2. What are the reasons for modulation?1. Frequency division multiplexing (To support multiple transmissions via a single channel) To avoid interference
M1(f) M(f) Multiplexed f signal0 + M2(f) 0 f1 f2 f f0
2. Practicality of Antennas Transmitting very low frequencies require antennas with miles in wavelength
3.What are the Different of Modulation Methods?
3. What are the Different of Modulation Methods?1. Analogue modulation- The modulating signal and carrier both are analogue signals Examples: Amplitude Modulation (AM) , Frequency Modulation (FM) , Phase Modulation (PM)2. Pulse modulation- The modulating signal is an analogue signal but Carrier is a train of pulses Examples : Pulse amplitude modulation (PAM), Pulse width modulation (PWM), Pulse position modulation (PPM)
3.What are the Different of Modulation Methods?3. Digital to Analogue modulation- The modulating signal is a digital signal , but the carrier is an analogue signal. Examples: Amplitude Shift Keying (ASK), FSK, Phase Shift Keying (PSK)4. Digital modulation - Examples: Pulse Code Modulation, Delta Modulation,Adaptive Delta Modulation
ANALOG AND DIGITALAnalog-to-analog conversion is the representation ofanalog information by an analog signal. One may askwhy we need to modulate an analog signal; it isalready analog. Modulation is needed if the medium isbandpass in nature or if only a bandpass channel isavailable to us.Topics discussed in this section:Amplitude ModulationFrequency ModulationPhase Modulation
NoteThe total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal: BPM = 2(1 + β)B.
4. What are the Basic Types ofAnalogue Modulation Methods ?
4. What are the Basic Types of Analogue Modulation Methods ?Consider the carrier signal below: sc(t ) = Ac(t) cos( 2πfc t + θ ) 1. Changing of the carrier amplitude Ac(t) produces Amplitude Modulation signal (AM) 2. Changing of the carrier frequency fc produces Frequency Modulation signal (FM) 3. Changing of the carrier phase θ produces Phase Modulation signal (PM)
5. What are the different Forms of Amplitude Modulation ?
5. What are the different Forms of Amplitude Modulation ?1. Conventional Amplitude Modulation (DSB-LC) (Alternatively known as Full AM or Double Sideband with Large carrier (DSB-LC) modulation2. Double Side Band Suppressed Carrier (DSB-SC) modulation3. Single Sideband (SSB) modulation4. Vestigial Sideband (VSB) modulation
6. Derive the Frequency Spectrum for Full-AM Modulation (DSB-LC)
6. Derive the Frequency Spectrum for Full-AM Modulation (DSB-LC)1 The carrier signal is sc (t ) = Ac cos(ω c t ) where ω c = 2πf c2 In the same way, a modulating signal (informationsignal) can also be expressed as sm (t ) = Am cos ωm t
3 The amplitude-modulated wave can be expressed as s (t ) = [ Ac + sm (t )] cos(ω c t )4 By substitution s (t ) = [ Ac + Am cos(ω mt )] cos(ω c t )5 The modulation index. Am m = Ac
6 Therefore The full AM signal may be written as s (t ) = Ac (1 + m cos(ω mt )) cos(ω c t ) cos A cos B = 1 / 2[cos( A + B ) + cos( A − B )] mAc mAcs (t ) = Ac (cos ω ct ) + cos(ω c + ω m )t + cos(ω c − ω m )t 2 2
7. Draw the Frequency Spectrum of the above AM signal and calculate the Bandwidth
7. Draw the Frequency Spectrum of the above AM signal and calculate the Bandwidth fc-fm fC fc+fm 2fm
8. Draw Frequency Spectrum for a complex input signal with AM
8. Draw Frequency Spectrum for a complex input signal with AM fc-fm fc fc+fm
Frequency Spectrum of an AM signalThe frequency spectrum of AM waveform containsthree parts: 1. A component at the carrier frequency fc 2. An upper side band (USB), whose highest frequency component is at fc+fm 3. A lower side band (LSB), whose highest frequency component is at fc-fmThe bandwidth of the modulated waveform is twice theinformation signal bandwidth.
• Because of the two side bands in the frequency spectrum its often called Double Sideband with Large Carrier.(DSB- LC)• The information in the base band (information) signal is duplicated in the LSB and USB and the carrier conveys no information.
ExampleWe have an audio signal with a bandwidth of 5 KHz.What is the bandwidth needed if we modulate the signalusing AM?
