![Analysis of the AM wave cont..
Vm
Vmin
Vc Vmax
Envelope = -[Vc + Vm sin mt]
Envelope = [Vc + Vm sin mt]](https://image.slidesharecdn.com/27662804lecture4nee2107-210919114356/85/2766280-4-lecture4nee2107-9-320.jpg)
![Analysis of the AM wave cont..
• Frequency spectrum of the AM wave.
eenv represents the peak envelope of the AM waveform at any instant in time ‘t’.
The carrier frequency c as a peak value represents an AM waveform using the
envelope.
Hence eAM = eenv sin = eenv sin c t
From earlier derivations, 𝑒𝐴𝑀 = 𝑉
𝑐 1 + m sin 𝜔𝑚𝑡 sin𝜔𝑐𝑡
𝑒𝐴𝑀 = 𝑉
𝑐𝑠𝑖𝑛 𝜔𝑐𝑡 + 𝑚𝑉
𝑐𝑠𝑖𝑛𝜔𝑐𝑡𝑠𝑖𝑛 𝜔𝑚t
The trigonometric identity sin sin =
1
2
[cos 𝛼 − 𝛽 − cos 𝛼 + 𝛽 ]
By applying the same identity to the eAM equation above we get
𝑒𝐴𝑀 = 𝑉
𝑐𝑠𝑖𝑛𝜔𝑐𝑡 + 𝑚
𝑉𝑐
2
cos(𝑐 − 𝑚)𝑡 − 𝑚
𝑉𝑐
2
cos(𝑐 + 𝑚)𝑡
Carrier frequency fc
Lower sideband (LSB) fc - fm Upper sideband (USB) fc + fm](https://image.slidesharecdn.com/27662804lecture4nee2107-210919114356/85/2766280-4-lecture4nee2107-13-320.jpg)
![Analysis of the AM wave cont..
Power distribution in the AM waveform
• The total effective(rms) power, PT in the AM wave is the sum of the effective carrier
power level Pc and the effective sideband power levels PLSB and PUSB
• Hence 𝑃𝑇 = 𝑃𝑐 + 𝑃𝐿𝑆𝐵 + 𝑃𝑈𝑆𝐵
• =
𝑉2
𝑐𝑟𝑚𝑠
𝑅
+
𝑉2
𝐿𝑆𝐵𝑟𝑚𝑠
𝑅
+
𝑉2
𝑈𝑆𝐵𝑟𝑚𝑠
𝑅
• =
(0.707𝑉𝑐)2
𝑅
+
(0.707𝑉𝐿𝑆𝐵 )2
𝑅
+
(0.707𝑉𝑈𝑆𝐵) 2
𝑅
• =
𝑉𝑐
2
2𝑅
+
𝑉𝐿𝑆𝐵
2
2𝑅
+
𝑉2
𝑈𝑆𝐵
2𝑅
Remember VLSB = VUSB =
𝑚
2
Vc hence PLSB = PUSB =
[0.707(
𝑚𝑉𝑐
2
)]2
𝑅
=
𝑚2𝑉2
𝑐
8𝑅
The total power in the carrier Pc is
𝑉𝑐
2
2𝑅
hence PLSB = PUSB =
𝑚2𝑃𝑐
4](https://image.slidesharecdn.com/27662804lecture4nee2107-210919114356/85/2766280-4-lecture4nee2107-17-320.jpg)
2766280 4 lecture4nee2107
- 1. Telecommunications NEE2107 Lecture 4 Dr Horace KING Office: G216 Phone:99194696 Email: horace.king@vu.edu.au
- 2. Modulation techniques • Modulation : A process through which audio, video, image or text information is added to an electrical or optical carrier signal to be transmitted over a telecommunication or electronic medium. Modulation enables the transfer of information on an electrical signal to a receiving device that demodulates the signal to extract the blended information. • Amplitude modulation (AM) Is defined as a carrier frequency whose amplitude is varied in proportion to the instantaneous amplitude of a modulating voltage. The modulating voltage is referred to as the intelligence. The carrier frequency, or carrier of the modulating voltage, is typically much higher than that of the modulating voltage o Usually a radio frequency(RF) signal in the Mid-Frequency range of 300KHz to 3MHz The frequency of the modulating voltage, is typically an audio frequency (AF) signal o Usually in the Frequency range of 20Hz to 20KHz
- 4. AM cont.. To produce an AM wave requires a modulator which produces the sum and difference products of the carrier and the modulation frequency. A modulator is a nonlinear device capable of producing a mixing action thus resulting in the sum and difference products. Transistors and diodes are nonlinear devices. Combining two frequencies through a nonlinear device produces a mixing action resulting in harmonics as well as the sum and difference frequencies of the original signals.
