Sample Problems for the 8th Quiz

Confusion in Notations and Amendaments

• As concerned by some students, the slides are indeed not consistent in the notations of w(t)and n(t) in Part V. For consistency, w(t) is used to denote specifically the white noisewith PSD

Sw(f) =N0

2for f ∈ R,

and n(t) denotes the filtered white noise with PSD

Sn(f) =

{N0

2, |f − fc| < BT

2or |f + fc| < BT

2

0, otherwise

By this notational convention, we should have the following corrections (and hope thisreduces the confusion caused by the notations).

– Slides 5-12, 5-16, 5-17: w(t), wI(t) and wQ(t) should be n(t), nI(t) and nQ(t).

– Slide 5-15: w(t), wI(t) and wQ(t) should be n(t), nI(t) and nQ(t) (except in the blockdiagram).

– Slides 5-20, 5-21, 5-22: w(t), w(t) wI(t) and wQ(t) should be n(t), n(t), nI(t) andnQ(t).

– Slide 5-23: “(namely, the power ratio between unmodulated carrier Ac cos(2πfct) andthe passband noise w(t), or sometimes denoted as n(t))” is now refined as “(namely,the power ratio between unmodulated carrier Ac cos(2πfct) and the passband noisen(t))”.

– Slides 5-36, 5-37, 5-38, 5-43, 5-60: In the block diagram, “Noise n(t)” should be “Noisew(t)”.

Corrections to Slides

• Slide 5-47: Change “3k2fP

W 2 =(BT,Carson

2W− 1)2, FM” to “

3k2fP

W 2 = 32

(BT,Carson

2W− 1)2, FM (D2 =

2k2fP

W 2 )”.

• Slide 5-47: “is nothing to do” should be “has nothing to do”.

• Slide 5-51: “Slide 5-151” should be “Slide 5-41”.

• Slides 5-61 & 5-62: “Pre-emphasis filter” on the right should be “De-emphasis filter”.

• Slide 5-64: “Hde(f) =1

1+f(f/f0)” should be “Hde(f) =

11+j(f/f0)

”.

• Slide 5-65: “A nonlinear pre-emphasis and de-emphasis filters” should be “Nonlinear pre-emphasis and de-emphasis filters”.

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1. For an FM receiver below, we have s(t) = Ac cos(2πfct+φ(t)), where φ(t) = 2πkf∫ t

0m(τ)dτ .

Denote the passband noise process (i.e., the noise process observed at the output of thebandpass filter) as

n(t) = nI(t) cos(2πfct)− nQ(t) sin(2πfct) = r(t) cos(2πfct+Ψ(t)),

where nI(t) = r(t) cos[Ψ(t)] and nQ(t) = r(t) sin[Ψ(t)].

(a) Due to the limiter, the amplitude of x(t) is of no influence on its output, and only thephase remains. Hence, we can assume the amplitude at the output of the limiter is 1for simplicity. Show that the output of the limiter xlimiter(t) is given by

xlimiter(t) = cos[2πfct+ θ(t)]

where

θ(t) = φ(t) + tan−1

(r(t) sin[Ψ(t)− φ(t)]

Ac + r(t) cos[Ψ(t)− φ(t)]

).

Hint: For simplicity during derivation, you can denote A = 2πfct + φ(t) and B =Ψ(t)− φ(t) and complete the following derivation:

x(t) = Ac cos[2πfct + φ(t)] + r(t) cos[2πfct +Ψ(t)]

= Ac cos[2πfct + φ(t)︸ ︷︷ ︸=A

] + r(t) cos[2πfct + φ(t)︸ ︷︷ ︸=A

+Ψ(t)− φ(t)︸ ︷︷ ︸=B

]

= Ac cos(A) + r(t) cos(A+B)

= · · ·(b) Since the pair of r(t) cos[Ψ(t) − φ(t)] and r(t) sin[Ψ(t) − φ(t)] has exactly the same

joint distribution as the pair of nI(t) = r(t) cos[Ψ(t)] and nQ(t) = r(t) sin[Ψ(t)] (andsince for the noise process, we only concern about its statistics), we can rewrite theresult in (a) as

θ(t) = φ(t) + tan−1

(nQ(t)

Ac + nI(t)

).

