ=
a
2
Pmn
1+a
2
Pmn
denotes the modulation e”ciency. Since
a
2
Pmn<1+a
2
Pmn, the SNR in conventional AM is always smaller
than the SNR in a baseband system!
14 / 35
-d
4
<1
-
>
Always
"¥
!let!mde
b!lg!n!n tas!d-!-
enerj!n!n toplameners!ye
orcen
/

Example 2
We assume that the message is a wide-sense stationary random
processM(t) with the autocorrelation function
RM(ϖ) = 16 sinc
2
(10,000ϖ)
We also know that all the realizations of the message process
satisfy the condition max|m(t)|= 6. We want to transmit this
message to a destination via a channel with a 50dBattenuation
and additive white noise with the power spectral density
Sn(f)=
N0
2
= 10
→12
W/Hz.WealsowanttoachieveanSNRat
the modulator output of at least 40dB.Whatistherequired
transmitter powerandchannel bandwidthif we employ the
following modulation schemes?
a. DSB-SC AM.
b. SSB AM.
c. ConventionalAMwith a modulation index equal to 0.8 .
15 / 35

E!ect of Noise on Angle Modulation
Recall that in AM, the message information is contained in the
amplitude of the modulated signal;sincenoiseisadditive,itisdirectly
added to the signal.
But, in a FM signal, the noise is added to the amplitude and the message
information is contained in thefrequency of the modulated signal.
Therefore, the message is contaminated by the noise to the extent that
the added noise changes the frequency of the modulated signal.
The frequency of a signal can be described by itszero crossings(as in
Exp. 6).
16 / 35

E!ect of Noise on Angle Modulation
The e!ect of additive noise on the demodulated FM signal can be
described by changes in the zero crossings of the modulated FM signal.
We observe that thee!ect of noise in a low-power FM system is more
severe than in a high-power FM system. In a low-power signal, noise
causes more changes in the zero crossings.
Recall the angle modulated signal is represented as
u(t)=Accos (2ωfct+ε(t))
=
!
Accos
'
2ωfct+2ωkf
(
t
→↑
m(ϑ)dϑ
)
FM
Accos (2ωfct+kpm(t)) PM
.
The block diagram of the receiver for a general angle-modulated signal is
Bandpass ﬁlter removes out-of-band noise (a noise-limiting ﬁlter) and its
bandwidth is equal to the bandwidth of the modulated signal.
17 / 35

E!ect of Noise on Angle Modulation
The output of the noise-limiting ﬁlter (BPF):
r(t)=u(t)+n(t)
=u(t)+nc(t) cos (2εfct)→ns(t)sin(2εfct)
Due to the nonlinearity of the demodulation process, a precise
analysis is quite involved!
Assume the signal power is much higher than the noise power.
The bandpass noise is represented as
n(t)=
%
n
2
c(t)+n
2
s(t) cos
#
2εfct+arctan
ns(t)
nc(t)
$
=Vn(t) cos (2εfct+#n(t))
whereVn(t)and# n(t) represent the envelope and the phase of
the bandpass noise process, respectively.
18 / 35

E!ect of Noise on Angle Modulation
Figure:Phasor diagram of an angle-modulated signal when the signal is
much stronger than the noise.
Noting thatε(t)=
!
kpm(t), PM
2ωkf
(
t
→↑
m(ϑ)dϑ,FM
The output of the angle demodulator:
y(t)=
!
kpm(t)+Yn(t), PM
kfm(t)+
1
2ω
d
dt
Yn(t),FM
whereYn(t)=
Vn(t)
Ac
sin (” n(t)→ε(t))
19 / 35

E!ect of Noise on Angle Modulation
y(t)=
&
kpm(t)+Yn(t), PM
kfm(t)+
1
2ω
d
dt
Yn(t),FM
whereYn(t)=
Vn(t)
Ac
sin (# n(t)→ω(t))
The ﬁrst term is the desired signal component, and the
second term is the noise component.
From this expression, we observe that the noise component is
inversely proportional to the signal amplitudeAc.
Hence, the higher the signal level, the lower the noise level.
Note that this is not the case with amplitude modulation
In AM systems, the noise component is independent of the
signal component, and scaling of the signal power does not
a!ect the received noise power.
20 / 35

