otomatik kontrol
description
Transcript of otomatik kontrol
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Otomatik Kontrol I
Laplace Dönüşümü
Vasfi Emre Ömürlü
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By Vasfi Emre Ömürlü, Ph.D., 2007 2
Laplace Dönüşümü:ÖzellikleriTeoremleri
Kısmî Kesirlere Ayırma
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By Vasfi Emre Ömürlü, Ph.D., 2007 3
Laplace TransformIt is advantageous to solveBy using, we can convert many common functions into
Operations like differentiation and integration can be replaced by algebraic equations.A linear differential equations can be transformed into an algebraic equation.If the algebraic equation in s is solved for the dependent variable, then the solution of the differential equation may be found by use of
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By Vasfi Emre Ömürlü, Ph.D., 2007 4
Laplace dönüşümünün avantajı
Grafik tekniklerin kullanımına imkan verir
Diferansiyel denklemlerin çözümünü kolaylaştırır
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By Vasfi Emre Ömürlü, Ph.D., 2007 5
Some dynamic systems and their mathematical representations
q1
Discharge valve
Discharge valve
Q4Q2 Q3
h1 h2
Automatic control valve to adjust the liquid levels of the tanks by controlling the flap angle ϕ
Valve to adjust the flow rate Q3 between tanks
Q1
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By Vasfi Emre Ömürlü, Ph.D., 2007 6
Kompleks DeğişkenBir kompleks sayı gerçek ve imajiner kısımlardan oluşur. Bu iki kısım değişken olduğundan kompleks değişken ismini alır.
G(s) kompleks fonksiyonu gerçek ve imajiner kısımlardan oluşur, Gx ve Gy.
Doğrusal kontrol sistemlerinde kompleks fonksiyonlara çokça rastlarız ki bunlar s cinsinden fonksiyonlardır.
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By Vasfi Emre Ömürlü, Ph.D., 2007 7
Euler`s Theorem
=θ=θ
sincos
Since
=θ+θ sincos j
...!4)(
!3)(
!2)()(1
432
+++++=xxxxex
Euler’s theorem
Also
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By Vasfi Emre Ömürlü, Ph.D., 2007 8
(Ters) Laplace Dönüşümü Tanımı ve varlığı
{ } dtetfsFf(t) st∫∞
−⋅==0
)()(L
f(t)=s=
L=F(s)=
Ters laplace dönüşümü de mevcuttur ve L-1 ile gösterilir.
Genellikle Laplace dönüşümünün integral fonksiyonu yerine daha basit yöntemleri kullanırız.
f(t) fonksiyonunun laplace dönüşümü laplace integrali yakınsarsa mevcuttur. Bu da ancak f(t) fonksiyonu t>0 için her sonlu aralıkta sürekli ise ve t sonsuza giderken fonksiyon üstel bir hal alıyorsa mümkündür.
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By Vasfi Emre Ömürlü, Ph.D., 2007 9
Bazı yaygın laplace dönüşümüörnekleri
Basamak fonksiyonu
⎩⎨⎧
≥<
=000
tforAtfor
f(t)
Yüksekliği bir olan basamak fonksiyonuna birim basamak fonksiyonu denir. t=to da gerçekleşen birim basamak fonksiyonu t-to ın fonksiyonu manasına 1(t-to) ile gösterilir.
