Math Review2

8/12/2019 Math Review2

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R6 Continuous-Time Signals 20

R6.1 Energy and Power . . . . . . . . . . . . . . . . . . . . . . . . 20R6.2 Continuous-Time Sinusoidal and Exponential Signals . . . . . 20R6.2.1 Denition . . . . . . . . . . . . . . . . . . . . . . . . . 20R6.2.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . 21

R6.3 Continuous-Time Eigenfunction . . . . . . . . . . . . . . . . . 22R6.4 Continuous-Time Fourier Series . . . . . . . . . . . . . . . . . 23

R6.4.1 Denition . . . . . . . . . . . . . . . . . . . . . . . . . 23R6.4.2 Dirichlet Conditions . . . . . . . . . . . . . . . . . . . 25

R6.5 Continuous-Time Fourier Transform . . . . . . . . . . . . . . . 26

R7 Discrete Fourier Series 27

R8 Matrix Algebra 29R8.1 Denition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29R8.2 Transpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29R8.3 Toeplitz Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 30R8.4 Circulant Matrix . . . . . . . . . . . . . . . . . . . . . . . . . 30R8.5 Determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30R8.6 Minor and Cofactor . . . . . . . . . . . . . . . . . . . . . . . . 32R8.7 Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . 33R8.8 Unitary Matrix and Orthogonal Matrix . . . . . . . . . . . . . 33R8.9 Cramers Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

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R1 Mathematical Formulas and Identities

R1.1 Finite and Innite Sums of Numbers 1

n

k=1

k = n(n + 1)

2 , (R1.1)

n

k=1

k2 = n(n + 1)(2 n + 1)

6 , (R1.2)

n

k=1

k3 = n2(n + 1) 2

4 , (R1.3)

k=1

1k2

= 2

6 , (R1.4)

k=1

1k4

= 4

90, (R1.5)

where n is a positive integer.

Note: The series

k=1

1

k (R1.6)

does not converge.

n 1

k=0

dk = 1 dn

1 d , (R1.7)

with d arbitrary integer and |d| = {1, 0}.

k=0

dk = 11 d

, (R1.8)

with 0 < |d| < 1.1 For test of the convergence of innite sums, a recommended reading is Table of Inte-

grals, Series, and Products, I.S. Gradshteyn and I.M. Ryzhik, c 2000, Academic Press.

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R1.2 Power Series

Binomial Series:

(x + y)n = xn + n1 xn 1y + n2 xn 2y2 + n3 xn 3y3 + + nn 1 xyn 1 + yn ,(R1.9)with n a positive integer.

Taylor Series:

If f (x) is an arbitrarily differentiable function, then it can be expressed inthe form

f (x) = f (a) + f (a)(x a) + f (a)2 (x a)2 + + f

(n )(a)n!

(x a)n + ,(R1.10)

where f (a) = df (x)dx x= a , f (a) = d2 f (x)

dx 2 x= a , and f (n )(a) = d

n f (x)dx n x= a .

Exponential Series:

ex

= 1 + x +

x2

2! +

x3

3! +

x4

4! + , (R1.11)ax = 1 + x loge a +

(x loge a)2

2! +

(x loge a)3

3! + . (R1.12)

Logarithmic Series:

loge(1+ x) = x12 x2+ 13 x314 x4+ , (Region of Convergence : 1 < x < 1).(R1.13)

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Trigonometric Series:

sin x = x x3

3! +

x5

5! x7

7! + , (R1.14)

cos x = 1 x2

2! +

x4

4! x6

6! + , (R1.15)

tan x = x + x3

3 +

2x5

15 +

17x7

315 +

62x9

2835 + , (Region of Convergence : x2 <

2

4 ),(R1.16)

cot x = 1x

x3

x3

45 2x5

945 + , (Region of Convergence : 0 < |x| < ).

(R1.17)

R1.3 Factorial

n! = n(n 1)(n 2) 2 1, with n a nonnegative integer , (R1.18)0! = (0 + 1) = 1 . (R1.19)

R1.4 Permutations and CombinationsThe number of permutations S of n things taken k at a time, with n and k

positive integers, is given by

S = n!

(n k)!. (R1.20)

The number of combinations S of n things taken k at a time, with n andk positive integers, is given by

S = nk = n!

k!(n k)!. (R1.21)

R1.5 Polynomial Factors and Productsxn yn = ( x y)(xn 1 + xn 2y + + yn 1), (R1.22)

with n a positive integer.

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xn + yn = ( x + y)(xn 1 xn 2y + xn 3y2 + yn 1), (R1.23)with n a positive and odd integer.

N

i=1

(x + i) = ( x + 1)(x + 2) (x + N )= 0 + 1x + 2x2 + + N 1xN 1 + N xN

(R1.24)

where

0 =

N

i=1 i , 1 =

N

i=1

0 i , 2 =

N

i= ji,j =1

0 i j , ,

N 1 = 1 + 2 + 3 + + N , N = 1 .

R1.6 Roots of Quadratic EquationThe roots x1, x2 of the quadratic equation

ax 2 + bx + c = 0,

with a,b, and c real numbers, are given by

x1 = b + b2 4ac2a , (R1.25)x2 = b b2 4ac2a . (R1.26)

Note:

x1 + x2 = ba , (R1.27)x1x2 =

ca

. (R1.28)

R1.7 Eulers Formula

e j = cos + j sin , (R1.29)

with a real number.

