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Frequency Modulation

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0% found this document useful (0 votes)
17 views63 pages

Frequency Modulation

Uploaded by

kavi20053878
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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AV314 - Communication Systems I

Frequency modulation and demodulation

Vineeth B. S.

Department of Avionics,
Indian Institute of Space Science and Technology.

October 21, 2020

VBS AV314 October 21, 2020 1 / 35


Possibilities
I The carrier signal is c(t) = Ac cos(2πfc t + θ)
I We can change two properties: the amplitude and the phase angle

VBS AV314 October 21, 2020 2 / 35


Possibilities
I The carrier signal is c(t) = Ac cos(2πfc t + θ)
I We can change two properties: the amplitude and the phase angle

I The modulated signal is s(t) = A(t)cos(θ(t))


I If A(t) is constant then we have constant envelope modulation
VBS AV314 October 21, 2020 2 / 35
Phase and frequency modulation

θ(t+∆t)−θ(t)
I The instantaneous frequency f (t) = 2π∆t
1 dθ(t)
I As ∆t → 0, we have f (t) = 2π dt

VBS AV314 October 21, 2020 3 / 35


Phase and frequency modulation

θ(t+∆t)−θ(t)
I The instantaneous frequency f (t) = 2π∆t
1 dθ(t)
I As ∆t → 0, we have f (t) = 2π dt
I s(t) = A(t)cos(θ(t))
I Suppose modulating signal is m(t)

VBS AV314 October 21, 2020 3 / 35


Phase and frequency modulation

θ(t+∆t)−θ(t)
I The instantaneous frequency f (t) = 2π∆t
1 dθ(t)
I As ∆t → 0, we have f (t) = 2π dt
I s(t) = A(t)cos(θ(t))
I Suppose modulating signal is m(t)
I Phase modulation: θ(t) = 2πfc t + kp m(t)
I kp : Phase (modulation) sensitivity in radians/volt

VBS AV314 October 21, 2020 3 / 35


Phase and frequency modulation

θ(t+∆t)−θ(t)
I The instantaneous frequency f (t) = 2π∆t
1 dθ(t)
I As ∆t → 0, we have f (t) = 2π dt
I s(t) = A(t)cos(θ(t))
I Suppose modulating signal is m(t)
I Phase modulation: θ(t) = 2πfc t + kp m(t)
I kp : Phase (modulation) sensitivity in radians/volt
I Frequency modulation (FM): f (t) = fc + kf m(t)
I kf : Frequency (modulation) sensitivity in Hz/volt

VBS AV314 October 21, 2020 3 / 35


Phase and frequency modulation

θ(t+∆t)−θ(t)
I The instantaneous frequency f (t) = 2π∆t
1 dθ(t)
I As ∆t → 0, we have f (t) = 2π dt
I s(t) = A(t)cos(θ(t))
I Suppose modulating signal is m(t)
I Phase modulation: θ(t) = 2πfc t + kp m(t)
I kp : Phase (modulation) sensitivity in radians/volt
I Frequency modulation (FM): f (t) = fc + kf m(t)
I kf : Frequency (modulation) sensitivity in Hz/volt
For FM, the phase angle θ(t) = 2πfc t + 2πkf 0t m(u).du
R
I

kp dm(t)
I For phase modulation, the frequency f (t) = fc + 2π dt

VBS AV314 October 21, 2020 3 / 35


Phase and frequency modulation

θ(t+∆t)−θ(t)
I The instantaneous frequency f (t) = 2π∆t
1 dθ(t)
I As ∆t → 0, we have f (t) = 2π dt
I s(t) = A(t)cos(θ(t))
I Suppose modulating signal is m(t)
I Phase modulation: θ(t) = 2πfc t + kp m(t)
I kp : Phase (modulation) sensitivity in radians/volt
I Frequency modulation (FM): f (t) = fc + kf m(t)
I kf : Frequency (modulation) sensitivity in Hz/volt
For FM, the phase angle θ(t) = 2πfc t + 2πkf 0t m(u).du
R
I

kp dm(t)
I For phase modulation, the frequency f (t) = fc + 2π dt
I We will consider only FM!

