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1

(Signals) (information) (data)

(

)

(Systems)

( inputs) ( outputs)

1-1

1.1.1

1.

(continuous-time signals,

1(a)) (discrete-time signals,

1(b))

1.

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2. (a) (b) (c) (d)

2.

()

(analog signals) (

) (digital signals) ( 2)

3.

(deterministic signals)

(stochastic or random signals)

4.

g(t) T0

0( ) ( ),= + "g t g t T t (1)

g(t) (periodic signals) (aperiodic signal)(1)

T0g(t)

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5.

(energy) (power)

g(t)

2

( ) .

-= gE g t dt (2)

/2 2

/2

1lim ( ) .

-=

T

gT T

P g t dtT

(3)

(energy signals)

(power signals)(2)(3)

()

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3.

1.1.2

1. (Time Shifting)

g(t)g(t T) T > 0 () T< 0

()

2.

(Time Scaling)

( 3)

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3.

(Time Reversal)

g(t)g(t)

4.

g(t)g(t) = g(t)g(t) (odd function)g(t) =g(t)g(t)

(even function)

[ ] [ ]1 1

( ) : ( ) ( ) , ( ) : ( ) ( ) .2 2

o eg t g t g t g t g t g t= - - = + -

go(t)g(t)ge(t)g(t)

1.1.3

(a)

(Sinusoidal Signal){ }0 0( 2 )0 0 0 0) cos(2 ) Re( ) cos( .j fx t A t A f t Ae p qw q p q ++ = + ==

(b) (Rectangular Pulse)

elsewhere

121,( )

0,( )x t

tt

= =-

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1.1.4 (Phasor Signals and Spectra)

A useful periodic signal in system analysis is the signal:

0( )( ) ,j tx t Ae tw q+= - < <

which is characterized by three parameters, amplitudeA, phase qin radians, and frequency

w0in radians per second orf0= w0/2pHz. We refer to ( )x t as a rotating phasor to distin-

guish it from the phasor jAe q , for which is 0j te w implicit. Using Eulers theorem, we may

readily show that 0( ) ( )x t x t T = + , where 00 /2T p w= . Thus ( )x t is a periodic signal withperiod 02 /p w .

The rotating phasor 0( )j tAe w q+ can be related to a real, sinusoidal signal 0 )cos(A tw q+ in two ways:

1. 0 )0 ) Re( ( )) Re( ),( ) cos( j tt xA tx t Ae w qw q ++ = ==

2. 0 0( ) ( )01 1 1 1

) ( ) (( ) cos( )2 2 2 2

j t j tt x t x t Aex t A Aew q w qw q + - +*+ = + == +

Frequency-Domain Representation of Signal: Line Spectra

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1-2 (Signal, Singularity Functions)

4.

1.2.1 (Unit Impulse Function)

( )td P.A.M. Dirac (Dirac delta function)

( ) 0, 0

( ) 1

t t

t dt

d

d

-

=

=

(singularity

function) (

4)

where0

1

( ) lim ( ), ( ) 2 2

t

t t td d d

= = P

where

2

0

1( ) lim ( ), ( ) sin

tt t t

t

pd d d

p

= =

(a) ( ) ( ) ( ) ( )t t T T t T f d f d - = - , f(t) t= T

(b) ( ) ( ) ( ) ( ) ( )t t T dt T t T dt T f d f d f

- -

= =- -

(sampling property)(sifting property)

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(c)1

( ) ( ),at ta

d d= a

(d)d(t) = d(t).

(e)

2

1

( ) ( )

1 2( ) ( ) ( 1) ( ),

n nnt

t t t T dt T t T tf d f- = - <

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Solution:

2Write the signal)( ) sin( 2 cos(10 )6x t tt

p p

+= as

(a) The real part of a sum of rotating phasors.

(b)A sum of rotating phasors plus their complex conjugates.

(c) From your results in parts (a) and (b), sketch the single-sided and double-sided amplitude

and phase spectra of x(t).

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Solution:

4For the signal g(t) shown in the following figure, sketch (a) g( t); (b) g(t+ 6); (c) g(3t);

(d) g(6 t). Also find their energies.

3For an energy signal g(t) with energy Eg, show that the energy of any one of the sig-

nals g(t), g(t), and g(t T) is Eg. Show also that the energy of g(at) as well as g(at b) is Eg/a.

This shows that time inversion and time shifting do not affect signal energy. On the other hand,

time compression of a signal by a factor areduces the energy by the factor a. What is the effect

on signal energy if the signal is (a) time-expanded by a factor a(a>1), and (b) multiplied by a

constant a?

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Solution:

1-3

1.3.1 Signals versus Vectors

x(t)[a, b]

N

where31 2 , 2 , ..., ( 1) ,1

, Nb a

t a t a N ba t aN

t -

= += = + = + - = =-

N x

[ ]1 2( ) ( ) ( )Nx t x t x t=x

N x Nx

x

l [ , ]im ( ), .N

a bx t t

= x

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This relationship clearly shows that continuous-time signals are straightforward generali-zations of finite dimension vectors. Thus, basic definitions and operations in a vector spacecan be applied to continuous-time signals as well.

