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Amplitude Modulation

Three signals are shown in Figure 1. The first is the carrier wave, which is the wave that the information is "carried on". The second signal is the modulating signal; this modifies or adjusts the carrier wave to represent the information to be transmitted. The final waveform is the resultant modulated signal. Note that the carrier frequency is modulated by the modulating signal to result in the modulated signal.

DOUBLE SIDEBAND/SUPPRESSED CARRIER (DSB-SC)

The amplitude modulation process can be represented by equation 1:

m(t) = a(t) cos(w_ct) (1)

where m(t) = modulated signal

a(t) = modulating signal

w_c= 2pf_c, where f_c = carrier frequency.

Note that m(t) is the modulated signal that results from a cosine wave being multiplied (or modulated) by a modulating baseband signal that contains the information to be transmitted, or a(t).

To being the exploration of equation 1, examine a(t) more closely:

a(t) = E_mcos(w_mt) (3)

where E_m = peak voltage of modulating signal.

w_m = modulating signal frequency, measured in radians/second.

The modulating signal a(t) contains all the information that is to be sent. Equation 3 represents the information signal with just one cosine wave oscillating at the modulating signal frequency ω_m. By inserting this expression for a(t) into equation 1, m(t) becomes

m(t) = E_mcos(ω_m t)cos(ω_c t) (4)

The product of two cosine waves can be written as the addition of half the amplitude of their sum and difference:

m(t) = (E_m/2) [cos(ω_c + ω_m)t + cos(ω_c - ω_m)t] (5)

Note that the carrier frequency has split into two pieces, each of half amplitude and each shifted in frequency by the modulating frequency. Figure 2 illustrates this phenomenon.

Because there are two pieces of sidebands; this type of AM is called double sideband (DSB) modulation. Also, because the carrier frequency is absent from the modulated signal, this type of AM is called suppressed carrier (SC) modulation. This type of modulation is therefore called double sideband, suppressed carrier amplitude modulation (DSB-SC AM). The two pieces of sidebands are named the upper sideband, USB, and the lower sideband, LSB.

In Figure 2, the carrier is assumed to have amplitude of 1 and the modulating signal is assumed to have the amplitude E_m. When these two are multiplied, or modulated, the amplitudes of the resulting signals are half of the product of the two individual amplitudes, or E_m/2.

The carrier is in a much higher frequency than the modulating signal, typically a factor of at least 100. For example, in broadcast AM, the carrier frequency is about 1000 kHz, and the modulating signal is about 5kHz.

Figure 3 shows all the frequencies that would exist between the lowest and the highest frequency of the modulating signal. In Figure 3, the spectrum does not extend right up to the carrier frequency. It stops a little short of it on both sides. This is because the lowest frequency in the modulating signal is not DC, but some slightly higher frequency. Because this is DSB-SC, there is no carrier present at w_c.

Sidebands

Because there are two sidebands and each is exactly the bandwidth of the original modulating signal, as shown in Figure 4, the resultant bandwidth is twice the original modulating signal's bandwidth. This is expressed in equation 6:

BW = 2ω_m or BW = 2f_m (6)

Additionally, the sidebands are always symmetric about the carrier frequency:

ω_usb = ω_c + ω_m or f_usb = f_c + f_m (7)

ω_lsb = ω_c - ω_m or f_lsb = f_c - f_m (8)

The sidebands have the same peak voltage given by E_m/2. Because they have the same peak voltage, the average powers will be the same as well. Power can be found by just squaring the rms voltage and dividing by the load resistance. Figure 5 shows how the DSB-SC signals would look on an oscilloscope.

DSB-SC Average Power

In Figure 4 and 5, the peak voltage of the modulated signal is half the peak voltage of the modulating signal. There are two sidebands and it is important to use rms voltages for computing power. The expression for the average power of a DSB-SC modulated signal is hence

P = 2^.(E_m/)²/R = E_m²/4R (11)

EXAMPLE 1:

This example will show sample calculations for a particular-DSB-SC modulator with a carrier frequency of 1000 rad/sec, a load resistance of 50 Ω, and a modulating signal a(t) given by a(t) = 10cos(20t). Find the following:

1. The peak voltage, E_m;

2. The carrier frequency, w_c;

3. The modulation frequency, w_m;

4. The DSB-SC bandwidth;

5. The USB bandwidth and frequencies;

6. The LSB bandwidth and frequencies;

The average power P.

