eec4 4a ss lecture 03
TRANSCRIPT
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Signals & Systems FEEE, HCMUT
Ch-2: Phn tch h thng LTI trong min thi gian
Lecture-3
2.1. Gii thiu
2.2. H thng LTI: tch chp
2.3. Cc tnh cht ca h thng LTI
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Signals & Systems FEEE, HCMUT
2.1. Gii thiu
Trong mn hc ny ta tp trung kho st h thng LTI:
Nhiu h thng vt l thc t c tnh LTI
H thng LTI tha nguyn l xp chng & bt bin: biu din
tn hiu vo thnh tng cc tn hiu c bn (hoc phin bn tr)
p ng ca h thng mt cch d dng.
Cc v d v biu din tnh hiu thnh tng cc tnh hiu c bn:
Biu din tnh hiu thnh tng ca cc xung n v
Biu din tnh hiu thnh tng cc tnh hiu hm m phc:
chui Fourier, bin i Fourier, bin i Laplace
Trong chng ny ta kho st vic biu din tn hiu thnh tng cc
xung n v tnh p ng ca h thng dng khi nim p ng
xung ca h thng v tch chp.
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Signals & Systems FEEE, HCMUT
2.2. H thng LTI: Tch chp
2.2.1. Biu din tn hiu thnh tng cc xung n v
2.2.2. p ng xung v biu din h thng LTI bng tch chp
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Signals & Systems FEEE, HCMUT
2.2.1. Biu din tn hiu thnh tng cc xung n v
nh ngha xung (t):
1 ; 0
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Signals & Systems FEEE, HCMUT
2.2.2. p ng xung v biu din h thng LTI bng tch chp
p ng xung ca h thng LTI: l p ng ca h thng vi (t)
V d: (a) h thng n v y(t)=f(t) h(t)= (t)
(b) h thng c phng trnh: t
-y(t)= f ( )d
t
-h(t)= ( )d u(t)
p ng ca h thng LTI vi xung (t):
0lim h (t)=h(t)
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Signals & Systems FEEE, HCMUT
2.2.2. p ng xung v biu din h thng LTI bng tch chp
p ng ca h thng LTI vi tn hiu gn ng ca f(t) ~
n
f (t)= f(n ) (t n )Vi:
Do h thng LTI nn: ~
n
y(t)= f(n )h (t n )
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Signals & Systems FEEE, HCMUT
2.2.2. p ng xung v biu din h thng LTI bng tch chp
p ng ca h thng LTI vi tn hiu vo f(t)
~
0f(t) lim f (t)= f( ) (t )dTa c:
Suy ra: ~
0 0n
y(t)= lim y(t)= lim f(n )h (t n )
y(t) f( )h(t )d y(t) f(t) h(t) (tch chp)
Trong phn tch v thit k ngi ta hay biu din m hnh h
thng LTI theo tch chp vi p ng xung h(t)
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Signals & Systems FEEE, HCMUT
2.2.2. p ng xung v biu din h thng LTI bng tch chp
Tnh tch chp: f(t) h(t)= f( )h(t )d
(Lu : ta s tnh tch phn trn tnh theo thang thi gian cn t l
tham s cng chnh l bin thi gian ca kt qu)
Xc nh h(t- ) theo bin :
Nhn f( ) vi h(t- )
Ly tch phn trn ton thang
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Signals & Systems FEEE, HCMUT
2.2.2. p ng xung v biu din h thng LTI bng tch chp
V d: cho f(t)=e-atu(t); a>0 l ng vo ca h thng LTI c p
ng xung h(t)=u(t). Xc nh p ng y(t) ca h thng?
f(t) h(t)=0
ta at1
a0
f(t) h(t)= e d (1-e )
at1a
y(t)=f(t) h(t)= (1-e )u(t)
t0
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Signals & Systems FEEE, HCMUT
2.3. Cc tnh cht ca h thng LTI
Tnh giao hon: +
y(t)=f(t) h(t)= f ( )h(t )d
t: 1 t 1t 1d d
1 1 1 1 1 1+
y(t)= f (t )h( )d h( )f(t ) =h(t) f(t)d
Tnh phn phi: 1 2 1 2y(t)=f(t) [h (t)+h (t)]=f(t) h (t)+f(t) h (t)
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Signals & Systems FEEE, HCMUT
2.3. Cc tnh cht ca h thng LTI
Tnh kt hp: 1 2 1 2y(t)=[f(t) h (t)] h (t)=f(t) [h (t) h (t)]
H thng LTI khng nh: h(t)=K (t)
y(t)=f(t) h(t)=f(t) K (t)=Kf(t)
Tnh kh ngch: ta d dng chng minh c h thng nghch o
ca mt h thng LTI cng l h thng LTI. Do h thng LTI
kh nghch khi tn ti hi(t) sao cho ih(t) h (t)= (t)
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Signals & Systems FEEE, HCMUT
2.3. Cc tnh cht ca h thng LTI
Tnh nhn qu: h thng LTI nhn qu khi h(t)=0 khi t