VBS AV314 October 21, 2020 3 / 35


The usual questions

I What is the spectrum of a FM signal?


I What is the transmit bandwidth of a FM signal?
I Condition for checking whether m(t) can be transmitted through a passband channel?
I What is the transmit power?

VBS AV314 October 21, 2020 4 / 35


Bandwidth of FM signals

I Recall DSBSC - the one sided channel bandwidth had to be larger than the two sided baseband bandwidth
I Why?

VBS AV314 October 21, 2020 5 / 35


Bandwidth of FM signals

I Recall DSBSC - the one sided channel bandwidth had to be larger than the two sided baseband bandwidth
I Why?
I Since we did only frequency translation, the bandwidth (one wided) of the modulated signal was equal to the
bandwidth (two sided) of the baseband signal

VBS AV314 October 21, 2020 5 / 35


Bandwidth of FM signals

I Recall DSBSC - the one sided channel bandwidth had to be larger than the two sided baseband bandwidth
I Why?
I Since we did only frequency translation, the bandwidth (one wided) of the modulated signal was equal to the
bandwidth (two sided) of the baseband signal
I Is FM a linear transformation?

VBS AV314 October 21, 2020 5 / 35


Bandwidth of FM signals

I Suppose m(t) is a single tone, i.e., m(t) = Am cos(2πfm t)

VBS AV314 October 21, 2020 6 / 35


Bandwidth of FM signals

I Suppose m(t) is a single tone, i.e., m(t) = Am cos(2πfm t)


I Then for FM signal s(t) the instantaneous frequency

f (t) = fc + kf Am cos(2πfm t)

VBS AV314 October 21, 2020 6 / 35


Bandwidth of FM signals

I Suppose m(t) is a single tone, i.e., m(t) = Am cos(2πfm t)


I Then for FM signal s(t) the instantaneous frequency

f (t) = fc + kf Am cos(2πfm t)

I Let ∆f = kf Am (frequency deviation, does not depend on fc )

VBS AV314 October 21, 2020 6 / 35


Bandwidth of FM signals

I Suppose m(t) is a single tone, i.e., m(t) = Am cos(2πfm t)


I Then for FM signal s(t) the instantaneous frequency

f (t) = fc + kf Am cos(2πfm t)

I Let ∆f = kf Am (frequency deviation, does not depend on fc )


I The phase θ(t) =
∆f
2πfc t + sin(2πfm t)
fm

VBS AV314 October 21, 2020 6 / 35


Bandwidth of FM signals

I Suppose m(t) is a single tone, i.e., m(t) = Am cos(2πfm t)


I Then for FM signal s(t) the instantaneous frequency

f (t) = fc + kf Am cos(2πfm t)

I Let ∆f = kf Am (frequency deviation, does not depend on fc )


I The phase θ(t) =
∆f
2πfc t + sin(2πfm t)
fm

∆f
I Let β = fm
(modulation index, measure of phase deviation)

VBS AV314 October 21, 2020 6 / 35


Bandwidth of FM signals

I The FM signal s(t) for single tone m(t) is therefore

s(t) = Ac cos(2πfc t + βsin(2πfm t))

I Intuitively, as ∆f or β increases, the amount by which the frequency changes is large


I Intuitively, then β should affect the bandwidth of the signal
I If β < 1, we have narrowband modulation
I If β > 1, we have wideband modulation

VBS AV314 October 21, 2020 7 / 35


Bandwidth of narrowband FM signals

I An approximate approach
I s(t) = Ac cos(2πfc t + βsin(2πfm t))
I s(t) = Ac cos(2πfc t)cos[βsin(2πfm t)] − Ac sin(2πfc t)sin[βsin(2πfm t)]
I Assume cos[βsin(2πfm (t))] = 1 and sin[βsin(2πfm t)] = βsin(2πfm t)
I Then
s(t) = Ac cos(2πfc t) − Ac βsin(2πfc t)sin(2πfm t)
I How to interpret this ?

