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**FREQUENCY MODULATION(FM )**

CHAPTER 3

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**FREQUENCY MODULATION (FM)**

Part 1

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FM :Introduction FM is the process of varying the frequency of a carrier wave in proportion to a modulating signal. The amplitude of the carrier wave is kept constant while its frequency and a rate of change are varied by the modulating signal.

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**FM :Introduction (cont…)**

Fig 3.1 : Frequency Modulated signal

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**FM :Introduction (cont…)**

The important features about FM waveforms are : The frequency varies. The rate of change of carrier frequency changes is the same as the frequency of the information signal. The amount of carrier frequency changes is proportional to the amplitude of the information signal. The amplitude is constant.

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**FM :Introduction (cont…)**

The FM modulator receives two signals,the information signal from an external source and the carrier signal from a built in oscillator. The modulator circuit combines the two signals producing a FM signal which passed on to the transmission medium. The demodulator receives the FM signal and separates it, passing the information signal on and eliminating the carrier signal. Federal Communication Coporation (FCC) allocation for a standard broadcast FM station is as shown in Fig.3.2

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**Figure 3.2: FM frequency allocation by FCC**

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**Analysis of FM Mathematical analysis: Let message signal: (3.1)**

And carrier signal: (3.2) Where carrier frequency is very much higher than message frequency.

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**Analysis of FM (cont’d)**

In FM, frequency changes with the change of the amplitude of the information signal. So the instantenous frequency of the FM wave is; (3.3) K is constant of proportionality

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**Analysis of FM(cont’d)**

Thus, we get the FM wave as: 3.4 Where modulation index for FM is given by

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**Analysis of FM(cont’d)**

Frequency deviation: ∆f is the relative placement of carrier frequency (Hz) w.r.t its unmodulated value. Given as:

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FM(cont’d) Therefore:

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Example 3.1 FM broadcast station is allowed to have a frequency deviation of 75 kHz. If a 4 kHz (highest voice frequency) audio signal causes full deviation (i.e. at maximun amplitude of information signal) , calculate the modulation index.

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Example 3.2 Determine the peak frequency deviation, , and the modulation index, mf, for an FM modulator with a deviation Kf = 10 kHz/V. The modulating signal to be transmitted is Vm(t) = 5 cos ( cos 10kπt).

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**Equations for Phase- and Frequency-Modulated Carriers**

Tomasi Electronic Communications Systems, 5e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.

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**FM&PM (Bessel function)**

Thus, for general equation:

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Bessel function

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B.F. (cont’d) It is seen that each pair of side band is preceded by J coefficients. The order of the coefficient is denoted by subscript m. The Bessel function can be written as N=number of the side frequency M=modulation index

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**Bessel Functions of the First Kind, Jn(m) for some value of modulation index**

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B.F. (cont’d)

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**Representation of frequency spectrum**

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**Angle Modulation Part 2 FM Bandwidth Power distribution of FM**

Generation & Detection of FM Application of FM

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FM Bandwidth Theoretically, the generation and transmission of FM requires infinite bandwidth. Practically, FM system have finite bandwidth and they perform well. The value of modulation index determine the number of sidebands that have the significant relative amplitudes If n is the number of sideband pairs, and line of frequency spectrum are spaced by fm, thus,, the bandwidth is: For n=>1

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**FM Bandwidth (cont’d) Estimation of transmission b/w;**

Assume mf is large and n is approximate mf +2; thus Bfm=2(mf +2)fm = (1) is called Carson’s rule

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Deviation Ratio (DR) The worse case modulation index which produces the widest output frequency spectrum. Where ∆f(max) = max. peak frequency deviation fm(max) = max. modulating signal frequency

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Example 3.3 An FM modulator is operating with a peak frequency deviation =20 kHz.The modulating signal frequency, fm is 10 kHz, and the 100 kHz carrier signal has an amplitude of 10 V. Determine : a) The minimum bandwidth using Bessel Function table. b)The minimum bandwidth using Carson’s Rule. Sketch the frequency spectrum for (a), with actual amplitudes.

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Example 3.4 For an FM modulator with a modulation index m=1, a modulating signal Vm(t)=Vm sin(2π1000t), and an unmodulated carrier Vc(t) = 10sin(2π500kt), determine : a) Number of sets of significant side frequencies b) Their amplitudes c) Draw the frequency spectrum showing their relative amplitudes.

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Example 3.4 (solution) a) From table of Bessel function, a modulation index of 1 yields a reduced carrier component and three sets of significant side frequencies. b) The relative amplitude of the carrier and side frequencies are Jo = 0.77 (10) = 7.7 V J1 = 0.44 (10) = 4.4 V J2 = 0.11 (10) = 1.1 V J3 = 0.02 (10) = 0.2 V

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**Example 3.4 (solution) : FREQUENCY SPECTRUM**

J J2 J JO J J J3 7.7 V 4.4V 1.1V 0..2

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FM Power Distribution As seen in Bessel function table, it shows that as the sideband relative amplitude increases, the carrier amplitude,J0 decreases. This is because, in FM, the total transmitted power is always constant and the total average power is equal to the unmodulated carrier power, that is the amplitude of the FM remains constant whether or not it is modulated.