- 5. AM cont.. • In a simple AM modulator, a summing amplifier is used to electrically sum the modulation and carrier frequencies. • The diode and the resistor are used to clip the negative half of the composite signal to produce a pulsating DC signal rich in harmonic content. • When a single pulse (vo ) is fed to a resonant tank circuit, it causes a damped oscillation to occur at the natural resonant frequency of the tank. • A damped oscillation is a sinusoid function with an exponentially decaying envelope. • Over time, decaying is caused by resistive losses in the tank components. V1 t V2 t vo Vo t vo t Gnd Ae-t
- 6. AM cont.. • If a parallel tank circuit is put at the diode output, the pulsating DC signal will sustain the oscillatory effect thereby producing complete sinusoids whose amplitudes are proportional to the amplitude of each pulse. • This is known as a flywheel effect and effectively results into an AM wave. • The resonant tank circuit filters out the undesirable harmonics and preserves the AM frequency components! • Since the diode is a passive device, it offers no gain to the AM wave! A transistor is a device that will add the element of gain to the signal. • Tank circuit vo t v
- 7. AM cont.. Replacing the diode with a transistor. • The amplitudes of the Carrier and modulating frequencies are adjusted to turn on the transistor in its nonlinear region during the positive peaks of the cycle. • During the negative peaks of the cycle, the transistor turns off, thus producing the pulsating DC signal at the tank circuit. • The AM wave is produced at the collector output of the transistor • Transistor modulator circuit em ec å - + - vcc Vo em ec Vo t
- 8. Analysis of the AM wave A carrier frequency, fc (angular velocity c ) is to be amplitude modulated with a sine wave fm (angular velocity m ). Therefore, the instantaneous voltage em of the sine wave used to modulate the carrier frequency is given as em = Vm sin m t where m is 2fm and Vm is the peak amplitude of the modulating signal. The instantaneous voltage of the carrier frequency is given as ec = Vc sin c t where c is 2fc and Vc is the peak amplitude of the carrier. The components of an AM signal are: • 𝑒𝐴𝑀 = 𝑉 𝑐𝑠𝑖𝑛𝑐𝑡 + 𝑚𝑉𝑐 2 cos 𝜔𝑐 + 𝜔𝑚 𝑡 − 𝑚𝑉𝑐 2 cos(𝜔𝑐 + 𝜔𝑚)𝑡 • carrier frequency Lower sideband Upper sideband • m in the equation is the modulation index or factor given by 𝑚 = 𝑉𝑚 𝑉𝑐 • Expressed as a percentage it becomes 𝑀 = 𝑉𝑚 𝑉𝑐 × 100
- 9. Analysis of the AM wave cont.. Vm Vmin Vc Vmax Envelope = -[Vc + Vm sin mt] Envelope = [Vc + Vm sin mt]