VBS AV314 October 21, 2020 8 / 35


Bandwidth of narrowband FM signals

I An approximate approach
I s(t) = Ac cos(2πfc t + βsin(2πfm t))
I s(t) = Ac cos(2πfc t)cos[βsin(2πfm t)] − Ac sin(2πfc t)sin[βsin(2πfm t)]
I Assume cos[βsin(2πfm (t))] = 1 and sin[βsin(2πfm t)] = βsin(2πfm t)
I Then
s(t) = Ac cos(2πfc t) − Ac βsin(2πfc t)sin(2πfm t)
I How to interpret this ?
I The second term is a DSB signal

VBS AV314 October 21, 2020 8 / 35


Narrowband FM modulator

m(t) s(t)
+
R
X

βAc sin(2πfc t)

-90 phase shift Ac cos(2πfc t)

I For general m(t) there will be some distortion in the envelope


I Also higher order harmonics at multiples of fm would be present

VBS AV314 October 21, 2020 9 / 35


General analysis

I The modulating signal m(t) = Am cos(2πfm t)


I Is s(t) = Ac cos(2πfc t + βsin(2πfm t)) periodic ?

VBS AV314 October 21, 2020 10 / 35


General analysis

I The modulating signal m(t) = Am cos(2πfm t)


I Is s(t) = Ac cos(2πfc t + βsin(2πfm t)) periodic ?
I s(t) = Re[s̃(t)e j(2πfc t) ]
I Here s̃(t) = Ac e jβsin(2πfm t) . Is s̃(t) periodic ?

VBS AV314 October 21, 2020 10 / 35


General analysis

I The modulating signal m(t) = Am cos(2πfm t)


I Is s(t) = Ac cos(2πfc t + βsin(2πfm t)) periodic ?
I s(t) = Re[s̃(t)e j(2πfc t) ]
I Here s̃(t) = Ac e jβsin(2πfm t) . Is s̃(t) periodic ?
Then s̃(t) = ∞ j2πnfm t
P
n=−∞ cn e
I

I What is cn ?

VBS AV314 October 21, 2020 10 / 35


General analysis

I The modulating signal m(t) = Am cos(2πfm t)


I Is s(t) = Ac cos(2πfc t + βsin(2πfm t)) periodic ?
I s(t) = Re[s̃(t)e j(2πfc t) ]
I Here s̃(t) = Ac e jβsin(2πfm t) . Is s̃(t) periodic ?
Then s̃(t) = ∞ j2πnfm t
P
n=−∞ cn e
I

I What is cn ?
R 1
I cn = fm 2fm1 s̃(t)e −j2πnfm t dt
− 2f
m

VBS AV314 October 21, 2020 10 / 35


Fourier series exp. for s̃(t)

1
2fm
s̃(t)e −j2πnfm t dt
R
I c n = fm
− 2f1
m

I With s̃(t) = Ac e jβsin(2πfm t) , we have


Z 1
2fm
c n = fm A c e jβsin(2πfm t)−j2πnfm t dt.
− 2f1
m

I Suppose x = 2πfm t
Ac
Z π
cn = e j(βsin(x)−πx) dx.
2π −π

VBS AV314 October 21, 2020 11 / 35


Fourier series exp. for s̃(t)

I We have Z π
Ac
cn = e j(βsin(x)−πx) dx.
2π −π
I nth order Bessel function of the first kind and argument β
Z π
1
Jn (β) = e j(βsin(x)−πx) dx.
2π −π
I Therefore, cn = Ac Jn (β).