Solution:

Because the carrier signal amplitude is assumed to be unity, the peak voltage is just the peak voltage of the modulating signal a(t), or 10V.

The carrier frequency is w_c = 1000 rad/sec. The modulation frequency is w_m = 20 rad/sec. The DSB-SC bandwidth is twice w_m, or 40 rad/sec. The USB and LSB are each 20 rad/sec in width.

The LSB starts at w_c - w_m = 980 rad/sec and extends to w_c = 1000 rad/sec. The USB starts at w_c = 1000 rad/sec and extends to w_c + w_m = 1020 rad/sec.

The average power of DSB-SC is given by

P = E_m²/4R = 10²/4(50) = 0.50W.

DOUBLE SIDEBAND / LARGE CARRIER (DSB-LC)

AMPLITUDE MODULATION

In DSB-SC the carrier frequency is suppressed, extra circuitry is required to locate and track the carrier and hence increase the cost of the receiver. The type of DSB-LC system is commonly known as AM radio, or broadcast AM.

DSB-LC is obtained by just adding the carrier amplitude E_c to equation 1, which was used to describe DSB-SC. With this change, m(t) becomes

m(t) = [a(t) + E_c]cos(ω_ct) = a(t)cos(ω_ct) + E_ccos(ω_ct) (12)

The only difference between DSB-SC and DSB-LC is that a DC voltage is added to the modulation signal, results in a cosine wave modulated by a signal given by [a(t) + E_c]. The spectrum diagrams are shown in Figure 6, and what a DSB-LC signal would look like on an oscilloscope is shown in Figure

In DSB-LC, there is a lot of energy at the carrier frequency to be used in the receiver. Even if a(t) = 0, there is still a component of m(t) = E_ccos(ω_ct). This energy comes from the amplitude E_c being modulated by the carrier frequency. This results in the modulated signal envelope.

In Figure 7, the modulated signal envelope opens and closes at the same frequency as the modulating signal. The peak voltage for the modulated waveform is just the sum of the modulating signal and carrier signal voltages. The voltage E_c must be greater than or equal to the smallest value that a(t) can obtain. Expressed mathematically, this becomes

E_c min[a(t)] (13)

Modulation Index

Figure 8 illustrates how the modulation index changes the DSB-LC envelope. The modulation index m is a dimensionless number, defined as the ratio of the sideband energy to the carrier energy:

m = modulation index = (14)

Alternatively, percentage modulation is obtained by multiplying m by 100% to obtain the number in percentage form:

percentage modulation = m x 100% (15)

The modulation index m should always > 0 and < 1. If m = 0, the resultant modulated waveform is just a constant envelope of amplitude E_c, the amplitude of the carrier frequency. There is no modulation of the carrier wave. If m > 1, the resultant waveform is overmodulated and is distorted. Figure 8 illustrates what the DSB-LC modulated signal looks like on an oscilloscope for various values of the modulation index.

DSB-LC Average Power

There are three components to any DSB-LC waveform, the upper and lower sidebands and the carrier frequency. If the powers in all three of these are added up, the total power in the modulated signal will be

P = P_c + P_usb + P_lsb (16)

= (19)

(21)

Through the relationship defining the modulation index, Equ. (14) can be substituted into Equ. (21) to result in

(22)

To compute the power in any system, rms values must be used. Considering this, equation 22 becomes

(23)

For the special case of zero percentage modulation (m = 0), the modulated signal should have the same power as the unmodulated carrier, results in

(26)

The modulated waveform can be developed directly from Equation 12 used to define DSB-LC:

m(t) = [a(t)+E_c]cos(ω_ct) = a(t)cos(w_ct)+E_ccos(w_ct) (12)

Substituting for a(t) = E_mcosw_mt will give an equation for the peak voltage of each component of the modulated signal envelope:

m(t) = [E_c + E_m cos(ω_mt)] cos(ω_ct) (28)

Substituting in Equ. (14) and relating the modulation index to the peak voltages of the carrier and modulating signal,

m(t) = [E_c + mE_ccos(ω_mt)] cos(ω_ct)

= [1 + m cos(ω_mt)] E_ccos(ω_ct), (29)

where the first bracketed term is a constant plus the modulating signal, and the second term is just the unmodulated carrier frequency.

