ECE 405
Continuous Wave Modulation
Z. Aliyazicioglu
Electrical & Computer Engineering Dept.
Cal Poly Pomona
Continuous Wave Modulation
A modulation is a process of transforming a baseband signal
(message) to another signal called modulated waveform.
The demodulation is the process of recovering the baseband signal
from the modulated waveform
In modulation, the baseband signal is referred to as the modulating
wave.
The result of modulation process is referred as the modulated wave
To shifting other frequency, we use carrier wave. Common form of
carrier is a sinusoidal wave.
Message
Signal
Modulation
Modulation
Modulated
signal
Sinusoidal
Carrier signal
1
Continuous Wave Modulation
Modulated signal is transmitted end of the communication system.
At the end of communication system, we have receiver. We require
the original baseband signal to be restored at the receiver. The
process at the receiver is called demodulation, which is the reverse
of the modulation process.
Received
signal
Demodulation
Demodulation
Estimated of
Message signal
The receiver receives the modulated signal with channel noise. The
performance of cannel noise may depend of the type of modulation
used.
Continuous Wave Modulation
Amplitude Modulation (AM):
The amplitude of the sinusoidal carrier wave is varied in
accordance with the baseband signal.
>> t=0:0.0001:.4;
>> y=cos(2*pi*10*t);
>> subplot(3,1,1)
>> plot(t,y)
>> ylabel('m(t)')
>> subplot(3,1,2)
>> x=cos(2*pi*100*t);
>> plot(t,x)
>>
s=1*(1+0.5*cos(2*pi*10*t)).*cos(2
*pi*100*t);
>> subplot(3,1,3)
>> plot(t,s)
>> ylabel('s(t)')
2
Continuous Wave Modulation
Amplitude Modulation (AM):
Consider a sinusoidal carrier wave c(t) as fallow
c(t ) = Ac cos(2π fct )
where Ac is carrier amplitude, fc is carrier frequency. We assume
the phase of carrier wave is zero.
The amplitude modulation signal is
s(t ) = Ac [1 + kam(t )] cos(2π fct )
= Ac cos(2π fct ) + Ac kam(t )cos(2π fct )
where m(t) is the message signal, ka is called the amplitude
sensitivity.
Continuous Wave Modulation
Amplitude Modulation (AM):
We have two requirements to have the baseband signal shape
over the envelope of the modulated signal.
1. The amplitude of is always les than one.
k am(t ) < 1 for all t
So that, 1 + kam(t ) stays positive. If the amplitude sensitivity is
large to make k am(t ) > 1 , then the carrier wave becomes
overmodulated when the factor crossses zero. This creates
envelope distortion in the modulated signal.
3
Continuous Wave Modulation
Amplitude Modulation (AM):
2. The carier frequency is much greater than the highest frequency
in baseband signal
f c >> W
where W is message bandwidth.
Continuous Wave Modulation
Amplitude Modulation (AM):
The Fourier Transform of AM signal is given by
S(f ) =
Ac
k A
δ (f − fc ) + δ (f + fc )] + a c [M (f − fc ) + M (f + fc )]
[
2
2
The message signal m(t) is band-limited to the interval c.
Ac/2
M(f)
Upper
sideband
M(0)
Lower
sideband
S(f)
ka Ac
2
M’(0)
Upper
sideband
Lower
sideband
f
-W
W
-fc-W
-fc
-fc+W
fc-W
The transmission bandwidth BT for AM wave is given
fc
fc+W
f
BT = 2W
4
Continuous Wave Modulation
Amplitude Modulation (AM):
AM wave has two main concern:
1. Amplitude modulation has wasteful of power. The carrier
wave c(t) is completely independent of the informationThen,
bearing signal m(t). Therefore, the transmission of the
carrier wave is waste of power.
2. Amplitude modulation is waste of bandwidth. The upper and
lower sideband of an AM wave are uniquely related each
other by virtue of symmetry about the carrier frequency.
Hence, given magnitude and phase spectra of either
sideband, we can determine the other.
These modifications naturally result in increased system complexity.