The discriminator will output the derivative of the phase, i.e.,

v(t) = θ′(t).

Show that if Ac � nI(t) and AC � nQ(t),

v(t) ≈ φ′(t) +n′Q(t)

Ac

Hint: (tan−1

(f(t)

g(t)

))′=f ′(t)g(t)− f(t)g′(t)

f 2(t) + g2(t).

2

(c) With

v(t) ≈ φ′(t) +n′Q(t)

Ac= 2πkfm(t) + 2πnd(t),

where nd(t) =1

2πAcn′Q(t), show that the SNRO of the FM system is given by

SNRO =3A2

ck2fP

2N0W 3,

provided E[m2(t)] = P and the bandwidth W of the (ideal) baseband lowpass filtersatisfies W < BT , where BT is the bandwidth of the (ideal) bandpass filter.

Hint: The PSD of nd(t) is equal to

PSDnd(f) =

∣∣∣∣ fAc

∣∣∣∣2 PSDnQ(f) =

f 2

A2c

N0 for |f | < BT

2.

(d) Based on (c), explain what the FM noise quieting effect is.

(e) Recall (from Slides 4-95 and 4-97) that for the discriminator to function properly, werequire ∣∣∣∣2kfBT

m(t)

∣∣∣∣ ≤ 1.

Justify that SNRO is proportional to B2T , provided that Ac, W , N0 and the average-

to-peak ratio E[m2(t)]maxt m2(t)

are given and fixed.

Hint: BT is not necessarily decided by the Carson’s rule or universal-curve rule. So,do not justify the statement by the two rules.

Solution.

(a)

x(t) = Ac cos(A) + r(t) cos(A+B)

= Ac cos(A) + r(t) cos(A) cos(B)− r(t) sin(A) sin(B)

= [Ac + r(t) cos(B)] cos(A)− r(t) sin(B) sin(A)

=√

[Ac + r(t) cos(B)]2 + [r(t) sin(B)]2

×(

[Ac + r(t) cos(B)]√[Ac + r(t) cos(B)]2 + [r(t) sin(B)]2

cos(A)

− r(t) sin(B)√[Ac + r(t) cos(B)]2 + [r(t) sin(B)]2

sin(A)

)

=√

[Ac + r(t) cos(B)]2 + [r(t) sin(B)]2 (cos(φnoise(t)) cos(A)− sin(φnoise(t)) sin(A))limiter→ cos(A+ φnoise(t)),

where

φnoise(t) = tan−1

(r(t) sin(B)

Ac + r(t) cos(B)

).

3

(b)

v(t) = φ′(t) +

(n′Q(t)(Ac + nI(t))− nQ(t)n

′I(t)

(Ac + nI(t))2 + n2Q(t)

)

≈ φ′(t) +(n′Q(t)Ac − nQ(t)n

′I(t)

A2c

)Because Ac � nI(t) and Ac � nQ(t)

= φ′(t) +n′Q(t)

Ac

−(nQ(t)

Ac

)︸ ︷︷ ︸Ac�nQ(t)

(n′I(t)

Ac

)

(Note thatn′Q(t)

Acand

n′I(t)

Acare about the same order.)

≈ φ′(t) +n′Q(t)

Ac.

(c) We can ignore the 2π factor and compute the SNRO based on kfm(t) + nd(t), fromwhich the average signal power is equal to

E[(kfm(t))2] = k2fE[m2(t)] = k2fP.

The average noise power is given by∫ W

−W

PSDnd(f)df =

∫ W

−W

f 2

A2c

N0df =2N0W

3

3A2c

.

This implies

SNRO =k2fP2N0W 3

3A2c

=3A2

ck2fP

2N0W 3.

(d) For fixed noise power level N0 and fixed message bandwidth W , increasing carrierpower A2

c/2 will decrease the effective noise power at the demodulator output (by afactor of 1/A2

c). This is called the noise quieting effect.

(e) 4k2fm2(t) ≤ B2

T implies

4k2f maxt∈R

m2(t) ≤ B2T ,

and hence the demodulation of an FM signal functions properly as long as

k2f ≤ B2T

4 ·maxt∈Rm2(t).