Rearranging equation 29, gives

m(t) = E_ccos(ω_ct) + mE_ccos(ω_ct)cos(ω_mt) (30)

Substituting for the product of the cosines, it is found that

m(t) = E_ccos(ω_ct)+ (31)

In equation 31, the first term is the carrier signal and the second and third terms are the sidebands. Note that for 100% modulation, m = 1, the sideband's amplitude is just half of the carrier frequencies amplitude.

EXAMPLE 2:

This example will show sample calculations for a particular broadcast AM radio station with a carrier frequency of 40cos(1000t), a load resistance of 50 Ω, and a modulating signal a(t) = 10cos(20t). Find the following:

1. The modulated signal m(t);

2. The peak carrier voltage, E_c;

3. The carrier frequency, ω_c;

4. The peak modulation voltage, E_m;

5. The modulation frequency, ω_m;

6. The LSB bandwidth and frequencies;

The USB bandwidth and frequencies;

8. The DSB-LC bandwidth;

9. The modulation index;

10. The average power, P.

Solution:

From m(t) = [E_mcos(ω_mt) + E_c]cos(ω_ct) = [10cos(20t) + 40]cos(1000t), we find the peak carrier voltage E_c = 40 V. The carrier frequency is 1000 rad/sec. The peak modulation voltage is E_m= 10 V. The modulation frequency is 20 rad/sec.

The LSB bandwidth is 20 rad/sec, starts at 980 rad/sec, and extends to 1000 rad/sec. The USB bandwidth is 20 rad/sec, starts at 1000 rad/sec, and extends to 1020 rad/sec. The DSB-LC bandwidth is 40 rad/sec.

The modulation index is 0.25, or a percentage modulation of 25%:

The average power is given by

The power of the carrier is given by equation 26 and is 16 W:

The total power was 16.5 W, so 97% of the total power is supplied by the carrier frequency. That means that only about 3% of the power is contained in the sidebands of the modulated signal.

Optimum transmitter use would be a modulation index of as close to 1.0 as possible. The modulating signal would need to be changed to something like

a(t) = 39cos(20t)

Now the modulation index soars to m = E_m/E_c = 39/40 = 0.975, or a percentage modulation of 95%. The power now contained in the sidebands also soars:

;

;

;

.

It should be clear that as the modulation index is increased, the amount of power in the sidebands also increases. It is determined that the smallest value that a(t) can obtain must be less than the peak carrier voltage. The smallest value that a(t) can obtain is -10V as it swings from +10 to-10. Because 10V is smaller than 40 V, the peak carrier voltage, the answer is yes.

DSB-LC Transmission Efficiency

In DSB-LC transmitter, a lot of power is placed in the carrier, and this depends on the value of the modulation coefficient m. Recall that

P = P_c +P_usb+P_lsb (16)

Because the sidebands have equal power,

P = P_c + 2P_sb (32)

Now define transmission efficiency α:

(33)

In terms of peak voltages and modulation index, it is found that

(34)

where α is the transmission efficiency of DSB-LC transmitter.

The maximum DSB-LC efficiency is obtained at m = 1. For a 100% percentage modulation, it is found that α = 33%. This means that the maximum transmission efficiency only results in one-third of the power being used to transmit the information. The remaining 66% of the power is "wasted" in sending the carrier.

A comparison of the two AM schemes for the power is of interest:

DSB-SC: , DSB-LC:

The only difference is in the modulation index term. For the maximum modulation index, m = 1, the power in a DSB-LC modulated waveform is only 1.5 times as large as the power in the unmodulated carrier. Clearly, the maximum power is contained in the sidebands for m = 1, and this still results in two-thirds of the power residing in the carrier:

DSB-LC, m = 0:

DSB-LC, m = 1:

EXAMPLE 3:

Suppose you are employed in an AM broadcast radio station that has an average carrier output power of 10 kW. The modulation index is 0.80. Find the total average power output, the peak amplitude of the output for a load resistance of 100 Ω, and the transmission efficiency.

