EI331 Signals and Systems Lecture 2 Bo Jiang John Hopcroft Center - - PowerPoint PPT Presentation

EI331 Signals and Systems

Lecture 2 Bo Jiang

John Hopcroft Center for Computer Science Shanghai Jiao Tong University

February 28, 2019

1/31

Contents

• 1. Complex Number

1.1 Arithmetic and conjugate 1.2 Trigonometric and exponential forms

• 2. Power, Root, and Exponential

2.1 Power 2.2 Root 2.3 Exponential

• 3. Convergence and Infinite Series

3.1 Infinite sequence 3.2 Infinite series

2/31

Complex Number

Complex number has form z = x + jy, with x, y ∈ R

• j imaginary unit: j2 = −1
• real part: Re z = x;

imaginary part: Im z = y

• z is real if y = 0, purely imaginary if x = 0
• equality: x1 + jy1 = x2 + jy2 ⇐⇒ x1 = x2 and y1 = y2
• no natural order
• set of all complex numbers: C

Geometric representation

• complex plane: x + jy ↔ (x, y)
• x axis: real axis
• y axis: imaginary axis
• free vector −

− → OP Re Im O z = x + jy P x y

3/31

Complex Arithmetic

Same rules for addition and multiplication as for real numbers. Given z1 = x1 + jy1 and z2 = x2 + jy2 Addition: z1 + z2 = (x1 + x2) + j(y1 + y2) Subtraction: z1 − z2 = (x1 − x2) + j(y1 − y2) Multiplication: z1z2 = (x1x2 − y1y2) + j(x1y2 + x2y1) Division: For z2 0, quotient of z1 and z2 is z satisfying z2z = z1 z1 z2 = x1x2 + y1y2 x2

2 + y2 2

+ j x2y1 − x1y2 x2

2 + y2 2

• Exercise. After slide 5, obtain above formula from

z1 z2 = z1¯ z2 |z2|2

4/31

Complex Arithmetic

Laws of complex arithmetic, same as for reals Commutative laws z1 + z2 = z2 + z1, z2z1 = z1z2 Associative laws (z1 + z2) + z3 = z1 + (z2 + z3), (z1z2)z3 = z1(z2z3) Distributive laws z1(z2 + z3) = z1z2 + z1z3, (z1 + z2)z3 = z1z3 + z2z3

5/31

Complex Conjugate

Complex conjugate of z = x + jy ¯ z = x − jy Properties

• (¯

z) = z

• z1 ± z2 = ¯

z1 ± ¯ z2

• z1z2 = ¯

z1¯ z2

• z1

z2

• =

¯ z1 ¯ z2

• z + ¯

z = 2Re z

• z − ¯

z = 2jIm z Re Im O z = x + jy x y ¯ z = x − jy −y

6/31

Trigonometric Form

Trigonometric form z = r(cosθ + j sinθ) Magnitude (modulus, absolute value) |z| = r =

• x2 + y2
• |z| = |¯

z| = r =

• x2 + y2
• z¯

z = |z|2 Phase (argument) Arg z = θ + 2πn for n ∈ Z

• undefined for z = 0

Principal value of phase: arg z = θ ∈ (−π, π] or [0, 2π) Re Im O r z = x + jy x y θ

7/31

Exponential Form

Euler’s formula ejθ cosθ + j sinθ Special cases

• ej0 = ej2π = 1
• ej π

2 = j

• ejπ = −1
• e−j π

2 = −j

Unit modulus: |ejθ | = 1 Periodicity: ej(θ+2π) = ejθ Exponential form z = rejθ, ¯ z = re−jθ

Leonhard Euler

(from Wikipedia)

Re Im O 1 ejθ θ cosθ sinθ e−jθ −θ − sinθ

8/31

Geometric Addition

Parallelogram law for addition and subtraction Re Im O z1 z2 z1 + z2 Re Im O z1 z2 −z2 z1 − z2 Triangle inequality |z1 ± z2| ≤ |z1| + |z2| =⇒ |z1 − z2| ≥

• |z1| − |z2|

9/31

Algebraic Proof of Triangle Inequality

|z1 + z2|2 = (z1 + z2)(z1 + z2) = (z1 + z2)(¯ z1 + ¯ z2) = |z1|2 + |z2|2 + z1¯ z2 + ¯ z1z2 = |z1|2 + |z2|2 + z1¯ z2 + z1¯ z2 = |z1|2 + |z2|2 + 2Re (z1¯ z2) ≤ |z1|2 + |z2|2 + 2|z1¯ z2| use |z| ≥ |Re z| = |z1|2 + |z2|2 + 2|z1| · |¯ z2| use |z1z2| = |z1| · |z2| (next slide) = |z1|2 + |z2|2 + 2|z1| · |z2| = (|z1| + |z2|)2

