eec4-4a-ss-lecture-01.pdf

Signals & Systems FEEE, HCMUT

404001 - Tn hiu v h thng


Chng 1. C bn v tn hiu v h thng

Chng 2. Phn tch HT tuyn tnh bt bin (LTI) trong min thi gian

Chng 3. Biu din tn hiu tun hon dng chui Fourier

Chng 4. Biu din tn hiu dng bin i Fourier

Chng 5. Ly mu

Chng 6. Phn tch h thng lin tc dng bin i Laplace

Chng 7. p ng tn s ca h thng LTI v thit k b lc tng t

404001 - Tn hiu v h thng


Ch-1: C bn v tn hiu v h thng

Lecture-1

1.1. C bn v tn hiu


1.1. C bn v tn hiu

1.1.1. Tn hiu v v d v tn hiu

1.1.2. Phn loi tn hiu

1.1.3. Nng lng v cng sut tn hiu

1.1.4. Cc php bin i thi gian

1.1.5. Cc dng tn hiu thng dng


nh ngha:

Tn hiu l hm ca mt hoc nhiu bin c lp (thi gian, khng

gian,) mang thng tin v hnh vi hoc bn cht ca cc hin

tng (vt l, kinh t, x hi,)

Tn hiu l hm theo 1 bin thi gian

V d 1: tn hiu in p uc(t) v dng in i(t) trong mch RC

c -t/RC

0; t


V d 2: Tn hiu thoi ghi li di dng in p u(t)



V d 3: Tn hiu in tim ghi li di dng in p u(t)



V d 4: The weekly Down-Jones stock market index



Tn hiu l hm nhiu bin:

nh tnh nh ng

f(x,y)f(x,y,t)



X l tn hiu: x l tng t & x l s tp trung XL tng t


Unfiltered signal

Filtered signal



C nhiu tiu ch phn loi tn hiu:

Tn hiu lin tc

Tn hiu tng t

Tn hiu khng tun hon

Tn hiu nng lng

Tn hiu xc nh

Tn hiu nhn qu

Tn hiu ri rc

Tn hiu s

Tn hiu tun hon

Tn hiu cng sut

Tn hiu ngu nhin

Tn hiu khng nhn qu

-

-

-

-

-

-

Trong , cch phn loi tn hiu lin tc v tn hiu ri rc l

thng dng nht (trong mn hc ny ta ch kho st tn hiu

lin tc)

Tn hiu thc - Tn hiu phc



V d: phn loi tin hiu lin tc & ri rc, tng t & s

(b)

t

f(t)(a) f(t)

t

f(t)

t t

f(t)(c) (d)

f(t)

t

Continuous-time

vs

discrete-time

Analog

vs

digital

time

am

pli

tud

e



Xt tn hiu dng in i(t) qua in tr R:

Cng sut tc thi trn R: p(t)=u(t)i(t)=Ri2(t)

Nng lng tn hao trong khong thi gian [t1t2]: 2 2

1 1

t t2

t tp(t)dt Ri (t)dt

Cng sut tn hao trung bnh trong khong thi gian [t1t2]:

2 2

1 1

t t2

t t2 1 2 1

1 1p(t)dt Ri (t)dt

t t t t

Nng lng & cng sut trn in tr R=1 c gi l nng

lng v cng sut ca tn hiu dng in i(t)

Nng lng tn hiu trong khong [t1t2]: 2

1

t2

it

E i (t)dt

Cng sut tn hiu khong thi gian [t1t2]: 2

1

t2

it

2 1

1P i (t)dt

t t



Nh vy nng lng tn hiu v cng sut tn hiu khng phi l

nng lng v cng sut v mt vt l thng s nh gi ln

ca tn hiu

Trn thc t xc nh ln tn hiu ta thng xem tng qut

l tn hiu phc tn ti trn ton thang thi gian. Khi nng

lng v cng sut ca tn hiu f(t) c vit li dng tng qut

nh sau:

* 2

fE f(t)f (t)dt |f(t)| dt

T/22

f-T/2

T

1P |f(t)| dt

Tlim

Nng lng:

Cng sut:



V d:

t

f(t)

-1 0

2 2e-t/2

0-t

f-1 0

E = 4dt+ 4e 8

ff

T

EP = lim 0

T

Tn hiu

nng lng

2

f-

E = |f(t)| dtTn hiu

cng sut

t

f(t)

