주기 [전력] 신호인경우, 기본주기 T0

정현 Fourier Series• 직교함수집합 : 구간 t0 ≤ t ≤ t0 + T0

복소지수함수 Fourier Series• 직교함수집합 : 구간 t0 ≤ t ≤ t0 + T0

푸리에급수(Fourier Series)

0

2( ) e , 0, 1, 2, ,ojn tn ot n

T

( ) 1, cos , sin , 1,2,3, ,n o ot n t n t n

복소지수함수근저함수 (basis function)집합에의한표현• 구간 t0 ≤ t ≤ t0 + T0에서임의의신호 x(t)는아래선형조합으로표현

• 선형조합의계수 은다음과같이구함

• 만일신호 x(t)가주기 T0를가진주기신호라면 basis함수들도같은주기의주기함수가되므로위의무한급수표현은한주기의구간을넘어서모든시구간 - < t < 로확장할수있음

복소지수함수 Fourier Series

0

2( ) e ,ojn tn o

n

x t CT

0

0

0 020

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

o

n oo o

o o o

o o

t T

t Tt jn tnn t T t T t

nt t

x t t dtx t tC x t e dtTt dt dt

nC

• Fourier Series 및 Fourier Series 계수

• FA Coefficient Cn은 n0주파수성분(nth harmonic)의전력량• 일명, 스펙트럼계수(spectrum coefficient)• 복소수(polar form) :

0

0

2 /

2 /

0 0

( ) (Fourier Series: Synthesis)

1 1( ) ( ) (FS Coefficients: Analysis)

o

o

o o

jn t j nt Tn n

n n

jn t j nt Tn T T

x t c e C e

c x t e dt x t e dtT T

arg nj Cn nC C e

대칭성• x(t)의값이 real인경우, 즉 인경우

• If x(t) is real & even (i.e. ), then cn is also real & even

• If x(t) is real & odd (i.e. ), then cn is imaginary & odd

푸리에급수의성질

( ) ( )x t x t

(conjugate symmetric)n nc c

(amplitude spectrum is even) arg arg (phase spectrum is odd)

n n

n n

c cc c

( ) ( )x t x t

( ) ( )x t x t

선형성• x(t)와 y(t)가동일한주기를갖는주기신호인경우

• 의푸리에급수는

0

0

( )

( )

jn tn

n

jn tn

n

x t c e

y t d e

( ) ( ) ( )z t x t y t

0( )

with

jn tn

n

n n n

z t g e

g c d

시간천이• x(t)의푸리에계수가 인경우

• x(t-)의푸리에계수 은

• 진폭스펙트럼은변화가없다.

nc

nd

0

0

0

( )

0 0

0

1 1( ) ( )

1 ( )

o o

o o

o

o

jn t jn t jnn T T

jn jn

T

jnn

d x t e dt x t e e dtT T

e x e dT

e c

n nd c

주기신호의평균전력

• Real valued signal의경우

•

Parseval의정리

0

2 2

0

1 ( ) nTn

P x t dt cT

2 20

12 n

nP c c

2 is called power spectrumnc

[proof]• 의푸리에계수

• z(t)의푸리에계수

• z(t)의직류값(k = 0에서의푸리에계수)

• x(t) = y(t)인경우

0 0

0 0

* *

( )* *

( ) ( ) ( ) jn t jm tn m

n m

j n m t jk tn m k m m

n m k m

z t x t y t c e d e

c d e c d e

*( ) ( ) ( )z t x t y t

0

0

* *

0

1 ( ) ( ) jk tk m m T

mc d x t y t e dt

T

0

* *

0

1 ( ) ( )m m Tm

c d x t y t dtT

0

2 2

0

1 ( )n Tn

c x t dtT

LTI system• 주파수가 인복소지수함수를입력하면• 동일한주파수의복소지수함수가출력되며,이득은 H()만큼곱해짐

• Complex gain H()

• Output can be written as

주기신호에대한 LTI 시스템의응답

( ) ( ) ( )j t j tx t e y t H e

arg ( )( ) ( ) Frequency responsej HH H e

( arg ( ))( ) ( ) j t Hy t H e

[Case 2] 의경우• 주파수가 인복소지수함수의선형조합을입력하면

• Linear system이므로출력은

• 주파수성분별로다른이득을갖고출력됨

1 21 2( ) j t j tx t c e c e

1 21 1 2 2( ) ( ) ( )j t j ty t H c e H c e

1 21 2

j t j tc e c e 1 21 1 2 2( ) ( )j t j tH c e H c e

( )H

[Case 3] 임의의주기신호를입력하는경우• 입력신호를푸리에급수로표현하면

• Linear system이므로출력은

• 출력의푸리에계수는

( ) ojn tn

n

x t c e

0( ) ( ) o ojn t jn tn n

n n

y t H n c e d e

0

0 0

( )( ) , arg arg ( ) arg

n n

n n n n

d H n cd H n c d H n c

ojn t

nn

c e

0( ) ojn t

nn

H n c e

( )H

주파수응답의대칭성• 실제시스템에서임펄스응답은실수값을갖는함수이므로, 즉

• 주파수응답은다음성질을갖는다.

• 따라서

정현파신호에대한 LTI 시스템의응답

*( ) ( )h t h t

* **

*

( ) ( ) ( ) ( )

( )

j t j t j tH h t e dt h t e dt h t e dt

H

( ) ( ) evenarg ( ) arg ( ) oddH H

H H

정현파신호에대한응답• For x(t)=cos(t)

• LTI 시스템의출력은

• 동일한주파수의정현파가출력되는데, 진폭은 배곱해지고

위상은 만큼더해진다.

