lecture 01

1Signal & Systems - Tran Quang Viet FEEE, HCMUT Semester: 02/10-11

404001 - Tn hiu v h thng

CBGD: Trn Quang Vit Lin h : B mn CSKT P.104 nh B3

Email : [email protected] ; [email protected]

Ti liu tham kho

[1] B. P. Lathi, Signal Processing and Linear Systems, Berkeley-Cambridge Press, 1998.

[2] A. V. Oppenheim, Signals and Systems, Prentice-Hall, 1983.[3] Phm Th C, L Thuyt Tn Hiu, Nh xut bn i Hc

Quc Gia TP. H Ch Minh, 2005.

Signal & Systems - Tran Quang Viet FEEE, HCMUT Semester: 02/10-11


nh gi: Bi tp (Quiz, In-Class) : 20%

- Quiz (15-30 pht): chim 80%; 5 bichn 4 maxTB- In-Class : chim 20% ; gi ln bng TB

Kim tra gia k (Mid) : 20% Thi cui k : 60%


Chng 1. C bn v tn hiu v h thngChng 2. Phn tch HT tuyn tnh bt bin (LTI) trong min thi gianChng 3. Biu din tn hiu tun hon dng chui FourierChng 4. Biu din tn hiu dng bin i FourierChng 5. Ly muChng 6. Biu din tn hiu dng bin i LaplaceChng 7. p ng tn s v b lc tng t



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 hiu1.1.2. Phn loi tn hiu1.1.3. Nng lng v cng sut tn hiu1.1.4. Cc php bin i thi gian1.1.5. Cc dng tn hiu thng dng


nh ngha:Tn hiu l hm ca mt hoc nhiu bin c lp (thi gian, khnggian,) mang thng tin v hnh vi hoc bn cht ca cc hintng (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)

1.1.1. Tn hiu v v d v tn hiu


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)



1.1.2. Phn loi tn hiu

C nhiu tiu ch phn loi tn hiu:

Tn hiu lin tcTn hiu tng t

Tn hiu khng tun honTn hiu nng lng

Tn hiu xc nhTn hiu nhn qu

Tn hiu ri rcTn hiu s

Tn hiu tun honTn hiu cng sutTn hiu ngu nhinTn hiu khng nhn qu

-

-

-

-

-

-

Trong , cch phn loi tn hiu lin tc v tn hiu ri rc lthng dng nht (trong mn hc ny ta ch kho st tn hiulin tc)

Tn hiu thc - Tn hiu phc


1.1.2. Phn loi tn hiu

V d:

(b)(a)

(c) (d)

f(t)

t

Continuous-time

vs

discrete-time

Analog

vs

digital

time

ampl

itude


1.1.3. Nng lng v cng sut tn hiu

Xt tn hiu in p u(t) trn in tr R:

Cng sut tc thi trn R: p(t)=u(t)i(t)=(1/R)u2(t)

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

2 2

1 1

t t 2

t t

1p(t)dt u (t)dtR

=

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

2 2

1 1

t t 2

t t2 1 2 1

1 1 1p(t)dt u (t)dtt t t t R

=

Nu R=1 nng lng & cng sut thc t c xem l nnglng v cng sut ca tn hiu in p u(t)

Nng lng tn hiu trong khong [t1t2]:

2

1

t 2u t

E u (t)dt=

Cng sut tn hiu khong thi gian [t1t2]:

2

1

t 2u t

2 1

1P u (t)dtt t

=



Nh vy nng lng tn hiu v cng sut tn hiu khng phi lnng lng v cng sut v mt vt l (c nhng tn hiu khngphi l tn hiu vt l) m ch n thun l thng s nh gi ln ca tn hiu.

Trn thc t xc nh ln tn hiu ta thng xem tng qutl tn hiu phc tn ti trn ton thang thi gian. Khi nnglng v cng sut tn hiu c vit li dng tng qut nh sau:

2fE |f(t)| dt

=

T/2 2f

-T/2T

1P |f(t)| dtTlim

=

Nng lng:

Cng sut:



V d:

0-t

f-1 0

E = 4dt+ 4e 8

= f

f T

EP = lim 0T

=

Tn hiunng lng

2f

-

E = |f(t)| dt

= Tn hiucng sut1 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 gianb) Php o thi gianc) Php t l thi giand) Kt hp cc php bin i


a) Php dch thi gian

f(t) (t)=f(t T) T>0 dch sang phi (delay) T0 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

10


b) Php o thi gian

f(t) (t)=f( t) i xng f(t) qua trc tung V d 1:

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

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

11


V d 3:-at

0; t0)

e ; t 0

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

Vi:

= +

at12

e-at1

2

e ; t0

at12

o-at1

2

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 bi mt h s l a

0

12


d) Kt hp cc php bin i

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

Bc 1 Bc 2



f(t) (t)=f(at b);a 0 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) V d: (t)=f(2t+1)

Bc 1 Bc 2

13



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

14


a) Hm bc n v u(t)

1; t>0u(t)=

0; t

15


b) Xung n v (t)

nh ngha :( ) 0; 0t t =

( ) 1t dt

=0

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

2

2 +1 1

( 1)= ( 1) +9 5

V d:


b) Xung n v (t)

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

=

V d:2

sin ( 2) sin 14 4

t

t tt dtpi pi

=

= =

Tnh cht 3:

du(t)(t)=

dt ( )d u(t)t

=

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

=

0( ) '( )f f t dt= 0( ) ( ) (0)f f t f

= = ( ) ( )f t t dt

=

16


c) Hm m

s=+j : Tn s phcst te =e (cost+jsint)s*t te =e (cost-jsint)

V d: st t st s*t1Re{e }=e cost= (e +e )2

t0 =

0 < 0 >

) 0b =) 0a =


c) Hm m

) 0; 0c < ) 0; 0d >

j

LHP RHP

ab

c d

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

17


Bi tp

Bi 1: Tnh nng lng ca cc tn hiu nh hnh 1

hnh 1

Bi 2: Hy v cc hm f(-2t), f(2t+1), f(-2t-3), sau vit hm m tca chng; vi f(t) c cho nh hnh v di y

lecture 01

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