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Amplitude Modulation
Ahmad Bilal
5.2
5-2 ANALOG AND DIGITAL
Analog-to-analog conversion is the representation of
analog information by an analog signal.
Amplitude Modulation
Frequency Modulation
Phase Modulation
Topics discussed in this section:
5.3
Types of analog-to-analog modulation
Basics of Modulation
•In modulation we have two signals
–Message signal.
•Low frequency
•Low Energy
•Example voice
–Carrier Signal
•High frequency andenergy
Basic of Modulation
•We know a signal with frequency can travel a largedistance with small antenna size , where as abaseband signal is low in frequency and energy
•Question is what to do if we want to send ourbaseband signal to large distance
Answer
Superimpose
it
•An AM signal is made up of a carrier (with constantfrequency) in which its amplitude is changed(modulated) with respect to the input signal(modulating signal).
•The modulating signal (message signal) causes thecarriers amplitude to change with time. This resultingshape of the carrier is called the envelope. Note theenvelope has the shape of a sine wave.
Message Signal •The modulating waveform can either be a single tone. This can be represented by a cosine waveform, or the modulating waveform could be a wide variety of frequencies
•Message signal may be represented as
m (t) = M sin (ωmt)Where:
• modulating signal frequency in Hertz is equal to ωm / 2 π
• M is the carrier amplitude• φ is the phase of the signal at the start of the reference time
•It is worth noting that normally the modulating signal frequency is well below that of the carrier frequency.
Carrier Signal
•Carrier signal is represented as
C (t) = C sin (ωct)Where:
carrier frequency in Hertz is equal to ωc / 2 πC is the carrier amplitude
The Modulated wave
M(t)
C(t)
M(t)
C(t)
M(t)*c(t)
The Time and Frequency Analysis
•We know
m(t)= M(w)
When the signal is modulated we get
m(t) X c(t) = Vm cos (ωmt) x cos (ωct)
=
½ Vm Cos ( (m - c) t ) + ½ Vm Cos ((m + c)t)
Things to Note
½ Vm Cos ( (m - c) t ) + ½ Vm Cos ((m + c)t))
Frequency range has shifted
Amplitude of signal have become half
•So cos(1000t) = pi (δ (w-1000)+ pi (δ (w+1000)
Representation of any sinusoidal wave
-1000Hz 1000Hz
π
Things to Note
½ Vm Cos ( (m - c) t ) + ½ Vm Cos ((m + c)t))
So lets say
M(t) =cos 1000t and c(t)=10000t : Modulated signal will be
½ 1 Cos ( (-9000) t ) + ½ 1Vm Cos ((11000))
Similarly in Fourier
Mt*cos(wct) = ½ [M(wm+wc)+M(wm-wc)]
Means in frequency
domain signal will exist
on -9000 hz and +9000 hz
infrequency domain
signal will exist on -
11000 hz and +11000 hz
Things to Note
M(t) =cos 1000t and c(t)=10000t
The modulated signal will be
½ 1 Cos ( (-9000) t ) + ½ 1Vm Cos ((11000))
Means in frequency
domain signal will exist
on -9000 hz and +9000 hz
infrequency domain
signal will exist on -
11000 hz and +11000 hz
-9000Hz 9000Hz-11000Hz +11000HZ
π/2
Things to Note
-9000Hz 9000Hz-11000 +11000
Final Result of Modulated Signal
π/2
Note the signal has exact two copies
one with positive frequency components and other
With negative frequency components
As the modulated signal consist of two copies of exact same signal. We may study any one , the properties for other will be same
•Modulated signal is centered around Wc
•If we add message signal frequency to carrier frequency , we get upper and lower limit of signal
•Let these limits be called upper side band and Lower side band
9000Hz +11000
π/2
Wc =10000 Hz
Wc+WmWc-WmUpper side band Lower Side band
The whole Picture
The Complete Picture
•As signal has two copies of low side band and upper side band . Thus the signal is called double side band with suppressed carrier
•Both copies contains exact information
•Upper side band and lower band also contain same information
Band width calculation
•The message signal had a bandwidth of 1000 hz
•Can u derive the bandwidth relation of message signal and modulated signals
9000Hz +11000Wc =10000 Hz
Wc+WmWc-Wm
•A standard AM broadcast station is allowed to transmit modulating frequencies up to 5 kHz. If the AM station is transmitting on a frequency of 980 kHz, compute the maximum and minimum upper and lower sidebands and the total bandwidth occupied by the AM station.