Continuous Wave Modulation
Amplitude Modulation (AM):
Average Transmitted Power of AM :
The normalized average power of an arbitrary signal x(t) is defined
as
_____
T /2
1
2
x(t ) dt
T →∞ T ∫
−T / 2
P = x 2 (t ) = lim
Using the definition, the average transmitted power of the AM
signal is
T /2
1
s 2 (t )dt
T →∞ T ∫
−T / 2
P = lim
5
Continuous Wave Modulation
Amplitude Modulation (AM):
Average Transmitted Power of AM :
T /2
P = lim
T →∞
1
2
Ac2 [1 + kam(t )] cos 2 (2π fct )dt
T −T∫/ 2
T /2
= lim
T →∞
T /2
1
1 2
1
1 2
A 1 + 2kam(t ) + ka2m 2 (t ) dt + lim
Ac 1 + 2kam(t ) + ka2m 2 (t ) cos(4π fct )dt
T →∞ T ∫ 2
T −T∫/ 2 2 c 
−T / 2
1
Where f c >> W. Since the integral Ac2 1 + 2kam(t ) + ka2m2 (t ) cos(4π fct ) of
2
is close to zero, Therefore, the average power is
T /2
1
1 2
Ac 1 + 2ka m(t ) + ka2m 2 (t )dt
T →∞ T ∫ 2
−T / 2
P = lim
Assume that the average value of message signal is zero
∞
M (0) =
∫ m(t )dt = 0
−∞
Continuous Wave Modulation
Amplitude Modulation (AM):
Average Transmitted Power of AM :
Therefore, the average power is
T /2
P = lim
T →∞
=
1
1 2
Ac 1 + ka2m 2 (t )dt
T −T∫/ 2 2 
T /2
_____
m 2 (t ) = lim
T →∞
1 2 1 2 2 _____
Ac + Ac ka m 2 (t )
2
2
Pc =
2PSB =
1
m 2 (t )dt
T −T∫/ 2
1 2 the power in the carrier
Ac
2
1 2 2 ______
Ac ka m 2 (t )
2
the power contained in the two side-band
The efficiency of the AM wave is
______
Eff =
ka2 m 2 (t )
______
x100%
1 + ka2 m 2 (t )
6
Continuous Wave Modulation
Example.
When the message signal is a singe-tone
m(t ) = Am cos(2π fmt )
The total power of message signal is ka = 1, Am = 1
T /2
_____
1
Am2 cos2 (2π fmt )dt
T →∞ T ∫
−T / 2
m 2 (t ) = lim
T /2
1
1
Am2 [1 + 2cos(4π fmt )]dt
T →∞ T ∫
2
−T / 2
= lim
=
Am2
2
Thus, the efficiency is
Am2
Eff = 2 2 x100% ≤ 33.33%
A
1+ m
2
The maximum efficienys is
That means 66.66% power is spent on carrier.
Continuous Wave Modulation
Double Sideband Suppressed Carrier (DSB-SC) Modulation
A double sideband suppressed carrier wave is obtained when a
message signal m(t) is multiplied by the carrier Ac cos(2π fct ) .
The DSB-SC wave is given by
sDSB−SC (t ) = Ac m(t )cos(2π fct )
m(t)
Acm(t)cos(ωct)
X
Accos(ωct)
Generation of DSB-SC signal
7
Continuous Wave Modulation
Double Sideband Suppressed Carrier (DSB-SC) Modulation
The FT of the DSB-SC wave in the frequency domain is
sDSB−SC (f ) =
Ac
[M (f − fc ) + M (f + fc )]
2
SDSB-SC (f)
M(f)
Upper
sideband
M(0)
-W
W
f
Lower
sideband
-fc-W -fc
Ac
M (0)
2
Lower
sideband
-fc+W
fc-W
Upper
sideband
fc
fc+W
f
The transof mission bandwidth the DSB-SC wave is BT=2W (Hz)
Continuous Wave Modulation
Example:
Sketch the waveform in the time domain and spectrum in
the frequency domain of the sinusoidal message
signal m(t ) = A cos(2π f t )
m
m
The carrier signal is
c (t ) = Ac cos(2π fct )
sDSB−SC (t ) = Ac Am cos(2π fct )cos(2π fat )
SDSB-SC(f)
M(f)
Upper
sideband
Am
2
-fm
fm
fc >> fm
f
Lower
sideband
-fc- fm -fc -fc+ fm
Lower
sideband
Upper
sideband
Am Ac
M (0)
4
fc- fm fc
fc+ fm
f
8
Matlab Example
Double Sideband Suppressed Carrier (DSB-SC) Modulation
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
t=0:0.0001e-3:4e-3;
m=cos(2*pi*1000*t);
subplot(3,1,1)
plot(t,m)
ylabel('m(t)')
subplot(3,1,2)
c=cos(2*pi*10000*t);
plot(t,c)
s=m.*c;
subplot(3,1,3)
plot(t,s)
ylabel('s(t)')
Continuous Wave Modulation
Example:
Let’s have a message signal as m(t ) = 2cos(2π 1000t )
The carrier signal is c(t ) = 100cos(2π 10000t )
a. Determine and sketch the modulated waveform sDSB−SC (t )
b. Determine and sketch the spectrum of the modulated waveform
c. What is the transmission bandwidth of the modulated signal?
