Continuous Wave Modulation

Home , Carrier wave, Continuous wave, Modulation

ECE 405

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 Modulated Signal signal ModulationModulation

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 Estimated of signal Message signal DemodulationDemodulation

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

ct( )= Acc cos(2π ft )

where Ac is carrier amplitude, fc is carrier frequency. We assume the phase of carrier wave is zero.

The amplitude modulation signal is

st()=+ Aca[ 1 kmt ()cos(2] π ft c )

=+Acccaccos(2ππft ) Akmt ( )cos(2 ft )

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.

kmta ( )< 1 for all t

So that, 1() + kmt a stays positive. If the amplitude sensitivity is large to make kmt a ()> 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

fc >> W where W is message bandwidth.

Continuous Wave Modulation

Amplitude Modulation (AM):

The Fourier Transform of AM signal is given by

AkA Sf()=−+++−++cac[]δδ ( ff ) ( ff )[] Mff ( ) Mff ( ) 22cc c c

The message signal m(t) is band-limited to the interval c.

S(f) M(f) Ac/2

kAac M(0) Upper M’(0) sideband 2 Lower Lower Upper sideband sideband sideband f f

-W W -fc-W-fc -fc+W fc-W fc fc+W

The transmission bandwidth BT for AM wave is given 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 information- Then,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 Pxt==2() lim xtdt () T →∞ ∫ T −T /2

Using the definition, the average transmitted power of the AM signal is

1 T /2 Pstdt= lim2 ( ) T →∞ ∫ T −T /2

5 Continuous Wave Modulation

Amplitude Modulation (AM):

Average Transmitted Power of AM :

T /2 1 222 PAkmtftdt=+limca[] 1 ( ) cos (2π c ) T →∞ ∫ T −T /2 TT/2 /2 11222222 11 =+++++limAca 1 2kmt ( ) km a ( t ) dt lim A ca 1 2 kmt ( ) km a ( t ) cos(4π ftdt c ) TT→∞∫∫ →∞ TT−−TT/222 /2

1 Where f >> W . Since the integral A2221++ 2kmt () km () t cos(4π ft ) of c caa2  c is close to zero, Therefore, the average power is

T /2 11222 PAkmtkmtdt=++limca 1 2 () a () T →∞ ∫   T −T /2 2 Assume that the average value of message signal is zero

∞ Mmtdt(0)= ∫ ( )= 0 −∞

Continuous Wave Modulation

Amplitude Modulation (AM): Average Transmitted Power of AM :

Therefore, the average power is

T /2 11222 _____ T /2 PAkmtdt=+limca 1 ( ) 221 T →∞ T ∫ 2 mt()= lim mtdt () −T /2 T →∞ ∫ T −T /2 11 _____ =+AAkmt2222() 22cca 1 PA= 2 the power in the carrier cc2

1 ______2()PAkmt= 22 2 the power contained in the two side-band SB2 c a

______The efficiency of the AM wave is 22 kmta () Exff = ______100% 22 1()+ kmta

6 Continuous Wave Modulation

Example. When the message signal is a singe-tone

mt( )= Amm cos(2π f t )

The total power of message signal is kAam= 1,= 1

_____ T /2 2221 mt( )= lim Amm cos (2π ftdt ) T →∞ ∫ T −T /2 T /2 112 =+limAmm[] 1 2cos(4πft ) dt T →∞ ∫ T −T /2 2 A2 = m 2

A2 Thus, the efficiency is m Ex=≤2 100% 33.33% ff A2 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 A cc cos(2 π ft ) . The DSB-SC wave is given by

stAmtftDSB− SC()= c ()cos(2π c )

m(t) A m(t)cos(ω t) X c c

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

A sf()=−++c [] MffMff ( ) ( ) DSB− SC2 c c

S (f) M(f) DSB-SC A Upper c M(0) M(0) 2 sideband Lower Lower Upper sideband sideband sideband f -W W f -fc-W-fc -fc+W fc-W fc fc+W

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 mt( )= Amm cos(2π f t ) The carrier signal is ct()= Acc cos(2π ft )

stAAftftDSB− SC( )= c m cos(2π c )cos(2π a ) ffcm>>

S (f) M(f) DSB-SC Lower Upper Lower sideband Am Upper sideband sideband AA 2 mcM(0) sideband 4 f f -fm fm -fc-fm -fc -fc+ fm fc-fm fc fc+ fm

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 mt( )= 2cos(2π 1000 t ) The carrier signal is ct( )= 100cos(2π 10000 t )

a. Determine and sketch the modulated waveform stDSB− SC () b. Determine and sketch the spectrum of the modulated waveform c. What is the transmission bandwidth of the modulated signal?

