Transcript EE302: Lesson 2 Gain and decibels

EET260
Frequency Modulation
Modulation

A sine wave carrier can be modulated by varying
its amplitude, frequency, or phase shift.
v  Vc sin  2 fct   
In AM, the amplitude of the carrier is modulated
by a low-frequency information signal.
2
2
1.5
1.5
1
1
0.5
0.5
Voltage (V)
Voltage (V)

0
0
-0.5
-0.5
-1
-1
-1.5
-2
0
-1.5
0.001
Information signal
0.002
0.003
0.004
0.005
0.006
T ime (sec)
0.007
0.008
0.009
0.01
-2
0
Amplitude modulated signal
0.001
0.002
0.003
0.004
0.005
0.006
T ime (sec)
0.007
0.008
0.009
0.01
Frequency modulation
In frequency modulation (FM) the instantaneous
frequency of the carrier is caused to deviate by
an amount proportional to the modulating signal
amplitude.
1
Voltage (V)
0.5
0
-0.5
-1
0
0.5
1
1.5
2
T ime (msec)
2.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
1
v  Vc sin  2 fct   
0.5
Voltage (V)

0
-0.5
-1
0
instantaneous frequency changed
in accordance with modulating signal
0.5
1
1.5
3
3.5
frequency modulated signal
4
-3
x 10
Phase modulation

In phase modulation (PM) the phase of carrier is
caused to deviate by an amount proportional to
the modulating signal amplitude.
v  Vc sin  2 fct   
carrier phase is changed in accordance
with modulating signal

phase modulated signal
Both FM and PM are collectively referred to as
angle modulation.
Frequency modulation

Consider the equation below for a frequency
modulated carrier.
center frequency
frequency deviation
modulating signal
vFM  Vc sin  2  f c  f d vm (t )  t 
defines the instantaneous frequency

We will begin with a simple binary input signal.
Frequency modulation

We will consider a 1-V, 1-kHz square wave as
an input.
Voltage (V)
1
0.5
0
-0.5
-1
0
1
1.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
Our input signal has 3 levels, and fd = 4-kHz
vFM 1Vc sin  2  f c  f d vm (t )  t 
0.5
Voltage (V)

0.5
input signal (vi )
instantaneous frequency of FM signal ( f )
no signal (0 V)
f  f c  f d (0)  f c  10-kHz
0
"high" signal (+1 V)
f  f c  f d (1)  10  4  14-kHz
-1
"low" signal
(-1 V)
f  f c  f d (1)  10  4  6-kHz
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
Frequency modulation
vFM  Vc sin  2  f c  f d vm (t )  t 
input signal (1-V, 1-kHz square wave)
Voltage (V)
1
0.5
0
-0.5
-1
0
0.5
1
1.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
FM signal (fc = 10-kHz, fd = 4-kHz)
Voltage (V)
1
0.5
0
-0.5
-1
0
0.5
f = 10-kHz
1
1.5
2
T ime (msec)
2.5
3
3.5
f = 6-kHz
f = 6-kHz
f = 14-kHz
f = 14-kHz
f = 14-kHz
4
-3
x 10
Fundamental FM concepts


The amount frequency deviation is directly
proportional to the amplitude of the modulating
signal.
The frequency deviation rate is determined by
the frequency of the modulating signal.
 The
deviation rate is the number of times per second
that carrier deviates above and below its center
frequency.
center frequency
frequency deviation
modulating signal
vFM  Vc sin  2  f c  f d vm (t )  t 
Fundamental FM concepts
input signal (2-V, 1-kHz square wave)
2
2
1
1
Voltage (V)
Voltage (V)
input signal (1-V, 1-kHz square wave)
0
-1
-1
-2
-2
0
0.5
1
1.5
2
T ime (msec)
2.5
3
3.5
4
0
1
1
0.5
0.5
0
-0.5
-1
-1

0.5
1
1.5
2
T ime (msec)
1
1.5
2.5
3
3.5
4
-3
x 10
2
T ime (msec)
2.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
0
-0.5
0
0.5
-3
x 10
Voltage (V)
Voltage (V)
0
0
0.5
1
1.5
3
3.5
Doubling the amplitude of the input doubles the
frequency deviation of the carrier.
4
-3
x 10
Fundamental FM concepts
input signal (1-V, 2-kHz square wave)
2
2
1
1
Voltage (V)
Voltage (V)
input signal (1-V, 1-kHz square wave)
0
-1
-1
-2
-2
0
0.5
1
1.5
2
T ime (msec)
2.5
3
3.5
4
0
1
1
0.5
0.5
0
-0.5
-1
-1