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R1.8 Trigonometric Functions and Formulas

sin = 12 j (e j e j ), (R1.30)

cos = 12 (e j + e j ), (R1.31)

tan = sin cos

= (e j e j ) j (e j + e j )

, (R1.32)

cot = 1tan

= j (e j + e j )(e j e j )

, (R1.33)

csc = 1sin

= 2 j

e j e j, (R1.34)

sec = 1cos = 2e j + e j , (R1.35)

sin = cos( 2 ) = sin( ), (R1.36)cos = sin( 2 ) = cos( ), (R1.37)tan = cot( 2 ) = tan( ), (R1.38)

sinh = 12 (e e ), (R1.39)

cosh = 12 (e + e ), (R1.40)

tanh = sinh cosh

= (e e )(e + e )

, (R1.41)

with a real number.

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sin(1 2) = sin 1 cos2 cos1 sin 2, (R1.42)cos(1 2) = cos 1 cos2 sin 1 sin 2, (R1.43)

sin2 1 sin2 2 = sin( 1 + 2) sin(1 2), (R1.44)cos2 1 cos2 2 = sin(1 + 2) sin(1 2), (R1.45)cos2 1 sin2 2 = cos(1 + 2) cos(1 2), (R1.46)cos2 1 + sin 2 2 = 1 (R1.47)

sin 1 sin 2 = 2 sin1 2

2 cos(1 2), (R1.48)cos 1 + cos 2 = 2 cos

1 + 2

2 cos

1 22

, (R1.49)

cos 1 cos2 = 2sin1 + 2

2 sin1 2

2, (R1.50)

sin2 = 2 sin cos , (R1.51)cos2 = cos2 sin2 , (R1.52)sin3 = 3 sin 4 sin3 , (R1.53)cos3 = 4 cos3 3 cos. (R1.54)

with , 1, and 2 real numbers.

R1.9 Newton-Raphson Method: Finding a root of apolynomial equation

The Newton-Raphson method is a numerical technique to determine approx-imately the root of the equation f (x) = 0. The procedure starts from ainitial guess of the root x = x1. Then using the recurrence relation

xn +1 = xn f (xn )f (xn )

, n = 1, 2, ,where

f (xn ) = df (x)

dx x= x n,

the successive approximations xn +1 , n 1, beginning with n = 1, can befound. The approximation is assumed to converge when the difference be-tween xn +1 and xn is below a prescribed small number, typically 10 6.

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The Newton-Raphson method converges fast to the actual root if the

initial guess of the root is close to the actual root. However, there are threemain drawbacks: (1) The method fails when f (xn ) = 0, (2) The method doesnot always converge, and (3) The method may converge to a root differentfrom that expected if the initial guess x1 is far from the actual root.

Example R1.1. In this example, we would like to show how the Newton-Raphson method is used to nd the root of f (x) = x3 3x2 + x 1 = 0.Assume the numerical resolution required is 14 decimal digits.

We start with an initial guess of the root x1 = 2.5:

x1 = 2.5,

x2 = x1 f (x1)

f (x1) = 2.84210526315789,

x3 = x2 f (x2)f (x2)

= 2.77282691999216,

x4 = x3 f (x3)f (x3)

= 2.76930129255045,

x5 = x4 f (x4)f (x4)

= 2.76929235429601,

x6 = x5

f (x5)

f (x5) = 2.76929235423863,

x7 = x6 f (x6)f (x6)

= 2.76929235423863.

The recurrence process stops as |x7 x6| 10 15 . Hence x = x7 is a root of f (x).

R1.10 H olders Inequality and Cauchy-SchwartzsInequality

The Holders inequality for integrals is given by

a1

a 0f (x)g(x) dx a

1

a 0 |f (x)| p dx 1/p a

1

a 0 |g(x)|q dx 1/q , (R1.55)

where1 p

+ 1q

= 1 .

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The equality holds when

f (x) = kg(x) p 1, with k any constant .

If p = q = 2, the inequality becomes Schwartzs inequality

a1

a 0f (x)g(x) dx

a1

a 0 |f (x)|2 dx

1/ 2 a1

a 0 |g(x)|2 dx

1/ 2. (R1.56)


f (x) = kg(x), with k any constant .

The Holders inequality for sums is given byN

i=1

xiyi N

i=1|xi| p

1/p N

i=1|yi|q

1/q, (R1.57)

where1 p

+ 1q

= 1 .


yi = kx p 1i , with k any constant .

If p = q = 2, the inequality becomes Cauchys inequality

N

i=1

xiyi N

i=1|xi|2

1/ 2 N

i=1|yi|2

1/ 2, (R1.58)


yi = kx i , with k any constant .

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R2 Useful Functions1. rect function

rect( x) = 1, |x| < 120, |x| > 12.

2. sinc functionsinc(x) =

sin xx

.

3. signum function

sgn(x) =+1 , x > 00, x = 0

1, x < 0.

4. ceiling function rounds the input x towards the closest integer largerthan or equal to x and is denoted as x .For example, 3.2 = 3.8 = 4 and 3.2 = 3.9 = 3.

