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# Chapter2 cont

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### Chapter2 cont

1. 1. Modulation Index & AM Power Signals
2. 2. MODULATION INDEX <ul><li>Modulation index, m is defined as a parameter, which determines the amount of modulation. </li></ul><ul><li>For AM, it is defined as a measure of the extent which a carrier voltage is varied by the modulating signal. </li></ul><ul><li>For proper AM to occur </li></ul><ul><li> V m  V c </li></ul><ul><li>Modulation index, or in percentage , </li></ul><ul><li>By this definition, we could distinguished three different types of amplitude modulation. </li></ul><ul><ul><li>Under modulated AM for m < 1 </li></ul></ul><ul><ul><li>Ideal AM for m = 1 </li></ul></ul><ul><ul><li>Over modulated AM for m > 1 </li></ul></ul>
3. 3. <ul><li>m < 1 : under modulation </li></ul><ul><ul><li>V m < V c </li></ul></ul><ul><ul><li>signal strength obtained at the receiver is not exactly the same as the signal strength at the transmitter. </li></ul></ul><ul><ul><li>No distortion to the signal, just reduced signal strength. </li></ul></ul>Figure 2.4 m < 1 , under modulation.
4. 4. <ul><li>m=1 : ideal modulation </li></ul><ul><ul><li>V m = V c </li></ul></ul><ul><ul><li>will produce greatest output at the receiver without distortion </li></ul></ul><ul><ul><li>maximum info signal amplitude is transmitted </li></ul></ul><ul><ul><li>more info signal power is transmitted  producing stronger, more intelligible signal </li></ul></ul><ul><ul><li>hard to achieve especially when the modulating signal amplitude varies randomly over a wide range – only the peak of the signal will produce 100% modulation. </li></ul></ul>Figure 2.5 m = 1 , ideal modulation.
5. 5. <ul><li>m > 1 : over modulation </li></ul><ul><ul><li>V m > V c </li></ul></ul><ul><ul><li>cause distortion </li></ul></ul><ul><ul><li>negative peaks have been clipped off. </li></ul></ul><ul><ul><li>The original shape of the signal is destroyed. </li></ul></ul>Figure 2.6 m > 1, over modulated AM
6. 6. <ul><li>Modulation index can be calculated directly from the AM wave. Figure 2.7(a) shows the measurement of modulation index using the minimum and maximum value while Figure 2.7(b) using peak-to-peak value. </li></ul>Figure 2.7(a) Measuring m using max and min value
7. 7. Figure 2.7(b) Measuring m using peak-to-peak value and the form of trapezoidal A B
8. 8. Example 2.3 <ul><li>A carrier signal with a peak voltage of 2.0V is amplitude modulated with a 10kHz sine wave. The modulation voltage has an effective value of 750mV. Compute the following: </li></ul><ul><ul><li>The percent modulation, m </li></ul></ul><ul><ul><li>The instantaneous voltage of the positive and negative envelope when the 10kHz sine has completed 68µs of its cycle. </li></ul></ul><ul><ul><li>Illustrate the resulting AM waveform </li></ul></ul>
9. 9. Answers <ul><ul><li>Modulation index </li></ul></ul><ul><ul><li>Voltage at t = 68µs </li></ul></ul><ul><ul><li>Therefore the negative envelope is -1.04V </li></ul></ul>
10. 10. POWER FOR AM <ul><li>The equation of amplitude modulation and the definition of modulation index can be combined to show the amount of power in a carrier and total signal: </li></ul>
11. 11. AM POWER TRANSMISSION <ul><li>We have seen that the power in AM is equal to the carrier power and the sideband power. </li></ul><ul><li>Total transmitted power, P T is equal to the sum of the carrier power ( P C ) and both of the sidebands ( P USB , P LSB ) </li></ul>
12. 12. <ul><li>The term R correspond to the fact that antenna (which is a load) will have its own impedance and dissipate power for a resistor is </li></ul><ul><li>In the previous discussion, we assume the impedance of the antenna is unity ( R = 1) . </li></ul><ul><li>From the sideband voltage </li></ul>
13. 13. <ul><li>Upper and lower sidebands power are equal, therefore </li></ul>
14. 14. <ul><li>If m = 1 , </li></ul><ul><li>In practice, carrier signal will be modulated by several signals simultaneously. The total modulation index or effective modulation index has to be calculated. </li></ul>for m < 1 , P LSB + P USB are even less !!
