HIGHER TECHNNOLOGICAL INSTITUTE
TENTH OF RAMADAN CITY
ELECTRONICS AND COMMUNICATION
ENGINEERING DEPARTMENT
Communication (1)
(EEC 213)
Prepared by :
Dr.Eng. Eslam Samy EL-Mokadem
Communication (1)
(EEC 213)
Lecture (6)
Angle Modulation
Angle Modulation
❑ Angle Modulation: Is the process in which the frequency or the
phase of the carrier signal is varied with respect to the
Modulating signal( message)
❑ Modulating signal m(t)
❑ carrier signal c(t) = 𝑨𝒄 𝒄𝒐𝒔 𝝎𝒄 𝒕 + 𝜽𝒄
phase
Amplitude Frequency
Phase angle ∅(𝒕)= 𝝎𝒄 𝒕 + 𝜽𝒄
❑ Modulated signal
𝒔𝟏 (t) = 𝑨𝒄 𝒎 𝒕 𝒄𝒐𝒔 𝝎𝒄 𝒕 + 𝜽𝒄 Amplitude Modulation
𝒔𝟐 (t) = 𝑨𝒄 𝒄𝒐𝒔 𝝎𝒄 𝒕 + 𝜽𝒄 + 𝒎 𝒕 Angle Modulation
Types of Angle Modulation
Frequency Phase
Modulation Modulation
(FM) (PM)
Frequency Modulation
❑ Is the process in which the frequency of the carrier signal Varies in
accordance with the instantaneous change in the amplitude of the modulating
signal (message )
Phase Modulation
❑ Is the process in which the phase of the carrier signal Varies in accordance
with the instantaneous change in the amplitude of the modulating signal
(message )
Properties of Angle modulation
1) Modulated signal has constant amplitude
2) The angle of the modulated signal is proportional to the message signal
3) Information is embedded in the Zero Crossing
4) Higher immunity to noise
❑ General form for angle Modulation 𝑺(t) = 𝑨𝒄 𝒄𝒐𝒔 ∅(𝒕)
∅(𝒕) is the instantaneous phase angle
Example 1
𝝅
Assume 𝒔𝟏 (t) = 𝑨𝒄 cos (𝝎𝒄 𝒕 + 𝟐)
𝝅
𝒔𝟐 (t) = 𝑨𝒄 cos (𝝎𝒄 𝒕𝟐 + 𝟑𝒕 + )
𝟐
1- Calculate instantaneous phase for 𝒔𝟏 (t) and 𝒔𝟐 (t)
2- Calculate initial phase for 𝒔𝟏 (t) and 𝒔𝟐 (t)
3- Calculate instantaneous frequency for 𝒔𝟏 (t) and 𝒔𝟐 (t)
solution
For 𝒔𝟏 (t)
For 𝒔𝟏 (t) 1- instantaneous phase (∅𝒊𝒏𝒔𝒕. (t))
𝝅 𝝅
1- instantaneous phase (∅𝒊𝒏𝒔𝒕. (t)) = 𝝎𝒄 𝒕 + = 𝝎𝒄 𝒕𝟐 + 𝟑𝒕 +
𝟐 𝟐
𝝅 𝝅
2- initial phase (∅𝒊 (t)) = at t =0 2- initial phase (∅𝒊 (t)) = at t =0
𝟐 𝟐
\
3- instantaneous frequency(𝝎𝒊𝒏𝒔 )= ∅. 𝒊𝒏𝒔𝒕 (t) 3- instantaneous frequency(𝝎𝒊𝒏𝒔 )=
𝒅 𝒅 𝝅 𝒅
= (∅𝒊𝒏𝒔𝒕. (t)) = (𝝎𝒄 𝒕 + ) = 𝝎𝒄 ∅. \𝒊𝒏𝒔𝒕 (t) = (∅𝒊𝒏𝒔𝒕. (t))=