step
0
1
2
3
4
5
6
7
0 2 4 6 8 10 12
time(sec)
sign
al s
tren
gth
(sig
nal u
nit)
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By Vasfi Emre Ömürlü, Ph.D., 2007 10
Bazı yaygın laplace dönüşümüörnekleri
Üstel fonksiyon
⎩⎨⎧
≥⋅<
= − 000
tforeAtfor
f(t) at
exp decay
0
1
2
3
4
5
6
0 2 4 6 8 10 12
time(sec)
sign
al s
tren
gth
(sig
nal u
nit)
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By Vasfi Emre Ömürlü, Ph.D., 2007 11
Bazı yaygın laplace dönüşüm örnekleri
Rampa fonksiyonu
{ } dtteAdtetAtf
tfortAtfor
f(t)
atst ∫∫∞
−∞
− ⋅⋅=⋅⋅=
⎩⎨⎧
≥⋅<
=
00
)(
000
L ramp
0
1
2
3
4
5
6
7
0 2 4 6 8 10 12
time(sec)
sign
al s
tren
gth
(sig
nal u
nit)
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By Vasfi Emre Ömürlü, Ph.D., 2007 12
Bazı genel laplace dönüşümüörnekleri
Sinüs fonksiyonu
( )tjtj eej
AtArecalltfortAtfor
f(t) ω−ω −=ω⎩⎨⎧
≥ω⋅<
=2
sin0sin00
sine
-6
-4
-2
0
2
4
6
0 2 4 6 8 10 12
time(sec)
sign
al s
tren
gth
(sig
nal u
nit)
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By Vasfi Emre Ömürlü, Ph.D., 2007 13
En çok kullanılan (kullanacağımız) dönüşümler
f(t) F(s)A
AtnAt
ateA −⋅
)sin( taA ⋅⋅)cos( taA ⋅⋅
)sin( taeA bt ⋅⋅ −
)cos( taeA bt ⋅⋅ −
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By Vasfi Emre Ömürlü, Ph.D., 2007 14
Sinyal şekilleri
-6
-4
-2
0
2
4
6
8
0 2 4 6 8 10 12
time(sec)
sign
al s
tren
gth
(sig
nal u
nit) A
At
nAt
ateA −⋅
)sin( taA ⋅⋅)cos( taA ⋅⋅
)sin( taeA bt ⋅⋅ −
)cos( taeA bt ⋅⋅ −
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By Vasfi Emre Ömürlü, Ph.D., 2007 15
Laplace dönüşümü özellikleri -süperpozisyon
)()(
)()(
21
21
sFsF
tftff(t)
⋅β+⋅α=
⋅β+⋅α=
Ölçekleme özelliği { } )()( 11 sFtf ⋅=⋅ ααL
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By Vasfi Emre Ömürlü, Ph.D., 2007 16
Laplace dönüşümü özelliği - gecikme
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By Vasfi Emre Ömürlü, Ph.D., 2007 17
Laplace dönüşümü özelliği - gecikme
{ } ∫∞
− ⋅⋅−=0
1 )()( dtetftf stλL
Suppose f(t) is delayed by λ>0. The Laplase transform of the function,
Define a new variable, t’=t- λ, and then, dt’=dt, f(t)=0 for t<0
{ } )()( sFetf s ⋅=− −λλLt00
A/t0
f(t)
tλ+t0λ
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By Vasfi Emre Ömürlü, Ph.D., 2007 18
Laplace dönüşümü özelliği – türev
)0()0(........)0()0()()(
)(
)(
)1()2(21
2
2
−−−− −−−−−=⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
nnnnnn
n
fsffsfssFstfdtd
tfdtd
tfdtd
&L
L
L
f(0) fonksiyonun başlangıç şartı ve df(0)/dt fonksiyonun türevinin başlangıç şartıdır.Mesela, fonksiyon mekanik sistemin konumunu veriyorsa, konum ve hız başlangıçşartları gibi.
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By Vasfi Emre Ömürlü, Ph.D., 2007 19
Bazı laplace dönüşümü örnekleriDarbe fonksiyonu
{ }
)1(
)(1)(1)(1)(1)(
)(1)(1)(,00
0
00
000
00 0
000
00
000
0
00
stst
ststst
est
Aest
Ast
A
dtettdtettAdtett
tAt
tAtf
tttAt
tAtf
tttfor
ttfortA
f(t)
−−
∞∞ −−
∞−
−=−=
⎥⎦
⎤⎢⎣
⎡⋅−−⋅⋅=⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛−⋅−⋅=
−⋅−⋅=⎭⎬⎫
⎪⎩
⎪⎨⎧
><
<<=
∫ ∫∫L
Burada, A ve t0 sabittir. Darbe fonksiyonu yüksekliği A/t0 olan, t=0 da başlayan bir basamak fonksiyonu ve t=t0 da negatif aynı şiddette bir basamak fonksiyonu ile birleşen bir toplam fonksiyon olarak düşünülebilir.