Solution:

Relating Oscilloscope Readings of the Modulated

Waveform to E_m and E_c

Take a close look at the waveforms that a DSB-LC amplitude modulator generates in Figure 9. Examining Figure 9, observe two things. First, the frequency and shape of the modulating signal's envelope are the same as the modulated signal. Second, V_max and V_min are the peak voltages of the modulated signal's envelope. It is much easier to measure V_max and V_min on an oscilloscope than it is to measure E_m or E_c once the modulation has occurred. Several other relationships can be generated from this observation and are shown below:

V_max = E_c + E_m (35)

V_min = E_c - E_m (36)

V_max = E_m/m + E_m = E_m(1+1/m)=E_c + mE_c =E_c(1+m) (38)

E_m = 1/2(V_max-V_min) (39)

E_c = 1/2(V_max+V_min) (40)

The peak voltage of any DSB modulated signal is just the addition of the peak carrier signal amplitude plus or minus the peak modulating signal amplitude. Figure 10 summarizes this statement, which is also captured by the equations above.

Figure 11 shows how the various voltages of the signals relate to each other. For case 1, where the percentage modulation is 0%, it is seen that the modulated signal is just the carrier signal. At any moment in time, the voltage is equal to the voltage of the carrier wave. Because there is no modulating signal, m = 0, the result is just the carrier.

In case 2, the maximum voltage occurs when the peak voltage of the modulating signal reaches a maximum. This voltage, for the special case of 100% percentage modulation, is equal to twice the modulation voltage or twice the carrier voltage. Only for this special case, the minimum voltage of the modulated signal is defined as 0V.

EXAMPLE 4:

You are examining a DSB-LC waveform on the oscilloscope and try to see if you can determine the modulating voltage, carrier voltage, modulation index, and modulation frequency from the readings shown in Figure 12.

From the figure, read the following values:

V_max = 12 V V_min = 6 V

f_m =1/(20×10^-3)=50 Hz

Now applying the formulas above,

E_m = 1/2(V_max-V_min) = 1/2(12-6) = 3 V

E_c = 1/2(V_max+V_min) = 1/2(12+6) = 9 V

The modulation index is

m = E_m/E_c = 3/9 = 0.33.

Assuming a value of 100 Ω load resistance, then you can calculate the average power to be

EXAMPLE 5:

These four thought experiments on modulation index apply to all types of amplitude modulation. Refer to equation 14.

If the voltage of a modulating signal is doubled, what happens to the modulation index? It doubles.

If the voltage of the carrier signal is doubled, what happens to the modulation index? It halves.

If the frequency of a modulating signal is doubled, what happens to the modulation index? Nothing.

If the frequency of the carrier signal is doubled, what happens to the modulation index? Nothing.

EXAMPLE 6:

Here are two more thought experiments, this time just for large carrier systems. Both have to do with power. Refer to equations 23 and 24.

If the voltage of the modulating signal is doubled, what happens to the average power? It goes up by 4.

If the voltage of the carrier signal is doubled, what happens to the average power? It goes up by 4.

EXAMPLE 7:

Now we have five more thought experiments, this time for all forms of AM. These have to do with bandwidth. Refer to Figure 4.

If the frequency of the modulating signal is doubled, what happens to the bandwidth? It doubles.

If the frequency of the carrier signal is doubled, what happens to the bandwidth? Nothing.

If the voltage of the modulating signal is doubled, what happens to the bandwidth. Nothing.

If the voltage of the carrier signal is doubled, what happens to bandwidth. Nothing.

If the modulation index is doubled, what happens to bandwidth. Nothing.

EXAMPLE 8:

These are the last thought experiments, they concern transmission efficiency a. Refer to equation 34.

If the frequency of the modulating signal is doubled, what happens to α? Nothing.

If the frequency of the carrier signal is doubled, what happens to α? Nothing.

