A First Course in Digital Communications - Solutions Manual

307 Pages·2009·5.041 MB·English

by  Ha H. Nguyen, Ed Shwedyk

k
y
d
e
w
h
Solutions Manual
“A First Course in Digital Communications”
S
Cambridge University Press
&
Ha H. Nguyen
Department of Electrical & Computer Engineering
University of Saskatchewan
Saskatoon, SK, CANADA S7N 5A9
n Ed Shwedyk
Department of Electrical & Computer Engineering
University of Manitoba
e
Winnipeg, MB, CANADA R3T 5V6
y
August 5, 2011
u
g
N
k
y
d
Preface
e
w
This Solutions Manual was last updated in August, 2011. We appreciate receiving comments
and corrections you might have on the current version. Please send emails [email protected]
or [email protected].
h
S
&
n
e
y
u
g
N
1
k
y
d
Chapter 2
e
Deterministic Signal Characterization
w
and Analysis
h
P2.1 First let us establish (or review) the reSlationships for real signals
(cid:80)
(a) Given {A ,B }, i.e., s(t) = ∞ [A cos(2πkf t)+B sin(2πkf t)], A ,B are real.
k (cid:113)k k=0 k r k r k k
Then C = A2 +B2, θ = tan−1 Bk, where we write s(t) as
k k k k Ak
&
(cid:88)∞
s(t) = C cos(2πkf t−θ )
k r k
k=0
If we choose to write it as
n (cid:88)∞
s(t) = C cos(2πkf t+θ )
k r k
k=0
e
then the phase becomes θ = −tan−1 Bk. Further
k Ak
y
A −jB
k k
D =
k
2
u D = D∗, k = 1,2,3,...
−k k
D = A (B = 0 always).
0 0 0
g
(b) Given {C ,θ } then A = C cosθ , B = C sinθ . They are obtained from:
k k k k k k k k
(cid:88)∞ (cid:88)∞
N
s(t) = C cos(2πkf t−θ ) = [C cosθ cos(2πkf t)+C sinθ sin(2πkf t)]
k r k k k r k k r
k=0 k=0
Now s(t) can written as:
(cid:40) (cid:41)
(cid:88)∞ ej[2πkfrt−θk]+e−j[2πkfrt−θk]
s(t) = C
k
2
k=0
(cid:183) (cid:184) (cid:183) (cid:184)
e−jθ0 +ejθ0 (cid:88)∞ C C
s(t) = C + ke−jθkej2πkfrt+ kejθke−j2πkfrt
0
2 2 2
(cid:124) (cid:123)(cid:122) (cid:125) k=1
cosθ0
1
Nguyen & Shwedyk A First Course in Digital Communications
k
Therefore
D = C cosθ where θ is either 0 or π
0 0 0 0 y
C
D = ke−jθk, k = 1,2,3,...
k
2
d
What about negative frequencies? Write the third term as Ckejθkej[2πk·(−fr)·t], where
2
(−fr) is interpreted as negative frequency. Therefore D−k = C2ke+jθk, i.e., D−k = Dk∗.
e
(c) Given {D }, then A = 2R{D } and B = −2I{D }. Also C = 2|D |, θ = −∠D ,
k k k k k k k k k
where Dk is in general complex and written as Dk = |Dk|ej∠Dk.
w
Remark: Even though given any set of the coefficients, we can find the other 2 sets, we
can only determine {A ,B } or {D } from the signal, s(t), i.e., there is no method to
k k k
determine {C ,θ } directly.
k k h
Consider now that s(t) is a complex, periodic time signal with period T = 1 , i.e.,
fr
s(t) = s (t)+js (t) where the rSeal and imaginary components, s (t), s (t), are each
R I R I
periodic with period T = 1 . Again we represent s(t) in terms of the orthogonal basis
fr
set {cos(2πkf t),sin(2πkf t)} . That is
r r k=1,2,...