0.5
1
1.5
2
T ime (msec)
1
1.5
2.5
3
3.5
4
-3
x 10
2
T ime (msec)
2.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
0
-0.5
0
0.5
-3
x 10
Voltage (V)
Voltage (V)
0
0
0.5
1
1.5
3
3.5
Doubling the frequency of the input doubles the
frequency deviation rate of the carrier.
4
-3
x 10
Example Problem 1
A transmitter operates on a carrier frequency of 915-MHz.
A 1-V square wave modulating signal produces 12.5kHz deviation the carrier. The frequency of the input
signal is 2-kHz.
a. Make a rough sketch of the FM signal.
b. If the modulating signal amplitude is doubled, what is
the resulting carrier frequency deviation?
c. What is the frequency deviation rate of the carrier?
FM with sinusoidal input

Consider a sinusoidal modulating input.
input signal (1-V, 500-Hz sine wave)
1
Voltage (V)
0.5
500-Hz modulating signal
0
-0.5
-1
0
0.5
1
1.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
1
f d  8-kHz
0.5
Voltage (V)


f
vFM  Vc cos  2 f ct  d sin(2 f mt ) 
fm


f c  16-kHz
fi  500-Hz
0
-0.5
-1
0
0.5
1
1.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
Frequency content of an FM signal
1
Voltage (V)
0.5
0

-0.5
-1
0
What does an FM signal look like in the
frequency domain?
0.5
1
1.5
2
T ime (msec)
2.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
1
Voltage (V)
0.5
0
-0.5
-1
0

0.5
1
1.5
3
3.5
4
-3
x 10
We will consider the case of a sinusoidal
modulating signal.


f
vFM  Vc cos  2 fct  d sin(2 f mt ) 
fm


Frequency content of an AM signal
Information signal vm
( fim= 500-Hz )
0.15
2
1.5
0

-0.05
-0.1
-0.15
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0.008
0.009
0.5
Voltage (V)
0.05
Voltage (V)
1
Modulator
or mixer
0.1
0
Amplitude modulated signal vAM
0.01
0
-0.5
-1
-1.5
Time (sec)
-2
0
Frequency domain
0.5
0.001
0.003
0.004
0.005
0.006
T ime (sec)
0.007
0.008
0.009
0.01
Frequency domain
0.5
1.5
0.4
1
0.3
0.5
Voltage (V)
0.2
0.1
0
Voltage (V)
0.4
0.3
0.2
0
0.1
-0.5
0
1000
2000
3000
Frequency (Hz)
4000
5000
6000
0
-1
-1.5
0.5
-2
0
0
1000
Frequency domain
0.001
0.002
0.003
0.4
Voltage (V)
Voltage (V)
0.002
2
0.004
0.005
0.006
T ime (sec)
0.007
0.008
0.009
0.01
0.3
0.2
0.1
0
0
1000
2000
3000
Frequency (Hz)
4000
5000
6000
Carrier signal vc (carrier frequency fc = 5-kHz)
2000
3000
Frequency (Hz)
4000
5000
6000
Frequency modulation index

The modulation index for FM is defined
m f  FM modulation index 
fd
fm
f d  frequency deviation
where
f m  modulating signal frequency
1
Voltage (V)

Just as in AM it is used to describe the depth of
modulation achieved.
From the previous example
0.5
0
-0.5
-1

0
0.5
1
1.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
vFM  Vc cos  2 16,000t  m f sin(2 500t ) 
1
Voltage (V)
0.5
0
thus m f 
-0.5
-1
0
0.5
1
1.5
2
T ime (msec)
2.5
3
3.5
4
-3
x 10
f d 8000

 16
f m 500
Frequency analysis of FM

In order to determine the frequency content of
vFM  Vc cos  2 fct  m f sin(2 f mt ) 
we can use the Fourier series expansion given
vFM (t ) 
where

 J n (m f )cos(c  nm )t
n 
c  2 fc  carrier frequency
m  2 f m  modulating signal frequency
where Jn(mf) is the Bessel function of the first kind
of order n and argument mf.
Frequency analysis of FM

Expanding the series,
we see that a single-frequency modulating
signal produces an infinite number of sets of
side frequencies.
Frequency analysis of FM

Each sideband pair includes an upper and lower
side frequency

The magnitudes of the side frequencies are
given by coefficients Jn(m).
 Although
there are an infinite number of side
frequencies, not all are significant.
FM spectrum for mf = 2.0

For the case for mf = 2.0, refer to the table in
Figure 5-2 to determine significant sidebands.
Bessel functions Jn(mf)
1
if mf = 2.0, then the side frequencies we
need to consider are J0, J1, J2, J3, J4
J0
0.8
J1
0.6
J2
J3
J4
0.4
J5
J6
J7
J8
J9
J10
0.2
0
-0.2
-0.4
0
1
2
3
4
5
6
7
8
9
10
11
12
FM spectrum for mf = 2.0

