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

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 1/31

Complex Number Complex number has form z = x + jy , with x , y ∈ R • j imaginary unit: j 2 = − 1 • real part: Re z = x ; imaginary part: Im z = y • z is real if y = 0 , purely imaginary if x = 0 • equality: x 1 + jy 1 = x 2 + jy 2 ⇐⇒ x 1 = x 2 and y 1 = y 2 • no natural order • set of all complex numbers: C Im Geometric representation z = x + jy y P • complex plane: x + jy ↔ ( x , y ) • x axis: real axis • y axis: imaginary axis x • free vector − − → O Re OP 2/31

Complex Arithmetic Same rules for addition and multiplication as for real numbers. Given z 1 = x 1 + jy 1 and z 2 = x 2 + jy 2 Addition: z 1 + z 2 = ( x 1 + x 2 ) + j ( y 1 + y 2 ) Subtraction: z 1 − z 2 = ( x 1 − x 2 ) + j ( y 1 − y 2 ) Multiplication: z 1 z 2 = ( x 1 x 2 − y 1 y 2 ) + j ( x 1 y 2 + x 2 y 1 ) Division: For z 2 � 0 , quotient of z 1 and z 2 is z satisfying z 2 z = z 1 z 1 x 1 x 2 + y 1 y 2 x 2 y 1 − x 1 y 2 + j = x 2 2 + y 2 x 2 2 + y 2 z 2 2 2 Exercise. After slide 5, obtain above formula from = z 1 ¯ z 1 z 2 | z 2 | 2 z 2 3/31

Complex Arithmetic Laws of complex arithmetic, same as for reals Commutative laws z 1 + z 2 = z 2 + z 1 , z 2 z 1 = z 1 z 2 Associative laws ( z 1 + z 2 ) + z 3 = z 1 + ( z 2 + z 3 ) , ( z 1 z 2 ) z 3 = z 1 ( z 2 z 3 ) Distributive laws z 1 ( z 2 + z 3 ) = z 1 z 2 + z 1 z 3 , ( z 1 + z 2 ) z 3 = z 1 z 3 + z 2 z 3 4/31

Complex Conjugate Im Complex conjugate of z = x + jy z = x + jy y ¯ z = x − jy Properties x O Re • ( ¯ z ) = z • z 1 ± z 2 = ¯ z 1 ± ¯ z 2 − y ¯ z = x − jy • z 1 z 2 = ¯ z 1 ¯ z 2 � � ¯ z 1 z 1 • z + ¯ z = 2 Re z • = ¯ z 2 z 2 • z − ¯ z = 2 j Im z 5/31

Trigonometric Form Trigonometric form Im z = x + jy y z = r ( cos θ + j sin θ ) r Magnitude (modulus, absolute value) θ � x O Re x 2 + y 2 | z | = r = � z = | z | 2 • z ¯ • | z | = | ¯ x 2 + y 2 z | = r = Phase (argument) Arg z = θ + 2 π n for n ∈ Z • undefined for z = 0 Principal value of phase: arg z = θ ∈ (− π , π ] or [ 0 , 2 π ) 6/31

Exponential Form Euler’s formula Leonhard Euler (from Wikipedia) e j θ � cos θ + j sin θ Special cases Im • e j 0 = e j 2 π = 1 e j θ • e j π 2 = j sin θ • e j π = − 1 • e − j π 2 = − j cos θ θ Unit modulus: | e j θ | = 1 O 1 Re − θ Periodicity: e j ( θ + 2 π ) = e j θ Exponential form − sin θ z = re − j θ z = re j θ , ¯ e − j θ 7/31

Geometric Addition Parallelogram law for addition and subtraction Im Im z 1 + z 2 z 2 z 2 z 1 z 1 O Re O Re z 1 − z 2 − z 2 Triangle inequality � � | z 1 ± z 2 | ≤ | z 1 | + | z 2 | = ⇒ | z 1 − z 2 | ≥ � | z 1 | − | z 2 | � 8/31

