Download Report

Fremgangsmåte (metode) for endringer på et system.
For et kontinuerlig system med en inngang og en utgang (SISO-system):
Inngang u(t)
Utgang x(t)
System
1.
Bestem balanseligning (diff.ligning) for systemet:
•
∆ x = − a∆x + b∆u (1.orden system)
Skisser matematisk blokkdiagram:
x(0)
u
d x/dt
b
x
a/b
••
•
∆ x = − a∆x − c∆ x + b∆u (2.orden system)
Skisser matematisk blokkdiagram:
dx(0)
dt
x(0)
2
u
b
d
dt 2
d x
dt
x
x
c/b
a/b
2.
•
Sett ∆x = s∆x(s ) (1.orden)
•
Transformer ∆x = −a∆x + b∆u ⇒ s∆x( s ) = −a∆x( s ) + b∆u ( s )
∆x ( s )( s + a ) = b∆u ( s )
Bestem transferfunksjon på standard form:
1
b
a
∆x( s)
K
b
=
=
⇒
∆u ( s) s + a 1
Ts + 1
s +1
a
b
1
K = og T =
a
a
h( s ) =
•
••
Sett ∆x = s∆x( s) og ∆ x = s 2 ∆x( s ) (2.orden)
••
•
Transformer ∆ x = −c∆ x − a∆x + b∆u ⇒ s 2 ∆x( s) = −cs∆x( s ) − a∆x( s ) + b∆u ( s )
∆x( s)( s 2 + cs + a ) = b∆u ( s)
Bestem transferfunksjon på standard form:
b
⋅a
K ⋅ ω 02
∆x ( s )
b
a
=
=
⇒ 2
h( s ) =
∆u ( s ) s 2 + cs + a s 2 + cs + a
s + 2ζω 0 s + ω 02
K=
b
c
c
=
, ω 0 = a og ζ =
a
2ω 0 2 a
ζ = 1 : p = p1 = p 2 = −ω 0 = − a og T1 = T2 = T = −
ζ = 1 : h( s ) =
b
s + 2 a ⋅s+a
2
=
b
(s − a )
ζ 〉1 : p1, 2 = −ζω 0 ± ω 0 ζ 2 − 1 = −
ζ 〉1 : T1, 2 = −
ζ 〉1 : h( s ) =
c
2 a
2
1
1
=
p
a
b
=
a(
1
a
a± a (
b
a
=
s + 1) 2
(
1
a
s + 1) 2
=
K
(Ts + 1) 2
c
1
) 2 − 1 = − (c ± c 2 − 4a )
2
2 a
1
1
=
p1, 2 1
(c ± c 2 − 4a )
2
b
=
s + cs + a
2
b
1
1
(c + c 2 − 4a ))( s + (c − c 2 − 4a ))
2
2
b
b
=
ζ 〉1 : h( s ) = 2
1
1
1
1
s + cs + a
s + 1)(
s + 1) (c + c 2 − 4a ) (c − c 2 − 4a )
(
1
1
2
2
(c − c 2 − 4 a )
(c + c 2 − 4 a )
2
2
(s +
2
ζ 〉1 : h( s ) =
ζ 〉1 : h( s ) =
b
=
s + cs + a (
2
1
1
(c + c 2 − 4a )
2
s + 1)(
b
1
1
(c − c 2 − 4a
2
1
s + 1) (c 2 − (c 2 − 4a ))
4
b
a
b
=
s + cs + a
2
(
1
1
(c + c 2 − 4 a )
2
0〈ζ 〈1 : h( s) =
K ⋅ ω 02
s 2 + 2ζω 0 s + ω 02
ζ = 0 : h( s ) =
K ⋅ ω 02
K ⋅ ω 02
=
s 2 + 2ζω 0 s + ω 02 s 2 + ω 02
s + 1)(
1
1
(c − c 2 − 4a
2
=
s + 1)
K
(T1 s + 1)(T2 s + 1)
3.
Skaff greie på ∆u (t ) .
Bestem ∆u (s ) .
4.