t00
A/t0
f(t)
t
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By Vasfi Emre Ömürlü, Ph.D., 2007 20
Bazı laplace dönüşümü örnekleriDarbe fonksiyonu
{ }
[ ]
( )AsAs
stdtd
eAdtd
est
Atf
tttfor
ttfortA
f(t)
st
t
st
t
t
==−
=
−=
⎪⎩
⎪⎨⎧
><
<<=
−
→
−
→
→
/)1(
lim
)1(lim)(
,00
0lim
00
0
0
00
0
00
0
0
0
0
0
0
L
Darbe fonksiyonun yüksekliği A/t0 ve süresi t0 olduğundan, bunun altındaki alan direk olarak A dır. t0 0 a yaklaştığında, alan A olarak kalır. Şu da hatırlanmalıdır ki darbe fonksiyonunun genliği altındaki alanla ölçülür.Darbe fonksiyonunun altındaki alan 1 e eşit ise buna birim darbe fonksiyonu veya Dirak Delta fonksiyonu denir.
0,t0
A/t0
f(t)
t
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By Vasfi Emre Ömürlü, Ph.D., 2007 21
Laplace dönüşümü teoremleri – son değer teoremi
)(lim)(lim0
ssFtfst →∞→
=
Example: aşağıdaki sistemin kalıcı hal değerini y(∞) bulunuz.
=
+++
=
∞→
⋅−±−
)(lim
)102()2(3)(
210442:
2
ty
sssssY
t
poles
44 344 21
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By Vasfi Emre Ömürlü, Ph.D., 2007 22
Laplace dönüşümü teoremleri – ilk değer teoremi ve DC kazanç
43421existshould
sssFf
−
∞→=+ )(lim)0(
)(lim0
sGGainDCs→
=−
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By Vasfi Emre Ömürlü, Ph.D., 2007 23
Kısmî kesirlere ayırma
}
{
=
−
−=
++⋅+++⋅+⋅
==
∏
∏
=
=−
+−
n
jpolescalled
j
m
i
zeroscalled
i
rootspnwithpolynomialreen
nnn
rootszmwithpolynomialreem
mmm
ps
zsK
asasbsbsb
sAsBsF
jth
ith
1
1
.....deg
11
.....deg
11
21
)(
)(
............
)()()(
4444 34444 21
44444 844444 76
Neden ihtiyaç duyuyoruz?
Fonksiyonun s-ortamında paydasının köklerine bağlı olarak kısmî kesirlere ayırma üçayrı şekilde yapılır.1. Payda ayrık gerçek köklere sahipse, 2. paydada kompleks kökler varsa, 3. paydada tekrar eden kökler varsa.
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By Vasfi Emre Ömürlü, Ph.D., 2007 24
Kısmî kesirlere ayırma – ayrık kökler
12
2
1
1 ,)(
...)()(
)( Cforps
Cps
Cps
CsF
n
n
−++
−+
−=
1)()( 11 ps
sFpsC→
⋅−=
npsnn sFpsC→
⋅−= )()(
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By Vasfi Emre Ömürlü, Ph.D., 2007 25
Kısmî kesirlere ayırma – ayrık kompleks kökler
{
( )
=
=
=
⎟⎟⎠
⎞⎜⎜⎝
⎛++
++⇒
=⇒−=−=⇒=++++
⇒++
++++=
+++
+=++
=++
++=
++=
)(
)(
23)2/1(
2/12/1
)(1,111)1()1(1
1)1()1(1
1)1(
1
,1)1(
1)(
2
2
3232
2
23
22
232
2
1232
.