15. 15. <ul><li>A 400 W carrier is modulated to a depth of 75%. Calculate the total power in the modulated wave </li></ul><ul><li>Solution: </li></ul>Example 2.4
16. 16. Example 2.5 <ul><li>An AM broadcast station’s peak carrier voltage of 2 kV has been amplitude modulated to an index of 75% with a 2 kHz test tone. The station broadcast frequency is 810 kHz . Compute the following: </li></ul><ul><ul><li>The lower and upper sidebands frequencies, f LSB and f USB </li></ul></ul><ul><ul><li>The peak modulation voltage, V m </li></ul></ul><ul><ul><li>The peak lower and upper sideband voltages, V LSB and V USB </li></ul></ul><ul><ul><li>The maximum signal amplitude, V max </li></ul></ul>
17. 17. Example 2.6 <ul><li>A spectrum analyzer with an input impedance of 50  is used to measure the power spectrum of an AM signal at the output of a preamplifier circuit. The AM signal has been modulated with a sine wave. The effective power P C is 745 mW , and each sideband, P USB and P LSB is 125 mW . Compute the following: </li></ul><ul><ul><li>The total effective power, P T </li></ul></ul><ul><ul><li>The peak carrier voltage, V C </li></ul></ul><ul><ul><li>The modulation index, m , and the percentage of modulation index M </li></ul></ul><ul><ul><li>The modulation voltage V m </li></ul></ul><ul><ul><li>The lower and upper sideband voltages, V LSB and V USB </li></ul></ul><ul><ul><li>Sketch the waveform that you would see with an oscilloscope if it were placed in parallel with the spectrum analyzer </li></ul></ul>
18. 18. MODULATION BY SEVERAL SINE WAVES <ul><li>In practice, modulation of a carrier by a several sine waves simultaneously is the rule rather than exception. </li></ul><ul><li>To calculate the resulting power </li></ul><ul><ul><li>Let V 1 , V 2 and V 3 etc. be the amplitude of the information signals, the resultant voltages, V T becomes </li></ul></ul><ul><ul><li>dividing both sides by V C </li></ul></ul>
19. 19. <ul><li>The equation of the total current and carrier current is derived from the total power equation :- </li></ul>(4.26)
20. 20. Example 2.7 <ul><li>Q. The antenna current of an AM transmitter is 12A when unmodulated but increases to 13A when modulated. Calculate the index percentage. </li></ul><ul><li>A. 59% </li></ul>
21. 21. Example 2.8 <ul><li>Q. The antenna of an AM transmitter is 8 A when only the carrier is sent, but it increases to 8.93 A when the carrier is modulated by a single sine wave. Find the percentage of modulation and the antenna current when the percentage of modulation changes to 0.8 . </li></ul>
22. 22. AM Circuits
23. 23. AM CIRCUITS <ul><li>2 ways to produce AM :- </li></ul><ul><li>Analog Multiplication </li></ul><ul><ul><li>multiply carrier by a gain or attenuation factor that varies with the modulating signal. </li></ul></ul><ul><ul><li>eg. : analog multiplier </li></ul></ul><ul><li>Non-linear Mixing </li></ul><ul><ul><li>linearly mix or algebraically add the carrier and modulating signal and then apply the composite signal to a nonlinear device or circuit </li></ul></ul><ul><ul><li>eg. : the simplest circuit is a resistive mixing network + diode rectifier + LC tuned circuit </li></ul></ul>
24. 24. Analog Multiplication <ul><li>AM is produced by multiplying the carrier by a factor = 1 + modulating sine wave. </li></ul><ul><li>Device like amplifiers or voltage dividers (PIN diodes) can create the gain or attenuation that varies with the modulating signal and hence produces the AM by passing the carrier through the device. </li></ul>
25. 25. Non-Linear Mixing <ul><li>Linearly mix the carrier and modulating signals. </li></ul><ul><li>A composite voltage is used to vary the current in the nonlinear device. </li></ul><ul><li>The current is proportional to the nonlinear device but not vary linearly with the applied voltage. </li></ul><ul><li>Nonlinear device </li></ul><ul><ul><li>FETs, diodes & BJTs </li></ul></ul><ul><ul><li>Square law response </li></ul></ul>
26. 26. Continue.. <ul><li>Figure 2.8 – Square law response curve </li></ul><ul><ul><li>Current in the device is proportional to the square of the input voltage. </li></ul></ul><ul><ul><li>Squaring the sum of the carrier and modulating signals produces the classic AM equation. </li></ul></ul><ul><li>Figure 2.9 – AM with diode </li></ul><ul><ul><li>the simplest circuit is a resistive mixing network + diode rectifier + LC tuned circuit. </li></ul></ul><ul><ul><li>LC filters the unwanted higher order harmonics. </li></ul></ul>a = constant i = av 2