Thus, for given W and N0,

SNRO =3A2

ck2fP

2N0W 3≤ 3A2

c

2N0W 3

P

4 ·max∈Rm2(t)B2

T .

The upper bound can be achieved by setting

k2f =B2

T

4 ·maxtm2(t).

Accordingly, SNRO can be made proportional to B2T .

4

2. Continuing from Problems 1(b) and 1(c), we know that the exact relation of v(t) and inputφ′(t) as well as noise term should be:

v(t) = φ′(t) +

(n′Q(t)(Ac + nI(t))− nQ(t)n

′I(t)

(Ac + nI(t))2 + n2Q(t)

)︸ ︷︷ ︸

=2π·nd(t)

.

(a) To examine the impact of the noise, we let nI(t) = λAc cos(ψ(t)) and nQ(t) =λAc sin(ψ(t)) (i.e., we let the instantaneous noise power be given by n2

I(t) + n2Q(t) =

λ2A2c). Show that

2πnd(t) = λψ′(t)cos(ψ(t)) + λ

1 + 2λ cos(ψ(t)) + λ2.

(b) Continue from (a). Find the value of nd(t) when ψ(t) = π, and argue that there is adiscontinuity at λ = 1.

(c) The previous sub-problem indicates that when r2(t) = n2I(t)+n2

Q(t) ≈ A2c and ψ(t) ≈

π, a noise spike will occur. Noting that E[n2(t)] = E[n2I(t)] = E[n2

Q(t)] = BTN0

(cf. the PSD of nQ(t) on Slide 5-43), we know from Slide 3-46 that

i) the pdf of X = r(t) is

fX(x) =x

BTN0e− x2

2BT N0 for x ≥ 0,

ii) ψ(t) is uniformly distributed over [0, 2π), and

iii) r(t) and ψ(t) are independent.

For 0 < ε < 1, show that

Pr

[∣∣∣∣r2(t)A2c

− 1

∣∣∣∣ ≤ ε and

∣∣∣∣ψ(t)π − 1

∣∣∣∣ ≤ ε

]= 2εe

− A2c

2BT N0 sinh

(A2

c

2BTN0ε

).

(d) A way to reduce the threshold 20 in A2c

2BTN0≥ 20 for “near-elimination” of clicking-

sound effect is to reduce the “effective noise power” from BTN0 down to (1−α)BTN0

by introducing a feedback as shown below:

Denote s(t) = cos(2πfct+φ(t)), and let svco(f) = 2 cos(2πfvcot+φvco(t)) with φvco(t) =αφ(t). Show that

x(t) = cos(2π(fc − fvco)t+ (1− α)φ(t)),

provided that the transfer function Hvco(f) of the ideal bandpass filter satisfies

Hvco(f) =

{1, |f − (fc − fvco)| < (1− α)BT

2

0, otherwise

where BT is the transmission bandwidth of s(t).

5

Solution.

(a) Noting that n′I(t) = −nQ(t)ψ

′(t), n′Q(t) = nI(t)ψ

′(t) and n2I(t) + n2

Q(t) = λ2A2c , we

have

2πnd(t) =n′Q(t)(Ac + nI(t))− nQ(t)n

′I(t)

(Ac + nI(t))2 + n2Q(t)

=nI(t)ψ

′(t)(Ac + nI(t)) + n2Q(t)ψ

′(t)

(Ac + nI(t))2 + n2Q(t)

=nI(t)Ac + λ2A2

c

A2c + 2nI(t)Ac + λ2A2

c

ψ′(t)

=λ cos(ψ(t))A2

c + λ2A2c

A2c + 2λ cos(ψ(t))A2

c + λ2A2c

ψ′(t)

= λψ′(t)cos(ψ(t)) + λ

1 + 2λ cos(ψ(t)) + λ2

(b) When ψ(t) = π,

2πnd(t) = λψ′(t)cos(ψ(t)) + λ

1 + 2λ cos(ψ(t)) + λ2= λψ′(t)

−1 + λ

1− 2λ+ λ2=

λ

λ− 1ψ′(t).

Since

limλ↓1

=λ

λ− 1= ∞ and lim

λ↑1=

λ

λ− 1= −∞,

a discontinuity appears at λ = 1.

