Frequency Modulation
Frequency Modulation
Vineeth B. S.
Department of Avionics,
Indian Institute of Space Science and Technology.
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
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)
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
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
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
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!
I Recall DSBSC - the one sided channel bandwidth had to be larger than the two sided baseband bandwidth
I Why?
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 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?
f (t) = fc + kf Am cos(2πfm t)
f (t) = fc + kf Am cos(2πfm t)
f (t) = fc + kf Am cos(2πfm t)
f (t) = fc + kf Am cos(2πfm t)
∆f
I Let β = fm
(modulation index, measure of phase deviation)
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 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
m(t) s(t)
+
R
X
βAc sin(2πfc t)
I What is cn ?
I What is cn ?
R 1
I cn = fm 2fm1 s̃(t)e −j2πnfm t dt
− 2f
m
1
2fm
s̃(t)e −j2πnfm t dt
R
I c n = fm
− 2f1
m
I Suppose x = 2πfm t
Ac
Z π
cn = e j(βsin(x)−πx) dx.
2π −π
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).
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
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
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 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 )]
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
m(t) s(t)
+
R
X
βAc sin(2πfc t)
Direct FM generation
I VCO is a system which generates a signal (sinusoid) with instantaneous frequency (ideal)
f (t) = fc + kf m(t)
Frequency multiplication
An example
Demodulation of FM signals
We have s(t) = Ac cos 2πfc t + 2πkf 0t m(u).du
R
I
Demodulation of FM signals
We have s(t) = Ac cos 2πfc t + 2πkf 0t m(u).du
R
I
Slope filter/circuit
dx(t)
I Suppose x(t) has a FT X (f ), then what is the FT of dt
?
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 .
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
Differentiator 1 ED 1
Differentiator 2 ED 2
Broadcast FM
Broadcast stereo FM
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
Broadcast stereo FM
I The “baseband format” of stereo FM is motivated by the following simple scheme for generating the signal
Demonstration of FM demodulation
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
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
Effect on AM signals
Effect on AM signals
Effect on AM signals
Effect on FM signals
Effect on FM signals
Effect on FM signals
Effect on FM signals