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