VBS AV314 October 21, 2020 12 / 35


Fourier series exp. for s̃(t)

I We have Z π
Ac
cn = e j(βsin(x)−πx) dx.
2π −π
I nth order Bessel function of the first kind and argument β
Z π
1
Jn (β) = e j(βsin(x)−πx) dx.
2π −π
I Therefore, cn = Ac Jn (β).
P∞ j2πnfm t
I Recall the Fourier series expansion s̃(t) = n=−∞ cn e
Then s̃(t) = ∞ j2πnfm t
P
n=−∞ Ac Jn (β)e
I

I Also recall that s(t) = Re[s̃(t)e j2πfc t ]

VBS AV314 October 21, 2020 12 / 35


An expansion for s(t)

P∞
I Then s̃(t) = n=−∞ Ac Jn (β)e j2πnfm t
I Also recall that s(t) = Re[s̃(t)e j2πfc t ]
s(t) = Ac ∞
P
n=−∞ Jn (β)cos(2π(fc + nfm )t)
I

VBS AV314 October 21, 2020 13 / 35


An expansion for s(t)

P∞
I Then s̃(t) = n=−∞ Ac Jn (β)e j2πnfm t
I Also recall that s(t) = Re[s̃(t)e j2πfc t ]
s(t) = Ac ∞
P
n=−∞ Jn (β)cos(2π(fc + nfm )t)
I

I What is S(f )?

VBS AV314 October 21, 2020 13 / 35


An expansion for s(t)

P∞
I Then s̃(t) = n=−∞ Ac Jn (β)e j2πnfm t
I Also recall that s(t) = Re[s̃(t)e j2πfc t ]
s(t) = Ac ∞
P
n=−∞ Jn (β)cos(2π(fc + nfm )t)
I

I What is S(f )?
P∞
I S(f ) = A2c n=−∞ Jn (β) [δ(f − fc − nfm ) + δ(f + fc + nfm )]

VBS AV314 October 21, 2020 13 / 35


The Bessel coefficients Jn (β)

I Jn (β) = (−1)n J−n (β)


I For β small, J0 (β) = 1, J1 (β) = β2 , Jn (β) ≈ 0, n ≥ 2
P 2
n Jn (β) = 1. What is the power in an FM signal?
I

VBS AV314 October 21, 2020 14 / 35


The Bessel coefficients Jn (β)

I Jn (β) = (−1)n J−n (β)


I For β small, J0 (β) = 1, J1 (β) = β2 , Jn (β) ≈ 0, n ≥ 2
P 2
n Jn (β) = 1. What is the power in an FM signal?
I

A2c
I Power = 2
VBS AV314 October 21, 2020 14 / 35
The effect of β and ∆f

∆f
I Recall ∆f = kf Am , β = fm
, and S(f ).
fm fixed, Am is varied fm is varied, Am is fixed

VBS AV314 October 21, 2020 15 / 35


Carson’s rule

I For wideband FM, the actual bandwidth is ∞


I Empirically, the bandwidth is 2∆f for wideband FM
I For narrowband FM, the bandwidth is 2fm
I Carson’s rule: Bandwidth is 2∆f + 2fm
I Read the textbook for an alternative method using Jn (β)

VBS AV314 October 21, 2020 16 / 35


Carson’s rule for general m(t)

I Let the maximum freq. component in m(t) be W


I Let ∆f = kf max|m(t)|
I Use Carson’s rule (is an underestimate!)

VBS AV314 October 21, 2020 17 / 35


FM modulation

Recall: Narrowband direct FM generation

m(t) s(t)
+
R
X

βAc sin(2πfc t)

-90 phase shift Ac cos(2πfc t)

VBS AV314 October 21, 2020 18 / 35


FM modulation

Direct FM generation

I VCO is a system which generates a signal (sinusoid) with instantaneous frequency (ideal)

f (t) = fc + kf m(t)

I The FM signal is generated from a VCO


I The frequency is directly controlled by m(t)
I kf is a parameter of the VCO
I Difficult to obtain wideband FM (why do we need wideband FM?) due to carrier drift

VBS AV314 October 21, 2020 19 / 35


FM modulation

Indirect FM generation (Armstrong modulator)

I Generate narrow band FM and then frequency multiply


I The narrow band FM signal is  Z t 
s(t) = Ac cos 2πfc t + 2πkf m(u)du
0
I If we frequency multiply by n  Z t 
s 0 (t) = Ac cos 2π(nfc )t + 2π(nkf ) m(u)du
0

VBS AV314 October 21, 2020 20 / 35


FM modulation

Frequency multiplication

I The device has IO characteristics given by:

v (t) = a1 s(t) + a2 s(t)2 + · · · + an s(t)n

I Would v (t) contain a component centered at nfc ?