12
tf
sF
s
s
sFCCsCsCsss
sCsCssCsC
ssss
CssCsC
sC
ssssF
usualassolve
Bazı kökler kompleks ise
-0,2
0
0,2
0,4
0,6
0,8
1
1,2
0 2 4 6 8 10
time(sec)
sign
al s
tren
gth
(sig
nal u
nit)
stepsine decaycosine decayf(t)
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By Vasfi Emre Ömürlü, Ph.D., 2007 26
Kısmî kesirlere ayırma – tekrar eden kökler
( ) ( )
( ) ( )( ) ( )
( ) ( )
( ) ( )[ ] ( ) ( )[ ]( )[ ]
( )[ ]
( ) ( )tt
s
etetfsss
sF
CsFsdsdagainatingdifferenti
sCsFsdsd
CCsCsdsdsFs
dsdalsoCsFs
CCsCssC
sC
sC
ss
sss
sC
sC
sC
ssssF
s
ss
−−
−→
+=⇒+
++
++
=
===⎭⎬⎫
⎩⎨⎧
+
=+==⎭⎬⎫
⎩⎨⎧ +
++++=+==+
++++=⎥⎦
⎤⎢⎣
⎡
++
++
++=
+++
+
++
++
+=
+++
=
−→
−→−→
232
13
2
2
23
32123
31
3
3212
33
2213
3
23
33
221
3
2
)(1
21
01
1)(
1221)(1
21
022)(1
11)(1,2)(1
11111
1)1(
321
111)1(32)(
1
11
Bazı kökler tekrar ediyorsa
0
0,2
0,4
0,6
0,8
1
1,2
0 2 4 6 8 10
time(sec)
sign
al s
tren
gth
(sig
nal u
nit) f(t)
e^-t
t^2*e^-t
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By Vasfi Emre Ömürlü, Ph.D., 2007 27
Örnek: tank dinamiğiProseste kullanıla tank dinamiği şöyle veriliyor:
Q(s)
-h(t) yi bulunuz-h(t) nin t = 105 teki genliğini bulunuz
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By Vasfi Emre Ömürlü, Ph.D., 2007 28
Örnek: tank dinamiği)(
101)( sQ
ssH ⋅
+=
I
II
III
IV
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By Vasfi Emre Ömürlü, Ph.D., 2007 29
Örnek: tank dinamiği
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By Vasfi Emre Ömürlü, Ph.D., 2007 30
Örnek: tank dinamiği=)(sH
=3C
=2C
=⎭⎬⎫
⎩⎨⎧=
=0
21 )(
ssHs
dsdC
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By Vasfi Emre Ömürlü, Ph.D., 2007 31
Örnek: tank dinamiğitetth 10
1001
101001)( −++
−=
Bu sonuç sadece 1/s2 girişi içindir, ama diğer cevaplar süperpozisyon ve ölçeklendirme özelliği kullanılarak elde edilebilir.
h(t) for only 1/s^2
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
0 0,2 0,4 0,6 0,8 1time (sec)
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By Vasfi Emre Ömürlü, Ph.D., 2007 32
Örnek: tank dinamiğiOverall system response is
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪
⎨
⎧
∞<≤
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
+−+−
++−+−
−+−+−
−++−
<≤
⎪⎪⎪
⎭
⎪⎪⎪
⎬
⎫
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
+−+−
−+−+−
−++−
<≤
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
+−+−
−++−
<⎭⎬⎫
⎩⎨⎧ ++−
=
−−
−−
−−
−
−−
−−
−
−−
−
−
tT
eTQ
TtTQTQ
eTQ
TtTQTQ
eTQ
TtTQTQ
eTQ
tTQTQ
TtT
eTQ
TtTQTQ
eTQ
TtTQTQ
eTQ
tTQTQ
TtTe
TQTtTQ
TQ
eTQ
tTQTQ
TtforeTQ
tTQTQ
th
Tt
Tt
Tt
t
Tt
Tt
t
Tt
t
t
3"
10/
)3)(/(10
/10
/)2)(/(
10/
10/
))(/(10
/10
/)/(
10/
32"
10/
)2)(/(10
/10
/))(/(
10/
10/
)/(10
/
2"
10/
))(/(10
/10
/)/(
10/
10/
)/(10
/
)(
)3(1000
0
)2(1000
0
)(1000
0
1000
0
)2(1000
0
)(1000
0
1000
0
)(1000
0
1000
0
1000
0
tank height, h(t)
0
0,05
0,1
0,15
0,2
0,25
0,3
0,35
0 0,5 1 1,5time (sec)
heig
ht (m
)
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By Vasfi Emre Ömürlü, Ph.D., 2007 33
Örnek: kütle-sönüm-yay sistemi
?