ExampleWe have an audio signal with a bandwidth of 5 KHz.What is the bandwidth needed if we modulate the signalusing AM?SolutionAn AM signal requires twice the bandwidth of theoriginal signal: BW = 2 x 5 KHz = 10 KHz
Modulation Index (m) 9. What is the significance of modulation index ?• m is merely defined as a parameter, which determines the amount of modulation.• What is the degree of modulation required to establish a desirable AM communication link? Answer is to maintain m<1.0 (m<100%).• This is important for successful retrieval of the original transmitted information at the receiver end.
9.Modulation Index (m) What is the significance of modulation index ?
• If the amplitude of the modulating signal is higher than the carrier amplitude, which in turn implies the modulation index m ≥ 1.0(100%) . This will cause severe distortion to the modulated signal.
Power distribution in full AM10. Calculate the power efficiency of AM signals
10. Calculate the power efficiency of AM signals• The ratio of useful power, power efficiency : sidebands power m2 / 2 m2 = = total power 1 + m / 2 2 + m2 2• In terms of power efficiency, for m=1 modulation, only 33% power efficiency is achieved which tells us that only one-third of the transmitted power carries the useful information.
Double Side Band Suppressed Carrier (DSB-SC) Modulation• The carrier component in full AM or DSB-LC does not convey any information. Hence it may be removed or suppressed during the modulation process to attain higher power efficiency.• The trade off of achieving a higher power efficiency using DSB-SC is at the expense of requiring a complex and expensive receiver due to the absence of carrier in order to maintain transmitter/receiver synchronization.
11. Derive the Frequency Spectrum for Double Sideband Suppressed Carrier Modulation (DSB-SC) 1 Consider the carrier sc (t ) = Ac cos(ω c t ) where ω c = 2πf c 2 modulated by a single sinusoidal signal sm (t ) = Am cos ω mt where ω m = 2πf m 3 The modulated signal is simply the product of these two s (t ) = Ac cos(ω t ) Am cos(ω t ) c m = Ac Am cos(ω t ) cos(ω t ) c m 1 since cos A cos B = (cos( A +B ) +cos( A −B ) ) 2 Am Ac Am Ac = cos(ω + m )t + c ω cos(ω − m )t c ω 2 2 USB LSB
sc (t ) = Ac cos ω c tsm (t ) = Am cos ω mt X s (t ) = Ac cos(ω c t ) Am cos(ω mt )Frequency Spectrum of a DSB-SC AM Signal fc-fm fc fc+fm
• All the transmitted power is contained in the two sidebands (no carrier present).• The bandwidth is twice the modulating signal bandwidth.• USB displays the positive components of sm(t) and LSB displays the negative components of sm(t).
Generation and Detection of DSB-SC• The simplest method of generating a DSB-SC signal is merely to filter out the carrier portion of a full AM (or DSB-LC) waveform.• Given carrier reference, modulation and demodulation (detection) can be implemented using product devices or balanced modulators.
BALANCED MODULATOR Sm(t) S1(t) AM Modulator 1Sm(t) Accos(ωct) S(t) Carrier Accos(ωct) DSB-SC AM Modulator 2 -Sm(t) S2(t)
• The two modulators are identical except for the sign reversal of the input to one of them. Thus, s1 (t ) = Ac (1 + m cos(ω mt )) cos(ω c t ) s2 (t ) = Ac (1 − m cos(ω mt )) cos(ω c t ) s (t ) = s1 (t ) − s2 (t ) = 2mAc cos(ω mt ) cos(ω c t )
COHERENT (SYNCHRONOUS) DETECTOR OR DSB-SC (PRODUCT DETECTOR) v(t) vo(t) DSB-SC Signal s(t) X LPF Cosωct Local Oscillator• Since the carrier is suppressed the envelope no longer represents the modulating signal and hence envelope detector which is of the non-coherent type cannot be used.
v(t ) = s (t ) cos(ω c t ) = [ 2mAc cos(ω mt ) cos(ω c t )] cos(ω c t ) Am =2 Ac cos(ω mt ) cos (ω c t ) 2 Ac 1 + cos 2ω c t = 2 Am cos(ω mt ) 2 = Am cos(ω mt ) + Am cos(ω mt ) cos(2ω c t )since sm (t ) = Am cos(ω mt ) = sm(t) + sm(t ) cos ( 2ω c t) Unwanted term(removed by LPF)
• It is necessary to have synchronization in both frequency and phase between the transmitter (modulator) & receiver (demodulator), when DSB-SC modulation ,which is of the coherent type, is used. Both phase and frequency must be known to demodulate DSB-SC waveforms.
