Detailed Lecture Notes on Amplitude Modulation

Amplitude Modulation Problems

Problem 6 discusses finding the percent modulation given a carrier power of 8 kilowatts and sideband powers of 2 kilowatts each. The total sideband power is calculated as the sum of the lower and upper sideband powers, resulting in 4 kilowatts. The total power is then the sum of the carrier power and total sideband power, equaling 12 kilowatts. Using the formula $P<em>T = P</em>c (1 + \frac{m^2}{2})$, the percent modulation is found to be 100%.

Problem 7 involves finding the percent modulation with a total power of 600 watts and sideband powers of 75 watts each. The carrier power is calculated as $P<em>c = P</em>T - 2 \cdot P_\text{sideband}$, resulting in 450 watts. The percent modulation is then calculated using the same formula as in Problem 6, yielding approximately 81.6%.

Problem 10 explores power in upper and lower sidebands. With a given percent modulation of 70% and carrier power of 1500 watts, the power in each sideband is calculated using $P<em>{\text{LSB}} = P</em>{\text{USB}} = \frac{m^2 P*c}{4}$, resulting in 183.75 watts. The problem then considers a second condition where the percent modulation is 50%, and the power in each sideband is recalculated to be 93.75 watts.

Problem 12 calculates the total output power given a percent modulation of 85%, antenna load impedance of 50 ohms, and carrier current of 10 amperes. Using the formula $I<em>T = I</em>c \sqrt{1 + \frac{m^2}{2}}$, the total current is found to be 11.667 amperes, and the total power is calculated as $P<em>T = I</em>T^2 R$, resulting in approximately 6805 watts or 6.805 kilowatts. An alternative solution involves calculating the carrier power first and then using the formula $P<em>T = P</em>c (1 + \frac{m^2}{2})$ to find the total power, yielding a similar result of 6.806 kilowatts.

Problem 13 details calculating carrier power, total power, and sideband power, given an antenna impedance of 40 ohms, an unmodulated AM signal current of 4.8 amperes, and a modulation of 90%. The carrier power is calculated as $P<em>c = I</em>c^2 R$, resulting in 921.6 watts. The total power is calculated using $P<em>T = P</em>c (1 + \frac{m^2}{2})$, resulting in 1295 watts or 1.295 kilowatts. The total sideband power is then found by subtracting the carrier power from the total power, yielding 373.4 watts. Additionally, the problem explores the scenario where the antenna current changes from 4.8 A to 5.1 A upon modulation, asking for the percentage of modulation. Using the formula $I<em>T = I</em>c \sqrt{1 + \frac{m^2}{2}}$, the percentage of modulation is calculated to be 51%.

The section on Modulation by Several Sine Waves gives equations for calculating total modulating voltage $V<em>t = V</em>1 + V<em>2 + V</em>3 + …$ and $V<em>m (total) = \sqrt{V</em>{m1}^2 + V<em>{m2}^2 + V</em>{m3}^2 + …}$. It also provides the modulation index relationship $\frac{V<em>m}{V</em>c} = m$ and formulates $m<em>t = \sqrt{m</em>1^2 + m<em>2^2 + m</em>3^2 + …}$.

Problem 14 covers calculating total radiated power with simultaneous modulation. Given a carrier power of 9 kilowatts, a total power with initial modulation of 10.125 kilowatts, and an additional modulation index of 0.4, the initial modulation index is calculated. Then, the total modulation index with simultaneous transmission is calculated using $m<em>t = \sqrt{m</em>1^2 + m_2^2}$, resulting in 0.64. Finally, the total transmitted power is calculated using $P<em>T = P</em>c (1 + \frac{m^2}{2})$, yielding 10.84 kilowatts.

The Percentage Efficiency in AM is defined as the ratio of power in sidebands to total power, with the formula $η = \frac{P<em>{\text{LSB}} + P</em>{\text{USB}}}{P*T} * 100$, simplified to $η = \frac{\frac{m^2}{2}}{1 + \frac{m^2}{2}} * 100 = \frac{m^2}{2 + m^2} * 100$. For conventional AM, the maximum percentage efficiency is 33.3% when m = 1. Examples are provided for calculating efficiency when m = 0.5 and m = 0.3, resulting in efficiencies of 11.11% and 4.3%, respectively.

The Drawbacks of AM include that the carrier signal does not carry information, it constitutes a large portion of the total transmitted power, both sidebands contain the same information (making transmission redundant), and conventional AM is both power and bandwidth inefficient.

Double Sideband Suppressed Carrier (DSBSC) suppresses the carrier, leaving only the upper and lower sidebands, thus no power is wasted on the carrier. Formulas include $V<em>{\text{max}} = Vm$, $V{\text{min}} = 0$, and bandwidth $BW = 2f</em>m$. The instantaneous voltage is given by $V<em>t = \frac{mV</em>c}{2} cos(2π(f<em>c - f</em>m)t) - \frac{mV<em>c}{2} cos(2π(f</em>c + f_m)t)$.

Power Relations in DSBSC indicate that the carrier component is suppressed, and $P<em>t = P{\text{LSB}} + P{\text{USB}} = \frac{m^2 V</em>c^2}{8R} + \frac{m^2 V<em>c^2}{8R} = \frac{m^2 V</em>c^2}{4R} = \frac{P_c m^2}{2}$. The power saving is $\% \text{Power Saving} = \frac{\text{Total Power(AM)} - \text{Power(DSB)}}{\text{Total Power(AM)}}*100 = \frac{2}{2 + m^2} * 100$, implying that if m = 1, the power saving is 66.7%.

Example 1 (DSBSC) calculates the total power in AM and DSBSC techniques given a carrier power of 400 watts and a modulation index of 1. It also considers how much power saving in watts is achieved for DSBSC if the depth of modulation is changed to 75%. Example 2 (DSBSC) calculates the carrier power in kilowatts if you want to transmit the same message by an AM transmitter given $P_T = 1000$ Watts and $m= 60\% = 0.6$. $Pt = \frac{P<em>c m^2}{2}$ therefore $P</em>c = \frac{2PT}{m^2} = \frac{2*1000}{0.6^2}$ Watts = 5555.55 = 5.55 kW.