10/31

Geometric Multiplication

Given zi = ri(cosθi + j sinθi) = ejθi z1z2 = r1r2(cosθ1 + j sinθ1)(cosθ2 + j sinθ2) = r1r2(cosθ1 cosθ2 − sinθ1 sinθ2) + jr1r2(cosθ1 sinθ2 + cosθ2 sinθ1) = r1r2(cos(θ1 + θ2) + j sin(θ1 + θ2)) = r1r2ej(θ1+θ2) |z1z2| = |z1|·|z2| Arg(z1z2) = Arg z1+Arg z2 (r1ejθ1)(r2ejθ2) = r1r2ej(θ1+θ2),

r1ejθ1 r2ejθ2 = r1 r2ej(θ1−θ2)

Re Im O z2 θ2 θ2 z1 θ1 z1z2 Similarly,

• z1

z2

• =

|z1| |z2| Arg

• z1

z2

• = Arg z1 − Arg z2

11/31

Calculation with Exponential Form

• Example. Simplify S = n

k=1 cos(nθ)

• Solution. If θ = 2mπ for m ∈ Z, then S = n. Now assume

θ 2mπ for any m ∈ Z. 1. S =

n

• k=1

cos(nθ) =

n

• k=1

Re ejnθ = Re n

• k=1

ejnθ

• 2.

n

• k=1

ejnθ =

n

• k=1
• ejθn

= ej(n+1)θ − ejθ ejθ − 1 = ej (n+2)θ

2

(ej nθ

2 − e−j nθ 2 )

ej θ

2 (ej θ 2 − e−j θ 2 )

= ej (n+1)θ

2

sin nθ

2

sin θ

2

3. S = cos (n+1)θ

2

sin nθ

2

sin θ

2

12/31

Contents

• 1. Complex Number

1.1 Arithmetic and conjugate 1.2 Trigonometric and exponential forms

• 2. Power, Root, and Exponential

2.1 Power 2.2 Root 2.3 Exponential

• 3. Convergence and Infinite Series

3.1 Infinite sequence 3.2 Infinite series

13/31

Power

For n ∈ N, n-th power of z is defined by zn z · z · · · z

• n times

z−n 1 zn Exponential and trigonometric forms

• rejθn

= rnejnθ = rn[cos(nθ) + j sin(nθ)] r = 1 =⇒ de Moivre’s formula (cosθ + j sinθ)n = cos(nθ) + j sin(nθ)

Abraham de Moivre

(from Wikipedia)

14/31

Root

For n ∈ N, n-th root of z = rejθ is w = ρejφ satisfying wn = z 1. wn = z ⇔ ρnejnφ = rejθ ⇔

• ρn = r ⇔ ρ = r

1 n

ejnφ = ejθ 2. ejnφ = ejθ ⇔ nφ = θ + 2kπ, (k ∈ Z) ⇔ φ = θ n + 2kπ n , (k ∈ Z) 3. w = r

1 nej( θ n + 2kπ n ),

(k ∈ Z)

• 4. ej( θ

n + 2kπ n ) is periodic in k with period n

• 5. n distinct n-th roots for z 0

wk = r

1 nej( θ n + 2kπ n ),

k = 0, 1, 2, . . . , n − 1

15/31

Root

• n-th root of unity: wn = 1

wk = ej 2kπ

n ,

k = 0, . . . , n−1

• let

ω = ej 2π

n

=⇒ wk = ωk

• equally spaced on unit

circle

• 1 + ω + · · · + ωn−1 = 0

Re Im O 1

2π 5

ω ω2 ω3 ω4

16/31

Root

Example Find all 4th roots of 1 + j Solution. Use exponential form 1 + j = √ 2ej π

4

For k = 0, 1, 2, 3, wk =

8

√ 2ej( π

16+ 2kπ 4 )

Note w1 = jw0, w2 = −w0, w3 = −jw0 w0 + w1 + w2 + w3 = 0 Re Im O

8

√ 2 1 + j w0 w1 w2 w3

17/31

Complex Exponential

ez = ex+jy exejy = ex(cos y + j sin y) Equivalently,

• |ez| = eRe z

Arg(ez) = Im z + 2kπ, k ∈ Z Properties

• ez 0, since |ez| = ex > 0
• ez1+z2 = ez1ez2
• 1/ez = e−z
• ez = e¯

z

• Periodicity: ez+j2π = ez

Re Im O ez ex y ex cos y ex sin y

18/31

Complex Exponential

Complex exponential vs. real exponential real ex complex ez with x ∈ R with z ∈ C always nonzero Yes Yes always positive Yes No monotonic Yes No periodic No Yes

19/31

Complex Exponential

Alternative definition ez =

∞

• n=0

zn n! = 1 + z + z2 2! + z3 3! + · · ·

• can do formal calculations as if z is real
• will revisit convergence issue later
• Example. Check Euler’s formula

ejθ =

∞

• n=0

(jθ)n n! =

∞

• k=0

(jθ)2k (2k)! +

∞

• k=0

(jθ)2k+1 (2k + 1)! =

∞

• k=0

(−1)kθ2k (2k)! + j

∞

• k=0

(−1)kθ2k+1 (2k + 1)! = cosθ + j sinθ

20/31

Complex Exponential

• Example. Check the addition formula ez1+z2 = ez1ez2

ez1+z2 =

∞

• k=0

(z1 + z2)k k! =

∞

• k=0

1 k!