- 10. Analysis of the AM wave cont.. • The modulation index m was given as 𝑚 = 𝑉𝑚 𝑉𝑐 • As a percentage M = 𝑉𝑚 𝑉𝐶 100 • From the AM wave, because ‘m’ is symmetrical about the carrier frequency, therefore Vmax – Vc = Vc - Vmin • And 𝑉 𝑚 = 𝑉𝑚𝑎𝑥−𝑉𝑚𝑖𝑛 2 and 𝑉 𝑐 = 𝑉𝑚𝑎𝑥+𝑉𝑚𝑖𝑛 2 dividing Vm by Vc gives us • 𝑚 = 𝑉𝑚𝑎𝑥−𝑉𝑚𝑖𝑛 𝑉𝑚𝑎𝑥+𝑉𝑚𝑖𝑛 and is equal to 𝑉𝑚𝑎𝑥𝑝−𝑝−𝑉𝑚𝑖𝑛𝑝−𝑝 𝑉𝑚𝑎𝑥𝑝−𝑝+𝑉𝑚𝑖𝑛𝑝−𝑝
- 11. Analysis of the AM wave cont.. Envelope of the AM waveform. • The amplitude of the AM waveform (eenv ) is the envelope (peak values of each cycle). • The envelope outlines the peaks and troughs of the waveform and hence the envelope can be a positive or negative signed value. • Hence for Sine wave modulation eenv = Vc + em = Vc + Vm sin m t • Remember 𝑚 = 𝑉𝑚 𝑉𝑐 ! Therefore Vm = mVc and by substitution in the equation above • eenv =Vc + mVc sin m t = Vc (1 + m sin m t) the positive envelope = -Vc (1 + m sin m t) the negative envelope
- 12. Worked example 1 A carrier signal with a peak voltage of 2V is amplitude modulated with a 10 KHz sine wave. The modulation voltage has an effective value(rms) of 750mV. Determine (i)the percent modulation index m (ii) the instantaneous voltage of the positive and negative envelope when the 10KHz sine wave has completed 68s of its cycle (iii) draw the resultant AM wave form. V t Vc=2Vp eenv=-1.04V eenv =+1.04V t=68s Vm =750mV
- 13. Analysis of the AM wave cont.. • Frequency spectrum of the AM wave. eenv represents the peak envelope of the AM waveform at any instant in time ‘t’. The carrier frequency c as a peak value represents an AM waveform using the envelope. Hence eAM = eenv sin = eenv sin c t From earlier derivations, 𝑒𝐴𝑀 = 𝑉 𝑐 1 + m sin 𝜔𝑚𝑡 sin𝜔𝑐𝑡 𝑒𝐴𝑀 = 𝑉 𝑐𝑠𝑖𝑛 𝜔𝑐𝑡 + 𝑚𝑉 𝑐𝑠𝑖𝑛𝜔𝑐𝑡𝑠𝑖𝑛 𝜔𝑚t The trigonometric identity sin sin = 1 2 [cos 𝛼 − 𝛽 − cos 𝛼 + 𝛽 ] By applying the same identity to the eAM equation above we get 𝑒𝐴𝑀 = 𝑉 𝑐𝑠𝑖𝑛𝜔𝑐𝑡 + 𝑚 𝑉𝑐 2 cos(𝑐 − 𝑚)𝑡 − 𝑚 𝑉𝑐 2 cos(𝑐 + 𝑚)𝑡 Carrier frequency fc Lower sideband (LSB) fc - fm Upper sideband (USB) fc + fm
- 14. Analysis of the AM wave cont. • The total composite AM signal comprised of two side bands and a carrier frequency, is normally referred to as double sideband full carrier (DSBFC) • The peak value terms of the LSB and the USB are equal i.e. 𝑚𝑉𝑐 2 = 𝑉𝐿𝑆𝐵 = 𝑉𝑈𝑆𝐵 • When all three frequency components are in phase they add together linearly and form the maximum signal amplitude, Vmax i.e Vmax = Vc +VLSB + VUSB • Vmax is also equal to the sum of Vc and Vm hence Vm = VLSB + VUSB • Because peak sideband voltages are equal Vm = 2 VLSB = 2 VUSB
- 15. Analysis of the AM wave cont.. Vm Vc Vmax Vmax =Vc + Vm =Vc +VLSB + VUSB Vm =VLSB + VUSB = 2VLSB =2VUSB V t fLSB fc fUSB 𝑉𝐿𝑆𝐵 = 𝑚 2 𝑉 𝑐 𝑉𝑈𝑆𝐵 = 𝑚 2 𝑉 𝑐 Vc Time domain Frequency domain