27. 27. Diode Modulator <ul><li>It is the oldest and the simplest modulator </li></ul><ul><li>consist of : </li></ul><ul><ul><li>resistive mixing network </li></ul></ul><ul><ul><li>diode rectifier </li></ul></ul><ul><ul><li>LC tuned circuit </li></ul></ul>Figure 2.10 Diode modulator Figure 2.11 Waveform in the diode modulator
28. 28. <ul><li>Carrier signal is applied to one input resistor </li></ul><ul><li>Modulating signal is applied to other input resistor </li></ul><ul><li>The mixed signal appear across R 3 </li></ul><ul><li>The signal are linearly mixed  algebraically added </li></ul><ul><li>Resultant wave : modulating signal seems riding on the carrier signal </li></ul><ul><li>The positive going pulses are applied to the parallel tune circuit, which will resonate at carrier frequency </li></ul><ul><li>L and C repeatedly exchange causing an oscillation at resonant frequency  it will create a –ve half cycle for every +ve input pulses </li></ul>
29. 29. <ul><li>Most commonly used in low power AM transmitter </li></ul><ul><li>Carrier signal is applied to the emitter of a transmitter through T 1 </li></ul><ul><li>Modulating signal is applied to the collector </li></ul><ul><li>Also called collector modulation </li></ul><ul><li>Q 1 is a class C base amplifier </li></ul><ul><li>Collector bias is developed by V cc through R 2 and the small DC resistance of L 1 . </li></ul><ul><li>Base bias is also developed by V cc and R 2 through the voltage divider network of R 1 and R 3 </li></ul><ul><li>C 3 compensate for temperature variation </li></ul>Transistor Modulator
30. 30. Figure 2.12 Transistor modulator
31. 31. <ul><li>C 4 prevents ac signal from reaching the power supply </li></ul><ul><li>The output signal is developed across the high ac resistance of L 1 </li></ul><ul><li>Since Q 1 is biased class C, only the amplified +ve pulses of the carrier signal will appear on the output </li></ul><ul><li>-ve pulses are developed in the same manner as in the diode modulator  a pie filter circuit consisting of L 2 and C 6 / C 7 is resonant at the carrier frequency and adds to the –ve pulses </li></ul><ul><li>C 5 is the coupling capacitor and prevents the dc bias on the collector from reaching the output </li></ul><ul><li>L 2 is in parallel to C 6 and C 7 making a tank circuit that is resonant at the carrier frequency </li></ul><ul><li>This circuit also can operate as an impedance matching circuit. </li></ul>
32. 32. Other Examples of Modulators Figure 2.13 Series Modulator Fig. 2.14 PIN diode modulator
33. 33. Linear Integrated Circuit AM Modulators <ul><li>Use a unique arrangement of transistors and FETs to perform signal multiplication  ideally suited to generate AM waveforms. </li></ul><ul><li>ICs can perfectly match current flow, amplifier voltage gain, and temperature variations. </li></ul><ul><li>Also offer excellent frequency stability, symmetrical modulation characteristic, circuit miniaturization, fewer components, temperature immunity and simplicity of design and troubleshooting. </li></ul><ul><li>Disadvantages of ICs on the other hand are low output power, relatively low usable frequency range, susceptibility to fluctuation in the dc power supply, eg. XR-2206 </li></ul>
34. 34. FIGURE 2.15 XR-2206: (a) Block diagram; (b) schematic diagram; (c) output voltage-versus-input voltage curve
35. 35. FIGURE 2.16 Linear integrated-circuit AM modulator
36. 36. SINGLE SIDEBAND SUPPRESSED CARRIERS
37. 37. Suppressed Carriers <ul><li>In Fig. 2.17, the center spectrum is removed or reduced, hence the whole signal becomes double sideband suppressed carrier or DSB-SC </li></ul><ul><li>Total transmitted power is in the sidebands. </li></ul><ul><li>Instead of 2/3 of power lost in the carrier, almost all being the available power is used in the sidebands. </li></ul>Fig. 2.17 DSB AM in the frequency domain LSB USB f A Suppressed carrier - absent
38. 38. <ul><li>The algebraic sum of the 2 sinusoidal sidebands. </li></ul><ul><li>During the modulation, the carrier is suppressed, but the 2 sidebands remains (refer Fig. 2.17). </li></ul><ul><li>Repetition rate is determined by the RF carrier wave. </li></ul><ul><li>Amplitude is controlled by the level of modulating signal. </li></ul><ul><li>Output takes the shape of the modulating signal except with alternating +ve and –ve polarities that corresponds to the polarity of the carrier signal. </li></ul><ul><li>Phase transition. </li></ul><ul><li>Generated using balance modulator. </li></ul>DSB-SC
39. 39. Figure 2.18 DSBSC generation Figure 2.19 Block diagram of balanced modulator
40. 40. Figure 2.20 (a) The (magnitude) spectrum of a DSB SC AM signal for a sinusoid message signal and (b) its lower and (c) upper sidebands.