)(tfkxxbxm =++ &&&
Sistem matematik modeli
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By Vasfi Emre Ömürlü, Ph.D., 2007 34
Örnek: dinamik sistem cevabının laplace dönüşümü
?t
1
u (t)=⎩⎨⎧01
∞<<<<
tt
110
G(s)=1/(s+2)
1 saniye süren bir darbe fonksiyonu için yukarıdaki sistemin cevabını bulunuz.
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By Vasfi Emre Ömürlü, Ph.D., 2007 35
Örnek: dinamik sistem cevabının laplace dönüşümü
=Ι )(sY
=Ι )(sY
=ΙΙ )(sY One-second delayed of YI(s).
Response of the System to a second long pulse
00,10,20,30,40,5
0 1 2 3
time (sec)sy
stem
out
put
t
1
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By Vasfi Emre Ömürlü, Ph.D., 2007 36
Ex-1
1 , 0 t 1u(t) 0 , 1 t 3
1 , 3 t
< <⎧ ⎫⎪ ⎪= < <⎨ ⎬⎪ ⎪− < < ∞⎩ ⎭
{ }u(t) ?=L
(Time delay)
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By Vasfi Emre Ömürlü, Ph.D., 2007 37
Ex-1
{ } )()( sFetf s ⋅=− −λλL
s
s
1 0s
1U(s) e 1s1 e 3
s
− λ
− λ
⎧ ⎫λ =⎪ ⎪⎪ ⎪−⎪ ⎪= ⋅ λ =⎨ ⎬
⎪ ⎪−⎪ ⎪⋅ λ =⎪ ⎪⎩ ⎭
s 3s1 1 1U(s) e es s s
− −⎡ ⎤ ⎡ ⎤= − ⋅ − ⋅⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
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By Vasfi Emre Ömürlü, Ph.D., 2007 38
Ex-2 (Differentation)
y 10 y 9y 5ty(0) 1
y(0) 2
•• •
•
− + == −
=
Find the Laplace Transform of thisequation…
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By Vasfi Emre Ömürlü, Ph.D., 2007 39
Ex-2
)0()0(........)0()0()()(
)0()0()()(
)0()()(
)1()2(21
22
2
−−−− −−−−−=⎭⎬⎫
⎩⎨⎧
−−=⎭⎬⎫
⎩⎨⎧
−=⎭⎬⎫
⎩⎨⎧
nnnnnn
n
fsffsfssFstfdtd
fsfsFstfdtd
fssFtfdtd
&
&
L
L
L
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By Vasfi Emre Ömürlü, Ph.D., 2007 40
Ex-2
{ }{ }{ }
{ }
2
2
y s Y(s) s y(0) y(0)
y s Y(s) y(0)
y Y(s)55ts
•• •
•
= ⋅ − ⋅ −
= ⋅ −
=
=
L
L
L
L2
2
2 2
5Y(s) s 10s 9 s 12s
5 s 12Y(s)s (s 9)(s 1) s (s 9)(s 1)
− + + − =⎡ ⎤⎣ ⎦
−= −
− − − −
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4-2
0
2
4
6
8
10
12
time (sec)
mag
nitu
de
System ResponseHO1
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By Vasfi Emre Ömürlü, Ph.D., 2007 41
Ex-3 (Distinct Poles)
s 2F(s)(s 9)(s 1)
+=
− −
Find the Inverse Laplace Transform of thisequation…