𝒅𝒕 𝒅𝒕 𝟐 𝒅𝒕
𝒅 𝝅
= (𝝎𝒄 𝒕𝟐 + 𝟑𝒕 + ) = 2𝝎𝒄 𝒕 + 𝟑
𝒅𝒕 𝟐
Relation between phase modulation
and frequency modulation
In Phase modulation (PM) In frequency modulation (FM)
∅. \𝒊𝒏𝒔𝒕 (t) 𝜶 𝒎 𝒕
∅𝒊𝒏𝒔𝒕. (t)) 𝜶 𝒎 𝒕
∅. \𝒊𝒏𝒔𝒕 (t)) = 𝒌𝒇 𝒎 𝒕
∅𝒊𝒏𝒔𝒕. (t)) = 𝒌𝒑 𝒎 𝒕 𝒌𝒇 : frequency deviation constant
𝒌𝒑 : phase deviation constant ∅𝒊𝒏𝒔𝒕. (t)) = 𝒌𝒇 𝒕𝒅 )𝒕(𝒎
S𝑷𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔[𝒘𝒄 𝒕 + 𝒌𝒑 𝒎(𝒕)] S𝑭𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔[𝒘𝒄 𝒕 + 𝒌𝒇 ]𝒕𝒅 )𝒕(𝒎
∅𝒊𝒏𝒔𝒕. (t) = 𝒘𝒄 𝒕 + 𝒌𝒑 𝒎(𝒕)
(𝝎𝒊𝒏𝒔 )= ∅. \𝒊𝒏𝒔𝒕 (t)= 𝒘𝒄 + 𝒌𝒑 𝒎\ (𝒕) ∅𝒊𝒏𝒔𝒕. (t) = 𝒘𝒄 𝒕 + 𝒌𝒇 )𝒕(𝒎
\
𝑷𝒉𝒂𝒔𝒆 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 (∆∅)=∅𝒊𝒏𝒔𝒕. - 𝒘𝒄 𝒕 (𝝎𝒊𝒏𝒔 )= ∅. 𝒊𝒏𝒔𝒕 (t)= 𝒘𝒄 + 𝒌𝒇 𝒎(𝒕)
∆∅=𝒘𝒄 𝒕 + 𝒌𝒑 𝒎(𝒕) - 𝒘𝒄 𝐭 𝐟𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝒚 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 (∆𝝎)= 𝝎𝒊𝒏𝒔 - 𝒘𝒄
𝐌𝐚𝐱. 𝑷𝒉𝒂𝒔𝒆 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 (∆∅𝒎𝒂𝒙 ) = ∆𝝎=𝒘𝒄 + 𝒌𝒇 𝒎(𝒕) - 𝒘𝒄
=𝒌𝑷 𝒎𝒂𝒙{𝒎 𝒕 } = 𝜷𝑷𝑴 𝐌𝐚𝐱. 𝐟𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝒚 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 (∆𝝎𝒎𝒂𝒙 ) =
Where 𝜷𝑷𝑴 is Modulation Index
=𝒌𝒇 𝒎𝒂𝒙{𝒎 𝒕 } = 𝜷𝑭𝑴
Where 𝜷𝑭𝑴 is Modulation Index
Phase modulation
For single tone
𝑨𝑺𝑺𝑼𝑴𝑬
𝒎 𝒕 = 𝑨𝒎 cos (𝝎𝒎 𝒕)
𝒄 𝒕 = 𝑨𝒄 cos (𝝎𝒄 𝒕)
S𝑷𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔 𝒘𝒄 𝒕 + 𝒌𝒑 𝑨𝒎 cos (𝝎𝒎 𝒕)
∅𝒊𝒏𝒔𝒕. (t)) = 𝒘𝒄 𝒕 + 𝒌𝒑 𝑨𝒎 cos (𝝎𝒎 𝒕)
𝝎𝒊𝒏𝒔 = ∅. \𝒊𝒏𝒔𝒕 (t)= 𝒘𝒄 + 𝒌𝒑 𝑨𝒎 𝝎𝒎 𝐬𝐢𝐧(𝝎𝒎 t)
𝝎𝒊𝒏𝒔 = ∅. \𝒊𝒏𝒔𝒕 (t)= 𝒘𝒄 + 𝒌𝒑 𝑨𝒎 𝝎𝒎 𝐬𝐢𝐧(𝝎𝒎 t)
𝐌𝐚𝐱. 𝑷𝒉𝒂𝒔𝒆 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 (∆∅𝒎𝒂𝒙 ) = ∅𝒊𝒏𝒔𝒕. (t)) - ∅𝒄 (t))
=𝒌𝒑 𝑨𝒎 𝒎𝒂𝒙{𝐜𝐨𝐬(𝝎𝒎 t)} = 𝜷𝑷𝑴
Modulation Index (𝜷𝑷𝑴 )= 𝒌𝒑 𝑨𝒎
Frequency Modulation
❑ For single tone
Assume 𝒎 𝒕 = 𝑨𝒎 cos (𝝎𝒎 𝒕)
𝒄 𝒕 = 𝑨𝒄 cos (𝝎𝒄 𝒕)
S𝑭𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔[𝒘𝒄 𝒕 + 𝒌𝒇 ]𝒕𝒅 )𝒕(𝒎
S𝑭𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔[𝒘𝒄 𝒕 + 𝒌𝒇 𝒎𝑨 cos (𝝎𝒎 𝒕) 𝒅𝒕]
𝒌𝒇 𝑨𝒎
S𝑭𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔[𝒘𝒄 𝒕 + sin (𝝎𝒎 𝒕)]