Bestem ∆x( s ) = ∆u ( s ) ⋅ h( s )
5.
Bestem x(t ) = x(0) + L−1 {∆x( s)} fra formelark.
For spesialtilfelle:
T
K
K
, ( 1 ≈ 10) ≈
ζ 〉1 : h( s ) =
⋅ e −T2 ⋅s
(T1 s + 1)(T2 s + 1) T2
T1 s + 1
Responser for endringer i systemet.
U
for 1.orden system:
s
U K
K ⋅U
∆x( s ) = ∆u ( s ) ⋅ h( s ) =
=
s Ts + 1 s (Ts + 1)
Sprangendring ∆u (t ) = U ⇒ ∆u ( s ) =
x(t ) = x(0) + K ⋅ U (1 − e
−
t
T
)
Merk av punktene:
x(t = 0) = x(0) + K ⋅ U (1 − e
−
x(t = T ) = x(0) + K ⋅ U (1 − e
0
T
−
T
T
x(t = 5T ) = x(0) + K ⋅ U (1 − e
) = x(0)
) = x(0) + 0,63K ⋅ U
5T
−
T
) = x(0) + K ⋅ U og skisser sprangresponsen.
3
For x(0) = 0, K = U = T = 1:
Impulsendring ∆u (t ) = δ (t ) ⇒ ∆u ( s ) = 1 for 1.orden system:
For enhetsimpulsen δ (t ) (Dirac-impulsen) er denne definert som en ”spiker” med stor,stor
A
A
amplitude med kort, kort varighet og et areal At1 →0 =
⋅ ∆t = ⋅ t1 = 1 eller
∆t
t1
⎡ t1
⎤
=1
⎢ ∫ δ (t )dt ⎥
⎢⎣ 0
⎥⎦ t1 →0
t
0
Svært stor amplitude
Pulsareal A=1
t
0 t1
4
∆x ( s ) = u ( s ) h ( s ) = 1 ⋅
K
K
=
Ts + 1 Ts + 1
t
x(t ) = x(0) +
K −T
e
T
Merk av punktene:
0
K −T
K
e = x ( 0) +
T
T
T
K −
K
x(t = T ) = x(0) + e T = x(0) + 0,37
T
T
5T
K −
x(t = 5T ) = x(0) + e T = x(0) og skisser impulsresponsen
T
x(t = 0) = x(0) +
For x(0) = 0, K = T = 1:
Rampendringendring ∆u (t ) = U ⋅ t ⇒ ∆u ( s ) =
∆x ( s ) = ∆u ( s ) h ( s ) =
U
for 1.orden system:
s2
U K
K ⋅U
= 2
2
s Ts + 1 s (Ts + 1)
x(t ) = x(0) + K ⋅ U (t − T + T ⋅ e
−
t
T
)
5
Merk av punktene:
x(t = 0) = x(0) + K ⋅ U (0 − T + T ⋅ e
−
0
T
) = x(0)
x(t = 0,5T ) = x(0) + K ⋅ U (0,5T − T + T ⋅ e
x(t = T ) = x(0) + K ⋅ U (T − T + T ⋅ e
−
T
T
x(t = 3T ) = x(0) + K ⋅ U (3T − T + T ⋅ e
−
0 , 5Tt
T
) = x(0) + 0,1T ⋅ K ⋅ U
) = x(0) + 0,37T ⋅ K ⋅ U
−
3Tt
T
x(t = 10T ) = x(0) + K ⋅ U (10T − T + T ⋅ e
) = x(0) + 2T ⋅ K ⋅ U
10Tt
−
T
) = x(0) + 9T ⋅ K ⋅ U
For x(0) = 0, K = U = T = 1:
U
for 2.orden system med relativ dempning ζ = 1 :
s
U
K
K ⋅U
∆x ( s ) = ∆u ( s ) ⋅ h ( s ) =
=
2
s (Ts + 1)
s (Ts + 1) 2
Sprangendring ∆u (t ) = U ⇒ ∆u ( s ) =
t
−
t
x(t ) = x(0) + K ⋅ U (1 − (1 + ) ⋅ e T )
T
Merk av punktene:
6
0
−
0
x(t = 0) = x(0) + K ⋅ U (1 − (1 + ) ⋅ e T ) = x(0)
T
2T
−
2T
) ⋅ e T ) ≈ x(0) + 0,63K ⋅ U
x(t = Tr = 2T ) = x(0) + K ⋅ U (1 − (1 +
T
10T
−
10T
) ⋅ e T ) = x(0) + K ⋅ U og skisser
x(t = 5Tr = 10T ) = x(0) + K ⋅ U (1 − (1 +
T
sprangresponsen.