( ) cos exp( ) exp( ) / 2x t t j t j t

( arg ( )) ( arg ( ))

( arg ( )) ( arg ( ))

1 1( ) ( ) ( )2 21 1( ) ( )2 2

( ) cos arg ( )

j t H j t H

j t H j t H

y t H e H e

H e H e

H t H

( )H

arg ( )H

[예제 4.10] • 그림에보인주파수응답을가진 LTI 시스템이있다고하자. 다음신호에대한출력을구하라.

LTI 시스템의주파수영역해석문제

(a) ( ) 2exp( 4 )(b) ( ) 2exp( 10 )

(c) ( ) cos 24

x t j tx t j t

x t t

0

( )H

88

( )H

1

88

2

2

[풀이] 4

( arg ( ))

(4 /4)

10

(a) ( ) 2 For ( ) , ( ) ( )

Since (4) 1, arg (4) / 4

( ) 2(b) ( ) 2 (10) 0 ( ) 0

(c) ( ) cos 24

j t

j t j t H

j t

j t

x t ex t Ae y t A H e

H H

y t ex t eH

y t

x t t

For ( ) cos( ), ( ) ( ) cos( arg ( ))

Since (2) 1, arg (2) / 8 ( ) cos(2 / 8)

x t A t y t A H t H

H Hy t t

[예제 4.11] • LTI 시스템의주파수응답이다음과같다고하자.

• 다음신호에대한출력을구하라.

(a) ( ) cos(b) ( ) cos 2(c) ( ) cos ,

x t tx t tx t t

1( )1

Hj

[풀이] Lowpass filter

1

2

1

1

1( ) , arg ( ) tan ( )1

(a) 11 (1) 0.707, arg (1) tan (1) 452

( ) (1) cos( arg (1)) 0.707 cos( 45 )(b) 2

1 (2) 0.447, arg (2) tan (2) 63.45

( ) (

H H

H H

y t H t H t

H H

y t H

2

2) cos(2 arg (2)) 0.447 cos(2 63.4 )(c)

1 lim ( ) lim 01

( ) 0

t H t

H

y t

[예제 4.12] • LTI 시스템의미분방정식이다음과같다고하자.

• 다음신호에대한출력을구하라.

( ) ( )( )dy t dx ty tdt dt

(a) ( ) cos(b) ( ) cos 2(c) ( ) cos ,

x t tx t tx t t

[풀이]

( ) e j tx t • 이면, 이다.

• 에 와 를대입하면,

( ) ( ) e j ty t H

1

2

( ) ( )

1 ( )

( )1

( ) , arg ( ) 90 tan ( ) for 01

j t j t j t

j t j t

d dH e H e edt dt

j H e j e

jHj

H H

( ) ( )( )dy t dx ty tdt dt

( )y t( )x t

[풀이] Highpass Filter

1

1

(a) 11 (1) 0.707, arg (1) 90 tan (1) 452

( ) (1) cos( arg (1)) 0.707 cos( 45 )(b) 2

2 (2) 0.894, arg (2) 90 tan (2) 26.65

( ) (2) cos(2 arg (2)) 0.894c

H H

y t H t H t

H H

y t H t H

2

os(2 26.6 )(c)

lim ( ) lim 1, lim arg ( ) 01

( ) ( )

t

H H

y t x t

[예제 4.14]• LTI system의 impulse response가다음과같다.

• 다음입력이주어지는경우

(a) 출력을구하라.(b) 입력과출력의푸리에급수를구하라.(c) 입력과출력의평균전력을구하라.

( ) 2exp( 2 ) ( )h t t u t

1 1( ) 1 cos 2 cos(4 30 ) cos62 3

x t t t t

[풀이]

(a) 입력의주파수성분은 0, 2, 4, 6 rad/sec이며, 0 = 2 rad/sec

2 (2 )

0 0

(2 )

0

1

2

( ) ( ) 2 2

2 2 2 2

2( ) , arg ( ) tan 24

j t t j t j t

j t

H h t e dt e e dt e dt

ej j

H H

1 1

1

1( ) (0) 1 (2) cos(2 arg (2)) (4) cos(4 30 arg (4))2

1 (6) cos(6 arg (6))3

2 1 21 cos 2 tan 1 cos 4 30 tan 228 20

2 1 cos 6 tan 3340

1 0.354cos 2 45 0.448cos 4 33.4 0.106cos 6 71.6

y t H H t H H t H

H t H

t t

t

t t t

[풀이](b) 입력과출력의푸리에계수를 cn, dn이라하자.

(c) Parseval의정리를이용하여전력을구하면

30 300 1 1 2 2 3 3

1 1 1 11, , , , , 0 for other 4 2 2 6

j jnc c c c e c e c c c n

45 45 33.4 33.40 1 1 2 2

71.6 71.63 3

1, 0.177 , 0.177 , 0.224 , 0.224 ,

0.053 , 0.053 , 0 for other

j j j j

j jn

d d e d e d e d e

d e d e d n

32 2

01

32 2 2 2 2 2

01

1 1 12 1 2 2 2 2.5564 2 36

2 1 2 0.177 2 0.224 2 0.053 1.169

x nn

y nn

P c c

P d d