5. floor function rounds the input x towards the closest integer less thanor equal to x and is denoted as x .For example, 3.2 = 3.8 = 3 and 3.2 = 3.9 = 4.

6. median of a set of real numbers {x1, x2,...,x N } is obtained by rankordering the numbers in the set and choosing the middle number in theordered set.

For example, the median of {7, 13, 1, 6, 3} is 6 and the median of {7, 13, 1, 6, 3, 9}is (6 + 7) / 2 = 6.5.7. Dirac delta function ( ) is a function of with innite height, zero

width, and unit area. It is the limiting form of a unit area pulse function

p ( ) = 12 , < 0, elsewhere

as goes to 0, i.e.,lim 0

p ( ) d =

( ) d = 1. (R2.1)

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Equation( R2.1) also holds even when we reverse the direction of axis

and shift ( ) by an amount t, i.e.,

(t ) d = 1. (R2.2)

Because of the above properties, we have

x( ) (t ) d = x( )| = t = x(t). (R2.3)

Equation( R2.3) holds for any value of t, and it is referred as the sifting property of the Dirac delta function.

8. The modulo operation of integer X over integer N is the residue of X divided by N :

X N = X kN, k = X/N .It can be veried that the modulo operation is linear . When nega-tive numbers are used, X N has the same sign as N . For example,67 13 = 67 5 13 = 2, 67 13 = 67 (13) (6) = 11 and

67 13 = 67 13 (6) = 11.

The statement X is congruent to Y , modulo N means that

X N = Y N .The notation

X 1 N denotes the multiplicative inverse of X evaluated modulo N , i.e., if X 1 N = , then X N = 1. For example, 3 1 4 = 3 because3 3 4 = 1, and 8 1 5 = 2 because 8 2 5 = 1.

In the case of polynomial, the operation a(z ) mod b(z ) is the residuer (z ) after the polynomial division a(z )/b (z ). For example, if a(z ) =4z 3 + 2 z 2 + 5 z 1 + 1 and b(z ) = z 2 + 3 z 1 + 4 then the residue after

the division a(z )b(z )

= 4z 1 10 + 19z 1 + 41z 2 + 3 z 1 + 4

is 19z 1 + 41 . Therefore, a(z ) mod b(z ) = r (z ) = 19z 1 + 41 .

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R3 Commonly Used Differentials andIntegrals

R3.1 Differentials

d(uv) = u dv + v du (R3.1)

d(uv

) = v du u dv

v2 (R3.2)

d(un ) = n un 1 du (R3.3)d eu = eu du (R3.4)

d au

= ( au

loge a) du (R3.5)d(loge u) = u

1 du (R3.6)d sin u = cos u du (R3.7)d cosu = sin udu. (R3.8)

R3.2 Integrals

f (g(x))g (x) dx = f (y) dy, y = g(x) and g (x) = dy/dx (R3.9)

u dv = uv v du (R3.10) f (x) dxf (x) = loge f (x) (R3.11) dxx = loge x (R3.12)

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xn dx = xn +1n + 1 (R3.13) ex dx = ex (R3.14) ax dx = a

x

loge a (R3.15)

abx dx = abx

bloge a (R3.16)

loge x dx = x loge x x (R3.17) sin x dx = cos x (R3.18) cosx dx = sin x (R3.19) tan x dx = log cosx. (R3.20)

R3.3 lHopitals RuleConsider a fraction f (x)/g (x) for which at x = x0, f (x0) = g(x0) = 0 (or

f (x0) = g(x0) = ). Thenlim

xx 0

f (x)g(x)

= limxx 0

f (x)g (x)

(R3.21)

as long as the limits on the right-hand side exist and are nite.

R3.4 ExamplesExample R3.1. Evaluate the integral

x[n] = 1

2

X ()e jn d,

whereX () = cos(), || 00, 0 < || .

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Answer:

x[n] = 12

0

0cos()e jn d

= 12

0

0

12

(e j + e j )e jn d

= 14

0

0e j e jn d +

0

0e j e jn d

= 14

1 j ( + n)

e j ( + n )0

0+

1 j ( + n)

e j ( + n )0

0

= 14

1 j ( + n)

2 j sin( + n)0 + 1 j ( + n)2 j sin( + n)0

= sin( + n)0

2( + n) +

sin( + n)02( + n)

.

Example R3.2. Evaluate the integral

x[n] = 12

X ()e jn d,

where

X () = , || 00, 0 < || .using integration by parts.

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Answer:

x[n] = 12

0

0e jn d

= 12

1 jn

0

0e jn d( jn )

= 12

1 jn

0

0 de jn

= 12

1 jn

e jn0

0 0

0e jn d

= 1

2

1

jn0e j 0 n

(

0)e j 0 n

1

jne jn

0

0

= 12

1 jn

0(e j 0 n + e j 0 n ) 1 jn

(e j 0 n e j 0 n )=

1( jn )

0 cos(0n) 1n

sin(0n) .

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R4 Complex Numbers

R4.1 DenitionA complex number z is represented in the Cartesian coordinate as

z = x + jy

where j = 1, and x and y are real numbers and denoted as the realand imaginary parts of z , respectively. The complex numbers can also berepresented in polar form as

z = |z |e j ,where |z | and are the magnitude and angle of z , respectively:

|z | = x2 + y2, = tan 1(yx

).