For x(0) = 0, K = U = T = 1:
Impulsendring ∆u (t ) = δ (t ) ⇒ ∆u ( s ) = 1 for 2.orden system med relativ dempning ζ = 1 :
∆x ( s ) = ∆u ( s ) h ( s ) = 1 ⋅
K
K
=
2
(Ts + 1)
(Ts + 1) 2
−
−
t ( n −1)
t
⋅ e T = x ( 0) + K ⋅ 2 ⋅ e T
n
T (n − 1)!
T
Merk av punktene:
t
t
x(t ) = x(0) + K ⋅
7
0
0 −T
⋅ e = x ( 0)
T2
T
−
T
K
x(t = T ) = x(0) + K ⋅ 2 ⋅ e T = x(0) + 0,37
T
T
4T
4T −
K
x(t = 2Tr = 4T ) = x(0) + K ⋅ 2 ⋅ e T = x(0) + 0,07
T
T
10T
10T −
x(t = 5Tr = 10T ) = x(0) + K ⋅ 2 ⋅ e T = x(0)
T
Skisser impulsresponsen.
x(t = 0) = x(0) + K ⋅
For x(0) = 0, K = T = 1:
Sprangendring ∆u (t ) = U ⇒ ∆u ( s ) =
∆x ( s ) = ∆u ( s ) h ( s ) =
K ⋅U
s (T1 s + 1)(T2 s + 1)
t
U
for 2.orden system med relativ dempning ζ 〉1 :
s
t
−
−
1
x(t ) = x(0) + K ⋅ U (1 −
⋅ (T1e T1 − T2 e T2 ))
T2 − T1
Merk av punktene:
8
0
0
−
−
1
x(t = 0) = x(0) + K ⋅ U (1 −
⋅ (T1e T1 − T2 e T2 )) = x(0)
T2 − T1
−
1
x(t = Tr = T1 + T2 ) = x(0) + K ⋅ U (1 −
⋅ (T1e
T2 − T1
T1 +T2
T1
−
1
x(t = 5Tr = 5(T1 + T2 )) = x(0) + K ⋅ U (1 −
⋅ (T1e
T2 − T1
Skisser sprangresponsen.
− T2 e
5 (T1 +T2 )
T1
−
T1 +T2
T2
)) ≈ x(0) + 0,63K ⋅ U
− T2 e
−
5 (T1 +T2 )
T2
)) = x(0) + K ⋅ U
For x(0) = 0, K = U = T1 = 1 og T2 = 3:
Impulsendring ∆u (t ) = δ (t ) ⇒ ∆u ( s ) = 1 for 2.orden system med relativ dempning ζ 〉1 :
∆x ( s ) = ∆u ( s ) h ( s ) =
K
(T1 s + 1)(T2 s + 1)
t
t
−
−
K
x(t ) = x(0) +
⋅ (e T1 − e T2 )
T1 − T2
9
Merk av punktene:
0
0
−
−
1
x(t = 0) = x(0) + K ⋅
⋅ (e T1 − e T2 )) = x(0)
T1 − T2
−
K
x(t = 0,4Tr = 0,4(T1 + T2 )) = x(0) +
⋅ (e
T1 − T2
−
K
x(t = 2Tr = 2(T1 + T2 )) = x(0) +
⋅ (e
T1 − T2
−
K
x(t = 5Tr = 5(T1 + T2 )) = x(0) +
⋅ (e
T1 − T2
0 , 4 (T1 +T2 )
T1
2 (T1 +T2 )
T1
5 (T1 +T2 )
T1
−e
−e
−
−
−e
−
0 , 4 (T1 +T2 )
T2
2 (T1 +T2 )
T2
5 (T1 +T2 )