a. sDSB−SC (t ) waveform is
sDSB−SC (t ) = Ac Am cos(2π fct )cos(2π fat )
= 200cos(2π 1000t )cos(2π 10000t )
b. Using trigonometric identity
sDSB−SC (t ) = 100 cos(2π 9000t ) + 100 cos(2π 11000t )
9
Continuous Wave Modulation
Example: (cont)
The Fourier transform of cosine signal is
cos(2π fct ) ⇔
1
[δ (f − fc ) + δ (f + fc )]
2
Then, Euler Fourier transform of
sDSB−SC (f ) = 50 [δ (f − 9000) + δ (f + 9000)] + 50 [δ (f − 11000) + δ (f + 11000)]
c. As we can see from the following figure the transmission
bandwidth is 2 KHz.
Continuous Wave Modulation
Example: (cont)
SDSB-SC(f)
Bandwidth
Lower
Lower
sideband
sideband
50
Upper
sideband
-11
-fc -9
9
Upper
sideband
fc
11
f (KHz)
Frequency Spectrum of a SDSB-SC signal
10
Continuous Wave Modulation
Demodulation of DSB-SC Modulation
A coherent detector, also called synchronous detector can be
used to recover the message m(t) from the modulated signal
v(t)
SDSB-SC (t)
X
Lowpass
Filter at W
v0(t)
cos(ωct)
v1(t ) =
A
A
1
1
Ac m(t ) + Ac m(t )cos(2ωct ) ⇔ c M (f ) + c [M (f − 2fc ) + M (f + 2fc )]
2
2
2
2
The output of the low-pass filter
v0 (t ) =
1
Ac m(t )
2
Continuous Wave Modulation
Demodulation of DSB-SC Modulation
If there is a frequency offset ∆ω between the two signal,
received signal and oscillator signal
v1(t ) =
1
1
Ac m(t )cos(∆ω t ) + Ac m(t )cos(2ωct + ∆ω t )
2
2
The output of the low-pass filter
v0 (t ) =
1
Ac m(t )cos(∆ω t )
2
Thus recovered signal is a distorted version of the original
message signal
11
Continuous Wave Modulation
Demodulation of DSB-SC Modulation
If there is a phase offset φ between the two signal, received
signal and oscillator signal
1
1
Ac m(t )cos(φ ) + Ac m(t )cos(2ωct + φ )
2
2
The output of the low-pass filter v0 (t ) = 1 Ac m(t )cos(φ )
2
v1(t ) =
Thus, unless φ is small, the output amplitude is attenuated by a
factor of cos (φ). If π/2≤ φ ≤ 3π/2 the output is inverted
If there are both the frequency and the phase offset , the
output voltage is
v0 (t ) =
1
Ac m(t )cos( ∆ω t + φ )
2
Continuous Wave Modulation
Demodulation of DSB-SC Modulation using Costas Loop
We need some kind of tracking system at the receiver
v1(t)
X
v3(t)
Lowpass
Filter at W
A1cos[ωct+θ(t)]
v6(t)
Voltage
controlled osc
SDSB-SC (t)
v5(t)
Lowpass
Filter at W
X
Phase Shifter
-900
X
v2(t)
Lowpass
Filter at W
v4(t)
Output
12
Continuous Wave Modulation
Demodulation of DSB-SC Modulation using Costas Loop
SDSB−SC (t ) = Ac m(t )sin[ωct + φ (t )]
v1(t ) = Ac m(t )sin[ωct + φ (t )]A1 cos[ωct + θ (t )]
Ac A1
AA