Class Task : Synchronous Detection and Coherent Detection
Draw Frequency Domain spectrum for
Modulated signal x cos 10000 t
Modulated signal= ½ 1 Cos ( (-9000) t ) + ½ 1Vm Cos ((11000))
Modulators
•Multiplier modulators
Two signals are directly multiplied together .
Very difficult to implement (non linear behavior )
Very expensive
Balanced Modulator
•Also called non liner modulator
•NL represents any non linear devices that behaves as ax(t) +bx2(t)
M(t) + cos(wct)
M(t) - cos(wct)
ax(t) +bx2(t)
aM(t)+acos(wct)+ b(M2(t)+cos2(wc(t)+2m(t)cos(wc(t)))
ax(t) +bx2(t)
aM(t)-acos(wct)+ b(M2(t)+cos2(wc(t)-2m(t)cos(wc(t)))
•Also called non liner modulator
•NL represents any non linear devices that behaves as ax(t) +bx2(t)
aM(t)+acos(wct)+ b(M2(t)+cos2(wc(t)+2m(t)cos(wc(t)))
-aM(t)-acos(wct)+ b(M2(t)+cos2(wc(t)-2m(t)cos(wc(t)))
aM(t)+acos(wct)+ b(M2(t)+bcos2(wc(t)+2bm(t)cos(wc(t))) -am(t)+acos(wct)-bm2(t)-bcos2(wc(t)+2bm(t)cos(wc(t)))
=final Out put 2acos(wct(t))+4bm(t)(cos(wc(t))
Amplitude Modulation DSB
1. In last method , there was no carrier wave sent to , receiver.
2. Hence if the receiver does not know about the modulating frequency , it can not recover original signal .
3. The other alterative is to send carrier signal along with modulated signal to the receiver , so there is no requirement of generating carrier at receiver side .
4. How ever for sending carrier wave along with modulated waves , needs lot of power and energy.
5. Where DSB-SC is used and Where DSB is used
DSB = AM
•DSB method is generally what people refer to when they are talking about AM . The modulated signal is represented by .
•What will be the frequency domain representation
Choosing Values
Envelop Detection Condition
A+m(t)>0
Modulation Index μ
Amplitude of Modulating Signal Amplitude of Carrier Signal
μ= VmVc
Amplitude of carrier signal should always be greater than message signal .The value of modulation index is always between zero and 1 If value of μ > 1. This is called over modulation : loss of data
Effect mo Modulation Index
•Preventing overmodulation is tricky. •For example, at different times during voice transmission voices will go from low amplitude to high amplitude. •Normally, the amplitude of the modulating signal is adjusted so that only the voice peaks produce 100 percent modulation.• This prevents overmodulation and distortion. Automatic circuits called compression circuits solve this problem by amplifying the lower-level signals and suppressing or compressing the higher-level signals. •The result is a higher average power output level without overmodulation. •Distortion caused by overmodulation also produces adjacent channel interference.
Other Formulas for Calculating Modulating Index
•For any modulated wave , we may find, Vm and Vc as
•Suppose that on an AM signal, the Vmax (p-p) value read from the graticule on the oscilloscope screen is 5.9 divisions and is Vmin (p-p) 1.2 divisions.