T2
))
))
)) = x(0)
For x(0) = 0, K = T1 = 1 og T2 = 3:
Sprangresponsen ∆u (t ) = U ⇒ ∆u ( s ) =
dempning 0〈ζ 〈1 :
U
for 2.orden system med relativ
s
K ⋅ U ⋅ ω 02
s( s 2 + 2ζω 0 s + ω 02 )
1
x(t ) = x(0) + K ⋅ U (1 −
e −ω0 ⋅ζ ⋅t ⋅ cos(ω 0 1 − ζ 2 ⋅ t − sin −1 ζ ))
2
1−ζ
∆x( s) = ∆u ( s)h( s) =
10
Merk av punktene:
x(t = 0) = x(0) + K ⋅ U lim s →∞ [ s
Tr ≈
ω 02
] = x(0)
s( s 2 + 2ζω 0 s + ω 02 )
1,5
ω0
x(t = Tr ) = x(0) + K ⋅ U (1 −
Tmaks =
1
1−ζ
2
e −ω0 ⋅ζ ⋅Tr ⋅ cos(ω 0 1 − ζ 2 ⋅ Tr − sin −1 ζ )) ≈ x(0) + 0,63K ⋅ U
π
ω0 1 − ζ 2
K ⋅ U ⋅ ω 02
∆x S = lim s →0 [ s
] = K ⋅U
s( s 2 + 2ζω 0 s + ω 02 )
x(t = Tmaks ) = x(0) + ∆x S (1 + e
Skisser sprangresponsen.
−
π ⋅ζ
1−ζ 2
)
For x(0) = 0, ω 0 = K = U = 1 og ζ = 0,6 :
Tr ≈ 1,5 , x(t = Tr = 1,5) = 0,63 , Tmaks =
π
1 − 0,6
2
= 3,9 og x(t = Tmaks = 3,9) = 1 + e
−
0 , 6π
1− 0 , 6 2
= 1,1
Impulsresponsen ∆u (t ) = δ (t ) ⇒ ∆u ( s ) = 1 for 2.orden system med relativ
dempning 0〈ζ 〈1 :
11
K ⋅ ω 02
s 2 + 2ζω 0 s + ω 02
∆x( s) = ∆u ( s)h( s) =
x(t ) = x(0) +
K ⋅ ω0
1−ζ
2
e −ζ ⋅ω0 ⋅t ⋅ sin(ω 0 1 − ζ 2 ⋅ t )
2π
Periodetid T p =
ω0 1 − ζ 2
Merk av punktene:
K ⋅ ω 0 −ζ ⋅ω ⋅0
x(t = 0) = x(0) +
e
⋅ sin(ω 0 1 − ζ 2 ⋅ 0) = x(0)
2
1− ζ
K ⋅ ω 0 −ζ ⋅ω ⋅t
d
d
e
⋅ sin(ω 0 1 − ζ 2 ⋅ t )] = 0 gir:
Vendepkt.: x(t ) = [ x(0) +
2
dt
dt
1−ζ
0
0
1−ζ 2
tg −1
t = Tmaks =
ζ2
ω0 1 − ζ 2
K ⋅ ω0
x(t = Tmaks ) = x(0) +
1−ζ
Første nullgjennomgang:
x(t = T0 ≈
π
ω0 1 − ζ
Første min.pkt:
x(t = Tmin ≈
2
2
e −ζ ⋅ω0 ⋅Tmaks ⋅ sin(ω 0 1 − ζ 2 ⋅ Tmaks )
) ≈ x(0) +
1,5π
ω0 1 − ζ 2
) ≈ x(0) +
K ⋅ ω0
1− ζ
2
e −ζ ⋅ω0 ⋅T0 ⋅ sin(ω 0 1 − ζ 2 ⋅ T0 )
K ⋅ U ⋅ ω0
1−ζ 2
e −ζ ⋅ω0 ⋅Tmin ⋅ sin(ω 0 1 − ζ 2 ⋅ Tmin )
Skisser impulsresponsen
Simulering for x(0) = 0, ω 0 = K = 1 og ζ = 0,6 :
12
U