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By Vasfi Emre Ömürlü, Ph.D., 2007 42
npsnn sFpsC=
⋅−= )()(
Ex-3
s 2 A BF(s)(s 9)(s 1) s 9 s 1
+= = +
− − − −
[ ]
[ ]
s 9
s 1
11A (s 9)F(s)83B (s 1)F(s)
8
=
=
= − =
−= − =
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By Vasfi Emre Ömürlü, Ph.D., 2007 43
Ex-3
{ }1
9 t t
3118 8F(s)
s 9 s 1f (t) F(s)
11 3f (t) e e8 8
−
−= +
− −=
= −
L
Impulse Response
Time (sec)
Ampl
itude
0 0.05 0.1 0.15 0.2 0.25 0.30
2
4
6
8
10
12
14
16
18
20
HO2
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By Vasfi Emre Ömürlü, Ph.D., 2007 44
Ex-4 (Repeated poles)
2
s 25F(s) )s (s 5)
+=
−
2 2
s 25 A B CF(s) )s (s 5) s s s 5
+= = + +
− −
Find the Inverse Laplace Transform of thisequation…
HO3
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By Vasfi Emre Ömürlü, Ph.D., 2007 45
Ex-4
[ ]s 5
2
s 0
2
s 0 s 0
2s 0 s 0
30 6C (s 5)F(s)25 5
B (s )F(s) 5
d d s 25A (s )F(s)ds ds s 5
d s 25 (s 5) (s 25) 30 6ds s 5 (s 5) 25 5
=
=
= =
= =
= − = =
= = −⎡ ⎤⎣ ⎦+⎧ ⎫ ⎧ ⎫= =⎨ ⎬ ⎨ ⎬−⎩ ⎭ ⎩ ⎭
⎧ ⎫+ − − + − −⎧ ⎫ = = =⎨ ⎬ ⎨ ⎬− −⎩ ⎭ ⎩ ⎭
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By Vasfi Emre Ömürlü, Ph.D., 2007 46
Ex-4
{ }2
1
5 t
6 655 5F(s)s s s 5
f (t) F(s)6 6f (t) 5t e5 5
−
= − −−
=
= − −
L
Impulse Response
Time (sec)
Ampl
itude
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
2
4
6
8
10
12
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By Vasfi Emre Ömürlü, Ph.D., 2007 47
Ex-5
L1/L2 =1/2
M=1 kg
R=60Nsec/m
K=800N/m
F(t)=1 N
- Find y(t)
12
2
Y(s) L 1F(s) L ms rs k
= ⋅+ +
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By Vasfi Emre Ömürlü, Ph.D., 2007 48
Ex-5
0 0 0 0
st 2st 3st2 2 2 2
F F F FT T T T
F(s) e e es s s s
− − −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠= − − +
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By Vasfi Emre Ömürlü, Ph.D., 2007 49
12
2
12
2
2 2
Y(s) L 1F(s) L ms rs k
L 1Y(s) F(s)L ms rs k1 1 1Y(s)2 s 60s 800 s
= ⋅+ +
= ⋅ ⋅+ +
= ⋅ ⋅+ +
Ex-5
Common term for every sub-input
31 2 42 2 2
C1 1 1 C C CY(s)2 s 60s 800 s s s (s 40) (s 60)
= ⋅ ⋅ = + + ++ + + +
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By Vasfi Emre Ömürlü, Ph.D., 2007 50
