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Continuous Wave Modulation 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 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 A2221++ 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).
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