for 2.orden system med relativ dempning ζ = 0 :
s
K ⋅ U ⋅ ω 02
K ⋅ U ⋅ ω 02
∆x( s ) = ∆u ( s)h( s) =
=
s ( s 2 + 2ζω 0 s + ω 02 ) s( s 2 + ω 02 )
K ⋅U
x(t ) = x(0) +
(1 − cos ω 0 t )
2
Sprangendring ∆u (t ) = U ⇒ ∆u ( s ) =
ω0
Simulering for x(0) = 0, ω 0 = K = U = 1 og ζ = 0 :
13
Impulsendring ∆u (t ) = δ (t ) ⇒ ∆u ( s ) = 1 for 2.orden system med relativ dempning ζ = 0 :
K ⋅ ω 02
K ⋅ ω 02
∆x( s ) = ∆u ( s)h( s) = 2
=
s + 2ζω 0 s + ω 02 s 2 + ω 02
x(t ) = x(0) + K ⋅ sin(ω 0 t )
Simulering for x(0) = 0, ω 0 = K = 1 og ζ = 0 :
Nullpunktledd i 1.orden system.
h( s ) =
∆x( s ) Tn s + 1
=
∆u ( s ) Ts + 1
∆x( s) = ∆u ( s)h( s) = ∆u ( s)
Tn s + 1
Ts + 1
U
for nullpunktledd i 1.orden system.
s
T s +1
T s +1
∆x ( s ) = ∆u ( s ) h ( s ) = ∆u ( s ) ⋅ K ⋅ n
= K ⋅U n
Ts + 1
s (Ts + 1)
Sprangendring ∆u (t ) = U ⇒ ∆u ( s ) =
19
T2 s + 1
s(T1 s + 1)
t
T −T −
1 + 2 1 ⋅ e T1
T1
For K = U = Tn = 1 , T = 2 og x(0) = 0 :
−
1 − 2 −T
x(t ) = 1 +
e = 1 − 0,5e 2
2
t
t
14
Nullpunktledd i 2.orden system.
K ⋅ (Tn s + 1)
∆x( s)
=
h( s ) =
∆u ( s) (T1 s + 1)(T2 s + 1)
K ⋅ (Tn s + 1)
∆x( s) = ∆u ( s)h( s) = ∆u ( s)
(T1 s + 1)(T2 s + 1)
U
for nullpunktledd i 2.orden system.
s
K ⋅ (Tn s + 1)
∆x( s) = ∆u ( s)h( s) = ∆u ( s)
(T1 s + 1)(T2 s + 1)
Sprangendring ∆u (t ) = U ⇒ ∆u ( s ) =
For K = U = Tn = 1 , T1 = 2, T2 = 4 og x(0) = 0 :
K ⋅U
1s + 1
1s + 1
∆x( s ) =
⋅
= K ⋅U
s
s(2s + 1)(4 s + 1)
(2s + 1)(4 s + 1)
27
T3 s + 1
, T1 ≠ T2
s(T1 s + 1)(T2 s + 1)
t
−
−
2 −1 −2 4 −1 −4
e +
e = 1 + 0,5e 2 − 1,5e 4
2−4
2−4
t
x(t ) = 1 −
t
T −T −
T − T3 − T2
1 − 1 3 e T1 + 2
e
T1 − T2
T1 − T2
t
t
t
15
x ( 0) = 0
x S = x (∞ ) = 1
Tr = T1 + T2 = 2 + 4 = 6[s]
x(t = Tr = 6) = 0,63 ⋅ x S = 0,63
For alle responser er det tillatt å bruke kalkulator for å skissere denne.
