Solution Manual For Signals and Systems Using MATLAB® 4th Edition by Akan , Aydin| Latest Version

SOLUTION MANUAL FOR
Addition Problems For
Signals And Systems Using
Matlab

Luis F. Chaparro And Aydin Akan
4th Edition

,Chapter 0

From The Ground Up

0.1 Basic Problems

0.1 Consider The Following Problems About Trigonometric And Polar Forms.
(a) Let Z = 6ejπ/4 Find (I) Re(Z), (Ii) Im(Z)
(b) If Z = 8 + J3 And V = 9 − J2, Is It True That

(i) Re(Z) = 0.5(Z + Z∗)? (Ii) Im(V) = −0.5j(V − V∗)?
∗ ∗
(Iii) Re(Z + V ) = Re(Z + V)? (Iv) Im(Z + V ) = Im(Z − V)?
√ √
Answers: (A) Re(Z) = 3 2; Im(Z) = 3 2; (B) Yes To All.
Solution
(a) Z = 6ejπ/4 = 6 Cos(Π/4) + J6 Sin(Π/4)
√
i. Re(Z) = 6 Cos(Π/4) = 3 2
√
ii. Im(Z) = 6 Sin(Π/4) = 3 2
∗
(b) I. Yes, Re(Z) = 0.5(Z + Z ) = 0.5(2Re(Z)) = Re(Z) = 8
∗
ii. Yes, Im(V) = −0.5j(V − V ) = −0.5j(2jim(V)) = Im(V) = −2
∗ ∗
iii. Yes, Re(Z + V ) = Re(Re(Z) + Re(V ) + Im(Z) − Im(V)) = Re(Z + V) = 17
∗
iv. Yes, Im(Z + V ) = Im(17 + J5) = Im(Z − V) = Im(−9 + J5) = 5




1

,Chaparro-Akan — Signals and Systems using MATLAB 0.2


0.2 Using The Vectorial Representation Of Complex Numbers It Is Possible To Get Some Interesting Inequalities.
(a) Is It True That For A Complex Number Z = X + Jy We Have That |X| ≤ |Z|? Show It
Geometrically By Representing Z As A Vector.
(b) The So Called Triangle Inequality Says That For Any Complex (Or Real) Numbers Z And V We Have That
|Z + V| ≤ |Z| + |V|. Show This Geometrically.
(c) If Z = 1 + J And V = 2 + J Is It True That

(I) |Z + V| ≤ |Z| + |V|? (Ii) |Z − V| ≤ |Z| + |V|?

Answer: (A) |X| = |Z|| Cos(Θ)| And Since | Cos(Θ)| ≤ 1 Then |X| ≤ |Z|; (C) Yes To Both.
Solution
(a) Representing The Complex Number Z = X + Jy = |Z|Ejθ Then |X| = |Z|| Cos(Θ)| And Since | Cos(Θ)| ≤ 1
Then |X| ≤ |Z|, The Equality Holds When Θ = 0 Or When Z = X, I.E., It Is Real.
(b) Adding Two Complex Numbers Is Equivalent To Adding Two Vectors To Create A Triangle With Two
Sides The Two Vectors Being Added And The Other Side The Vector Resulting From The Addition.
Unless The Two Vector Being Added Have The Same Angle, In Which Case |Z| + |V| = |Z + V|, It Holds
That |Z| + |V| > |Z + V|.




Figure 1: Problem 2: Addition Of Two Vectors Illustrating The Triangular Inequality.

(c) The Answer To Both Is Yes. Indeed,
√ √ √
(A) |Z + V| = 13 ≤ |Z| + |V| = 2 + 5
√ √
(B) |Z − V| = | − 1| = 1 ≤ |Z| + |V| = 2 + 5




Copyright 2018, Elsevier, Inc. All rights reserved.

, Chaparro-Akan — Signals and Systems using MATLAB 0.3


0.3 Use Euler’s Identity In The Following Problems.
(a) Find Trigonometric Identities In Terms Of Sin(Α), Sin(Β), Cos(Α), Cos(Α) For

(I) Cos(Α + Β) (Ii) Sin(Α + Β)

(b) Is It True That
∫ 1
Ej2πt Dt = 0 ?
0

(c) Is It True That
i. (−1)N = Cos(Πn) For Any Integer
N?
ii. Ej0 + Ejπ/2 + Ejπ + Ej3π/2 = 0 ? (Sketch A Figure)
Answer: (A) Cos(Α + Β) = Cos(Α) Cos(Β) − Sin(Α) Sin(Β); (C) Yes To Both.

Solution
(a) I. 2 Cos(Α
Jα Jβ
+ Β) = Ej(Α+Β) + E−J(Α+Β) = (Ejαejβ) + (Ejαejβ)∗ = 2R E(Ejαejβ) And
Re[E E ] = Re[(Cos(Α) + J Sin(Α))(Cos(Β) + J Sin(Β))] = Cos(Α) Cos(Β) − Sin(Α) Sin(Β) So That

Cos(Α + Β) = Cos(Α) Cos(Β) − Sin(Α) Sin(Β)

Ii. 2j Sin(Α + Β) = Ejαejβ − (Ejαejβ)∗ = 2jim[Ejαejβ], And The Imaginary Is
Sin(Α) Cos(Β) + Cos(Α) Sin(Β) = Sin(Α + Β)
(b)
∫ 1
Ej2πt 1 Ej2π — 1
Ej2πt Dt = |0 = =0
0 J2π J2π
Also ∫ ∫ ∫
1 1 1
Ej2πt Dt = Cos(2πt)Dt + Sin(2πt)Dt = 0 + J0
0 0
J
0
Since The Integrals Of The Sinusoids Are Over A Period.
(c) I. Yes, (−1)N = (Ejπ)N = Ejnπ = Cos(Nπ) + J Sin(Nπ) = Cos(Nπ) Since Sin(Nπ) = 0 For Any Integer
N.
Ii. Yes, Ej0 = −Ejπ And Ejπ/2 = −Ej3π/2 So They Add To Zero.




Copyright 2018, Elsevier, Inc. All rights reserved.