41. 41. Example 2.9 <ul><li>For a 100W total available power in the sidebands, compare the power in the sidebands when the modulation is standard AM with m = 100%, vs a SC design where 90% of the carrier power is suppressed. How many times greater is the sideband power in the suppressed carrier case? </li></ul><ul><li>For m = 1, P T = P C (1+m 2 /2) = P C (3/2) </li></ul><ul><li>P C = 66.7W & P SB =100 – 66.7 =33.3W </li></ul><ul><li>DSB-SC P C = 66.7W x 0.9 = 60.0W (reduced) </li></ul><ul><li>New P SB =33.3W +60.0W = 93.3W </li></ul><ul><li>The power ratio = P SBnew / P SBold = 2.8 </li></ul><ul><li>In dB = 4.5 dB </li></ul><ul><li>Means in AM info transmitted only 33.3% but in DSB-SC is 93.3%. </li></ul>
42. 42. Example 2.10 <ul><li>A 500 W DSB-SC system with 100% modulation suppresses 50% of the carrier and the suppressed carrier power goes to the sidebands. How much power is in the sidebands and how much is in the carrier? By how many dB has the sideband power increased? </li></ul><ul><li>With 100% modulation, final Pc = 500W and total Psb = 250W, if Pc is diverted to SB, the new Pc = 500 – 250 = 250W and the new Psb = 250 +250 = 500W. The increase in power is 500/250 = 2 = 3dB. </li></ul>
43. 43. Figure 2.21(a) A time-domain display of a DSB SC AM signal (b) A frequency-domain display of a DSB AM signal
44. 44. <ul><li>Advantages of DSBSC compared to AM </li></ul><ul><ul><li>power conservation  can be allocated to both sidebands  longer distance </li></ul></ul><ul><ul><li>easy to generate </li></ul></ul><ul><li>Disadvantages of DSBSC compared to AM </li></ul><ul><ul><li>Rarely used because the signal is difficult to recover at the receiver  requires carrier reinsertion  complex circuitry </li></ul></ul><ul><ul><li>Take the same BW as in AM  not efficient </li></ul></ul>
45. 45. Single Sideband <ul><li>In DSB, info is transmitted twice, once in each sideband </li></ul><ul><li>One sideband may be suppressed  SSBSC / SSB </li></ul><ul><li>In SSB, when no info or modulating signal is present, no RF signal is transmitted </li></ul><ul><li>In a standard AM transmitter, carrier is still transmitted even though it may not be modulated </li></ul><ul><li>Refer Figure 2.22 and 2.23. </li></ul>
46. 46. Figure 2.22 SSB signal with carrier
47. 47. Figure 2.23 SSB signal without carrier
48. 48. Example 2.11 <ul><li>Q. Given f c = 14.3 MHz and f m = 2 kHz, what is the frequency of the signal if it is being SSB modulated? </li></ul><ul><li>A. After SSB modulation, the modulated signal is at </li></ul><ul><li>f = f c + f m </li></ul><ul><li> = 14.302 MHz  USB </li></ul><ul><li> or </li></ul><ul><li>f = f c – f m </li></ul><ul><li> = 14.298 MHz  LSB either one. </li></ul><ul><li>*most signals are not pure sine wave, eg. Voice,  it will create a complex SSB signal which varies in frequency and amplitude over a narrow spectrum defined by the voice signal BW. </li></ul>