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EET260 Frequency Modulation. Modulation A sine wave carrier can be modulated by varying its amplitude, frequency, or phase shift. In AM, the amplitude.

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Presentation on theme: "EET260 Frequency Modulation. Modulation A sine wave carrier can be modulated by varying its amplitude, frequency, or phase shift. In AM, the amplitude."— Presentation transcript:

1 EET260 Frequency Modulation

2 Modulation A sine wave carrier can be modulated by varying its amplitude, frequency, or phase shift. In AM, the amplitude of the carrier is modulated by a low-frequency information signal. Information signal Amplitude modulated signal

3 Frequency modulation In frequency modulation (FM) the instantaneous frequency of the carrier is caused to deviate by an amount proportional to the modulating signal amplitude. instantaneous frequency changed in accordance with modulating signal frequency modulated signal

4 Phase modulation In phase modulation (PM) the phase of carrier is caused to deviate by an amount proportional to the modulating signal amplitude. Both FM and PM are collectively referred to as angle modulation. carrier phase is changed in accordance with modulating signal phase modulated signal

5 Frequency modulation Consider the equation below for a frequency modulated carrier. We will begin with a simple binary input signal. defines the instantaneous frequency center frequency frequency deviation modulating signal

6 Frequency modulation We will consider a  1-V, 1-kHz square wave as an input. Our input signal has 3 levels, and f d = 4-kHz

7 Frequency modulation input signal (  1-V, 1-kHz square wave) FM signal ( f c = 10-kHz, f d = 4-kHz) f = 10-kHz f = 14-kHz f = 6-kHz f = 14-kHz f = 6-kHz f = 14-kHz

8 Fundamental FM concepts The amount frequency deviation is directly proportional to the amplitude of the modulating signal. The frequency deviation rate is determined by the frequency of the modulating signal.  The deviation rate is the number of times per second that carrier deviates above and below its center frequency. center frequency frequency deviation modulating signal

9 Fundamental FM concepts input signal (  1-V, 1-kHz square wave)input signal (  2-V, 1-kHz square wave) Doubling the amplitude of the input doubles the frequency deviation of the carrier.

10 Fundamental FM concepts input signal (  1-V, 1-kHz square wave)input signal (  1-V, 2-kHz square wave) Doubling the frequency of the input doubles the frequency deviation rate of the carrier.

11 A transmitter operates on a carrier frequency of 915-MHz. A  1-V square wave modulating signal produces  12.5- kHz deviation the carrier. The frequency of the input signal is 2-kHz. a. Make a rough sketch of the FM signal. b. If the modulating signal amplitude is doubled, what is the resulting carrier frequency deviation? c. What is the frequency deviation rate of the carrier? Example Problem 1

12 FM with sinusoidal input Consider a sinusoidal modulating input. input signal (  1-V, 500-Hz sine wave) 500-Hz modulating signal

13 Frequency content of an FM signal What does an FM signal look like in the frequency domain? We will consider the case of a sinusoidal modulating signal.

14 Frequency content of an AM signal Amplitude modulated signal v AM Carrier signal v c (carrier frequency f c = 5-kHz)  Modulator or mixer Information signal v m ( f im = 500-Hz ) Frequency domain

15 Frequency modulation index The modulation index for FM is defined Just as in AM it is used to describe the depth of modulation achieved. From the previous example

16 Frequency analysis of FM In order to determine the frequency content of we can use the Fourier series expansion given where J n ( m f ) is the Bessel function of the first kind of order n and argument m f.

17 Expanding the series, we see that a single-frequency modulating signal produces an infinite number of sets of side frequencies. Frequency analysis of FM

18 Each sideband pair includes an upper and lower side frequency The magnitudes of the side frequencies are given by coefficients J n (m).  Although there are an infinite number of side frequencies, not all are significant. Frequency analysis of FM

19 For the case for m f = 2.0, refer to the table in Figure 5-2 to determine significant sidebands. FM spectrum for m f = 2.0

20 Bessel functions J n ( m f ) if m f = 2.0, then the side frequencies we need to consider are J 0, J 1, J 2, J 3, J 4

21 For the case for m f = 2.0, the series can be rewritten Substituting the values for J 0 (2), J 1 (2),…, J 4 (2) FM spectrum for m f = 2.0

22 From the equation, the spectrum can be plotted. What is the bandwidth of this signal? FM spectrum for m f = 2.0 fcfc f c + f m f c + 2f m f c + 4f m f c + 3f m f c - 3f m f c - f m f c - 2f m f c - 4f m

23 The bandwidth of the previous signal is More generally, the bandwidth is given where N is the number of significant sidebands. FM spectrum for m f = 2.0 f c + 4f m f c - 4f m

24 A signal v m (t) = sin (2  1000t) is frequency modulates a carrier v c (t) = sin (2  500,000t). The frequency deviation of the carrier is f d = 1000 Hz. a.Determine the modulation index. b.The number of sets of significant side frequencies. c.Draw the frequency spectrum of the FM signal. Example Problem 2

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26 FM bandwidth increases with modulation index. FM bandwidth as function of m f

27 Note the case m f = 0.25 In this special case, FM produces only a single pair of significant sidebands, occupying no more bandwidth than an AM signal. This is called narrowband FM. FM bandwidth

28 FM systems with m f <  /2 are defined as narrowband.  This is true despite the fact that only values of m f in the range of 0.2 to 0.25 have a single pair of sidebands. The purpose of NBFM is conserve spectrum and they are widely used in mobile radios. Narrowband FM

29 An approximation for FM bandwidth is given by Carson’s rule: The bandwidth given by Carson’s rule includes ~98% of the total power. Carson’s Rule

30 What is the maximum bandwidth of an FM signal with a deviation of 30 kHz and maximum modulating signal of 5 kHz as determined the following two ways: a.Using the table of Bessel functions. b.Using Carson’s rule. Example Problem 3

31 BW = 90 kHz (Bessel functions) BW = 70 kHz (Carson’s rule)


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