Ex-5
[ ]
[ ]
22 s 0
21 2
s 0
52 2
s 0
53 2s 40
s 40
4 2s 60s 40
1C s Y(s)1600
d d 1C s Y(s)ds ds s 60s 800
2s 60 9 10(s 60s 800)
1C (s 40)Y(s) 1,56 102s (s 60)
1C (s 60)Y(s) 6, 92s (s 40)
=
=
−
=
−
=−=−
=−=−
= =⎡ ⎤⎣ ⎦
⎧ ⎫ ⎧ ⎫= ⋅ ⋅ = ⋅⎨ ⎬ ⎨ ⎬+ +⎩ ⎭ ⎩ ⎭
⎧ ⎫− −= = − ⋅⎨ ⎬+ +⎩ ⎭
⎡ ⎤= + = = ⋅⎢ ⎥+⎣ ⎦
⎡ ⎤= + = = − ⋅⎢ ⎥+⎣ ⎦
610−
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By Vasfi Emre Ömürlü, Ph.D., 2007 51
Ex-5
5 t 6 t
5 4 5 6
2
5 4 1,56 10 6,9 10
9 10 6, 25 10 1,56 10 6, 9 10Y(s)s s s 40 s 60
y(t) 9 10 6, 25 10 t e e− −
− − − −
− − − ⋅ − ⋅
− ⋅ ⋅ ⋅ − ⋅= + + +
+ += − ⋅ + ⋅ ⋅ + +
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By Vasfi Emre Ömürlü, Ph.D., 2007 52
Ex-5
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
5 t 6 t
5 t T 6 t T
5 t 2 T 6 t 2 T
5 t 3T
5 4 1,56 10 6,9 10
5 4 1,56 10 6,9 10
5 4 1,56 10 6,9 10
5 4 1,56 10
I 9 10 6, 25 10 t e e
II 9 10 6, 25 10 t T e e
III 9 10 6, 25 10 t 2T e e
IV 9 10 6, 25 10 t 3T e e
− −
− − − −
− − − −
− −
− − − ⋅ − ⋅
− − − ⋅ − ⋅
− − − ⋅ − ⋅
− − − ⋅ −
= − ⋅ + ⋅ ⋅ + +
= − ⋅ + ⋅ ⋅ − + +
= − ⋅ + ⋅ ⋅ − + +
= − ⋅ + ⋅ ⋅ − + +( )6 t 3T6,9 10− −⋅
Overall system response:
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By Vasfi Emre Ömürlü, Ph.D., 2007 53
Ex-5
Overall system response:
( )( )
( )
0
0
0
0
t T y(t) (F / T) IT t 2T y(t) (F / T) I II2T t 3T y(t) (F / T) I II III3T T y(t) (F / T) I II III IV
< = ⋅⎧ ⎫⎪ ⎪≤ ≤ = ⋅ −⎪ ⎪⎨ ⎬≤ ≤ = ⋅ − −⎪ ⎪⎪ ⎪≤ ≤ ∞ = ⋅ − − +⎩ ⎭
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By Vasfi Emre Ömürlü, Ph.D., 2007 54
Ex-5
)(lim)(lim0
ssFtfst →∞→
=
Final value teorem:
{ }
{ }
5 t 6 t5 4 1,56 10 6,9 10
5
t
2 2
2s 0
t s 0
y(t) 9 10 6, 25 10 t e elim y(t) 9 10 1 1
1 1 1Y(s)2 s 60s 800 s
1 1 1 1lim sY(s)2 s 60s 800 s 0
lim y(t) lim sY(s)
− −− − − ⋅ − ⋅
−
→∞
→
→∞ →
= − ⋅ + ⋅ ⋅ + +
= − ⋅ + ∞ + + = ∞
= ⋅ ⋅+ +
= ⋅ ⋅ = = ∞+ +
=
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By Vasfi Emre Ömürlü, Ph.D., 2007 55
Ex-5
Initial value teorem:
43421existshould
sssFf
−
∞→=+ )(lim)0(
{ }
2 2
2s
1 1 1Y(s)2 s 60s 800 s
1 1 1 1y(0 ) lim sY(s) 02 s 60s 800 s→∞
= ⋅ ⋅+ +
+ = = ⋅ ⋅ = =+ + ∞