Har vi responsen x(t) og kjenner pådrag u(t) skal vi bruke ”vedlegg” for å bestemme
systemets transferfunksjon h(s) = x(s)/u(s).
16
Endringer i systemer med flere inngangsvariable (MISO-systemer).
SYSTEM
x
h1
y
T
h2
z
h3
Transferfunksjoner:
∆T ( s )
∆x( s)
∆T ( s)
∆x = ∆z = 0 : h2 ( s ) =
∆y ( s )
∆T ( s)
∆x = ∆y = 0 : h3 ( s) =
∆z ( s)
∆y = ∆z = 0 : h1 ( s) =
Superposisjon:
∆T ( s) ∆T ( s ) ∆T ( s)
+
+
∆x( s) ∆y ( s) ∆z ( s)
∆y ≠ 0, ∆z ≠ 0, ∆x ≠ 0 : ∆T ( s) = ∆x( s) ⋅ h1 ( s) + ∆y ( s) ⋅ h2 ( s) + ∆z ( s) ⋅ h3 ( s)
∆y ≠ 0, ∆z ≠ 0, ∆x ≠ 0 : h( s ) = h1 ( s) + h2 ( s ) + h3 ( s) =
∆y ≠ 0, ∆z ≠ 0, ∆x ≠ 0 : T (t ) = T (0) + L−1 {∆x( s )h1 ( s ) + ∆y ( s )h2 ( s) + ∆z ( s )h3 ( s )} = T (0) + ∑ ∆T
Systemer med negativ tilbakekobling.
Vi tar som eksempel en elmaskin med last som skal turtallsreguleres ved settpunktendring:
17
Nett
Frekvens-
Frekvens f
El.maskin
w
3
omformer
Last
Pådragsignal u
Turtalls-
Målt turtall y
regulator
Turtallsregulator
r(s)
e(s)
Transmitter
Frekvensomformer
hr(s)
u(s)
hf(s)
Turtall w
Turtalls-
Elmaskin+Last
f(s)
hp(s)
Regulert størrelse w
Turtalls-Transmitter
hm(s)
Målt turtall y
Når vi tegner blokkdiagram for reguleringssystemet starter vi med regulator
(turtallsregulator).
Deretter følger vi signalgangen for systemet fra u til y.
Alle blokker må modelleres:
hr ( s ) =
∆u ( s)
∆f ( s)
∆ω ( s)
∆y ( s )
, h f (s) =
, hP ( s ) =
, hm ( s) =
∆e( s)
∆u ( s)
∆f ( s)
∆ω ( s)
For blokkdiagram får vi:
∆ω ( s ) = ∆e ( s ) ⋅ hr ( s ) ⋅ h f ( s ) ⋅ h P ( s )
∆ e ( s ) = ∆r ( s ) − ∆y ( s ) :
∆ω ( s ) = ( ∆r ( s ) − ∆y ( s )) ⋅ hr ( s ) ⋅ h f ( s ) ⋅ hP ( s ) = ∆r ( s ) hr ( s ) h f ( s ) hP ( s ) − ∆y ( s )hr ( s ) h f ( s ) hP ( s )
∆y ( s ) = ∆ω ( s ) ⋅ hm ( s) :
∆ ω ( s ) = ∆ r ( s ) hr ( s ) h f ( s ) h P ( s ) − ∆ ω ( s ) h r ( s ) h f ( s ) h P ( s ) h m ( s )
∆ ω ( s ) + ∆ ω ( s ) h r ( s ) h f ( s ) h P ( s ) hm ( s ) = ∆ r ( s ) h r ( s ) h f ( s ) h P ( s )
∆ω ( s )(1 + ∆ω ( s )hr ( s ) h f ( s ) hP ( s ) hm ( s )) = ∆r ( s ) hr ( s ) h f ( s ) hP ( s )
Transferfunksjon for hele reguleringssystemet (ingen lastendring):
h( s ) =
hr ( s )h f ( s )hP ( s )
∆ω ( s )
=
∆r ( s ) 1 + hr ( s )h f ( s )h p ( s )hm ( s )
18
19