(cid:88)∞
s(t) =&[A cos(2πkf t)+B sin(2πkf t)] (2.1)
k r k r
k=0
(cid:82) (cid:82)
where A = 2 s(t)cos(2πkf t)dt; B = 2 s(t)sin(2πkf t)dt are now complex
k T t∈T r k T t∈T r
numbers.
n
One approach to finding A ,B is to express s (t),s (t) in their own individual Fourier
k k R I
series, and then combine to determine A ,B . That is
k k
e (cid:88)∞ (cid:104) (cid:105)
(R) (R)
s (t) = A cos(2πkf t)+B sin(2πkf t)
R k r k r
k=0
y
(cid:88)∞ (cid:104) (cid:105)
(I) (I)
s (t) = A cos(2πkf t)+B sin(2πkf t)
I k r k r
u k=0
 
(cid:88)∞ (cid:179) (cid:180) (cid:179) (cid:180) 
g⇒ s(t) =  A(R)+jA(I) cos(2πkf t)+ B(R)+jB(I) sin(2πkf t)
 k k r k k r 
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
k=0
=Ak =Bk
N
(d) So now suppose we are given {A ,B }. How do we determine {C ,θ } and D ?
k k k k k
Again, (2.1) can be written as
(cid:183)(cid:181) (cid:182) (cid:181) (cid:182) (cid:184)
(cid:88)∞ A −jB A +jB
s(t) = k k ej2πkfrt+ k k e−j2πkfrt
2 2
k=0
(cid:80)
As before, define D as D ≡ Ak−jBk and note that the term ∞ Ak+jBke−j2πkfrt can
k k 2 k=0 2
be written as
(cid:181) (cid:182)
(cid:88)0 A +jB (cid:88)0 A −jB (cid:88)0
k −kej2πkfrt = k k ej2πkfrt = D ej2πkfrt
k
2 2
k=−∞ k=−∞ k=−∞
Solutions Page 2–2
Nguyen & Shwedyk A First Course in Digital Communications
(cid:80) k
Therefore s(t) = ∞ D ej2πkfrt, where D = Ak−jBk, k = ±1,±2,... and D =
k=−∞ k k 2 0
A −jB (note that B is not necessary equal to zero now).
0 0 0
y
Equation (2.1) can also be written as:
(cid:88)∞ (cid:113) (cid:181) d B (cid:182)
s(t) = A2 +B2cos 2πkf t−tan−1 k
k k r A
k
k=0
e
(cid:113)
from which it follows that C ≡ A2 +B2 and θ = −tan−1 Bk.
k k k k Ak
w
Remarks:
(i) D (cid:54)= D∗ if s(t) is complex.
k k
(ii) The amplitude, C , and θ are in general complex quantities and as such lose their
k k h
usual physical meanings.
(iii) A complex signal, s(t) would arise not from the physical phenomena but from our
S
analysis procedure. This occurs for instance when we go to an equivalent baseband
model.
(cid:110) (cid:111)
Considernowthat D ≡ Ak−jBk aregiven. ThensinceD = Ak+jBk, wehave, upon
k 2 −k 2
adding and subtracting D , D : A = D +D ; B = j(D −D ).
k&−k k k −k k k −k
{C ,θ } can be determined from {A ,B }, which in turn can be determined from D
k k k k k
as above.
(cid:82) (cid:82)
P2.2 (a) A = 2 s(t)cos(2πkf t)dt = 2 s(t)cos(2π(−k)f t)dt = A , i.e., even.
k T (cid:82)t∈T r T t(cid:82)∈T r −k
(b) B = 2 s(t)sin(2πkf t)dt = −2 s(t)sin(2π(−k)f t)dt = −B , i.e., odd.
k (cid:113)T t∈T nr T t∈T r −k
(cid:112) (cid:112)
(c) C = A2 +B2; C = (A )2+(B )2 = (A )2+(−B )2 = C , i.e., even (note
k k k −k −k −k k k k
that squaring an odd real function always gives an even function).
e(cid:179) (cid:180) (cid:179) (cid:180) (cid:179) (cid:180) (cid:179) (cid:180)
(d) θ = −tan−1 Bk ; θ = −tan−1 B−k = −tan−1 −Bk = tan−1 Bk = −θ , i.e.,
k Ak −k A−k Ak Ak k
odd (noteythat tan−1(·) is an odd function).