VBS AV314 October 21, 2020 21 / 35


FM modulation

An example

VBS AV314 October 21, 2020 22 / 35


FM demodulation

Demodulation of FM signals

 
We have s(t) = Ac cos 2πfc t + 2πkf 0t m(u).du
R
I

I We need to extract the instantaneous frequency or phase


ds(t)
I What about dt
?

VBS AV314 October 21, 2020 23 / 35


FM demodulation

Demodulation of FM signals

 
We have s(t) = Ac cos 2πfc t + 2πkf 0t m(u).du
R
I

I We need to extract the instantaneous frequency or phase


ds(t)
I What about dt
?
 
ds
= −2πAc (fc + 2πkf m(t))sin 2πfc t + 2πkf 0t m(u).du
R
I
dt
I The envelope of the signal contains the message signal
I FM demodulation in principle using a differentiator followed by an envelope detector

VBS AV314 October 21, 2020 23 / 35


FM demodulation

Limiter Discriminator Method

VBS AV314 October 21, 2020 24 / 35


FM demodulation

Slope filter/circuit

dx(t)
I Suppose x(t) has a FT X (f ), then what is the FT of dt
?

VBS AV314 October 21, 2020 25 / 35


FM demodulation

Slope filter/circuit

dx(t)
I Suppose x(t) has a FT X (f ), then what is the FT of dt
?
I j2πfX (f )
I So H(f ) for a differentiator should be j2πf
I But we only need to differentiate our signal within the FM transmission bandwidth BT .

VBS AV314 October 21, 2020 25 / 35


FM demodulation

Slope filter/circuit

x
I Consider the filter with response H1 (f ) defined as
BT
)), for fc − B2T ≤ f ≤ fc + B2T ,

j2πa(f − (fc −
 2
BT
H1 (f ) = j2πa(f + (fc − 2
)), for − fc − B2T ≤ f ≤ −fc + BT
2
,

0, otherwise

VBS AV314 October 21, 2020 26 / 35


FM demodulation

Balanced FM demodulator ...

VBS AV314 October 21, 2020 27 / 35


FM demodulation

... Balanced FM demodulator

Differentiator 1 ED 1

Differentiator 2 ED 2

VBS AV314 October 21, 2020 28 / 35


FM demodulation

Broadcast FM

I Broadcast FM (mono) - some points to keep in mind


I Frequency deviation ≈ 75kHz
I One sided bandwidth ≈ 15kHz
I Modulation index ≈ 5
I FM channel bandwidth (allocated) is 200kHz
I Approximate transmit bandwidth is 180kHz
I Recall the transmission scheme that we had looked at

VBS AV314 October 21, 2020 29 / 35


FM demodulation

Broadcast stereo FM

I Broadcast FM (stereo) transmits a stereo signal consisting of a L channel and R channel


I But it transmits it as L + R and L - R (so that mono channel receivers can demodulate the L + R)
I The baseband signal looks like (and by Carson’s formula has a bandwidth > 200kHz)

I A stereo FM receiver needs to demodulate the FM signal and further process the baseband signal to get the L and R
channels separately

VBS AV314 October 21, 2020 30 / 35


FM demodulation

Broadcast stereo FM

I The “baseband format” of stereo FM is motivated by the following simple scheme for generating the signal

VBS AV314 October 21, 2020 31 / 35


FM demodulation

Demonstration of FM demodulation

VBS AV314 October 21, 2020 32 / 35


Effect of channel non-linearities

Effects of channel non-linearities

I Suppose vi (t) and vo (t) are the input and output signals resp.
I The input output relationship is non-linear and an example model is given by

vo (t) = a1 vi (t) + a2 vi (t)2 + a3 vi (t)3

VBS AV314 October 21, 2020 33 / 35


Effect of channel non-linearities

Effects of channel non-linearities

I Suppose vi (t) and vo (t) are the input and output signals resp.
I The input output relationship is non-linear and an example model is given by

vo (t) = a1 vi (t) + a2 vi (t)2 + a3 vi (t)3

I What happens if we send an AM or FM vi (t) = s(t) through this channel?

