Amplitude Modulation. Ahmad Bilal
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1 Amplitude Modulation Ahmad Bilal
2 5-2 ANALOG AND DIGITAL Analog-to-analog conversion is the representation of analog information by an analog signal. Topics discussed in this section: Amplitude Modulation Frequency Modulation Phase Modulation 5.2
3 Types of analog-to-analog modulation 5.3
4 Basics of Modulation In modulation we have two signals Message signal. Low frequency Low Energy Example voice Carrier Signal High frequency and energy
5 Basic of Modulation We know a signal with frequency can travel a large distance with small antenna size, where as a baseband signal is low in frequency and energy Question is what to do if we want to send our baseband signal to large distance
6 Answer Superimpose it
7
8
9 An AM signal is made up of a carrier (with constant frequency) in which its amplitude is changed (modulated) with respect to the input signal (modulating signal). The modulating signal (message signal) causes the carriers amplitude to change with time. This resulting shape of the carrier is called the envelope. Note the envelope has the shape of a sine wave.
10 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.
11 Carrier Signal Carrier signal is represented as C (t) = C sin (ωct) Where: carrier frequency in Hertz is equal to ωc / 2 π C is the carrier amplitude
12 The Modulated wave M(t) M(t)*c(t) C(t) C(t)
13 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 V m Cos ( ( m - c ) t ) + ½ V m Cos (( m + c )t)
14 Things to Note ½ V m Cos ( ( m - c ) t ) + ½ V m Cos (( m + c )t)) Frequency range has shifted Amplitude of signal have become half
15 Representation of any sinusoidal wave So cos(1000t) = pi (δ (w-1000)+ pi (δ (w+1000) π -1000Hz 1000Hz
16 Things to Note ½ V m Cos ( ( m - c ) t ) + ½ V m 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 1V m Cos ((11000)) Means in frequency domain signal will exist on hz and hz infrequency domain signal will exist on hz and hz Similarly in Fourier Mt*cos(wct) = ½ [M(wm+wc)+M(wm-wc)]
17 Things to Note M(t) =cos 1000t and c(t)=10000t The modulated signal will be ½ 1 Cos ( (-9000) t ) + ½ 1V m Cos ((11000)) Means in frequency domain signal will exist on hz and hz infrequency domain signal will exist on hz and hz π/ hz -9000Hz 9000Hz HZ
18 Things to Note π/ Hz 9000Hz Final Result of Modulated Signal Note the signal has exact two copies one with positive frequency components and other With negative frequency components
19 As the modulated signal consist of two copies of exact same signal. We may study any one, the properties for other will be same π/2 Wc-Wm Lower Side band Wc+Wm Upper side band Wc =10000 Hz 9000Hz 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
20 The whole Picture
21 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
22
23 Band width calculation Wc-Wm Wc+Wm 9000Hz Wc =10000 Hz The message signal had a bandwidth of 1000 hz Can u derive the bandwidth relation of message signal and modulated signals
24 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.
25
26 Class Task : Synchronous Detection and Coherent Detection Modulated signal= ½ 1 Cos ( (-9000) t ) + ½ 1V m Cos ((11000)) Draw Frequency Domain spectrum for Modulated signal x cos t
27 Modulators Multiplier modulators Two signals are directly multiplied together. Very difficult to implement (non linear behavior ) Very expensive
28 Balanced Modulator
29 Also called non liner modulator NL represents any non linear devices that behaves as ax(t) +bx 2 (t) M(t) + cos(wct) ax(t) +bx 2 (t) am(t)+acos(wct)+ b(m 2 (t)+cos 2 (wc(t)+2m(t)cos(wc(t))) M(t) - cos(wct) ax(t) +bx 2 (t) am(t)-acos(wct)+ b(m 2 (t)+cos 2 (wc(t)-2m(t)cos(wc(t)))
30 Also called non liner modulator NL represents any non linear devices that behaves as ax(t) +bx 2 (t) am(t)+acos(wct)+ b(m 2 (t)+cos 2 (wc(t)+2m(t)cos(wc(t))) - am(t)-acos(wct)+ b(m 2 (t)+cos 2 (wc(t)-2m(t)cos(wc(t))) am(t)+acos(wct)+ b(m 2 (t)+bcos 2 (wc(t)+2bm(t)cos(wc(t))) -am(t)+acos(wct)-bm 2 (t)-bcos 2 (wc(t)+2bm(t)cos(wc(t))) =final Out put 2acos(wct(t))+4bm(t)(cos(wc(t))
31 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
32 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
33 Choosing Values
34 Envelop Detection Condition A+m(t)>0
35 Modulation Index μ Amplitude of Modulating Signal Amplitude of Carrier Signal μ= Vm Vc 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
36 Effect mo Modulation Index
37
38
39 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.
40 Other Formulas for Calculating Modulating Index For any modulated wave, we may find, Vm and Vc as
41 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.
42 Power of AM Waves 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
43 For Power calculation, we need Peak Values RMS values. Can be obtained multiplying the voltage value to Using Power formula
44 Power of AM wave
45
46 if the carrier of an AM transmitter is 1000 W and it is modulated 100 percent (m = 1), the total AM power i
47 Solve for 70 percent modulated 250-W carrier, the total power in the composite AM signal is
48 Power Factor in Terms Of I 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=I 2 R : So our equation will become P t =I t2 R Where I t is
49 Quick Example The total output power of an 85 percent modulated AM transmitter, whose unmodulated carrier current into a 50 Ohm antenna load is 10 Ampere Hmm easy Just Apply the formula I t =I c (1+(m 2 /2)) Where Ic = 10 m =0.85 I t =10 (1.36) = Amp NOW CALCULATE POWER
50 Find Modulation Factor
51 Find Modulation Factor I t =I c (1+(m 2 /2)) Where Ic = 10 I t =10 (1.36) = Amp NOW CALCULATE Modulation Index
52 Question
53 Question Remember you can also calculate via
54
55 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?
56
57 Single Side Band In DSB the basic information is transmitted twice. (Practically speaking there is no advantage of doing so) So one side band may be suppressed Advantages 1. Occupy Less spectrum (more signals can be transmitted) 2. Strong signal (No power for double bands and carrier ) 3. Less fading
58 Single Side Band Typical AM Suppressed LSB FC USB
59 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.
60 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.
61 SSB Power 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.
62 SSB Power
63 Applications SSB Telephone systems Two way radio (for military apps) DSB FM and TV Audio Broad casting
64 ,Assume that a voice signal produces a 360-V, peakto-peak signal across a 50-V load. The rms voltage is times the peak value, and the peak value is onehalf the peak-to-peak voltage.
65 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
66
67 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 onethird of the PEP value with typical human speech
68 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.
69 An SSB transmitter produces a peak-to-peak voltage of 178 V across a 75-V antenna load. What is the PEP
70 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
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