E!ect of Noise on Angle Modulation
After some math operations, the
demodulated-noise power
spectral density (for|f|<W)
Sn0
(f)=
&
N0
A
2
c
,PM
N0
A
2
c
f
2
,FM
PM has a ﬂat noise spectrum and FM has a parabolic noise
spectrum.
Therefore, in FM,the e!ect of noise on higher-frequency
components is much higher than the e!ect of noise on
lower-frequency components.
21 / 35

E!ect of Noise on Angle Modulation
The noise power at the output of LPF:
Pno
=
'
+W
→W
Sno
(f)df=
&
2N0W
A
2
c
,PM
2N0W
3
3A
2
c
,FM
The output signal power
Pso
=
&
k
2
pPM,PM
k
2
f
PM,FM
The SNR is deﬁned as
(
S
N
)
o
=
Pso
Pno
.Thus,
#
S
N
$
o
=
&
k
2
pA
2
c
2
PM
N0W
,PM
3k
2
f
A
2
c
2W
2
PM
N0W
,FM
22 / 35

E!ect of Noise on Angle Modulation
Noting that the received signal powerPR=
A
2
c
2
and
!
ϖp=kpmax|m(t)|,PM
ϖf=
k
fmax|m(t)|
W
, FM
We may express the output SNR as
#
S
N
$
o
=
*
+
,
PR
'
ε
P
max|m(t)|
)
2
P
M
N0W
,PM
3PR
'
ε
f
max|m(t)|
)
2
P
M
N0W
,FM
If we denote
P
R
N0W
by
%
S
N
&
b
,theSNRofabasebandsystemwiththesame
received power, we obtain
#
S
N
$
o
=
*
+
,
P
Mε
2
P
(max|m(t)|)
2
%
S
N
&
b
PM
3P
Mε
2
f
(max|m(t)|)
2
%
S
N
&
b
FM
P
M
(max|m(t)|)
2is power of the normalized message,PMn,sowehave
#
S
N
$
o
=
!
ϖ
2
pPMn
%
S
N
&
b
PM
3ϖ
2
fPMn
%
S
N
&
b
FM
23 / 35

E!ect of Noise on Angle Modulation
Observations:
In both PM and FM, the outputSNRis proportional toϖ
2
increasingϖincreases the output SNR.
The increase in the received SNR is obtained by increasing the
bandwidth. Therefore angle modulation provides a way to trade o!
bandwidth for transmitted power.
Increasing the transmitter power will increase the output SNR due to a
decrease in the received noise power.
In AM, any increase in the received power directly increases the
signal power at the output of the demodulator. This is basically
because the message is in the amplitude of the transmitted signal
and an increase in the transmitted power directly a!ects the
demodulated signal power. However, in angle modulation, the
message is in the phase of the modulated signal, and increasing the
transmitter power does not increase the demodulated message
power. The output SNR is increased by a decrease in the received
noise power.
In FM, the e!ect of noise is higher at higher frequencies. This means that
signal components at higher frequencies will su!er more from noise than
signal components at lower frequencies.
24 / 35

Example 3
What is the required received power in an FM system withϱ=5if
W= 15kHzandN0= 10
→14
W/Hz? The power of the
normalized message signal is assumed to be 0.1 Watt and the
required SNR after demodulation is 60dB.
25 / 35

Example 4
Design an FM system that achieves anSNRat the receiver equal
to 30dBand requires the minimum amount of transmitter power.
The bandwidth of the channel is 120kHz; the message bandwidth
is 10kHz; the average-to-peak power ratio for the message,
PMn
=
PM
(max|m(t))
2is
1
2
; and the (one-sided) noise power spectral
density isN0= 10
→8
W/Hz.Whatistherequiredtransmitter
power if the signal is attenuated by 40dBin transmission through
the channel?
26 / 35

Preemphasis and Deemphasis Filtering for FM
As observed in previous slides, the noise power spectral
density at the output of the FM demodulator has a parabolic
shape within the message signal bandwidth.
This parabolic increase in the noise power spectral density is
due to the use of a di!erentiator in the FM demodulator.
As a consequence, the higher-frequency components of the
message signal are degraded by this increase in noise power.
27 / 35