a. stDSB− SC () waveform is

stAAftftDSB− SC( )= c m cos(2ππ c )cos(2 a ) = 200cos(2ππ 1000tt )cos(2 10000 ) b. Using trigonometric identity

stDSB− SC ( )=+ 100cos(2ππ 9000 t ) 100cos(2 11000 t )

9 Continuous Wave Modulation

Example: (cont)

The Fourier transform of cosine signal is

1 cos(2πδδftccc )⇔−++[] ( f f ) ( f f ) 2

Then, Euler Fourier transform of

sfDSB− SC ( )=−+++−++ 50[δδ ( f 9000) ( 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 Upper Lower sideband Upper sideband sideband 50 sideband

f (KHz) -11-fc -9 9 fc 11

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) v (t) X Lowpass 0 Filter at W

cos(ωct)

11 AAcc vt1()=+ Amtcc () Amt ()cos(2ω c t ) ⇔+−++ Mf ()[] Mf ( 2 f c ) Mf ( 2 f c ) 22 22 The output of the low-pass filter

1 vt()= Amt () 0 2 c

Continuous Wave Modulation

Demodulation of DSB-SC Modulation

If there is a frequency offset ∆ω between the two signal, received signal and oscillator signal

11 vt()=∆++∆ Amt ()cos(ω t ) Amt ()cos(2ωω t t ) 1 22ccc The output of the low-pass filter

1 vt()=∆ Amt ()cos(ω t ) 0 2 c 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

11 vt( )=+ Amt ( )cos(φ ) Amt ( )cos(2ωφ t + ) 1 22ccc 1 The output of the low-pass filter vt( )= Amt ( )cos(φ ) 0 2 c 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

1 vt()=∆+ Amt ()cos(ω t φ ) 0 2 c

Continuous Wave Modulation

Demodulation of DSB-SC Modulation using Costas Loop

We need some kind of tracking system at the receiver

v1(t) v3(t) X Lowpass Filter at W

A cos[ω t+θ(t)] v5(t) 1 c v6(t) Voltage Lowpass X SDSB-SC (t) controlled osc Filter at W Phase Shifter -900 Output v2(t) v4(t) X Lowpass Filter at W

12 Continuous Wave Modulation

Demodulation of DSB-SC Modulation using Costas Loop

StAmtttDSB− SC()=+ c ()sin[ω c φ ()]

vt11( )= Amtcc ( )sin[ω t++φωθ ( tA )] cos[ c t ( t )] AA AA =−+++cc11mt()sin[()φθ t ()] t mt ()sin[2 ωφθ t () t ()] t 22c AA 11 vt()=−c 1 mt ()sin[()φθ t ()] t sin(xy )cos( )= sin( xy++ ) sin( xy − ) 3 2 22

vt11()=++ Amtcc ()sin[ω tφωθ ()] tA sin[ c t ()] t AA AA =−−++cc11mt()cos[()φθ t ()] t mt ()cos[2 ωφθ t () t ()] t 22c

AAc 1 22 vt( )=− mt ( )cos[φθ ( t ) ( t )] AA 2 4 vt()=−−c 1 mttt ()sin[()φθ ()]cos[() φθ tt ()] 2 5 4 22 AAc 1 2 11 =−mt()sin2[()[]φθ t ()] t sin(xy )sin( )=−−+ cos( xy ) cos( xy ) 8 22

Continuous Wave Modulation

Demodulation of DSB-SC Modulation using Costas Loop

22 AAc 1 2 vt6()=− mt ()sin2[()[]φθ t ()] 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()tt≈ φ ()

Thus, the demodulated output signal when the loop is in lock

AA AA vt( )=−=cc11 mt ( )cos[φθ ( t ) ( t )] mt ( ) 4 22

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 1 2 Pxt==2() lim xtdt () T →∞ ∫ T −T /2 The average transmit ed power of the DSB-Sc signal is

T /2 1 2 PStdtT= lim DSB− SC ( ) T →∞ ∫ T −T /2 T /2 1 22 2 = limAmCc ( t )cos ( wtdt ) T →∞ ∫ T −T /2 T /2 1122 =+limACcmt ( )[] 1 cos(2 wtdt ) T →∞ ∫ T −T /2 2

Continuous Wave Modulation

Demodulation of DSB-SC Modulation using Costas Loop

2 TT/2 /2 AC 1122 PmtmtwtdtT =+lim ( ) lim ( )cos(2c ) 2 TT→∞TT∫∫ →∞ −−TT/2 /2

Usually the carrier frequency fc is much greater than the baseband bandwidth W. fc>>W. The second term is close to zero 22T /2 _____ AACC1 22 PmtmtT ==lim ( ) ( ) 22T →∞ T ∫ −T /2 The mean-squared value of the message signal