(e) D = Ak−jBk; D = A−k−jB−k = Ak+jBk = D∗.
k 2 −k 2 2 k
u
Remarks: Properties (a), (b) are true for complex signals as well. But not (c), (d), (e),
at least not always.
g
P2.3 f = 1 = 1 (Hz).
r T 8
A = 2R(D ) = 0; B = −2I(D ) = −2.
1N 1 1 1
A = 2R(D ) = 4; B = −2I(D ) = 0.
5 5 5 5
C = 2|D | = 2; θ = ∠D = π.
1 1 1 1 2
C = 2|D | = 4; θ = 0.
5 5 5
(cid:161) (cid:161) (cid:162)(cid:162) (cid:161) (cid:161) (cid:162)(cid:162) (cid:161) (cid:161) (cid:162)(cid:162) (cid:161) (cid:161) (cid:162)(cid:162)
∴ s(t) = −2sin 2π 1t +4cos 2π 5t = 2cos 2π 1t+ π +4cos 2π 5t .
8 8 8 2 8
P2.4 The fundamental period, f , is f = f (Hz).
r r c
V V
(a) Write s(t) as s(t) = ejαej2πfct+ e−jαe−j2πfct.
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
2 2
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
k=1 k=−1
D1 D−1
Solutions Page 2–3
Nguyen & Shwedyk A First Course in Digital Communications
k
(i) D = V ejα, D = V e−jα = D∗.
1 2 −1 2 1
(ii) α = 0 ⇒ D = V = D .
1 2 −1 y
(iii) α = π ⇒ D = −V = D .
1 2 −1
(iv) α = −π ⇒ same as (iii), not much difference between +π and −π.
(v) α = π ⇒ D = jV , D = −jV . d
2 1 2 −1 2
(vi) α = −π ⇒ D = −jV , D = jV .
2 1 2 −1 2
(b) The magnitude spectrum is the same for all theecases, as expected, since only the phase
of the sinusoid changes. It looks like:
w
| | (volts)
V • •V
2 2
h
- f 0 f f (Hz)
c c
S
Figure 2.1
The phase spectra of cosines change as shown in Fig. 2.2. For (iii) and (iv), which
plot(s) do you prefer and w&hy?
Note: In all cases, the magnitude and phase spectra are even and odd functions, respec-
tively, as they should be.
(c) s(t) = V cosαcos(2πf t)−V sinαsin(2πf t)
c c
(i) A = V cosα, B = −V sinα.
1 n1
(ii) A = V, B = 0.
1 1
(iii) A = −V, B = 0.
1 1
e
(iv) same as (iii).
(v) A = 0, B = −V.
1 1
y
(vi) A = 0, B = V.
1 1
For {C ,θ } we have C = V always and θ = +α.
k k 1 1
u
P2.5 s(t) = 2sin(2πt−3)+6sin(6πt).
(a) Tghe frequency of the first sinusoid is 1 Hz and that of the second one is 6π = 3 Hz. The
2π
fundamental frequency is f = GCD{1,3} = 1 Hz. Rewrite sin(x) as cos(x− π). Then
r 2
   
N
 (cid:179) π(cid:180)  π 
s(t) = 2 cos2πt− 3+ + 1 cos2π(3)t− .
(cid:124)(cid:123)(cid:122)(cid:125)   (cid:124)(cid:123)(cid:122)(cid:125)  
2 2
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124)(cid:123)(cid:122)(cid:125)
C1 C3
θ1 θ3
Use |D | = Ck, ∠D = θ , D = D∗ relationships to obtain:
k 2 k k −k k
∴ D1 = 1ej(3+π2), D3 = 12ejπ2.
∴ D−1 = 1e−j(3+π2), D−3 = 12e−jπ2.
(b) See Fig. 2.3.
