E4702 HW#3 solutions
by Anmo Kim (ak2263@columbia.edu)
1. (P2.29) The given phase modulated signal is described as
s(t) = Ac cos[2πfc t + kp Am cos(2πfm t)]
(a) the spectrum of the signal when βp = kp Am ≤ 0.3
(Solution.)
Rewriting the signal,
s(t) = Ac cos[2πfc t + kp Am cos(2πfm t)]
= Ac cos(2πfc t) cos(βp cos(2πfm t)) − Ac sin(2πfc t) sin(βp cos(2πfm t))
For a small value of x, we can use approximation rules, sin(x) ≈ x and cos(x) ≈ 1.
s(t) ≈ Ac cos(2πfc t) − Ac sin(2πfc t)βp cos(2πfm t)
1
1
= Ac cos(2πfc t) − Ac βp sin(2π(fc + fm )t) − Ac βp sin(2π(fc − fm )t)
2
2
In the frequency domain,
S(f ) =
1
Ac [δ(f + fc ) + δ(f − fc )]
2
1
1
− Ac βp [δ(f + fc + fm ) − δ(f − fc − fm )] − Ac βp [δ(f + fc − fm ) − δ(f − fc + fm )]
4j
4j
(a) phasor diagram of (3)
(b) phasor diagram of the narrowband FM signal
Figure 1: phasor diagrams
1
(1)
(2)
(b) phasor diagram
(Solution.)
From (2),
1
1
s(t) = Ac cos(2πfc t) + Ac βp cos(2π(fc + fm )t + π/2) + Ac βp cos(2π(fc − fm )t + π/2)
2
2
(3)
The phasor diagrams of 3 can be drawn with three vectors whose (amplitude, phase) pairs are:
1
π 1
π
(Ac , 0), ( Ac βp , 2πfm t + ), ( Ac βp , −2πfm t + )
2
2
2
2
The phasor diagram of these vectors are shown in fig.1(a). Fig.1(b) is the phasor diagram of the
narrowband FM signal, which is from the figure 2.22 of Haykin.
Two phasor diagrams show the exactly same behavior except its phase. (Notice that the direction of two
small phasors are different in (a) and (b).) Here, narrowband FM is lagged by π/2 in time compared to
the phase modulated signal in (a).
2. (P2.40)
Figure 2: demodulation diagram
s(t) = Ac cos[2πfc t + β sin(2πfm t)]
T is so small as cos(2πfm T ) ≈ 1 and sin(2πfm T ) ≈ 2πfm T
(Solution.)
We define v(t) and w(t) as the signal before the envelope detector and demodulator output signal respectively.
v(t) = s(t) − s(t − T )
= Ac cos[2πfc t + β sin(2πfm t)] − Ac cos[2πfc (t − T ) + β sin(2πfm (t − T ))]
= Ac cos[2πfc t + φ(t)] − Ac cos[2πfc (t − T ) + φ(t − T )]
where φ(t) = β sin(2πfm t).
·
v(t) = −2Ac sin
¸
·
¸
1
1
(2πfc (2t − T ) + φ(t) + φ(t − T ) sin (2πfc T + φ(t) − φ(t − T ))
2
2
2
Here, we can simplify φ(t) − φ(t − T ) by approximations .
φ(t) − φ(t − T ) = β sin(2πfm t) − β sin(2πfm (t − T ))
= β sin(2πfm t) − β sin(2πfm t) cos(2πfm T ) + β cos(2πfm t) sin(2πfm T )
≈ β sin(2πfm t) − β sin(2πfm t) + β cos(2πfm t) · (2πfm T )
= 2πβfm T cos(2πfm t)
= 2π∆f T cos(2πfm t)
where ∆f = βfm . Then, considering the given condition, 2πfc T = π/2,
¸
·
1
sin (2πfc T + φ(t) − φ(t − T ))
2
·
¸
1
= sin (2πfc T + 2π∆f T cos(2πfm t)
2
= sin[πfc T + π∆f T cos(2πfm t)]
= sin[π/4 + π∆f T cos(2πfm t)]
= sin(π/4) cos[π∆f T cos(2πfm t)] + cos(π/4) sin[π∆f T cos(2πfm t)]
1
1
= √ cos[π∆f T cos(2πfm t)] + √ sin[π∆f T cos(2πfm t)]
2
2
1
≈ √ [1 + π∆f T cos(2πfm t)]
2
Then the composite signal becomes
·
¸
·
¸
1
1
v(t) = −2Ac sin (2πfc (2t − T ) + φ(t) + φ(t − T ) sin (2πfc T + φ(t) − φ(t − T ))
2
2
·
¸
√
1
= − 2Ac sin (2πfc (2t − T ) + φ(t) + φ(t − T ) [1 + π∆f T cos(2πfm t)]
2
In the above expression of v(t), the first bracket is near fc in frequency domain and the second bracket is the
low frequency part. Thus, after going through the envelop detector, we will get
√
v(t) = 2Ac [1 + π∆f T cos(2πfm t)]
3. (P2.48)
Figure 3: DSB-SC signal demodulator with noise added OSC
3
(Solution.)