m(t )sin[φ (t ) − θ (t )] + c 1 m(t )sin[2ωct + φ (t ) + θ (t )]
2
2
=
Ac A1
m(t )sin[φ (t ) − θ (t )]
2
v1(t ) = Ac m(t )sin[ωct + φ (t )]A1 sin[ωct + θ (t )]
v3 (t ) =
=
sin( x )cos( y ) =
1
1
sin( x + y ) + sin( x − y )
2
2
Ac A1
AA
m(t )cos[φ (t ) − θ (t )] − c 1 m(t )cos[2ωct + φ (t ) + θ (t )]
2
2
Ac A1
2 2
m(t )cos[φ (t ) − θ (t )] v (t ) = Ac A1 m 2 (t )sin[φ (t ) − θ (t )]cos[φ (t ) − θ (t )]
5
2
4
Ac2 A12 2
=
m (t )sin [ 2[φ (t ) − θ (t )]]
1
1
sin( x )sin( y ) = cos( x − y ) − cos( x + y )
8
v 4 (t ) =
2
2
Continuous Wave Modulation
Demodulation of DSB-SC Modulation using Costas Loop
v 6 (t ) =
Ac2 A12 2
m (t )sin [ 2[φ (t ) − θ (t )]]
8
The output frequency of VCO is fc when the input amplitude is zero
When input amplitude is positive, the output frequency of VCO
increases from fc in proportion to the voltage
When input amplitude is negative, the output frequency of VCO
decreases from fc in proportion to the voltage
That means
θ (t ) ≈ φ (t )
Thus, the demodulated output signal when the loop is in lock
v 4 (t ) =
Ac A1
AA
m(t )cos[φ (t ) − θ (t )] = c 1 m(t )
2
2
13
Continuous Wave Modulation
Demodulation of DSB-SC Modulation using Costas Loop
The normalized average power of an arbitraray signal x(t) is
T /2
_____
P = x 2 (t ) = lim
T →∞
1
2
x(t ) dt
T −T∫/ 2
The average transmit ed power of the DSB-Sc signal is
T /2
PT = lim
T →∞
1
(t )dt
S2
T −T∫/ 2 DSB−SC
T /2
1
AC2 m 2 (t )cos2 (wct )dt
T →∞ T ∫
−T / 2
= lim
T /2
1
1
AC2 m 2 (t ) [1 + cos(2wct )] dt
T →∞ T ∫
2
−T / 2
= lim
Continuous Wave Modulation
Demodulation of DSB-SC Modulation using Costas Loop
PT =
AC2
2
T /2
T /2


1
1
m 2 (t ) + lim
m 2 (t )cos(2wct )dt 
Tlim
T →∞ T ∫
→∞ T ∫
−T / 2
−T / 2


Usually the carrier frequency fc is much greater than the baseband
bandwidth W. fc>>W. The second term is close to zero
PT =
T /2
 A2 _____
AC2 
1
m 2 (t ) = C m 2 (t )
Tlim
∫
→∞
T −T / 2
2 
 2
The mean-squared value of the message signal
T /2
____
m 2 (t ) = lim
T →∞
1
m 2 (t )dt +
T −T∫/ 2
14
Continuous Wave Modulation
Demodulation of DSB-SC Modulation using Costas Loop
When the message signal is a single-tone
m(t ) = Am cos(ωmt )
The mean-squared value of the message signal
______
m 2 (t ) =
Am2
2
The average transmitted power is
PT =
Am2 AC2
4
Handy Equations
cos(a + b) = cos(a)cos(b) − sin(a)sin(b)
cos(a − b) = cos(a)cos(b) + sin(a)sin(b)
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin(a − b) = sin(a)cos(b) − cos(a)sin(b)
cos(a)cos(b) =
1
1
cos(a − b) + cos(a + b)
2
2
sin(a)sin(b) =
1
1