(c)

Pr

[∣∣∣∣ψ(t)π − 1

∣∣∣∣ ≤ ε and

∣∣∣∣r2(t)A2c

− 1

∣∣∣∣ ≤ ε

]

= Pr

[∣∣∣∣ψ(t)π − 1

∣∣∣∣ ≤ ε

]· Pr

[∣∣∣∣r2(t)A2c

− 1

∣∣∣∣ ≤ ε

](r(t) and ψ(t) are independent.)

= Pr [(1− ε)π ≤ ψ(t) ≤ (1 + ε)π] · Pr[Ac

√1− ε ≤ r(t) ≤ Ac

√1 + ε

]= ε ·

(∫ Ac√1+ε

Ac√1−ε

x

BTN0e− x2

2BT N0 dx

)

= ε

(−e− x2

2BT N0

∣∣∣∣Ac√1+ε

Ac√1−ε

)

= ε

(e−A2

c(1−ε)

2BT N0 − e−A2

c(1+ε)

2BT N0

)

= 2εe− A2

c2BT N0 sinh

(A2

c

2BTN0ε

)

Hence, this probability is a function of carrier-to-noise power ratio A2c/(2BTN0).

Note: When A2c

2BTN0= 10 and ε = 0.1, this probability becomes

2× 0.1× e−10sinh (10 · 0.1) ≈ 10−5

6

As a result, the probability of occurrence of such an “extreme spike situation” seemssmall. However, as long as

λ cos(ψ(t)) + λ2

1 + 2λ cos(ψ(t)) + λ2

is adequately large in comparison with φ′(t) = 2πkfm(t), and ψ′(t) changes sign in avery short period of time, a clicking sound (that reflects the situation that the noisesuddenly jumps from a very negative value to a very positive value, or vice versa) will

be heard. Thus, experiments show that it requires A2c

2BTN0≥ 20 to guarantee that the

clicking sound almost disappears.

(d)

s(t)svco(t) = 2 cos(2πfct + φ(t)) cos(2πfvcot + φvco(t))

= cos(2π(fc − fvco)t+ (1− α)φ(t)) + cos(2π(fc + fvco)t + (1 + α)φ(t))

The frequency deviation of s(t)svco(t) is given by

(∆f)vco = maxt

∣∣∣∣(1− α)

2πφ′(t)

∣∣∣∣ = (1− α) ·∆f,

where ∆f is the frequency deviation of s(t), given by

∆f = maxt

1

2π|φ′(t)|.

As the frequency deviation of s(t)svco(t) is (1− α) of the frequency derivation of s(t),

the effective bandwidth is reduced to (1− α)BT , where BT is the transmission band-width of s(t). Consequently, the bandpass filter Hvco(f) is adequate to pass the firstterm of s(t)svco(t), which is(

s(t)svco(t))bandpass

= cos(2π(fc − fvco)t + (1− α)φ(t))

Note: For your reference, we analyze the noise process due to feedback as follows.

Rewrite n(t) = nI(t) cos(2πfct)−nQ(t) sin(2πfct), where the PSDs of nI(t) and nQ(t)are equal to

PSDnI(f) = PSDnQ

(f) =

{N0, |f | < BT

2

0, otherwise

Then,

n(t)svco(t) = 2nI(t) cos(2πfct) cos(2πfvcot + φvco(t))

−2nQ(t) sin(2πfct) cos(2πfvcot+ φvco(t))bandpass H(f)−→ nI,vco(t) cos(2π(fc − fvco)t− φvco(t))

−nQ,vco(t) sin(2π(fc − fvco)t− φvco(t))

where the PSDs of nI,vco(t) and nQ,vco(t) become{N0, |f | < (1−α)BT

2

0, otherwise

7

This reduces to the threshold to:

A2c

2(1− α)BTN0≥ 20 ⇔ A2

c

2BTN0≥ (1− α)20.

Accordingly, in principle, the feedback reduces the “noise-spike-free” threshold from20 down to (1− α)20.

3. (a) For the linearization approximation of phase-locked loop below, show that

Φe(f)

Φ1(f)=

jf

jf + k0H(f)

where H(f), Φe(f) and Φ1(f) are Fourier transforms of h(t), φe(t) and φ1(t), respec-tively.

