Class notes
A First Course in Digital Communications Solution Manual PDF
- Course
- Institution
- Book
Complete Solutions Manual for A First Course in Digital Communications by Ha H Nguyen and Ed Shwedyk.
[Show more]
Connected book
Book Title:
Author(s):
- Edition:
- ISBN:
- Edition:
Seller
Follow
SolutionsWizard
Reviews received
Content preview
kdy
we
Solutions Manual
“A First Course in Digital Communications”
Sh
Cambridge University Press
Ha H. Nguyen
Department of Electrical & Computer Engineering
&
University of Saskatchewan
Saskatoon, SK, CANADA S7N 5A9
Ed Shwedyk
Department of Electrical & Computer Engineering
University of Manitoba
en
Winnipeg, MB, CANADA R3T 5V6
August 5, 2011
uy
Ng
, k
dy
Preface
we
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 to ha.nguyen@usask.ca
or shwedyk@ee.umanitoba.ca.
Sh
&
en
uy
Ng
1
, k
dy
Chapter 2
Deterministic Signal Characterization
we
and Analysis
(a) Given {Ak , qBk }, i.e., s(t) =
Then Ck = Ak + Bk , θk = tan−1 B
2 2 Sh
P2.1 First let us establish (or review) the relationships for real signals
P∞
k=0 [Ak cos (2πkfr t) + Bk sin (2πkfr t)], Ak , Bk are real.
Ak , where we write s(t) as
k
s(t) =
∞
X
Ck cos (2πkfr t − θk )
&
k=0
If we choose to write it as
∞
X
s(t) = Ck cos (2πkfr t + θk )
k=0
en
Bk
then the phase becomes θk = − tan−1 Ak . Further
Ak − jBk
Dk =
2
uy
D−k = Dk∗ , k = 1, 2, 3, . . .
D0 = A0 (B0 = 0 always).
(b) Given {Ck , θk } then Ak = Ck cos θk , Bk = Ck sin θk . They are obtained from:
∞
X ∞
X
Ng
s(t) = Ck cos (2πkfr t − θk ) = [Ck cos θk cos(2πkfr t) + Ck sin θk sin(2πkfr t)]
k=0 k=0
Now s(t) can written as:
∞
( )
X ej[2πkfr t−θk ] + e−j[2πkfr t−θk ]
s(t) = Ck
2
k=0
· ¸ X∞ · ¸
e−jθ0 + ejθ0 Ck −jθk j2πkfr t Ck jθk −j2πkfr t
s(t) = C0 + e e + e e
2 2 2
| {z } k=1
cos θ0
1
, Nguyen & Shwedyk A First Course in Digital Communications
Therefore
D0 = C0 cos θ0 where θ0 is either 0 or π
k
Ck −jθk
Dk = e , k = 1, 2, 3, . . .
2
dy
What about negative frequencies? Write the third term as C2k ejθk ej[2πk·(−fr )·t] , where
(−fr ) is interpreted as negative frequency. Therefore D−k = C2k e+jθk , i.e., D−k = Dk∗ .
(c) Given {Dk }, then Ak = 2R{Dk } and Bk = −2I{Dk }. Also Ck = 2|Dk |, θk = −∠Dk ,
where Dk is in general complex and written as Dk = |Dk |ej∠Dk .
we
Remark: Even though given any set of the coefficients, we can find the other 2 sets, we
can only determine {Ak , Bk } or {Dk } from the signal, s(t), i.e., there is no method to
determine {Ck , θk } directly.
Consider now that s(t) is a complex, periodic time signal with period T = f1r , i.e.,
Sh
s(t) = sR (t) + jsI (t) where the real and imaginary components, sR (t), sI (t), are each
periodic with period T = f1r . Again we represent s(t) in terms of the orthogonal basis
set {cos (2πkfr t), sin (2πkfr t)}k=1,2,... . That is
s(t) =
∞
X
k=0
[Ak cos (2πkfr t) + Bk sin (2πkfr t)] (2.1)
2
R 2
R
&
where Ak = T t∈T s(t) cos (2πkfr t)dt; Bk = T t∈T s(t) sin (2πkfr t)dt are now complex
numbers.
One approach to finding Ak , Bk is to express sR (t), sI (t) in their own individual Fourier
series, and then combine to determine Ak , Bk . That is
en
∞ h
X i
(R) (R)
sR (t) = Ak cos (2πkfr t) + Bk sin (2πkfr t)
k=0
∞ h
X i
(I) (I)
sI (t) = Ak cos (2πkfr t) + Bk sin (2πkfr t)
uy
k=0
X ³ (R)
∞ ´ ³ ´
⇒ s(t) = A + jA(I) cos (2πkfr t) + B (R) + jB (I) sin (2πkfr t)
k k k k
k=0 | {z } | {z }
=Ak =Bk
Ng
(d) So now suppose we are given {Ak , Bk }. How do we determine {Ck , θk } and Dk ?
Again, (2.1) can be written as
∞ ·µ
X ¶ µ ¶ ¸
Ak − jBk j2πkfr t Ak + jBk −j2πkfr t
s(t) = e + e
2 2
k=0
Ak −jBk P∞ Ak +jBk −j2πkfr t
As before, define Dk as Dk ≡ 2 and note that the term k=0 2 e can
be written as
0
X 0
X µ ¶ 0
X
Ak + jB−k j2πkfr t Ak − jBk
e = ej2πkfr t = Dk ej2πkfr t
2 2
k=−∞ k=−∞ k=−∞
Solutions Page 2–2
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller SolutionsWizard. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $10.49. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
73314 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now