- 16. Worked example 2 • Given a station’s peak carrier voltage of 2KV has been modulated to an index of 75% with a 2 KHz test tone. The station’s broadcasting frequency is 810 KHz. Determine • (i) the LSB and the USB • (ii) the peak modulation voltage • (iii) the peak VLSB and VUSB • (iv) Vmax
- 17. Analysis of the AM wave cont.. Power distribution in the AM waveform • The total effective(rms) power, PT in the AM wave is the sum of the effective carrier power level Pc and the effective sideband power levels PLSB and PUSB • Hence 𝑃𝑇 = 𝑃𝑐 + 𝑃𝐿𝑆𝐵 + 𝑃𝑈𝑆𝐵 • = 𝑉2 𝑐𝑟𝑚𝑠 𝑅 + 𝑉2 𝐿𝑆𝐵𝑟𝑚𝑠 𝑅 + 𝑉2 𝑈𝑆𝐵𝑟𝑚𝑠 𝑅 • = (0.707𝑉𝑐)2 𝑅 + (0.707𝑉𝐿𝑆𝐵 )2 𝑅 + (0.707𝑉𝑈𝑆𝐵) 2 𝑅 • = 𝑉𝑐 2 2𝑅 + 𝑉𝐿𝑆𝐵 2 2𝑅 + 𝑉2 𝑈𝑆𝐵 2𝑅 Remember VLSB = VUSB = 𝑚 2 Vc hence PLSB = PUSB = [0.707( 𝑚𝑉𝑐 2 )]2 𝑅 = 𝑚2𝑉2 𝑐 8𝑅 The total power in the carrier Pc is 𝑉𝑐 2 2𝑅 hence PLSB = PUSB = 𝑚2𝑃𝑐 4
- 18. Worked example 3 A spectrum analyser with an input impedance of 50Ω 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 carrier power, PC is 750 mW and each sideband is 120 mW. Calculate (i) The total effective power, (ii) The peak carrier voltage, (iii) the modulation index m and the percent modulation index M, (iv) The modulation voltage (v) LSB and USB voltages. (vi) Draw the waveform.
- 19. Worked example cont.. Vm Vc Vmax Vmax =Vc + Vm = 8.66 + 6.93 = 15.59 𝑚 = 𝑉𝑚 𝑉𝑐 = 6.93 8.66 = 0.8 V t fc
- 20. Analysis of the AM wave cont.. The total effective power PT representation • Can be represented using the carrier power in the sidebands where PT = PC + PLSB + PUSB • Such that 𝑃𝑇 = 𝑃𝑐 + 𝑚2𝑃𝑐 4 + 𝑚2𝑃𝑐 4 = 𝑃𝑐 + 𝑚2𝑃𝑐 2 = 𝑃𝑐(1 + 𝑚2 2 ) • Hence only knowing the carrier power and the modulation index we can compute the total effective power of an AM wave form Vc
- 21. Analysis of the AM wave cont.. Considering an AM waveform in terms of power efficiency • The condition when 100% modulation is used m = 1 and this sets the condition for maximum power utilisation in the sidebands. • From 𝑃𝑇 = 𝑃𝑐 + 𝑚2𝑃𝑐 4 + 𝑚2𝑃𝑐 4 = 𝑃𝑐 + 𝑚2𝑃𝑐 2 = 𝑃𝑐 1 + 𝑚2 2 when m = 1, PT = 1.5Pc • The equation states that at 100% modulation, the total effective power in the sidebands of an AM waveform is half the carrier power.. • Or we can say that 2/3 of the total power is dissipated by the carrier frequency Pc = PT /1.5
- 22. Worked example 4 • An AM transmitter has an effective carrier power level of 50KW. If the carrier is modulated with a sine wave, determine (i) the total effective power for a percent modulation index of 50%. (ii) The effective power in each sideband for a modulation index of 50%. (iii) The total effective transmitted power for 100% modulation. (iv) The effective power in each sideband for 100% modulation.