____ 1 T /2 mt22()=+ lim mtdt () T →∞ ∫ T −T /2

14 Continuous Wave Modulation

Demodulation of DSB-SC Modulation using Costas Loop

When the message signal is a single-tone

mt( )= Amm cos(ω t )

The mean-squared value of the message signal ______A 2 mt2()= m 2 The average transmitted power is

A22A P = mC T 4

Handy Equations

cos(ab+= ) cos( a )cos( b ) − sin( a )sin( b ) cos(ab−= ) cos( a )cos( b ) + sin( a )sin( b ) sin(ab+= ) sin( a )cos( b ) + cos( a )sin( b ) sin(ab−= ) sin( a )cos( b ) − cos( a )sin( b ) 11 cos(ab )cos( )= cos( ab−+ ) cos( ab + ) 22 11 sin(ab )sin( )= cos( ab−− ) cos( ab + ) 22 11 sin(ab )cos( )= sin( abab−+ ) sin( + ) 22 1 cos2 (aa )=+[] 1 cos(2 ) 2 1 sin2 (aa )=−[] 1 cos(2 ) 2

15 PSpice Example

AM 4. 0V AM signal in time domain

2. 0V

0.8

V 0V

V1 VOFF = 0 V2 VAMPL = 1 VOFF = 0 FREQ = 1000 VAMPL = 2 FREQ = 10000 -2.0V

-4.0V 0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms 6. 0ms V( S UM1 : OUT) Ti me 2. 0V AM signal in Frequency domain

1. 0V

0V 0Hz 5 KHz 10KHz 15KHz 20KHz 25KHz V( S UM1 : OUT ) Fr equency

PSpice Example

AM 1. 0V Message signal

0.9

V 0. 5V

V1 = 0 V3 V2 = 1 V2 TD = 0 VOFF = 0 TR = 0.000001ns VAMPL = 1 TF = 0.000001ns FREQ = 10000 PW = 1ms PER = 20ms 0 0V 0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms 6. 0ms V( V3 : + ) Ti me

AM signal in time domain 1. 2V AM signal in Frequency domain

2. 0V

0. 8V

1. 0V

0V 0. 4V

-1.0V

0V -2.0V 0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms 6. 0ms 0Hz 5KHz 1 0 KHz 15KHz 20KHz 25KHz V( S UM1: OUT) V( S UM1: OUT) Ti me Fr e quenc y

16 PSpice Example

1. 0V AM Message signal 0. 5V

0.9

0V V V

V1 = -1 V3 V2 = 1 V2 TD = 0 VOFF = 0 -0.5V TR = 0.000001ns VAMPL = 1 TF = 0.000001ns FREQ = 10000 PW = 0.5ms PER = 1ms -1.0V 0 0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms 6. 0ms V( V3 : + ) Ti me

AM signal in time domain AM signal in Frequency domain 1. 0V 2. 0V

1. 0V

0. 5V 0V

-1.0V

-2.0V 0V 0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms 6. 0ms 0Hz 5KHz 10KHz 15KHz 20KHz 25KHz 3 0 KHz V( S UM1: OUT) V( S UM1 : OUT) Ti me Fr equency

PSpice Example

1. 0V DSB-SC DSB-SC signal in time domain

0. 5V

0V Vm VOFF = 0 V Ct VAMPL = 1 VOFF = 0 FREQ = 1000 VAMPL = 1 FREQ = 10000 -0.5V

0 -1.0V 0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms V( MULT1 : OUT) Ti me Message signal 500mV

1. 0V AM signal in Frequency domain

0. 5V

250mV

-0.5V

-1.0V 0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms 0V V( Vm: + ) Ti me 0Hz 5KHz 10KHz 15KHz 20KHz 25KHz V( MUL T1 : OUT) Fr equency

17 PSpice Example

DSB-SC 1. 0V DSB-SC signal in time domain

0. 5V

V V

V1 = -1 V3 0V V2 = 1 V2 TD = 0 VOFF = 0 TR = 0.000001ns VAMPL = 1 TF = 0.000001ns FREQ = 10000 -0.5V PW = 0.5ms PER = 1ms 0 -1.0V 0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms V( MULT1 : OUT) Message signal Ti me

1. 0V AM signal in Frequency domain

800mV

0. 5V

400mV

-0.5V

-1.0V 0s 1. 0ms 2. 0ms 3. 0ms 4. 0ms 5. 0ms V( V3 : + ) 0V Ti me 0Hz 5KHz 10KHz 15KHz 20KHz 25KHz 30KHz V( MULT 1 : OUT) Fr equency