Solutions Page 2–4
Nguyen & Shwedyk A First Course in Digital Communications
k
— —
(radians) (radians)
(i) (ii)
•a
y
- f
c • •
0 f f (Hz) - f d0 f f (Hz)
c c c
• (Phase spectrum of acos((cid:215)))
- a e
w
(iii), (iv) Or
— —
(radians) (radians)
•p p •
h
- f f
c c
0 f f (SHz) - f 0 f (Hz)
C c
- p • •- p
&
Or Or
— —
(radians) (radians)
• •
p
n - f f
c c
- f 0 f f (Hz) 0 f (Hz)
c c
(Phase sepectrum of a- cos(cid:215)()) • - p •- p
y
(v) — (vi) —
u (radians) (radians)
•p p •
2 2
g
- f f
c c
0 f f (Hz) - f 0 f (Hz)
c c
N
- p • •- p
2 2
(Phase spectrum of a- sin(cid:215)()) (Phase spectrum of asin((cid:215)))
Figure 2.2
P2.6 Classification of signals is as follows:
Solutions Page 2–5
Nguyen & Shwedyk A First Course in Digital Communications
k
|D |(volts)
k
1
1 y
2
- 3 - 2 - 1 0 1 2 d3 f (Hz)
— D (radians)
k e
( )
3+p
2
p
w
2
- 3 - 2 - 1
- p 0 1 2 3 f (Hz)
2
h
Figure 2.3
S
(a) It has odd (but not halfwave) symmetry.
(b) Neither even, nor odd but show a halfwave symmetry.
&
(c) Even symmetry, also halfwave symmetry, therefore even quarterwave symmetry.
(d) (i) odd symmetry, also halfwave ⇒ odd quarterwave symmetry.
(ii) neither even, nor odd, nor halfwave.
(iii) even but not halfwave.
(e) Neither even, nor odd, nor halfwave.
n
(f) Any even signal still remains even since s(cid:48)(t) = DC+s(t) and s(cid:48)(−t) = DC+s(−t) =
DC+s(t) = s(cid:48)(t).
e
Any odd signal is no longer odd. Note that a DC is an even signal.
y
Any halfwave symmetric signal is no longer halfwave symmetric. Again a DC signal is
not halfwave symmetric and this component destroys the halfwave symmetry. Note that
u
any signal with a nonzero DC component cannot be halfwave symmetric.
g
Regarding the values of the Fourier series coefficients, all that changes is the D , C or
0 0
A value, i.e., the DC component value. All other coefficients are unchanged.
0
(gN) In general the classification changes completely. But for certain specific time shift the
classification may remain the same. To see the effect of a time shift on the coefficients
consider first
(cid:90) (cid:183) (cid:90) (cid:184)
1 1
D(shifted) = s(t−τ)e−j2πkfrtdt λ==t−τ e−j2πkfrτ s(λ)e−j2πkfrλdλ
k T T
t∈T (cid:124) λ∈T (cid:123)(cid:122) (cid:125)
D
k
∴ D(shifted) = e−j2πkfrτD
k k
(cid:183) (cid:184)
(shifted) (shifted)
A −jB A −jB
D(shifted) = k k = [cos(2πkf τ)−jsin(2πkf τ)] k k
k 2 r r 2
Solutions Page 2–6
Nguyen & Shwedyk A First Course in Digital Communications
k
(shifted)
∴ A = A cos(2πkf τ)−B sin(2πkf τ).
k k r k r
(shifted)
∴ B = A sin(2πkf τ)+B cyos(2πkf τ).
k k r k r
Now
(cid:189) (cid:190)
(shifted)
D(shifted) = Ck ejθk(shifted) = e−dj2πkfrτ Ckejθk
k 2 2
(cid:124) (cid:123)(cid:122) (cid:125)
e Dk
(shifted)
∴ Ck =wCk,
(shifted)
∴ θ = θ −j2πkf τ.
k k r
Basically a time shift leaves the amplitude of the sinusoid unchanged and results in a
h
linear change in the phase.
P2.7 (a) s(−t) = s (−t)+s (−t) = s (t)+s (t) = s(t) ⇒ even.
1 2 1 S2
(b) s(−t) = s (−t)+s (−t) = s (t)−s (t), no symmetry.