k

• m=0

k! m!(k − m)!zm

1 zk−m 2

=

∞

• k=0

k

• m=0

1 m!(k − m)!zm

1 zk−m 2

=

∞

• m=0

∞

• k=m

1 m!(k − m)!zm

1 zk−m 2

=

∞

• m=0

∞

• ℓ=0

1 m!ℓ!zm

1 zℓ 2

= ∞

• m=0

1 m!zm

1

∞

• ℓ=0

1 ℓ!zℓ

2

• = ez1ez2

(to be justified by absolute convergence)

21/31

Contents

• 1. Complex Number

1.1 Arithmetic and conjugate 1.2 Trigonometric and exponential forms

• 2. Power, Root, and Exponential

2.1 Power 2.2 Root 2.3 Exponential

• 3. Convergence and Infinite Series

3.1 Infinite sequence 3.2 Infinite series

22/31

Distance

Distance between z1, z2 ∈ C d(z1, z2) = |z1 − z2| Same as Euclidean distance in R2 |z1 − z2| = |(x1 + jy1) − (x2 + jy2)| =

• (x1 − x2)2 + (y1 − y2)2

Re Im O z1 z2 −z2 z1 − z2

23/31

Convergence of Sequence

• Definition. A sequence {zn : n ≥ 1} ⊂ C converges to z ∈ C

if lim

n→∞ |zn − z| = 0.

We say z is the limit of the sequence {zn} and write z = lim

n→∞ zn,

• r

zn → z, as n → ∞

• Theorem. Let zn = xn + jyn and z = x + jy. Then

lim

n→∞ zn = z ⇐⇒ lim n→∞ xn = x

and lim

n→∞ yn = y

• Proof. Use

|xn − x| |yn − y|

• ≤ |zn − z| ≤ |xn − x| + |yn − y|

24/31

Cauchy Sequence

• Definition. {zn} ⊂ C is a Cauchy sequence if

|zn − zm| → 0, as n, m → ∞.

• Theorem. {zn = xn + jyn} is Cauchy iff both

{xn} and {yn} are Cauchy.

• Proof. Use

|xn − xm| |yn − ym|

• ≤ |zn − zm| ≤ |xn − xm| + |yn − ym|.

Augustin-Louis Cauchy (from Wikipedia)

25/31

Completeness of C

• Theorem. C is complete, i.e. every Cauchy sequence in C

converges to a point in C.

• Proof. Consider a Cauchy sequence zn = xn + jyn, n ≥ 1.
• 1. {xn} and {yn} are Cauchy
• 2. R is complete =⇒ xn → x and yn → y for some x, y ∈ R
• 3. zn → z = x + jy

26/31

Series

• Definition. An infinite series

∞

• n=1

zn = z1 + z2 + z3 + · · · converges if the sequence of its partial sums sk = k

n=1 zn

converges, and we call the limit of sk the sum of the series ∞

n=1 zn, i.e.

s =

∞

• n=1

zn ⇐⇒ lim

k→∞ sk = s

If sk does not converge, we say the series ∞

n=1 zn

diverges.

27/31

Series

Theorem (Cauchy’s convergence test). The series ∞

n=1 zn

converges iff given any ϵ > 0, there exists N ∈ N such that for any n > N and p ≥ 1 |zn+1 + zn+2 + · · · + zn+p| < ϵ.

• Proof. Let sk = k

n=1 zn. Note

sn+p − sn = zn+1 + zn+2 + · · · + zn+p. Use Cauchy’s convergence test for the sequence {sk}.

28/31

Series

• Theorem. Let zn = xn + jyn. The series ∞

n=1 zn converges

iff both ∞

n=1 xn and ∞ n=1 yn converge.

• Proof. Let sk = k

n=1 zn, σk = k n=1 xn, and τk = k n=1 yn.

Then sk = σk + jτk.

• n

zn converges ⇐⇒ {sk} converges ⇐⇒ {σk} and {τk} converge ⇐⇒

• n

xn and

• n

yn converge

29/31

Series

• Theorem. If ∞

n=1 zn converges, then zn → 0 as n → ∞.

• Proof. Let zn = xn + jyn
• n

zn converges ⇐⇒

• n

xn and

• n

yn converge =⇒ xn → 0 and yn → 0 as n → ∞ =⇒ zn → 0 as n → ∞

30/31

Absolute Convergence

• Theorem. If ∞

n=1 |zn| converges, then ∞ n=1 zn converges

• Proof. Let zn = xn + jyn. For any ϵ > 0, when n is large

enough,

• n+p
• k=n

zk

• ≤

p

• k=n

|zn| < ϵ. Definition.

• If ∞

n=1 |zn| converges, we say ∞ n=1 zn converges

absolutely.

• If ∞

n=1 zn converges but does not converge

absolutely, we say it converges conditionally.

31/31

Complex Exponential Revisited

Recall ez =

∞

• k=0

zk k! The series converges absolutely, as

n

• k=0
• zk

k!

• =

n

• k=0

|z|k k! → e|z|.