fm example

8/6/2019 FM Example

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Angle Modulation vs Amplitude Modulation

carrier signal

message signal

(modulating signal)

AM wave

PM wave

FM wave

Note: frequency constant,

envelope varies

Note: envelope constant,phase/frequency varies

8/6/2019 FM Example


Angle Modulation: Waveforms

• Note: FM and PM areidentical except for atime (phase shift)

• How do you know ifyou’re dealing with FMor PM?

– must know themodulating signal, andcheck if the

instantaneousfrequency is proportionto the amplitude of themessage (FM) or if theinstantaneous phase is

proportional to theamplitude of themessage (PM)

• Now we concentrate on

FM only

FIGURE 6-3 Phase and frequency modulation of a sine-

wave carrier by a sine-wave signal: (a) unmodulatedcarrier; (b) modulating signal; (c) frequency-modulatedwave; (d) phase-modulated wave

8/6/2019 FM Example


EE354:Bessel Plot

( )*

( ) ( )

1 1( ) ( ) ( )

2 2

c n mn

c c

G f A J f nf

S f G f f G f f

β δ ∞

=−∞

= −

= − + − −

∑

J 0( β )

J 1( β )

J 2( β )

J 3( β )

J 4( β ) J 5( β ) J 6 ( β )

J 7 ( β ) J 8( β )

J n

(β

)

Modulation Index, β

β = 0.2

• How do you determinespectrum using theBessel plot?

• Knowing the value of

β , enter each curve tofind J 0 ( β ), J 1( β ), etc, togive the coefficients ofeach cosine in theinfinite sum

• Example: β =0.2 – J 0 (0.2 )= 0.98

– J 1(0.2 )= 0.08

– J 2 (0.2 ) = ~0

– …

• Note: J 0 ( β )corresponds to thecarrier frequency (n=0)

– J 0 ( β ) disappears at β =2.4, 5.4, etc

Spectrum ofcomplex

envelope

Spectrum ofFM signal

8/6/2019 FM Example


EE354: Bessel Table• This table gives the values of the Bessel Functions of the first kind

• note: only the significant side frequencies are shown: those whoseamplitudes are ≥ 1% of the unmodulated carrier (i.e., J n ( β ) ≥ 0.01) areshown

β

8/6/2019 FM Example


EE354: FM Spectrum Example• For an FM modulator with modulation index β = 1, a modulating signal v m (t)=V m

cos(2 π 1000t),

and an unmodulated carrier ofv

c (t)=10cos(2

π 500,000t),

Find:• (a) Bandwidth by Carson’s Rule = ?

– Answer:

• (b) Bandwidth using Bessel Table.

– Answer: 3 sets of side frequencies

• (c) Total normalized average power.

– Answer:

• (d) (a) The spectrum of the FM signal, S(f ) (show positive frequencies only).

– Answer: see next slide

2( 1) 2(1 1)1000 4 kHzT

B B Hz β = + = + =

2 2 3 1 kHz = 6 kHzT m

B nf = = ⋅ ⋅

2 2

10 502 2

c AP W = = =

3 85

8/6/2019 FM Example


S( f )

0

110 (1) 3.85

2 J ⋅ =

1

110 (1) 2.2

2 J −⋅ = −

1

110 (1) 2.2

2 J ⋅ =

2

110 (1) 0.55

2

J ⋅ =

2110 (1) 0.552 J −⋅ =

3

110 (1) 0.1

2 J ⋅ =

3

110 (1) 0.1

2 J −⋅ = −

f (kHz)

0.10.55

2.2

3.85

0.55

- 0.1

497 498 499 500 501 502 5030 ( ) J β 1( ) J β 2 ( ) J β 3( ) J β

1( ) J β −2 ( ) J β −3( ) J β −

Notes:

• δ functions are separated by f m = 1 kHz

• # of δ functions depends on value of β (see Bessel table)• some δ functions are positive, some negative, but their magnitude is symmetric about fc

fm example

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