Solution 3

7.41.

xp(t) =∞∑

n=−∞x(nTs)p(t− nTs)

= p(t) ⋆∞∑

n=−∞x(nTs)δ(t− nTs)

= p(t) ⋆ x(t)∞∑

n=−∞δ(t− nTs)

Thus

Xp(f) = P (f) · F

[x(t)

∞∑n=−∞

δ(t− nTs)

]

= P (f)X(f) ⋆ F

[ ∞∑n=−∞

δ(t− nTs)

]

= P (f)X(f) ⋆1

Ts

∞∑n=−∞

δ(f − n

Ts)

=1

TsP (f)

∞∑n=−∞

X(f − n

Ts).

2. In order to avoid aliasing, Fs = 1Ts

> 2W . Furthermore, the spectrum P (f)should be invertible for |f | < W .

3. X(f) can be recovered using the reconstruction filter∏( f2W ′ ) with W < W ′ <

1Ts

−W . In this case

X(f) = Xp(f)TsP−1(f)

∏(

f

2W ′ ).

7.181.

RX(t; τ) = E[x(t)x(t+ τ)]

= E[Y 2 cos(2πf0t+ θ) cos(2πf0(t+ τ) + θ)

]= E[Y 2]E [cos(2πf0t+ θ) cos(2πf0(t+ τ) + θ)]

=3

2cos(2πf0t)

SX(f) = F (RX(t; τ)) =3

4[δ(f − f0) + δ(f + f0)]

1

2.

PX =

∫ ∞

−∞SX(f)df =

3

2.

SQNR =PX

E[e2]=

3× 4vPX

x2max

= 22v−1 ≥ 40dB = 104.

We have v ≥ 8, thus the required number of bits/sample and the bandwidthrequirement are 8 and 8f0, respectively.

3.

SQNR =PX

E[e2]=

3× 4vPX

x2max

dB= 10log10

PX

a2+ 6v + 4.8.

Thus, one extra bit increases the SQNR by 6dB. If the SQNR is to be increasedby 24dB, 4 more bits/sample must be introduced. The new minimum bandwidthrequirement is 12f0.

7.201.∫∞−∞ fX(x)dx = 1, then 3a = 1 which leads to a = 1

3 .

2.

PX = E[X2]

=

∫ 0

−2

x2(1

6x+

1

3)dx+

∫ 2

0

1

3x2dx

=10

9.

3. 2v = 32 leads to v = 5. Thus

SQNR = 10 log10PX

x2max

+ 6v + 4.8 = 10 log1010

9/4 + 30 + 4.8 ≈ 29.2dB.

4. BWreq = vW = 5× 12KHz = 60KHz.

5. ∆SQNR = 6∆v ≥ 20 leads to ∆vmin = 4.Then ∆BWreq = ∆vW = 4× 12KHz = 48KHz.

9.41. Taking the inverse Fourier transform of H(f), we obtain:

h(t) = F−1 [H(f)] = δ(t) +α

2δ(t− t0) +

α

2δ(t+ t0).

Hencey(t) = s(t) ⋆ h(t) = s(t) +

α

2s(t− t0) +

α

2s(t+ t0).

2

2. If the signal s(t) is used to modulate the sequence In, then the transmitted signalis

u(t) =∞∑

n=−∞Ins(t− nT ).

The received signal is the convolution of u(t) with h(t). Hence,

y(t) = u(t) ⋆ h(t) =

( ∞∑n=−∞

Ins(t− nT )

)⋆(δ(t) +

α

2δ(t− t0) +

α

2δ(t+ t0)

)=

∞∑n=−∞

Ins(t− nT ) +α

2

∞∑n=−∞

Ins(t− t0 − nT ) +α

2

∞∑n=−∞

Ins(t+ t0 − nT )

Thus, the output of the matched filter s(−t) at the time instant t1 is:

w(t1) =

∞∑n=−∞

In

∫ ∞

−∞s(τ − nT )s(τ − t1)dτ

+α

2

∞∑n=−∞

In

∫ ∞

−∞s(τ − t0 − nT )s(τ − t1)dτ

+α

2

∞∑n=−∞

In

∫ ∞

−∞s(τ + t0 − nT )s(τ − t1)dτ

If we denote the signal s(t) ⋆ s(t) by x(t), then the output of the matched filterat t1 = kT is:

w(kT ) =∞∑

n=−∞Inx(kT − nT )

+α

2

∞∑n=−∞

Inx(kT − t0 − nT ) +α

2

∞∑n=−∞

Inx(kT + t0 − nT )

3. With t0 = T and k = n in the previous equation, we obtain:

wk = Ikx0 +∑n̸=k

Inxk−n

+α

2Ikx−1 +

α

2

∑n̸=k

Inxk−n−1 +α

2Ikx1 +

α

2

∑n̸=k

Inxk−n+1

= Ik

(x0 +

α

2x−1 +

α

2x1

)+∑n ̸=k

In

[xk−n +

α

2xk−n−1

α

2xk−n+1

].

The terms under the summation is ISI introduced by the channel. If the signals(t) is designed so as to satisfy the Nyquist criterion, then:

xk = 0, k ̸= 0

and the above expression simplified to

wk = Ik +α

2(Ik+1 + Ik−1) .

3

9.5The pulse x(t) having the raising cosine spectrum given by (9.2.20/21) is:

x(t) = sinc(t/T )cos(πβt/T )

1− 4β2t2/T 2

The function sinc(t/T ) is 1 when t = 0 and 0 when t = nT . Therefore, the Nyquistcriterion will be satisfied as long as the function g(t) is:

g(t) =cos(πβt/T )

1− 4β2t2/T 2=

{1 t = 0bounded t ̸= 0

The function g(t) needs to be checked only for those values of t such that 4β2t2/T 2 =1 or βt = T

2 . However:

limβt→T

2

cos(πβt/T )

1− 4β2t2/T 2= lim

x→1

cos(π2x)

1− x2= lim

x→1− π

4xsin(

π

2x) = −π

4< ∞

Hence:

x(nT ) =

{1, n = 00, n ̸= 0

meaning that the pulse x(t) satisfied the Nyquist criterion.

9.12

9.241. The output of the zero-force equalizer is:

qm =1∑

n=−1

cnxm−n

4

With q0 = 1 and qm = 0 for m ̸= 0, we obtain the system: 1.0 0.1 −0.5−0.2 1.0 0.10.05 −0.2 1.0

c−1

c0c1

=

010

Solving the previous system in terms of the equalizer’s coefficients, we obtain: c−1

c0c1

=

0.0000.9800.196

2. The output of the equalizer is

qm =

0 m ≤ −4c−1x−2 = 0 m = −3c−1x−1 + c0x−2 = −0.49 m = −20 m = −11 m = 00 m = 1c0x2 + x1c1 = 0.0098 m = 2c1x2 = 0.0098 m = 30 m ≥ 4

Hence, the residual ISI sequence is

residual ISI = {..., 0,−0.49, 0, 0, 0, 0.0098, 0.0098, 0, ...}

and its span is 6 symbols.

5