The principal value of the angle of z is given by

< .Figure R4.1 shows the representation of a complex number z in the com-

plex plane. By using Euler s Formula (see Eq.(R1.29)) we can nd therepresentation of a complex number in Cartesian form from its polar form:

x = |z |cos , y = |z |sin .It should be pointed out here that negative real numbers have angle

= (2 k + 1)

with k any integer.

Example R4.1. Let z = 2 + j 3, we can express it in polar form bycalculating its magnitude and angle:

|z | = 22 + 3 = 7, = tan 1( 32 ).Therefore,

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Figure R4.1: Representation of a complex number z in Cartesian form andpolar form.

The conjugate of a complex number in Cartesian form is obtained bynegating the imaginary part:

z = ( x + jy )= x+ ( jy )= x jy.In polar form, the conjugate is obtained by changing the sign of the angle:

z = ( re j )= r e j .

R4.2 Complex Arithmetic(1) Addition and Subtraction

z 1 = x1 + jy1 and z 2 = x2 + jy2 be two complex numbers. Then

z 1 + z 2 = ( x1 + x2) + j (y1 + y2)

where (x1 + x2) are (y1 + y2) are the real and imaginary parts of the sum

z 1 + z 2, respectively. Similarly,z 1 z 2 = ( x1 x2) + j (y1 y2)

where (x1 x2) are (y1y2) are the real and imaginary parts of the differencez 1 z 2, respectively.16


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Example R4.2. Let z 1 = 1.3 + j 5.2 and z 2 = 2.7 j 3.6 thenz 1 + z 2 = (1 .3 + j 5.2) + (2 .7 j 3.6) = 4 + j 1.6z 1 z 2 = (1 .3 + j 5.2) (2.7 j 3.6) = 1.4 + 8.8 j.

(2) Multiplication

Let z 1 = x1 + jy 1 and z 2 = x2 + jy 2 then

z 1 z 2 = ( x1 + jy 1)(x2 + jy 2)= x1x2 + jx 1y2 + jx 2y1 + j 2y1y2= ( x1x2 y1y2) + j (x1y2 + x2y1).

Example R4.3. Let z 1 = 1 + j 3, z 2 = 2

j 2. The product of z 1, z 2calculated in polar form is given by

(1 + j 3)(2 j 2) = 2e j/ 3 2 2e j/ 4 = 4 2e j/ 12 = 5 .4641 + j 1.4641.Calculating in the Cartesian form we get

(1 + j 3)(2 j 2) = (2 + 2 3) + j (2 3 2) = 5.4641 + j 1.4641.(3) Division

The division of two complex numbers z 0 and z 1 can be carried out either inpolar form or in Cartesian form. In the former case

w = z 0z 1 =

r0e j 0

r 1e j 1 = r0r 1 e j (

0

1

) .

In the latter case

w = z 0z 1

= x0 + jy 0x1 + jy 1

= (x0 + jy 0)(x1 jy 1)(x1 + jy 1)(x1 jy 1)

= (x0x1 + y0y1) + j (x1y0 x0y1)

x21 + y21.

Example R4.4. To divide 2 + j 2 by 1 j , we calculate in polar form asfollows:2 + j 2

1 j =

2 2e j 4

2e j 4

= 2 e j (4 ) (

4 ) = 2 e j

2 = j 2.

Calculating in the Cartesian form we get2 + j 21 j

= (2 + j 2)(1 + j )

(1 j )(1 + j ) =

(2 2) + j (2 + 2)12 + 1 2

= j 2.

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(4) Inverse

The inverse of a complex number is a special case of division where thenumerator is 1. In polar form we have

z 1 = 1z

= 1re j

= 1r

e j .

Equivalently, in the Cartesian form we have

z 1 = 1z

= 1

x + jy =

x jy(x + jy )(x jy )

= x jyx2 + y2

.

R5 Complex Variables

R5.1 Function of a Complex VariableA function of the complex variable z can be written as

f (z ) = u(z ) + jv (z )

where u(z ) and v(z ) are real functions of z . In the Cartesian form, we denez = x + jy for real x and y. Therefore, the values of u(z ) and v(z ) dependon x and y, and we can express the complex function f (z ) as

f (z ) = u(x, y) + jv (x, y).

If z = re j , then f (z ) can be expressed as

f (z ) = u(r, ) + jv (r, ),where u(r, ) and v(r, ) are the real and imaginary parts of f (z ).

R5.2 Analytic Function of a Complex VariableDenition R5.1. A function f (z ) is said to be differentiable at a point z 0in the z -plane if the limit

f (z 0) = limz0

f (z 0 + z ) f (z 0)z

exists. Note that f (z 0 + z ) can approach f (z

0) along any path. This limit

is called the derivative of f (z ) at point z 0.Denition R5.2. A function f (z ) of a complex variable z is analytic in theregion R in the complex z -plane if and only if all the derivatives of f (z ) existat all points inside the region R.

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R5.3 Analytic Continuation

If the values of a function f (z ) of a complex variable are known everywhereon a closed contour C inside a region R where f (z ) is analytic, then thevalues of f (z ) at all points in R can be found by mapping from the contourC to any point in R.