(b) If H(f) = 1 and

φ1(t) = u(t) =

{0, t < 0

1, t > 0

determine φe(t) and φ2(t).

Hint:

TABLE A6.2 Summary of properties of the Fourier transformProperty Mathematical Description8. Differentiation in the time domain d

dtg(t) � j2πfG(f)

9. Integration in the time domain∫ t

−∞ g(τ)dτ � 1j2πf

G(f) + G(0)2δ(f)

TABLE A6.3 Fourier-transform pairsTime Function Fourier Transformexp(−at)u(t), a > 0 1

a+j2πf

exp(−a|t|), a > 0 2aa2+(2πf)2

Notes: u(t) = unit step function

(c) How long it takes to have |φe(t)| = |φ1(t) − φ2(t)| ≤ e−3 ≈ 0.05? Will a larger k0increase the time to reach |φe(t)| ≤ e−3?

8

(d) Continue from (b) and (c). In the system, we actually have φ1(t) = φ2(t) = φe(t) = 0for t < 0 and φ1(t) = 1 when t > 0, and therefore φe(t) can be larger than 0.5 atsome t. It is thus inaccurate to simplify sin(φe(t)) as φe(t). Since 0 < sin(θ) < θfor 0 < θ < π, will the actual time to achieve |φe(t)| = |φ1(t) − φ2(t)| ≤ e−3 ≈ 0.05longer than the time obtained from linearization approximation, or shorter than thetime obtained from linearization approximation? Justify your answer.

Solution.

(a) See Slide 5-71.

(b)

Φe(f) =j2πf

j2πf + 2πk0Φ1(f)

implies

φe(t) =d

dtφ1(t) F−1

{1

j2πf + 2πk0

}= δ(t) e−2πk0tu(t)

= e−2πk0tu(t)

Hence,

φ2(t) = φ1(t)− φe(t) = (1− e−2πk0t)u(t).

(c) e−2πk0t ≤ e−3 implies t ≥ 32πk0

; hence, a larger k0 will reduce the time to achieve

e−2πk0t ≤ e−3.

(d) Since sin(φe(t)) < φe(t), φ2(t) =∫ t

02πk0 sin(φe(s))ds should be smaller than

∫ t

02πk0φe(s)ds.

As a result, φe(t) = φ1(t) − φ2(t) shall take longer time to achieve e−3 that the timeobtained from the linearization approximation.

4. When performing sampling on a signal g(t), we obtain the sampled signal

gδ(t) =

∞∑n=−∞

g(nTs) · δ(t− nTs),

where Ts is the sampling period. Show that the Fourier transform of gδ(t) is

1

Ts

∞∑n=−∞

G

(f − n

1

Ts

).

9

Solution. First, we note that

Gδ(f) = F{gδ(t)}

= F{ ∞∑

n=−∞g(nTs) · δ(t− nTs)

}

=∞∑

n=−∞g(nTs) · F {δ(t− nTs)}

=∞∑

n=−∞g(nTs) · e−j2πnTsf (1)

Second, it is clear that 1Ts

∑∞n=−∞G

(f − n

Ts

)is a periodic function with period 1/Ts. By

Fourier series expansions,

1

Ts

∞∑n=−∞

G

(f − n

Ts

)=

∞∑n=−∞

cn ej2πnTsf (2)

where

cn = Ts

∫ 12Ts

− 12Ts

(1

Ts

∞∑n=−∞

G

(f − n

Ts

))e−j2πnTsfdf

=

∫ 12Ts

− 12Ts

( ∞∑n=−∞

G

(f − n

Ts

))e−j2πnTsfdf

=

∞∑n=−∞

∫ 12Ts

− 12Ts

G

(f − n

Ts

)e−j2πnTsfdf (s = f − n

Ts)

=

∞∑n=−∞

∫ 12Ts

− nTs

− 12Ts

− nTs

G (s) e−j2πnTs(s+nTs

)ds

=∞∑

n=−∞

∫ 12Ts

− nTs

− 12Ts

− nTs

G (s) e−j2πnTssds

=

∫ ∞

−∞G (s) e−j2πnTssds

= g(−nTs)

Consequently, equating (1) and (2) completes the proof.

10