LACK OF PHASE SYNCHRONISATIONLet the received DSB-SC signal be s DSB − SC (t ) = sm (t ) cos( ω c t + θ ) Acif θ is unknown, v(t ) = s DSB − SC (t ) cos ω c t = Ac sm (t ) cos( ω c t + θ ) cos ω c t Ac = sm (t )[ cosθ + cos( 2ω c t + θ ) ] 2Output of LPF Ac vo (t ) = sm (t ) cosθ 2
But we want just Ac vo (t ) = sm (t ) 2Due to lack of phase synchronization, we will see that thewanted signal at the output of LPF will be attenuated by anamount of cosθ.In other words, phase error causes an attenuation of theoutput signal proportional to the cosine of the phase error.The worst scenario is when θ=π/2, which will give rise tozero or no output at the output of the LPF.
LACK OF FREQUENCY SYNCHRONISATIONSuppose that the local oscillator is not stable at fc but at DSB − SC (t ) cos( ω c + ∆ω ) tfc+∆ f, then v(t ) = s = Ac sm (t ) cos ω c t cos( ω c + ∆ω ) t Ac = sm (t )[ cos ∆ωt + cos( 2ω c t + ∆ω ) ] 2Output of LPF Ac vo (t ) = sm (t ) cos ∆ωt 2Thus, the recovered baseband information signal will varysinusoidal according to cos ∆ ωt
This problem can be overcome by adding an extrasynchronization circuitry which is required to detect θ and∆ ωt and by providing the carrier signal to the receiver.A synchronizer is introduced to curb the synchronizationproblem exhibited in a coherent system.Let the baseband signal be sm (t ) = Am cos ω mt Received DSB-SC signal s (t ) = Ac sm (t ) cos ω c t
SYNCHRONISER ( )2 PLL BPF ÷2 Mathematical analysis of the synchronizer is shown below:s 2 (t ) = Ac2 Am cos 2 ω mt cos 2 ω c t 2 Ac2 Am 2 = [1 + cos 2ω mt ][1 + cos 2ω ct ] 4 Ac2 Am 2 = [1 + cos 2ω mt + cos 2ω ct + cos 2ω mt cos 2ω ct ] 4 Ac2 Am 2 1 1 = 1 + cos 2ω mt + cos 2ω c t + cos 2( ω c − ω m ) t + cos 2( ω c + ω m ) t 4 2 2 Output of BPF Ac2 Am 2 cos 2ω c t 4
Output of frequency divider k cos ω c t where k is a constant of proportionality. DISADVANTAGE OF USING COHERENT SYSTEMS• The frequency and phase of the local oscillator signal must be very precise which is very difficult to achieve. It requires additional circuitry such as synchronizer circuit and hence the cost is higher.
Single-SidebandBand Modulation (SSB) Single Side ModulationHow to generate SSB signal?• Generate DSB-SC signal• Band-pass filter to pass only one of the sideband and suppress the other.For the generation of an SSB modulated signalto be possible, the message spectrum must havean energy gap centered at the origin.
• Example of signal with -300 Hz ~ 300 Hz energy gap Voice : A band of 300 to 3100 Hz gives good articulation• Also required for SSB modulation is a highly selective filter
• Vestigial SidebandBand Modulation (VSB) Vestigial Side Modulation Instead of transmitting only one sideband as SSB, VSB modulation transmits a partially suppressed sideband and a vestige of the other sideband.
Comparison of Amplitude Modulation methods Full AM (or DSB-LC)- Sidebands are transmitted in full with the carrier.- Simple to demodulate / detect- Poor power efficiency- Wide bandwidth ( twice the bandwidth of the information signal)- Used in commercial AM radio broadcasting, one transmitter and many receivers.
Comparison of Amplitude Modulation methods DSB-SC- Less transmitted power than full AM and all the transmitted power is useful.- Requires a coherent carrier at the receiver; This results in increased complexity in the detector(i.e. synchroniser)- Suited for point to point communication involving one transmitter and one receiver which would justify the use of increased receiver complexity.
Comparison of Amplitude Modulation methods SSB- Good bandwidth utilization (message signal bandwidth = modulated signal bandwidth)- Good power efficiency- Demodulation is harder as compares to full AM; Exact filter design and coherent demodulation are required- Preferred in long distance transmission of voice signals
Comparison of Amplitude Modulation methods VSB- Offers a compromise between SSB and DSB-SC- VSB is standard for transmission of TV and similar signals- Bandwidth saving can be significant if modulating signals are of large bandwidth as in TV and wide band data signals.• For example with TV the bandwidth of the modulating signal can extend up to 5.5MHz; with full AM the bandwidth required is 11MHz