1 2 1 2
(cid:161) (cid:162) (cid:161) (cid:162) (cid:161) (cid:162)
(c) s t± T = s t± T +s t± T = −s (t)−s (t) = −s(t). Therefore s(t) is halfwave
2 1 2 2 2 1 2
symmetric.
&
(d) Can’t say anything.
(e) Again cannot say anything.
(cid:161) (cid:162) (cid:161) (cid:162) (cid:161) (cid:162)
(f) Consider s t± T = s t± T +s t± T = −s (t)−s (t) = −s(t). Therefore s(t) is
2 1 2 2 2 1 2
halfwave symmetric.
n
Now if both s (t), s (t) are even then s(t) is even and therefore it is even quarterwave
1 2
symmetric. If both are odd, s(t) is odd ands(t) is odd quarterwave symmetric. However
e
if one is even and the other is odd then s(t) is only halfwave symmetric.
(g) s(t) is even but not halfwave symmetric.
y
P2.8 s(t) = s (t)s (t)
1 2
(a) s(−t)u= s (−t)s (−t) = s (t)s (t) = s(t) ⇒ even.
1 2 1 2
(b) s(−t) = s (−t)s (−t) = −s (t)s (t) = −s(t) ⇒ odd.
1 2 1 2
(cid:161) (cid:162) (cid:161) (cid:162) (cid:161) (cid:162)
(c) sgt± T = s t± T s t± T = [−s (t)][−s (t)] = s(t). Therefore s(t) is not
2 1 2 2 2 1 2
halfwave symmetric. Note that the fundamental period changes to T.
2
(dN) Neither even nor halfwave symmetric.
(e) Neither odd nor halfwave symmetric.
(f) Have 3 possibilities: (i) s (t), s (t) are both even quarterwave; (ii) s (t), s (t) are both
1 2 1 2
odd quarterwave; (iii) one is odd quarterwave, the other is even quarterwave.
(i) From (a) it follows that s(t) is even. From (c) it follows that it is not halfwave
symmetric.
(cid:161) (cid:162) (cid:161) (cid:162) (cid:161) (cid:162)
(ii) Easytoshowthats(t)iseven. Buts t± T = s t± T s t± T = [−s (t)][−s (t)] =
(cid:161) (cid:162) 2 1 2 2 2 1 2
s (t)s (t) = s(t) (cid:54)= −s t± T , therefore not halfwave symmetric.
1 2 2
Solutions Page 2–7
Nguyen & Shwedyk A First Course in Digital Communications
(cid:161) (cid:162) k(cid:161) (cid:162)
(iii) s(t) is odd (follows from (b)) but again s t± T = s(t) (cid:54)= −s t± T , therefore not
2 2
halfwave symmetric.
y
Again note that in each case the fundamental period changes to T.
2
(g) s(t) is even. Is it halfwave symmetric? d
(cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182)
T T T T
s t± = s t± s t± = −s t± s (t) (cid:54)= −s(t)
2 1 2 2 2e1 2 2
No.
w
P2.9 (a) Check for 2 conditions: that the magnitude is even and the phase is odd. If both are
satisfied then the signal is real. If one or the other or neither condition is satisfied then
the signal is complex.
h
(i) real, (ii) real (note the phase of π is the same as −π), (iii) complex (phase spectrum
is not odd).
S
(b) If a harmonic frequency has a phase of π (or 0) then there is only a cos(·) term at this
frequency. Similarly if the phase is π (or 3π) then there is only a sin(·) term at that
2 2
frequency. Finally if the signal is even then it only has cos(·) terms in the Fourier series,
if odd then only sin(·) terms.
&
Therefore the signal in (ii) is even (and real). No signal is odd.
P2.10 The two spectra are
D coefficients of V cos(2p f t)
n k m m
V V
e m2 m2
- f 0 f f (Hz)
m m
y
D coefficients of V cos(2p f t)
k c c
uV V
c2 c2
g
- fc 0 - fc f (Hz)
N
V V V V V V V V
m c m c m c m c
4 4 4 4
- f- f - f+ f 0 f - f f + f f (Hz)
c m c m c m c m
Figure 2.4
Solutions Page 2–8