Preemphasis and Deemphasis Filtering for FM
To compensate for the increase in noise power at the higher
frequencies of the message signal, we can boost the
high-frequency components prior to the FM modulator at the
transmitter and, thus, transmit these frequencies at a higher
power level.
This can be easily accomplished by using a highpass ﬁlter at
the transmitter, called apreemphasis ﬁlter
Thus, the degradation of the high-frequency components of
the message due to the large noise power spectral density at
the demodulator is reduced.
28 / 35

Preemphasis and Deemphasis Filtering for FM
Having boosted the high frequencies of the message signal at
the transmitter, we need to restore these frequency
components in the demodulated message signal to their
original form.
This is accomplished by performing the inverse operation of
preemphasis, i.e., passing the demodulated signal through a
lowpass ﬁlter, called adeemphasis ﬁlter.
29 / 35

Preemphasis and Deemphasis Filtering for FM
the cascade of the preemphasis and the deemphasis ﬁlters has
reciprocal frequency response characteristics within the
bandwidth occupied by the message signal.
Use of pre-emphasis and de-emphasis in an FM system:
Hp(f)=
1
Hd(f)
|f|<W
The net e!ect of these ﬁlters should be a ﬂat-frequency
response
30 / 35

Comparison of Analog-Modulation Systems
Now, we are at a point where we can present an overall
comparison of di!erent analog communication systems. The
systems that we have studied include linear modulation (DSB-SC,
Conventional AM, SSB, and VSB) and nonlinear modulation
systems (FM and PM).
The comparison can be done based on three important practical
criteria:
Bandwidth E”ciency
Power E”ciency
Ease of Implementation (Tx and Rx)
31 / 35

Comparison of Analog-Modulation Systems
Bandwidth E#ciency
The most bandwidth-e”cient analog communication system is
the SSB system
This system is widely used in bandwidth-critical applications,
such as voice transmission over microwave and satellite links
and some point-to-point communication systems in congested
areas.
Since SSB cannot e!ectively transmit DC, it cannot be used
for the transmission of signals that have a signiﬁcant DC
component, such as image signals. A good compromise is the
VSB system, which has a bandwidth slightly larger than SSB
and is capable of transmitting DC values. VSB is used in TV
broadcasting and in some data communication systems.
PM, and particularly FM, are the least favorable systems when
bandwidth is a major concern, and their use is only justiﬁed
by their high level of noise immunity.
32 / 35

Comparison of Analog-Modulation Systems
Power E#ciency
A criterion for comparing the power e”ciency of various
systems is the comparison of their output signal-to-noise ratio
at a given received signal power. We have already seen that
angle-modulation schemes, and particularly FM, provide a
high level of noise immunity and, therefore, power e”ciency.
FM is widely used on power-critical communication links, such
as point-to-point communication systems and high-ﬁdelity
radio broadcasting. It is also used for the transmission of
voice (which has been already SSB/FDM multiplexed) on
microwave line-of-sight and satellite links.
Conventional AM and VSB+C are the least power-e”cient
systems and are not used when the transmitter power is a
major concern. However, their use is justiﬁed by the simplicity
of the receiver structure.
33 / 35

Comparison of Analog-Modulation Systems
Ease of Implementation
The simplest receiver structure is the receiver for conventional AM, and
the structure of the receiver for VSB+C system is only slightly more
complicated.
FM receivers are also easy to implement. These three systems are widely
used for AM, TV, and high-ﬁdelity FM broadcasting (including FM
stereo).
The power ine#ciency of the AM transmitter is compensated by the
extremely simple structure of literally hundreds of millions of receivers.
DSB-SC and SSB require synchronous demodulation and, therefore, their
receiver structure is much more complicated. These systems are,
therefore, never used for broadcasting purposes. Since the receiver
structure of SSB and DSB-SC have almost the same complexity and the
transmitter of SSB is only slightly more complicated than DSB-SC,
DSB-SC is hardly used in analog signal transmission, due to its relative
bandwidth ine#ciency.
34 / 35

End of Semester
35 / 35

EEM 308 Lecture telecommunications introduction

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EEM 308 Lecture telecommunications introduction

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