R5.4 Cauchys Integral FormulaIf a function f (z ) is analytic both on and inside a counterclockwise closedcontour C and if z 0 is any point inside C , then

f (z 0) = 1

2j C f (z )

1

z z 0dz, (R5.1)

f (z 0) = 12j C f (z ) 1(z z 0)2 dz (R5.2)

f (z 0) = 22j C f (z ) 1(z z 0)3 dz (R5.3)... (R5.4)

f (n )(z 0) = n!2j C f (z ) 1(z z 0)n +1 dz. (R5.5)

where f (z 0) = d f (z)dzz= z0

, f (z 0) = d2 f (z)dz 2

z= z0, and f (n )(z 0) = d

n f (z)dz n

z= z0.

Eq.( R5.1) is often referred as the Cauchys integral formula.

By combining Eq.( R5.1) - Eq.(R5.5) we arrive at an useful relation:

12j C z k 1 dz = 1 , k = 00 , k = 0 (R5.6)

where C is a counterclockwise closed contour encircling z = 0.

R5.5 Cauchys Residue TheoremIf a function f (z ) is analytic both on and inside a counterclockwise closedcontour C except at poles z k , k = 1, 2,...,n , then

12j C f (z ) dz = k residue of f (z ) at pole z k inside C . (R5.7)

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In the case when f (z ) is a rational function of z and has pole at z = z k

of multiplicity m, we can express f (z ) as

f (z ) = (z )(z z k)m

,

where (z ) does not have any pole at z = z k . Thus the residue of f (z ) atthe pole z k inside C is given by

1(m 1)!

dm 1(z )d z m 1 z= zk

. (R5.8)

R6 Continuous-Time SignalsR6.1 Energy and PowerThe total energy of a continuous-time signal x(t) is given by

E x = limT

T

T |x(t)|2 dt.

The average power of a continuous-time x(t) is given by

P x = limT 1

2T T

T |x(t)|2

dt.

The denition of total energy can explained as the area under the squaredsignal |x(t)|2, and it is a measurement of the strength of the signal x(t) overinnite time. However, there are signals with innite energy so we needto evaluate the average power of the signal x(t) as a measurement of thestrength over one unit time.

R6.2 Continuous-Time Sinusoidal and Exponential Sig-nals

R6.2.1 Denition

The continuous-time real sinusoidal signal with constant amplitude is of the form

x(t) = A cos(0t + ), (R6.1)

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where A, 0 and are real numbers. The parameters A, 0 and are called,

respectively, the amplitude , the angular frequency , and the phase of thesinusoidal signal x(t).The complex exponential signal is expressed in the form

x(t) = A t , (R6.2)

where

= e0 + j 0 , A = |A|e j . (R6.3)If A and are both real, the signal of Eq. ( R6.2) reduces to real

exponential signal . For t

0 such a signal with

|

| < 1 decays expo-

nentially as t increases and with || > 1 grows exponentially as t increases.In addition, we can rewrite Eq. ( R6.2) as

x(t) = Ae(0 + j 0 )t = |A|e0 t e j ( 0 t+ ) (R6.4)= |A|e0 t cos(0t + ) + j |A|e0 t sin(0t + ). (R6.5)

Thus the real and imaginary parts of a complex exponential signal are realsinusoidal signals.

The fundamental period T 0 of a complex exponential signal (Eq. ( R6.4))with

0 = 0 is dened to be the smallest positive T

0 satisfying

|A|e j (0 t+ ) = |A|e j (0 (t+ T 0 )+ ) , (R6.6)or equivalently,

e j 0 T 0 = 1 . (R6.7)

Therefore,T 0 =

2

|0|. (R6.8)

R6.2.2 Properties

The properties of continuous-time sinusoidal and exponential signals andcomparisons with discrete-time sinusoidal and exponential sequences are dis-cussed as follows.

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1. Periodicity for any choice of 0

Note that the continuous-time sinusoidal signal A cos(0t + ) (Eq.(R6.1)) and the continuous-time complex exponential signal |A|e j (0 t+ )are periodic signals of any choice of 0. However, discrete-time se-quences are not always periodic with any choice of 0. Discrete si-nusoidal sequence A cos(0n + ) and discrete complex exponential se-quence |A|e j (0 n + ) are periodic with period N only if 0N is an integermultiple of 2, i.e., 0N = 2r where N and r are positive integers.For example, cos( n4 ) is a periodic sequence while cos(

n4 ) is not periodic.

2. Distinctness for different 0, 1Any two continuous-time sinusoidal signals

A cos(0t + ), A cos(1t + ), 0 = 1have different waveforms. Similarly, any two continuous-time exponen-tial signals with 0 = 1 also have different waveforms. Unlike thecontinuous-time case, discrete-time sinusoidal sequences

A cos(0n + ), A cos(1n + ), 0 = 1 + 2 k

have the same sequence values. Similarly, any two discrete-time ex-ponential sequences with 0 = 1 + 2k also have the same sequence

values.

R6.3 Continuous-Time EigenfunctionIf the input signal of any LTI system has output signal being the input signalmultiplied by a complex constant, this certain type of input signal is calledthe eigenfunction and the complex constant is called the eigenvalue .