Algebraic Proof of Triangle Inequality | z 1 + z 2 | 2 = ( z 1 + z 2 )( z 1 + z 2 ) = ( z 1 + z 2 )( ¯ z 1 + ¯ z 2 ) = | z 1 | 2 + | z 2 | 2 + z 1 ¯ z 2 + ¯ z 1 z 2 = | z 1 | 2 + | z 2 | 2 + z 1 ¯ z 2 + z 1 ¯ z 2 = | z 1 | 2 + | z 2 | 2 + 2 Re ( z 1 ¯ z 2 ) ≤ | z 1 | 2 + | z 2 | 2 + 2 | z 1 ¯ z 2 | use | z | ≥ | Re z | = | z 1 | 2 + | z 2 | 2 + 2 | z 1 | · | ¯ z 2 | use | z 1 z 2 | = | z 1 | · | z 2 | (next slide) = | z 1 | 2 + | z 2 | 2 + 2 | z 1 | · | z 2 | = (| z 1 | + | z 2 |) 2 9/31

Geometric Multiplication Given z i = r i ( cos θ i + j sin θ i ) = e j θ i z 1 z 2 Im z 1 z 2 = r 1 r 2 ( cos θ 1 + j sin θ 1 )( cos θ 2 + j sin θ 2 ) z 1 = r 1 r 2 ( cos θ 1 cos θ 2 − sin θ 1 sin θ 2 ) θ 2 + jr 1 r 2 ( cos θ 1 sin θ 2 + cos θ 2 sin θ 1 ) z 2 θ 1 = r 1 r 2 ( cos ( θ 1 + θ 2 ) + j sin ( θ 1 + θ 2 )) θ 2 = r 1 r 2 e j ( θ 1 + θ 2 ) O Re | z 1 z 2 | = | z 1 |·| z 2 | Arg ( z 1 z 2 ) = Arg z 1 + Arg z 2 � � � � | z 1 | � z 1 � z 1 � � Similarly, Arg = Arg z 1 − Arg z 2 � = � � | z 2 | z 2 z 2 � r 1 e j θ 1 r 2 e j θ 2 = r 1 ( r 1 e j θ 1 )( r 2 e j θ 2 ) = r 1 r 2 e j ( θ 1 + θ 2 ) , r 2 e j ( θ 1 − θ 2 ) 10/31

Calculation with Exponential Form Example. Simplify S = � n k = 1 cos ( n θ ) Solution. If θ = 2 m π for m ∈ Z , then S = n . Now assume θ � 2 m π for any m ∈ Z . � n � n n � � � Re e jn θ = Re e jn θ cos ( n θ ) = 1 . S = k = 1 k = 1 k = 1 n n = e j ( n + 1 ) θ − e j θ � e j θ � n � � e jn θ = 2 . e j θ − 1 k = 1 k = 1 = e j ( n + 1 ) θ = e j ( n + 2 ) θ ( e j n θ 2 − e − j n θ sin n θ 2 ) 2 2 2 e j θ 2 ( e j θ 2 − e − j θ sin θ 2 ) 2 S = cos ( n + 1 ) θ sin n θ 2 2 3 . sin θ 2 11/31

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 12/31

Power For n ∈ N , n -th power of z is defined by z n � z · z · · · z � ��� �� ��� � n times z − n � 1 z n Exponential and trigonometric forms � re j θ � n = r n e jn θ = r n [ cos ( n θ ) + j sin ( n θ )] r = 1 = ⇒ de Moivre’s formula Abraham de Moivre ( cos θ + j sin θ ) n = cos ( n θ ) + j sin ( n θ ) (from Wikipedia) 13/31

Root For n ∈ N , n -th root of z = re j θ is w = ρ e j φ satisfying w n = z 1. � 1 ρ n = r ⇔ ρ = r w n = z ⇔ ρ n e jn φ = re j θ ⇔ n e jn φ = e j θ 2. e jn φ = e j θ ⇔ n φ = θ + 2 k π , ( k ∈ Z ) ⇔ φ = θ n + 2 k π n , ( k ∈ Z ) 3. 1 n e j ( θ n + 2 k π n ) , w = r ( k ∈ Z ) n ) is periodic in k with period n 4. e j ( θ n + 2 k π 5. n distinct n -th roots for z � 0 1 n e j ( θ n + 2 k π n ) , w k = r k = 0 , 1 , 2 , . . . , n − 1 14/31

Root Im • n -th root of unity: w n = 1 ω w k = e j 2 k π n , k = 0 , . . . , n − 1 ω 2 • let 2 π ω = e j 2 π 5 n = ⇒ w k = ω k O 1 Re • equally spaced on unit ω 3 circle • ω 4 1 + ω + · · · + ω n − 1 = 0 15/31