1. (Complex Exponential Fourier Series)

x(t) (t0, t0+T) 0 00

22 f

T

pw p= = x(t)

0

0 0 0( ) ,jn

ntx t X t te Ttw

-

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(Euler formula)

0 0( )cos( ( )sin) )( Re( ) Im( ) ,nj X

n n n nX x t n j x t n X j X X t t ew w= - = + =

x(t)Re(Xn)Im(Xn)

and ,n n n nX X X X - -= = -

*n nX X- = x(t) 0( )cos( )nX x t n tw= x(t) Xn=

0( )sin( )j x t n tw-

(a)

(Linearity)

x(t) y(t) Xn YnAx(t)+By(t)AXn+BYn

A, B(b)

(Time Shifting)

x(t) Xn x(t-a) 0jnaX e w-

(c)

(Frequency Shifting)

x(t) Xn 0( ) j tkx t e w Xn-k k

(d) (Conjugate)

x(t) Xn x*(t)*

nX-

(e)

(Time Reversal) x(t) Xn x(t) Xn

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(f) Parsevals Theorem0 0

0

2 2

0

1( ) :

t T

n avt

n

x t dt X PT

+

=-

= = Pav

1.

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2.

(Line Spectra)

x(t)(t0, t0+T0)

0 0( )cos( ( )sin) )( Re( ) Im( ) ,nj X

n n n nX x t n j x t n X j X X t t ew w= - = + =

nj X

nX e

Xn (phasor)

Xnw= nw0(n)

(line spectra)X0 x(t) DCXn x(t)

n (n-th harmonic)

Solution:

1-4 (Fourier Transform)

x(t) x(t) ( )x t dt-

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{ } 21 ( ) ( ) ( ) .ftjX f x t X f e df p--

= = F

X(f)

( )) )( ) ( )cos(2 ( )sin(2 Re( ( )) Im( ( )) ( ) .j X fX f x t dt j x t dt X f j X f X f eft ftp p - -

= - = + =

x(t) Re(X(f)) Im(X(f)) 2 2 2( ) Re( ( )) Im( ( ))X f X f X f = + tan ( ) Im( ( )) /Re( ( ))X f X f X f = ( )X f- =

( )X f ( ) ( )X f X f - = - X(f) = X*(f) () x(t)X(f) =

Re(X(f)) x(t) X(f) =jIm(X(f)) ( )X f

( )X f

x(t)x(t)

2 2*: ( ) ( ) ( ) ,j ftE x t dt x t X f e df dtp

- - -

= =

*

2 2*

*

( ) ( ) ( ) ( )

( ) ( )

ft fj j tE X f x t e dt df X f x t e dt df

X f X f df

p p

- - - -

-

= =

=

22

( ) ( )E x t dt X f df

- -= =

Rayleighs energy theorem Parseval 2

( )X f jouls/hertz (energy density)

(energy spectral density)2

( ) : ( )G f X f =

G(f)

()

1. (Superposition Theorem)

1 1 2 2 1 1 2 2( ) ( ) ( ) ( )a x t a x t a X f a X f + + ()Proof.

2. (Time-Delay Theorem)02

0( ) ( ) ftjx t t X f e p--

Proof.

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3. (Scale-Change Theorem)

1( )

fx at X

a a

Proof.

4. (Duality Theorem)

( ) ( )X t x f -

x(t) X(f)X(t) ( X t X(t))

x(f)

Proof.

5. (Frequency Transition Theorem)02

0( ) ( )f tjx t e X f f p -

Proof.

6. (Modulation Theorem)

0 0 0( )1 1

) ( ) (co 2 )(2 2

s f t X f t f fx f Xp - + +

7.

(Differentiation Theorem)( )

( 2 ) ( )n

n

nd x

df f

tj X

tp

8. (Integration Theorem)

Proof.

1( ( 21

) ) ( ) (0) ( )

2

t

d f X f x f Xjl l p d -

-

+

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9.

(Convolution Theorem)

Proof.

10. (Multiplication Theorem)

Proof.

Solution:

1 2 1 1 12 2 2( ) ( ) : ( ) ( ) ) ( )( ( ) ( )x tx t x t x x t X X d d fx fl l l l l l

-

-* = =- -

1 2 1 2 1 2( ) ( ) ( ) ( )) ( ) (x t x t X f X f f X X dl l l

-

* = -

4. Obtain the following Fourier transform pairs:

(1)

( )A t Ad (2) 020( )

j ftt t AA e pd -- (3)

( )A A fd

(4) 0 02 ( )f tjAe A f fp d -

Ziemer

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Solution:

ys(t)

2 1( ( )) , ,sjns n sm n

fs

s

tt mTy t Y e f T

pd

=- =-

= = =-

2)1 .( ss

n sT

jn f t

s

Y fT

t e dtpd -= =

2( ) ,sjns sn

f ty t f e p

=-

=

4(4)

2( .() )1 sjs s s sn

nf t

n

f nY f f e f f p d

=- =-

= = - F

( ) ( )s ssm n

ft mT t nf d d=-

-

=

- - (5)

(5)

2( ) ( ) ftj

m ms st mT t mT e dt

pd d --=

-

=-

=

- -

F

2 22( ) ( ) sft j ft mj j T

m

f

m s m st mT e t mT e edtdtp p pd d- -

- -

-

=- = - - =

- - = =

(5)mm

2 ( )sj sm

m f

n

Tse f t nf

p d=

- -=

= -

5. As another example of obtaining Fourier transform of signals involving impulses,

let us consider the signal

( ) ( .)s sm

t my Tt d=

-

= - It is a periodic waveform referred to as the ideal sampling waveform and consists of adoubly infinite sequence of impulses spaced by Tss. Find the Fourier transform of ys(t).

(Ziemer)

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4

x(t)p(t) ( /2 /2s sT t T-