Example R6.1. We want to show that the complex exponential signal de-ned in Eq. (R6.2)

x(t) = A t

is an eigenfunction of an LTI continuous-time system with an impulse re-sponse h(t).

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By using the convolution integral, we have

y(t) = h( )A (t ) d =

h( ) d A t .

Since the integral inside the brackets is independent of t, we can thereforesay that the input signal A t is an eigenfunction.

Example R6.2. We want to show that the sum of any two complex expo-nential signals

x(t) = A t + B t

is not an eigenfunction of an LTI continuous-time system with an impulseresponse h(t).

By using the convolution integral, we have

y(t) =

h( ) A (t ) + B (t ) d

=

h( ) d A t +

h( ) d B t .

Since the input signal x(t) cannot be extracted from the summation above,we can therefore say that the input signal A t + B t is not an eigenfunction.

R6.4 Continuous-Time Fourier SeriesR6.4.1 Denition

Given a periodic continuous-time signal x(t) with period T 0 and fundamentalfrequency 0 = 2/T 0, the Fourier series expansion of x(t) is given by thelinear combination of the set of harmonically related complex exponentials

e jk 0 t = e jk2 T 0

t , k = 0, 1, 2,...,i.e.,

x(t) =

k=

ake jk 0 t =

k=

ake jk 2 T 0 t (R6.9)

ak = 1T 0 T 0 x(t)e jk 0 t dt = 1T 0 T 0 x(t)e jk

2 T 0

t dt. (R6.10)

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Note that the notation

T 0 denotes the integration over any interval of

length T 0. The Eq.( R6.9) is referred to as the synthesis equation and theEq.( R6.10) is referred to as the analysis equation. The coefficient ak is calledthe Fourier series coefficient .

Table R6.1: Properties of Continuous-Time Fourier Series.

Type of Periodic Signal with Fourier SeriesProperty frequency 0 = 2/T Coefficients

g(t) akh(t) bk

Linearity g(t) + h (t) a k + bkTime Shifting g(t t0) ake jk 0 t 0

Frequency Shifting e jM 0 t g(t) ak M Multiplication g(t)h(t)

l= a lbk l

Time Reversal g(t) a kConjugation g(t) a kTime Scaling g(t ), > 0 ak

Periodic Convolution

T g( )h(t

) d Takbk

Example R6.3. Find the Fourier series coefficients of the continuous-timesignal

x(t) = 1 + cos( 0t) + 2 cos(20t + 3

) + 4 sin(30t 4

)

with fundamental frequency 0.By using the Eulers Formula, it can be shown that

x(t) = 1 + 12(e

j 0 t

+ e j 0 t

) + 22(e

j (2 0 t+ 3 )

+ e j (2 0 t+ 3 )

) + 42 j (e

j (3 0 t 4 )

e j (3 0 t 4 )

)

= 1 + 12

e j 0 t + 12

e j 0 t + e j3 e j 20 t + e j

3 e j 20 t +

2 j

e j4 e j 30 t

2 j

e j4 e j 30 t .

Therefore, the Fourier series coefficients are

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a0 = 1,

a1 = 12

, a 1 = 12

,

a2 = e j3 =

1 + j 32

, a 2 = e j3 =

1 j 32

,

a3 = 2 j

e j4 = 2(1 j ), a 3 =

2 j

e j4 = 2(1 + j ),

ak = 0, |k| > 3.Example R6.4. Find the Fourier series coefficients of the impulse train

x(t) =

k=

(t kT 0)

with period T 0.By calculating Eq.( R6.10) in the interval T 0/ 2 t T 0/ 2, we can get

ak = 1T 0

T 0 / 2

T 0 / 2 (t)e jk

2 T 0

t dt = 1T 0

.

Therefore, all the Fourier series coefficients of the impulse train have thesame value 1/T 0.

Some important properties of continuous-time Fourier series are listed inTable R6.1 for quick reference.

R6.4.2 Dirichlet Conditions

In order to verify the existence of Fourier series representation for a peri-odic signal x(t), we need to examine the Dirichlet conditions. The Dirichletconditions are given by:

1. x(t) must be absolutely integrable, i.e.,

T 0 |x(t)|dt < 25


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2. In any nite interval of time, x(t) has a nite number of local maxima

and local minima.3. In any nite interval of time, x(t) has a nite number of discontinuities.

The Dirichlet conditions guarantee that x(t) equals its Fourier series rep-resentation

k=

ake jk 0 t

at all values of t except at discontinuities of x(t). Note that Dirichlet condi-tions are only sufficient but not necessary conditions.

R6.5 Continuous-Time Fourier TransformGiven an aperiodic continuous-time signal x(t), the continuous-time Fouriertransform of x(t) is given by

X ( j) =

x(t)e j t dt (R6.11)

x(t) = 12

X ( j)e j t d. (R6.12)

The transform X ( j ) is referred to as the spectrum of x(t) because it pro-vides the information of x(t) when evaluated by complex exponential signalsat different frequencies.

Some important properties of continuous-time Fourier transform are listedin Table R6.2 for quick reference.

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Table R6.2: Properties of Continuous-Time Fourier Transform.