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**FM Power Distribution (cont’d)**

In effect, in FM, the total power that is originally in the carrier is redistributed between all components of the spectrum, in an amount determined by the modulation index, mf, and the corresponding Bessel functions. At certain value of modulation index, the carrier component goes to zero, where in this condition, the power is carried by the sidebands only.

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**Average Power The average power in unmodulated carrier**

The total instantaneous power in the angle modulated carrier. The total modulated power

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Example 3.5 a) Determine the unmodulated carrier power for the FM modulator and condition given in example 3.4, (assume a load resistance RL = 50 Ώ) b) Determine the total power in the angle modulated wave.

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Example 3.5 : solution a) Pc = (10)(10)/(2)(50) =1 W b)

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**Generation of FM Two major FM generation: Direct method:**

straight forward, requires a VCO whose oscillation frequency has linear dependence on applied voltage. Advantage: large frequency deviation Disadvantage: the carrier frequency tends to drift and must be stabilized. example circuit: Reactance modulator Varactor diode

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**Generation of FM (cont’d)**

ii) Indirect method: Frequency-up conversion. Two ways: Heterodyne method Multiplication method One most popular indirect method is the Armstrong modulator

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**Armstrong modulator Integrator Balanced modulator Down converter**

Frequency multiplier (x n) Crystal oscillator Phase shifter Vc(t) fc Vm(t) fm

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**Armstrong modulator For example: Let fm =15Hz and fc= 200kHz**

At frequency deviation= 75kHz,it need a frequency multiplication by a factor, n, n=75000/15=5000; So it need a chain of four triplers (34) and six doublers (26), ie:n= (34) x (26)=5184, But, n x fc=5000 x 200kHz=1000MHz So, down converter with oscillating frequency=900MHz is needed to put fc in the FM band of 88MHz-108MHz

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**FM Detection/Demodulation**

FM demodulation is a process of getting back or regenerate the original modulating signal from the modulated FM signal. It can be achieved by converting the frequency deviation of FM signal to the variation of equivalent voltage. The demodulator will produce an output where its instantaneous amplitude is proportional to the instantaneous frequency of the input FM signal.

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FM detection (cont’d) To detect an FM signal, it is necessary to have a circuit whose output voltage varies linearly with the frequency of the input signal. The most commonly used demodulator is the PLL demodulator. Can be use to detect either NBFM or WBFM.

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**PLL Demodulator fVc0 V0(t) VCO FM input Phase detector Low pass filter**

Amplifier fVc0 VCO Vc(t)

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PLL Demodulator The phase detector produces an average output voltage that is linear function of the phase difference between the two input signals. This low frequency component is selected by LPF. After amplification, part of the signal is fed back through VCO where it results in frequency modulation of the VCO frequency. When the loop is in lock, the VCO frequency follows or tracks the incoming frequency.

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**PLL Demodulator Let instantaneous freq of FM Input, fi(t)=fc +k1vm(t),**

and the VCO output frequency, f VCO(t)=f0 + k2Vc(t); f0 is the free running frequency. For the VCO frequency to track the instantaneous incoming frequency, fvco = fi; or

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**PLL Demodulator f0 + k2Vc(t)= fc +k1vm(t), so,**

If VCO can be tuned so that fc=f0, then Where Vc(t) is also taken as the output voltage, which therefore is the demodulated output

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Comparison AM and FM Its the SNR can be increased without increasing transmitted power about 25dB higher than in AM Certain forms of interference at the receiver are more easily to suppressed, as FM receiver has a limiter which eliminates the amplitude variations and fluctuations. The modulation process can take place at a low level power stage in the transmitter, thus a low modulating power is needed. Power content is constant and fixed, and there is no waste of power transmitted There are guard bands in FM systems allocated by the standardization body, which can reduce interference between the adjacent channels.

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Application of FM used by most of the field VHF portable, mobile and base radios in exploration use today. It is preferred because of its immunity to noise or interference and at the frequencies used the antennas are of a reasonable size.

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**Summary of angle modulation -what you need to be familiar with**

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Summary (cont’d)

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**Summary (cont’d) Bandwidth:**

Actual minimum bandwidth from Bessel table: b) Approximate minimum bandwidth using Carson’s rule:

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Summary (cont’d) Multitone modulation (equation in general):

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Summary (cont’d)

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**Summary (cont’d)- Comparison NBFM&WBFM**

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**Part 3 Advantages Disadvantages**

ANGLE MODULATION Part 3 Advantages Disadvantages

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Advantages Wideband FM gives significant improvement in the SNR at the output of the RX which proportional to the square of modulation index. Angle modulation is resistant to propagation-induced selective fading since amplitude variations are unimportant and are removed at the receiver using a limiting circuit. Angle modulation is very effective in rejecting interference. (minimizes the effect of noise). Angle modulation allows the use of more efficient transmitter power in information. Angle modulation is capable of handing a greater dynamic range of modulating signal without distortion than AM.

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Disadvantages Angle modulation requires a transmission bandwidth much larger than the message signal bandwidth. The capture effect where the wanted signal may be captured by an unwanted signal or noise voltage. Angle modulation requires more complex and inexpensive circuits than AM.

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**END OF ANGLE MODULATION**

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