The s(t) and n(t) are
s(t) = Ac m(t) cos(2πfc t)
n(t) = nI (t) cos(2πfc t) − nQ (t) sin(2πfc t)
Then, v(t) becomes
v(t) = [s(t) + n(t)] cos(2πfc t + θ(t))
= [Ac m(t) cos(2πfc t) + nI (t) cos(2πfc t) − nQ (t) sin(2πfc t)] cos(2πfc t + θ(t))
= [Ac m(t) + nI (t)] cos(2πfc t) cos(2πfc t + θ(t)) − nQ (t) sin(2πfc t) cos(2πfc t + θ(t))
1
1
1
1
= [Ac m(t) + nI (t)] cos(4πfc t + θ(t)) + [Ac m(t) + nI (t)] cos(θ(t)) − nQ (t) sin(4πfc t + θ(t)) + nQ (t) sin(θ(t))
2
2
2
2
After going through a low-pass filter, high frequency components will be filtered out.
w(t) =
1
1
[Ac m(t) + nI (t)] cos(θ(t)) + nQ (t) sin(θ(t))
2
2
The message signal component of w(t) is 12 Ac m(t) given zero oscillator noise. Thus, error signal is
1
e(t) =w(t) − Ac m(t)
2
1
1
1
= [Ac m(t)][cos(θ(t)) − 1] + nI (t) cos(θ(t)) + nQ (t) sin(θ(t))
2
2
2
2
Here, E[nQ (t)2 ] = E[nI (t)2 ] = σN
. Assuming nI (t), nQ (t) θ(t) and m(t) are all independent from each other,
the mean-square error is
1 2
A E[m(t)2 ]E[(cos2 (θ(t)) − 1)] +
4 c
1
= A2c E[m(t)2 ]E[(cos(θ(t)) − 1)2 ] +
4
E[e(t)2 ] =
1 2
1 2
σN E[cos2 (θ(t))] + σN
E[sin(θ(t)2 )]
4
4
1 2
σ
4 N
Notice that all cross-product terms disappear when expanding e(t)2 because E[nI (t)] = 0 and E[nQ (t)] = 0.
Using the approximation 1 − cos θ ≈ θ2 /2 and setting P = E[m(t)2 ],
E[e(t)2 ] =
1 2 2
1 2
A P E[θ(t)4 ] + σN
16 c
4
E[θ(t)4 ] can be computed by integration by parts, which gives E[θ(t)4 ] = 3σθ4 .
In an alternative way, characteristic function can be used. The characteristic function, ψ(u), of Gaussian
random variable Θ is
ψΘ (u) = E[eiuθ ]
Z ∞
=
eiuθ fΘ (θ)dθ
−∞
=e
imθ u− 12 σθ2 u2
4
1
2
2
where mθ = E[θ] and σθ2 = V AR[θ]. Since mθ = 0, ψΘ (u) = e− 2 σθ u .
nth moment of a random variable Θ is
·
¸
1 dn ψ(u)
E[θ ] = n
i
dun u=0
¸
· n
1 d − 1 σθ2 u2
2
= n
e
i dun
u=0
n
After several differentiations, we can get E[θ(t)4 ] = 3σθ4 .
As a conclusion, the final mean square error comes to
E[e(t)2 ] =
3 2 2 4
1 2
Ac P σΘ + σN
16
4
5