• An AM signal is said to be made up of 3 components
– Carrier + LSB+ USB
• The Power of sideband depends on Modulation index
• Greater is modulation index, More is side band Power
Power of AM Waves
For Power calculation , we need
Peak Values
RMS values .
Can be obtained multiplying the voltage value to 0.707
Using Power formula
Power of AM wave
if the carrier of an AM transmitter is 1000 W and it is modulated 100 percent (m = 1), the total AM power i
•Solve for 70 percent modulated 250-W carrier, the total power in the composite AM signal is
• Power can also be calculated in terms of current , as it is easy to measure the current across the know resistor/Load . As we all know that
P=I2R : So our equation will become
Pt=It2R
Where It is
Power Factor in Terms Of I
• The total output power of an 85 percent
modulated AM transmitter, whose
unmodulated carrier current into a 50
Ohm antenna load is 10 AmpereHmm easy …Just Apply the formula
It=Ic √(1+(m2/2))
Where Ic = 10
m =0.85
It=10√(1.36) = 11.67 Amp
NOW CALCULATE POWER
Quick Example
Find Modulation Factor
It=Ic √(1+(m2/2))
Where Ic = 10
It=10√(1.36) = 11.67 Amp
NOW CALCULATE Modulation Index
Find Modulation Factor
Question
Question
Remember you canalso calculate via
•The transmitter in Example 3-4 experiences an antenna current change from 4.8 A unmodulated to 5.1 A. What is the percentage of modulation?
• In DSB the basic information is transmitted twice. (Practically speaking there is no advantage of doing so)
• So one side band may be suppressed • Advantages1. Occupy Less spectrum (more signals can be
transmitted)2. Strong signal (No power for double bands and
carrier )3. Less fading
Single Side Band
• Typical AM
Single Side Band
LSB FC USB
Suppressed
Disadvantage
•Demodulation depends upon the carrier being present.• If the carrier is not present, then it must be regenerated at the receiver and reinserted into the signal. •To faithfully recover the intelligence signal, the reinserted carrier must have the same phase and frequency as those of the original carrier. •This is a difficult requirement. When SSB is used for voice transmission, the reinserted carrier can be made variable in, frequency so that it can be adjusted manually• This is not possible with some kinds of data signals.
•To solve this problem, a low-level carrier signal is sometimes transmitted along with the two sidebands in DSB or a single sideband in SSB.
• Because the carrier has a low power level, the essential benefits of SSB are retained, but a weak carrier is received so that it can be amplified and reinserted to recover the original information.
•Such a low-level carrier is referred to as a pilot carrier.
In SSB one of the band is transmitted, either upper or lower.
It has a ratio of 3:1 over AM
Mean A 50W SSB transmitter will have same performance as of 150 W AM
In SSB the power is expressed in terms of PEP , Peak Envelope Power .
SSB Power
•
SSB Power
Applications
• SSB
– Telephone systems
– Two way radio (for military apps)
• DSB
– FM and TV Audio Broad casting
•,Assume that a voice signal produces a 360-V, peak-to-peak signal across a 50-V load. The rms voltage is 0.707 times the peak value, and the peak value is one-half the peak-to-peak voltage.
PEP in terms of I
where Vs = amplifier supply voltage
Imax =current peak
A 450-V supply with a peak current of 0.8 A produces a what PEP in watts
Voice amplitude peaks are produced only when very loud sounds are generated during certain speech patterns or when some word or sound is emphasized.
During normal speech levels, the input and output power levels are much less than the PEP level.
The average power is typically only one-fourth to one-third of the PEP value with typical human speech
•With a PEP of 240 W, the average power is only 60 to 80 W. Typical SSB transmitters are designed to handle only the average power level on a continuous basis, not the PEP.
•An SSB transmitter produces a peak-to-peak voltage of 178 V across a 75-V antenna load. What is the PEP
•An SSB transmitter has a 24-V dc power supply. On voice peaks the current achieves a maximum of 9.3 A.
•What is PEP
•What is average power of transmitter