𝝎𝒎 To find frequency Deviation
❑ To find instantaneous phase
𝒌𝒇 𝑨𝒎
∅𝒊𝒏𝒔𝒕. (t)) = 𝒘𝒄 𝒕 + sin (𝝎𝒎 𝒕)] (∆𝝎)= 𝝎𝒊𝒏𝒔 - 𝒘𝒄
𝝎𝒎 𝒌 𝒇 𝑨𝒎
= 𝒘𝒄 + 𝝎𝒎 cos (𝝎𝒎 𝒕)] − 𝒘𝒄
To find instantaneous phase deviation 𝝎𝒎
𝒌𝒇 𝑨𝒎 = 𝒌𝒇 𝑨𝒎 cos (𝝎𝒎 𝒕)
∆∅𝒎𝒂𝒙 = ∅𝒊𝒏𝒔𝒕. (t)) - ∅𝒄. (t)) = 𝝎𝒎 To find Max frequency Deviation
To find instantaneous frequency ∆𝝎𝒎𝒂𝒙 =𝒌𝒇 𝑨𝒎 at cos (𝝎𝒎 𝒕) =1
\ 𝒌𝒇 𝑨𝒎
(𝝎𝒊𝒏𝒔 )= ∅. 𝒊𝒏𝒔𝒕 (t)= 𝒘𝒄 + 𝝎𝒎 cos (𝝎𝒎 𝒕) To find Modulation index
𝝎𝒎
𝒌 𝒇 𝑨𝒎 ∆𝝎𝒎𝒂𝒙 ∆𝒇𝒎𝒂𝒙
𝜷𝑭𝑴 = ∆∅𝒎𝒂𝒙= == =
𝝎𝒎 𝝎𝒎 𝒇𝒎
Modulation Index
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
𝜷=
𝒎𝒂𝒙 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝒇𝒐𝒓 𝒊𝒏𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝒔𝒊𝒈𝒏𝒂𝒍
Relationship between FM and PM
S𝑷𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔[ 𝒘𝒄 𝒕 + 𝒌𝒑 𝑨𝒎 cos (𝝎𝒎 𝒕) (1)
S𝑭𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔[𝒘𝒄 𝒕 + 𝒌𝒇 𝒎𝑨 cos (𝝎𝒎 𝒕) 𝒅𝒕]
𝒌𝒇 𝑨𝒎
S𝑭𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔[𝒘𝒄 𝒕 + sin (𝝎𝒎 𝒕)]
𝝎𝒎 (2)
❑ From eq(1) & eq(2) Frequency modulation (FM) and phase modulation
(PM) are closely related to each other because in both the cases there is a
variation in total phase angle
Relationship between FM and PM
Types of FM Modulation:
Narrow Band Wide Band FM
FM (NBFM) (WBFM)
Narrow Band FM (NBFM)
❑ Occurred when 𝜷 ≪ 𝟏
Assume 𝒎 𝒕 = 𝑨𝒎 cos (𝝎𝒎 𝒕)
𝒄 𝒕 = 𝑨𝒄 cos (𝝎𝒄 𝒕)
S𝑭𝑴 (t) = 𝑨𝒄 𝒄𝒐𝒔[𝒘𝒄 𝒕 + 𝒌𝒇 ]𝒕𝒅 )𝒕(𝒎
Narrow Band FM (NBFM)
FM similar to DSB-TC
Band width for NBFM= 2 𝒇𝒎
FM Bandwidth calculation
methods for WBFM:
Actual Method Approximation
Method
(Bessel Function
Table ) (Carson's Rule)
Finding bandwidth using
Bessel's Function Table
❑ Bessel Function : is an equation that expresses the phase angle
in terms of the sine wave modulating signal is solved with a
complex mathematical process
❑ Minimum bandwidth is determined mathematically as
BW = 2 (n 𝒇𝒎 )
Where
n : number of significant sidebands
𝒇𝒎 ∶ is maximum modulating signal frequency
Note : The number of significant sidebands (n)
depends on the value of modulation index(𝜷𝒇 ).