Property Signal Fourier Transformg(t) G( j )h(t) H ( j )

Linearity g(t) + h (t) G( j ) + H ( j )Time Shifting g(t t0) G( j)e j t 0

Frequency Shifting e j 0 t g(t) G( j ( 0))Multiplication g(t)h(t) 12 G( j)

H ( j )

Time Reversal g(t) G( j )Conjugation g(t) G( j )Time Scaling g(t ) 1| | G

j

Convolution g(t) h(t) G( j )H ( j )Differentiation in Time dd t g(t) j G( j)

Integration t

g( ) d 1

j G( j ) + G(0) ()Real and Even in Time g(t) real and even G( j ) real and evenReal and Odd in Time g(t) real and odd G( j ) purely imaginary and odd

R7 Discrete Fourier SeriesGiven a periodic sequence x[n] with period N , the fundamental period isdened to be the smallest integer N such that x[n] = x[n + N ] is satised,and the fundamental frequency is dened to be 0 = 2/N . The harmonicsare sequences whose frequencies are integer multiples of the fundamentalfrequency. For discrete complex exponential signals, the k th harmonic isexpressed as

e jk 0 n = e jk2 N n , k = 0, 1, 2, .

Note that there are only N distinct harmonics for discrete complex ex-ponential signals with fundamental frequency 0 = 2/N because every twosignals with frequencies which differ in 2m have the same waveform, i.e,

e jk (0 +2 m )n = e jk (2 N +2 m )n = e jk (

2 N )n e jk 2mn = e jk (

2 N )n .

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The discrete Fourier series expansion of periodic signal x[n] is the expres-

sion in form of a linearly weighted combination of a fundamental and a seriesof harmonic complex exponential signals.

x[n] =N 1

k=0

ake jk 0 n =N 1

k=0

ake jk2 N n ,

where

ak = 1N

N 1

n =0

x[n]e jk 0 n = 1N

N 1

n =0

x[n]e jk2 N n .

Example R7.1. Calculate the Fourier series coefficients ak for the followingperiodic signal

{x[n]} = {..., 1, 1, 1, 1, 0, 0,...}.

We can observe that N = 6 so

ak = 165n =0 x[n]e

jk ( 2 6 )6

= 16 (e jk ( 2 6 )0 + e jk (

2 6 )1 + e jk (

2 6 )2 + e jk (

2 6 )3 + 0 + 0)

= 16 (1 + e jk 3 + e jk

2 3 + e jk )

= 16 (1 + e jk 3 + (1)ke jk

3 + (1)k).

Example R7.2. Calculate signal x[n] from the following Fourier series co-

efficients ak

{ak} = {..., 1/ 4, 1/ 2, 1, 1/ 2, 1/ 4, 0, 1/ 4, 1/ 2, 1, 1/ 2...}. .

We can observe that N = 6 so

x[n] =k=

ake jk2

6 n = 1 + 12

e j2

6 n + 14

e j2

6 2n + 0 + 14

e j2

6 4n + 12

e j2

6 5n

= 1 + 12

e jn

3 + 14

e j2 n

3 + 0 + 14

e j4 n

3 + 12

e j5 n

3

= 1 + 1

2e j

n3 +

1

4e j

2 n3 + 0 +

1

4e j (2n

2 n3 ) +

1

2e j (2n

n3 )

= 1 + 12

e jn

3 + 14

e j2 n

3 + 0 + 14

e j2 n

3 + 12

e jn

3

= 1 + cos(n3

) + 12

cos(2n

3 ).

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R8 Matrix Algebra

R8.1 DenitionA matrix is a rectangular array of real or complex numbers enclosed in brack-ets; for instance,

3 41 2 ,

3 5 62 1 3

,3 j 5

2 4 j 1 + j7 5 3 j.

A matrix with K rows and M columns is called a K M matrix. Forexample, the matrix1 0 00 1 00 0 1

is a 3 3 matrix. The matrix1 42 6

3 j 11 + j 1is a 4

2 matrix. The matrix

U =

a11 a12 a1M a21 a22 a2M ... ... . . . ...aK 1 aK 2 aKM

(R8.1)

is a K M matrix and the number ars , r = 1, 2,..., K , and s = 1, 2,..., M is called the entry of U.

R8.2 TransposeThe transpose, UT , of a K M matrix U is the M K matrix formed byinterchanging the rows and columns of U . For example, the transpose of the

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matrix U given in Eq.(R8.1) is a M K matrix given by

U T =

a11 a21 aK 1a12 a22 aK 2... ... .. . ...a1M a2M aKM

. (R8.2)

R8.3 Toeplitz MatrixThe N N matrix U is a Toeplitz matrix if all entries along the line parallelto the main diagonal are the same. For example,

U =a0 a 1 a 2 a 3a1 a0 a 1 a 2a2 a1 a0 a 1a3 a2 a1 a0

is a 4 4 Toeplitz matrix.

R8.4 Circulant MatrixThe N N matrix U is a circulant matrix if each row equals the right circularshift of the previous row by one entry. For example,

U =

a0 a1 a2 a3a3 a0 a1 a2a2 a3 a0 a1a1 a2 a3 a0

is a 4 4 Circulant matrix.