- 23. Analysis of the AM wave cont.. • AM waveform with 100% modulation • 𝑚 = 𝑉𝑚𝑎𝑥𝑝−𝑝− 𝑉𝑚𝑖𝑛𝑝−𝑝 𝑉𝑚𝑎𝑥𝑝−𝑝+ 𝑉𝑚𝑖𝑛𝑝−𝑝 = 𝑉𝑚𝑎𝑥𝑝−𝑝− 0 𝑉𝑚𝑎𝑥𝑝−𝑝+ 0 = 1 • Carrier modulated by a sine wave. • At 100% modulation there is no Vmin p-p • Labelled diagram is the time domain of the AM waveform with 100% modulation. • Frequency domain of the AM wave form showing the amount of carrier power in the sidebands and the centre frequency • 𝑃𝑇 = 𝑃𝑐 + 1 4 𝑃𝑐 + 1 4 𝑃 𝑐 = 𝑃𝑐 + 1 2 𝑃𝑐 Vmax p-p Vmin p-p fLSB fc fUSB t V f P 1 4 𝑃𝑐 Pc 1 4 𝑃𝑐
- 24. Analysis of the AM wave cont.. Single sideband There are inherent problems with AM more specifically double sideband full carrier (DSBFC). The amount of power contained in the carrier frequency is enormous. Modulating the carrier frequency with intelligence, or modulating voltage does not alter the amplitude or the frequency of the carrier component itself. It remains independent of the modulating voltage! It is the sidebands that change in proportion to the modulating voltage and the depth of modulation. So the intelligence resides entirely in the sidebands! Hence the power transmitted in the AM carrier frequency is essentially wasted! o Remember that the power in each sideband is one-fourth the power in the carrier for 100% modulation! Or we can say that 2/3 of the total power is dissipated by the carrier frequency Pc = PT /1.5! For modulation indices less than 100% even more power is wasted by the carrier component! LSB and USB are mirror images of each other and contain the same information with the same power levels. There is no need to transmit both! By suppressing one sideband, 50% of the power is saved! Thus overall for 100% modulation a saving of 83.33% is realised i.e. 2/3 for the carrier plus ½ of 1/3 for one of the sidebands 2/3 + 1/6 = 5/6 translating into 0.833333 and is 83.33% Therefore suppressing the carrier frequency and one of the sidebands is referred to as single sideband suppressed carrier (SSBSC) or simply SSB. Suppressing the carrier component of the AM signal so that the transmitted wave only consists of the lower and Upper sidebands is known as Double sideband Suppressed Carrier (DSBSC)
- 25. Analysis of the AM wave cont.. • Double sideband suppressed carrier (DSBSC) with sine wave modulation in time domain. • The carrier frequency of the AM wave is suppressed by a balanced modulator fc fUSB fLSB fc - fm fc + fm Suppressed carrier f P t v Suppressed carrier modulator
- 26. Analysis of the AM wave cont.. Advantages • SSB requires only half the bandwidth of the AM signal • SSB is power efficient Selection of transmitting of LSB or USB requires high Q (Q = resonant frequency/bandwidth) filters with steep attenuation characteristics
- 27. Analysis of the AM wave cont.. Filtering the SSB LSB or USB • Suppose a SSB transceiver operating in the CB band has a carrier frequency in the 7 MHz region. If the carrier frequency is modulated with a low frequency tone of 100Hz the USB and LSB are separated by only 200 Hz. • Selecting the USB for transmission would require the filter to pass the USB frequency of 7.0001MHz (fc +100Hz) and totally reject 6.9999 MHz (fc – 100Hz) • Over a range of 200Hz selectivity requirements for the filter would be a Q of 35,000! Remember Q = resonant frequency/bandwidth) or (7MHz /200Hz) and for higher carrier frequencies higher would be necessary!. • This is highly impractical as multi stage LC filters with Qs beyond 200 are difficult and impractical to design ! • For the same 200Hz range, if we lower the fc to say 100KHz our Q becomes 500, the USB can be selected and up converted to the transmitting frequency (a process known as dual-conversion). • The fm is mixed with low carrier frequency fc1 and the LSB and USB signals at the output of the balanced modulator 1 are at a frequency better suited for SSB filtering. • The USB is filtered out and mixed again with a higher carrier frequency fc2 . • The output of balanced modulator 2 now contains sideband frequencies that are separated by a frequency range compatible with SSB filters Audio Amplifier Balanced Modulator 1 Carrier Oscillator1fc1 =100KHz Bandpass Filter1 USB Balanced Modulator 2 Carrier Oscillator2, fc2=7MHz Bandpass Filter2 USB Power Amplifier Antenna Output1 Output2 100.2KHz USB 7MHz+100.1KHz f fc2 7MHz LSB 7MHz–100KHz USB 200Hz LSB 100.1KHz fc1 100 KHz 99.9KHz Output1 Output2 fm=100Hz
- 28. Quadrature Amplitude Modulation (QAM) • The technique of combining Amplitude and phase modulation is called quadrature amplitude modulation (QAM). • What is Phase modulation? • See you in the next lecture!!!!!!!!!!!!!!