If the voltage of the modulating signal is doubled, what happens to α?

It gets larger.

If the voltage of the carrier signal is doubled, what happens to α?

It gets smaller.

If the modulation index is doubled, what happens to α? It gets larger.

SINGLE SIDEBAND (SSB) MODULATION

The SSB signal is just the upper or lower sideband only of the DSB signal. SSB can be generated either with a suppressed carrier, SSB-SC, or with carrier energy present, SSB-LC. Figure 13 illustrates the frequency domain generation process to produce SSB-LC.

SSB has one major advantage over DSB: The bandwidth is not doubled. On the other hand, the power of the SSB modulated signal is less than what would be obtained using DSB modulation:

P_SSB-LC = P_c + P_usb (41)

P_SSB-SC = P_usb (42)

The equations for the power in the SSB modulated signals can be written by just removing that component added for the LSB:

, (43)

This can be rearranged to give an expression with the modulating voltage instead of the carrier voltage:

(44)

By just considering only the single sideband, it is possible to find an expression for the SSB-SC case as well:

(45)

This is exactly half of what was obtained in equation 11 for the DSB-SC situation. Figure 14 illustrates just how the SSB-LC signal would look like on an oscilloscope.

One way of looking at SSB-LC is to realize it is the same as DSB-LC except the maximum modulation index is 0.50, or 50% percent modulation. The equations below relate the peak voltages of the SSB modulated waveform to the peak voltages of the modulating and carrier signal, as well as the modulation index:

(46)

(47)

(49)

(50)

VESTIGIAL SIDEBAND (VSB) MODULATION

The vestigial sideband (VSB) modulation works by combining some small part of one of the sidebands together with the entire other one. DSB transmits the both entire sidebands, SSB transmits only one entire sideband, and VSB transmits one entire sideband and some vestige of the other. The bandwidth used by VSB will lie between that calculated for DSB and SSB.

The situations where VSB is desirable are where the input signal is very wide and there is a need to conserve frequency or spectrum space. The bandwidth of a VSB signal is between 1 and 2 times the original modulating signal bandwidth, depending on how much of the vestige sideband is transmitted. VSB is usually chosen when bandwidth conservation is critical and SSB is not considered appropriate because it is too expensive to implement.

VSB modulation does not have a symmetric spectrum about the carrier frequency. In the application in which VSB transmits with a large carrier, envelope detection can be used, as is the case for broadcast TV, shown by the very large video carrier energy. Other applications might not transmit the carrier, and the techniques used with DSB-SC demodulation are required.

SIGNAL-TO-NOISE (S/N) COMPARISON

DSB-SC S/N

The expression for calculating signal-to-noise ratio is:

The above relationship defines signal-to-noise ratio for any modulation scheme. By modifying it to reflect the relative value of the signal, we can find a specific expression for each of the amplitude modulations considered. For DSB-SC, this results in the following expression

(52)

DSB-LC S/N

The signal-to-noise comparison will be made only for the case of 100% modulation. In the 100% case, the signal envelope of the DSB-LC signal is twice the amplitude of an equivalent DSB-SC signal. This means that the signal-to-noise ratio for a DSB-LC signal is twice that of the SC signal. Therefore, with equivalent transmit powers, the 100% modulated DSB-LC approach will yield twice the signal-to-noise performance of a DSB-SC signal. That is expressed below by the addition of 3 dB to the S/N value found for the DSB-SC situation:

(53)

SSB S/N

In the SSB-SC case, the signal strength is twice what it was in the DSB case because only one side of the band is being transmitted. Therefore, the signal-to-noise performance of an SSB-SC will be twice that of a DSB-SC signal.

In the SSB-LC case, the performance should be somewhat better than the DSB case, because of the savings of transmitting the extra sideband, but this factor depends on the modulation coefficient m. This is seen in the equations below:

(54)

For equivalent transmit powers, SSB-SC will exhibit half the performance that DSB-SC will. Although SSB-LC will exhibit equivalent performance for equivalent transmit powers, this result comes about because although both SSB forms of modulation feature a peak signal voltage of half the DSB-LC, each also has only one sideband, halving the total signal strength:

(55)