Root Example Find all 4 th roots of 1 + j Im Solution. w 1 1 + j Use exponential form √ 2 e j π 1 + j = 4 w 0 Re For k = 0 , 1 , 2 , 3 , √ O 8 √ 2 w 2 2 e j ( π 16 + 2 k π 8 4 ) w k = Note w 1 = jw 0 , w 2 = − w 0 , w 3 = − jw 0 w 3 w 0 + w 1 + w 2 + w 3 = 0 16/31

Complex Exponential e z = e x + jy � e x e jy = e x ( cos y + j sin y ) Equivalently, Im � | e z | = e Re z e z Arg ( e z ) = Im z + 2 k π , k ∈ Z e x sin y e x Properties y • e z � 0 , since | e z | = e x > 0 e x cos y O Re • e z 1 + z 2 = e z 1 e z 2 • 1 / e z = e − z • e z = e ¯ z • Periodicity: e z + j 2 π = e z 17/31

Complex Exponential Complex exponential vs. real exponential real e x complex e z with x ∈ R with z ∈ C always nonzero Yes Yes always positive Yes No monotonic Yes No periodic No Yes 18/31

Complex Exponential Alternative definition ∞ z n n ! = 1 + z + z 2 2 ! + z 3 � e z = 3 ! + · · · n = 0 • can do formal calculations as if z is real • will revisit convergence issue later Example. Check Euler’s formula ∞ ∞ ∞ ( j θ ) n ( j θ ) 2 k ( j θ ) 2 k + 1 � � � e j θ = = ( 2 k ) ! + ( 2 k + 1 ) ! n ! n = 0 k = 0 k = 0 ∞ ∞ (− 1 ) k θ 2 k (− 1 ) k θ 2 k + 1 � � + j = ( 2 k ) ! ( 2 k + 1 ) ! k = 0 k = 0 = cos θ + j sin θ 19/31

Complex Exponential Example. Check the addition formula e z 1 + z 2 = e z 1 e z 2 ∞ ∞ k ( z 1 + z 2 ) k 1 k ! � � � e z 1 + z 2 = m ! ( k − m ) ! z m 1 z k − m = 2 k ! k ! k = 0 k = 0 m = 0 ∞ k 1 � � 1 z k − m m ! ( k − m ) ! z m = 2 k = 0 m = 0 ∞ ∞ 1 � � (to be justified by m ! ( k − m ) ! z m 1 z k − m = 2 absolute convergence) m = 0 k = m ∞ ∞ � � 1 m ! ℓ ! z m 1 z ℓ = 2 m = 0 ℓ = 0 � ∞ � � ∞ � 1 1 � � m ! z m ℓ ! z ℓ = e z 1 e z 2 = 1 2 m = 0 ℓ = 0 20/31

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 21/31

Distance Distance between z 1 , z 2 ∈ C Im d ( z 1 , z 2 ) = | z 1 − z 2 | z 2 z 1 Same as Euclidean distance in R 2 O Re z 1 − z 2 | z 1 − z 2 | = |( x 1 + jy 1 ) − ( x 2 + jy 2 )| − z 2 � ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 = 22/31

Convergence of Sequence Definition. A sequence { z n : n ≥ 1 } ⊂ C converges to z ∈ C if n →∞ | z n − z | = 0 . lim We say z is the limit of the sequence { z n } and write z = lim or z n → z , as n → ∞ n →∞ z n , Theorem. Let z n = x n + jy n and z = x + jy . Then and n →∞ z n = z ⇐⇒ lim lim n →∞ x n = x n →∞ y n = y lim Proof. Use � | x n − x | ≤ | z n − z | ≤ | x n − x | + | y n − y | | y n − y | 23/31

Cauchy Sequence Definition. { z n } ⊂ C is a Cauchy sequence if | z n − z m | → 0 , as n , m → ∞ . Theorem. { z n = x n + jy n } is Cauchy iff both { x n } and { y n } are Cauchy. Augustin-Louis Cauchy Proof. Use (from Wikipedia) � | x n − x m | ≤ | z n − z m | ≤ | x n − x m | + | y n − y m | . | y n − y m | 24/31