Bessel's Function Table
Finding bandwidth using
Carson’s Rule
∆𝒇
BW = 2 (∆𝒇+ 𝒇𝒎 ) = 𝟐𝒇𝒎 ( +1) = 𝟐𝒇𝒎 (𝜷𝒇 +1)
𝒇𝒎
Where
∆𝒇 : is maximum frequency deviation
𝒇𝒎 ∶ is maximum modulating signal frequency
Note : Carson’s BW is less than actual transmitted BW
Example 2
For an FM modulator with peak frequency deviation 10
KHz, a modulating signal frequency 10 KHz, peak
modulating signal amplitude 10 V, and a 500 KHz carrier,
load resistance= 50Ω
determine:
a) Modulation Index
b) Actual minimum bandwidth from the Bessel table.
c) Approximation minimum bandwidth using Carson’s
rule.
d) Plot the output frequency spectrum for the Bessel
approximation.
e) Calculate total power distribution .
Solution
∆𝒇 𝟏𝟎 𝒌𝑯𝒛
𝒂) 𝜷𝒇 = = =𝟏 𝜷𝒇 =1
𝒇𝒎 𝟏𝟎 𝒌𝑯𝒛
b) From the Bessel function table at 𝜷𝒇 =1
n=3
BW = 2 (n 𝒇𝒎 ) = 2*3*10 kHz= 60 k Hz
c) minimum bandwidth using Carson’s rule.
BW = 2 (∆𝒇+ 𝒇𝒎 ) = 2(10 kHz+ 2(10 kHz) = 40 k Hz
Bessel's Function Table
Solution
d) Plot the output frequency spectrum for the Bessel
approximation.
The relative amplitudes of carrier and sidebands are
7.7 V
The corresponding spectrum
will be as shown in figure : 4.4 V 4.4 V
1.1 V 1.1 V
0.2 V 0.2 V
f KHz
470 480 490 500 510 520 530
26/37
Example 3
FM wave is given by s(t) = 10 cos(8𝝅𝟏𝟎𝟔 𝒕 + 𝟗 𝒔𝒊𝒏 2𝝅𝟏𝟎𝟑 𝒕)
load resistance= 50Ω
determine:
a) frequency deviation
b) bandwidth.
c) Calculate total power distribution .
Solution
𝑨𝒄 = 𝟏𝟎, 𝒇𝒄 = 4 * 𝟏𝟎𝟔 Hz , 𝜷 = 𝟗 , 𝒇𝒎 = 𝟏𝟎𝟑 𝑯𝒛
∆𝒇𝒎𝒂𝒙
𝜷𝑭𝑴 = ∆𝒇𝒎𝒂𝒙 = 𝜷𝑭𝑴 𝒇𝒎 =9k Hz
𝒇𝒎
BW= 2 𝒇𝒎 (𝜷𝑭𝑴 +1) = 2* 𝟏𝟎𝟑 (9+1)= 20k Hz
𝑨𝒄 𝟐 (𝟏𝟎)𝟐 𝟏𝟎𝟎
𝑷𝒄 = = = = 1 watt
𝟐𝑹 𝟐 (𝟓𝟎) 𝟏𝟎𝟎