VBS AV314 October 21, 2020 33 / 35


Effect of channel non-linearities AM

Effect on AM signals

I Suppose vi (t) = Ac Am cos(2πfm t)cos(2πfc t); Ac Am = 1


I Then we have that

vo (t) = a1 cos(2πfm t)cos(2πfc t) + a2 cos 2 (2πfm t)cos 2 (2πfc t)


+a3 cos 3 (2πfm t)cos 3 (2πfc t)

VBS AV314 October 21, 2020 34 / 35


Effect of channel non-linearities AM

Effect on AM signals

I Suppose vi (t) = Ac Am cos(2πfm t)cos(2πfc t); Ac Am = 1


I Then we have that

vo (t) = a1 cos(2πfm t)cos(2πfc t) + a2 cos 2 (2πfm t)cos 2 (2πfc t)


+a3 cos 3 (2πfm t)cos 3 (2πfc t)

I What all frequencies would vo (t) contain?

VBS AV314 October 21, 2020 34 / 35


Effect of channel non-linearities AM

Effect on AM signals

I Suppose vi (t) = Ac Am cos(2πfm t)cos(2πfc t); Ac Am = 1


I Then we have that

vo (t) = a1 cos(2πfm t)cos(2πfc t) + a2 cos 2 (2πfm t)cos 2 (2πfc t)


+a3 cos 3 (2πfm t)cos 3 (2πfc t)

I What all frequencies would vo (t) contain?


I What is the desired signal at center frequency of fc ?
I How to extract this desired signal?

VBS AV314 October 21, 2020 34 / 35


Effect of channel non-linearities FM

Effect on FM signals

I Suppose vi (t) = Ac cos(2πfc t + θ(t)); Ac = 1


I Then we have that

vo (t) = a1 cos(2πfc t + θ(t)) + a2 cos 2 (2πfc t + θ(t))


+a3 cos 3 (2πfc t + θ(t))

VBS AV314 October 21, 2020 35 / 35


Effect of channel non-linearities FM

Effect on FM signals

I Suppose vi (t) = Ac cos(2πfc t + θ(t)); Ac = 1


I Then we have that

vo (t) = a1 cos(2πfc t + θ(t)) + a2 cos 2 (2πfc t + θ(t))


+a3 cos 3 (2πfc t + θ(t))

I What all frequencies would vo (t) contain?

VBS AV314 October 21, 2020 35 / 35


Effect of channel non-linearities FM

Effect on FM signals

I Suppose vi (t) = Ac cos(2πfc t + θ(t)); Ac = 1


I Then we have that

vo (t) = a1 cos(2πfc t + θ(t)) + a2 cos 2 (2πfc t + θ(t))


+a3 cos 3 (2πfc t + θ(t))

I What all frequencies would vo (t) contain?


I What is the desired signal at center frequency of fc ?
I How to extract this desired signal?

VBS AV314 October 21, 2020 35 / 35


Effect of channel non-linearities FM

Effect on FM signals

I Suppose vi (t) = Ac cos(2πfc t + θ(t)); Ac = 1


I Then we have that

vo (t) = a1 cos(2πfc t + θ(t)) + a2 cos 2 (2πfc t + θ(t))


+a3 cos 3 (2πfc t + θ(t))

I What all frequencies would vo (t) contain?


I What is the desired signal at center frequency of fc ?
I How to extract this desired signal?
I fc > 3∆f + 2W

VBS AV314 October 21, 2020 35 / 35

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