cos(a − b) − cos(a + b)
2
2
sin(a)cos(b) =
1
1
sin(a − b) + sin(a + b)
2
2
1
[1 + cos(2a)]
2
1
sin2 (a) = [1 − cos(2a)]
2
cos2 (a) =
15
PSpice Example
AM
AM signal in time domain
4. 0V
2. 0V
0.8
0V
V
VOFF = 0
VAMPL = 1
FREQ = 1000
V1
VOFF = 0
VAMPL = 2
FREQ = 10000
V2
- 2. 0V
0
- 4. 0V
0s
1 . 0 ms
2 . 0 ms
3 . 0 ms
4 . 0 ms
5 . 0 ms
6 . 0 ms
V( SUM1 : OUT)
Ti me
AM signal in Frequency domain
2. 0V
1. 0V
0V
0 Hz
5 KHz
1 0 KHz
1 5 KHz
2 0 KHz
2 5 KHz
V( S UM1 : OUT)
Fr e q u e n c y
PSpice Example
AM
Message signal
1. 0V
0.9
V
V1 = 0
V3
V2 = 1
TD = 0
TR = 0.000001ns
TF = 0.000001ns
PW = 1ms
PER = 20ms
VOFF = 0
VAMPL = 1
FREQ = 10000
0. 5V
V2
0
0V
0s
1 . 0 ms
2 . 0 ms
3 . 0 ms
4 . 0 ms
5 . 0 ms
6 . 0 ms
V( V3 : +)
Ti me
AM signal in time domain
AM signal in Frequency domain
1. 2V
2. 0V
0. 8V
1. 0V
0V
0. 4V
- 1. 0V
- 2. 0V
0s
1 . 0 ms
2 . 0 ms
3 . 0 ms
V( S UM1 : OUT)
Ti me
4 . 0 ms
5 . 0 ms
6 . 0 ms
0V
0 Hz
5 KHz
1 0 KHz
1 5 KHz
2 0 KHz
2 5 KHz
V( SUM1 : OUT)
Fr e q u e n c y
16
PSpice Example
1. 0V
AM
Message signal
0. 5V
0.9
V
0V
V
V1 = -1
V3
V2 = 1
TD = 0
TR = 0.000001ns
TF = 0.000001ns
PW = 0.5ms
PER = 1ms
VOFF = 0
VAMPL = 1
FREQ = 10000
V2
- 0. 5V
- 1. 0V
0s
0
1 . 0 ms
2 . 0 ms
3 . 0 ms
4 . 0 ms
5 . 0 ms
6 . 0 ms
V( V3 : +)
Ti me
AM signal in Frequency domain
AM signal in time domain
1. 0V
2. 0V
1. 0V
0. 5V
0V
- 1. 0V
- 2. 0V
0s
1 . 0 ms
2 . 0 ms
3 . 0 ms
4 . 0 ms
5 . 0 ms
6 . 0 ms
V( SUM1 : OUT)
0V
0 Hz
5 KHz
1 0 KHz
1 5 KHz
2 0 KHz
2 5 KHz
3 0 KHz
V( S UM1 : OUT)
Ti me
Fr e q u e n c y
PSpice Example
1. 0V
DSB-SC
DSB-SC signal in time domain
0. 5V
V
0V
VOFF = 0
VAMPL = 1
FREQ = 1000
Vm
V
VOFF = 0
VAMPL = 1
FREQ = 10000
Ct
- 0. 5V
0
- 1. 0V
0s
1 . 0 ms
2 . 0 ms
3 . 0 ms
4 . 0 ms
5 . 0 ms
V( MULT1 : OUT)
Ti me
Message signal
5 0 0 mV
AM signal in Frequency domain
1. 0V
0. 5V
2 5 0 mV
0V
- 0. 5V
- 1. 0V
0s
1 . 0 ms
2 . 0 ms
3 . 0 ms
V( Vm: +)
Ti me
4 . 0 ms
5 . 0 ms
0V
0 Hz
5 KHz
1 0 KHz
1 5 KHz
2 0 KHz
2 5 KHz
V( MULT1 : OUT)
Fr e q u e n c y
17
PSpice Example
DSB-SC signal in time domain
1. 0V
DSB-SC
0. 5V
V
V
V1 = -1
V3
V2 = 1
TD = 0
TR = 0.000001ns
TF = 0.000001ns
PW = 0.5ms
PER = 1ms
0V
VOFF = 0
VAMPL = 1
FREQ = 10000
V2
- 0. 5V
0
- 1. 0V
0s
1 . 0 ms
2 . 0 ms
3 . 0 ms
4 . 0 ms
5 . 0 ms
V( MULT1 : OUT)
Message signal
Ti me
AM signal in Frequency domain
1. 0V
8 0 0 mV
0. 5V
0V
4 0 0 mV
- 0. 5V
- 1. 0V
0s
1 . 0 ms
2 . 0 ms
3 . 0 ms
V( V3 : +)
Ti me
4 . 0 ms
5 . 0 ms
0V
0 Hz
5 KHz
V( MULT1 : OUT)
1 0 KHz
1 5 KHz
2 0 KHz
2 5 KHz
3 0 KHz
Fr e q u e n c y
18