R8.5 DeterminantIf the 2 2 matrix U is

U = a11 a12a21 a22 , (R8.3)

then the determinant of the U is given by

det( U ) = a11a22 a12a21 . (R8.4)30


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Example R8.1. The determinant of the matrix

U = 3 41 2

is det(U ) = 3 2 4 1 = 6 4 = 2.If the 3 3 matrix U is

U =a11 a12 a13a21 a22 a23a31 a32 a33

, (R8.5)

then the determinant of U is given by

det( U ) = a11 a22a33 + a21a32a13 + a12a23a31 a13a22a31 a11a32a23 a12a21a33 .(R8.6)Example R8.2. The determinant of the matrix

1 4 62 1 34 5 2

is

det( U ) = 1

1

(

2) + 2

5

(

6) + 4

3

4

(

6)

1

4

1

5

3

4

2

(

2)

= ( 2) + ( 60) + 48 (24) 15 (16) = 11 .If the N N matrix U is

U =

a11 a12 a1N a21 a22 a2N ... ... . . . ...aN 1 aN 2 aNN

, (R8.7)

then the determinant of U is

det( U ) = ar 1(1)r +1 M r 1 + ar 2(1)r +2 M r 2 + + a rN (1)r + N M rN (R8.8)= a1s (1)1+ sM 1s + a2s (1)2+ s M 2s + + aNs (1)N + sM Ns (R8.9)where r, s = 1 or 2 or 3 , or N and M rs is the minor of ars (see SectionR8.6).

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Example R8.3. To calculate the determinant of the matrix

U =

1 6 4 67 1 3 62 3 6 52 4 2 6

, (R8.10)

we rst calculate the minors

M 11 =1 3 63 6 54 2 6

= 320, M 12 =7 3 62 6 52 2 6

= 344

M 13 = 7 1

6

2 3 52 4 6

= 232, M 14 = 7 1 3

2 3 62 4 2

= 200.

The determinant is therefore given by

det( U ) = 1 (1)1+1 320 + 6 (1)1+2 (344) + (4) (1)1+3 232 + 6 (1)1+4 200= 320 + 2064 928 1200 = 256.

R8.6 Minor and Cofactor

From U given in Eq.(R8.7), the minor M rs of ars in U is dened to be thedeterminant of the ( N 1)(N 1) matrix formed by deleting the r -th rowand s-th column of U . For example,

M 11 =


, M 12 =


.

The cofactor C rs of ars in U given in Eq.(R8.7) is dened to be

C rs = (1)r + sM rs . (R8.11)

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R8.7 Inverse of a Matrix

By Eq.(R8.7), if det(U ) = 0, then the inverse of U exists and is uniquelygiven by

U 1 = 1

det( U )

C 11 C 21 C N 1C 12 C 22 C N 2... ... . . . ...C 1N C 2N C NN

, (R8.12)

where C rs = (1)r + s M rs is the cofactor of a rs in U given in Eq.(R8.7).Example R8.4. In this example we want to nd the inverse of the matrixgiven in Eq.(R8.10). The cofactors are calculated as follows:

C 11 = (1)1+1 M 11 = 320, C 12 = (1)1+2 M 12 = 344,C 13 = (1)1+3 M 13 = 232, C 14 = (1)1+4 M 14 = 200C 21 = (1)2+1 M 21 = 176, C 22 = (1)2+2 M 22 = 210,C 23 = (1)2+3 M 23 = 158, C 24 = (1)2+4 M 24 = 134C 31 = (1)3+1 M 31 = 240, C 32 = (1)3+2 M 32 = 234,C 33 = (1)3+3 M 33 = 198, C 34 = (1)3+4 M 24 = 142C 41 = (1)4+1 M 41 = 344, C 42 = (1)4+2 M 42 = 329,C 43 = (1)4+3 M 43 = 239, C 44 = (1)4+4 M 44 = 227.

Therefore, the inverse is

U 1 = 1256

320 176 240 344344 210 234 329232 158 198 239200 134 142 227

.

R8.8 Unitary Matrix and Orthogonal MatrixThe N N matrix U is said to be unitary if

U H U = UU H = kI, (R8.13)

where k is any nonzero constant and U H = ( U T ) is the conjugate-transposeof U. Note that the unitary matrix is always invertible and U H = U 1.

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A real unitary matrix U is also called an orthogonal matrix, i.e.,

U T U = UU T = kI, (R8.14)where k is any nonzero constant and UT is the transpose of U. Similarly,the orthogonal matrix is always invertible and UH = U 1. If k = 1, thenthe matrix U is said to be orthonormal.

R8.9 Cramers RuleConsider the set of N linear equations

a11x1 + a12x2 + + a1N xN = b1,a

21x

1 + a

22x

2 +

+ a

2N x

N = b

2,

...aN 1x1 + aN 2x2 + + aNN xN = bN ,

(R8.15)

writing in matrix form yields

a11 a12 a1N a21 a22 a2N ... ... .. . ...aN 1 aN 2 aNN

x1x2...

xN

=

b1b2...

bN

. (R8.16)

Let D be the determinant of the coefficient matrix

D =


. (R8.17)

If D = 0, then the system( R8.15) has the unique solutionx1 =

D1D

, x2 = D2D

, , xN = DN

D , (R8.18)

where

D1 =b1 a12 a1N b2 a22 a2N ... ... . . . ...bN aN 2 aNN

, D 2 =a11 b1 a1N a21 b2 a2N ... ... . . . ...aN 1 bN aNN

, , etc.

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