-1 0

1

1 2 3-2-3

-1

1 12 2

f-1 -1

1 1 1P = |f(t)| dt= t dt=

2 2 3


1.1.4. Cc php bin i thi gian

a) Php dch thi gian

b) Php o thi gian

c) Php t l thi gian

d) Kt hp cc php bin i


a) Php dch thi gian

f(t) (t)=f(t T)

T>0 dch sang phi (tr) T giy

T


V d 2: tn hiu tun hon & tn hiu khng tun hon

f(t) l tun hon nu vi T>0 f(t) = f(t+T) vi mi t

Gi tr nh nht ca T c gi l chu k ca f(t)

f(t) l tn hiu khng tun hon nu khng tn ti gi tr ca T tha tnh cht trn

t

f(t)

a) Php dch thi gian


b) Php o thi gian

f(t) (t)=f( t)

i xng f(t) qua trc tung

V d 1:

f(t)

t10 3

f(-t)

t-1 0-3

V d 2: Tn hiu chn v l

Hm chn: fe(-t)=fe(t); i xng qua trc tung

Hm l: fo(-t)=-fo(t); i xng ngc qua trc tung


Phn tch tn hiu thnh thnh phn chn v l

t

fe(t)

t

fo(t)

e of(t)=f (t)+f (t)

e

1f (t)= [f(t)+f(-t)]

2

o

1f (t)= [f(t)-f(-t)]

2

Thnh phn chn

Thnh phn l

b) Php o thi gian


V d 3: -at

0; t0)

e ; t 0

t

f(t)

1

t

fe(t)

1/2

t

fo(t)

1/2

-1/2

e o=f (t)+f (t)

Vi:

= +

at12

e -at12

e ; t0

at12

o -at12

e ; t0

b) Php o thi gian


c) Php t l thi gian

f(t) (t)=f(at); a>0

a>1 : co thi gian a ln

0



f(t) (t)=f(at b);a 0

Trng hp a>0:

Phng php 1:

Bc 1: Php dch thi gian g(t)=f(t-b)

Bc 2: Php t l (t)=g(at)

f(t)

t-2 4 -3 3

g(t)=f(t+1)

t

V d: (t)=f(2t+1)

t g(2t)=f(2t+1)

t-3/2 3/2

Bc 1 Bc 2




Trng hp a>0:

Phng php 2:

Bc 1: Php t l g(t)=f(at)

Bc 2: Php dch thi gian (t)=g(t-b/a)

f(t)

t-2 4

V d: (t)=f(2t+1)

t g(t+0.5)=f(2t+1)

t-3/2 3/2

Bc 1 Bc 2 g(t)=f(2t)

t-1 2




Trng hp a


1.1.5. Cc tn hiu thng dng

a) Hm bc n v u(t)

b) Xung n v (t)

c) Hm m


a) Hm bc n v u(t)

u(t)

t

1 1; t>0u(t)=

0; t


a) Hm bc n v u(t)

V d 2:

t; 0


b) Xung n v (t)

nh ngha : ( ) 0; 0t t

( ) 1t dtt

/2 /2t

(t)

0

Tnh cht 1: Nu f(t) lin tc ti t0 th: 0 0 0f(t)(t t )=f(t )(t t )

f(t)

tt0

t-t0

f(t0) (t-t

0)

tt0

2

2

+1 1( 1)= ( 1)

+9 5V d:


b) Xung n v (t)

Tnh cht 2: 0 0f(t)(t t )dt f(t )

V d: t=2

t tsin (t 2)dt=sin =1

4 4

Tnh cht 3:

du(t)(t)=

dt

t

()d u(t)

'du(t)f(t)dt= u(t)f(t) u(t)f (t)dtdt

'

0f ( ) f (t)dt

0f ( ) f(t) f(0) f(t)(t)dt


c) Hm m

s= +j : Tn s phc

st te =e (cost+jsint)

s*t te =e (cost-jsint)

V d: st t st s*t1

Re{e }=e cost= (e +e )2

t

0

0 0

t

) 0b) 0a


c) Hm m

t t

) 0; 0c ) 0; 0d

j

LHP RHP

a b

c d

V tr ca bin phc s= +j trong cc v d a, b, c, v d

eec4-4a-ss-lecture-01.pdf

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