For the case for mf = 2.0, the series can be
rewritten

Substituting the values for J0(2), J1(2),…, J4(2)
FM spectrum for mf = 2.0

From the equation, the spectrum can be plotted.
1
Amplitude (V)
0.8
0.58
0.6
0.58
0.35
0.4
0.35
0.22
0.2
0.13
0.13
0.03
0.03
0
494
496
498
fc - 4fm
500
fc - 2fmFrequency
fc (Hz)
fc - 3fm

fc - fm
502
504
fc + 2fm
fc + f m
fc + 4fm
fc + 3fm
What is the bandwidth of this signal?
506
FM spectrum for mf = 2.0

The bandwidth of the previous signal is
1
Amplitude (V)
0.8
0.58
0.6
BW  ( f c  4 f m )  ( f c  4 f m )
 8 fm
0.35
0.4
0.35
0.22
0.2
0.13
0.13
0.03
0.03
0
494
496
fc - 4fm

0.58
498
500
Frequency (Hz)
502
504
506
fc + 4fm
More generally, the bandwidth is given
BW  2 fm N
where N is the number of significant sidebands.
Example Problem 2
A signal vm(t) = sin (21000t) is frequency modulates a
carrier vc(t) = sin (2500,000t). The frequency deviation
of the carrier is fd = 1000 Hz.
a. Determine the modulation index.
b. The number of sets of significant side frequencies.
c. Draw the frequency spectrum of the FM signal.
Example Problem 2
FM Spectrum with m f=1.0
1
0.77
Amplitude (V)
0.8
0.6
0.44
0.44
0.4
0.2
0.11
0.11
0.02
0
494
496
0.02
498
500
Frequency (kHz)
502
504
506
FM bandwidth as function of mf
FM spectrum with m f=1.00
FM spectrum with m f=0.25
0.98
1
1
Amplitude (V)
Amplitude (V)
0.6
0.4
0.2
0
490
0.77
0.8
0.8
0.12
492
494
496
0.6
0.44
0.2
0.12
498
500
502
Frequency (kHz)
504
506
508
0
490
510
0.11
0.02
492
494
496
0.8
0.8
0.58
0.58
0.35
0.4
0.35
0.22
0.2
0
490

0.13
0.03
492
494
496
0.13
0.03
498
500
502
Frequency (kHz)
0.11
0.02
498
500
502
Frequency (kHz)
504
506
508
510
FM spectrum with m f=4.00
1
Amplitude (V)
Amplitude (V)
FM spectrum with m f=2.00
1
0.6
0.44
0.4
504
0.6
0.43
0.36
0.28
0.4
0.2
506
508
510
0
490
0.13
0.05
0.02
492
494
0.40
0.07
496
0.43
0.36
0.28
0.13
0.050.02
0.07
498
500
502
Frequency (kHz)
504
506
508
FM bandwidth increases with modulation index.
510
FM bandwidth
Note the case mf = 0.25
0.98
1
Amplitude (V)

FM spectrum with m f=0.25
1.2
0.8
0.6
0.4
0.2
0
490


0.12
492
494
496
0.12
498
500
502
Frequency (kHz)
504
506
508
In this special case, FM produces only a single
pair of significant sidebands, occupying no more
bandwidth than an AM signal.
This is called narrowband FM.
510
Narrowband FM

FM systems with mf < /2 are defined as
narrowband.
 This
is true despite the fact that only values of mf in
the range of 0.2 to 0.25 have a single pair of
sidebands.

The purpose of NBFM is conserve spectrum and
they are widely used in mobile radios.
Carson’s Rule

An approximation for FM bandwidth is given by
Carson’s rule:
BW  2  f d (max)  f m(max) 

The bandwidth given by Carson’s rule includes
~98% of the total power.
Example Problem 3
What is the maximum bandwidth of an FM signal with a
deviation of 30 kHz and maximum modulating signal of
5 kHz as determined the following two ways:
a. Using the table of Bessel functions.
b. Using Carson’s rule.
Example Problem 3
FM spectrum with m f=6.00
1
Amplitude (V)
0.8
0.6
0.36 0.36
0.4
0.36 0.36
0.25
0.2
0.13
0.02
0
950
0.24
0.11
0.28
0.28
0.15
0.24
0.25
0.13
0.11
0.06
960
0.06
970
980
990
1000
1010
Frequency (kHz)
1020
BW = 90 kHz (Bessel functions)
BW = 70 kHz (Carson’s rule)
1030
0.02
1040
1050