Nyquistov dijagram
Nyquistov dijagram
Nyquistov dijagram
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NYQUIST-ovi DIJAGRAMI<br />
AUDITORNE VJEŽBE
ZADATAK 1.<br />
PRIKAZ OSNOVNIH ČLANOVA U NYQUIST-ovom DIJAGRAMU<br />
s jω<br />
P 0 član<br />
Im<br />
G(s) =<br />
G( jω ) =<br />
K P<br />
K P<br />
G = K P<br />
ϕ = ϕ B<br />
− ϕ N<br />
= Im Im<br />
arctg( )<br />
B<br />
− arctg( ) = 0<br />
N<br />
Re Re<br />
P 1 član<br />
K P<br />
Re<br />
K<br />
P<br />
G(s) =<br />
Ts + 1<br />
KP<br />
G( jω ) =<br />
1+ Tjω<br />
G =<br />
K<br />
P<br />
1 + (T ω)<br />
2<br />
G( jω ) =<br />
G =<br />
KP<br />
1+ Tjω<br />
1−<br />
Tjω<br />
1−<br />
Tjω<br />
KP<br />
=<br />
2 2<br />
1 + T<br />
+ −ωTKP<br />
ω 1 + T ω<br />
2 2<br />
⎛ KP<br />
⎞ ⎛ −K PωT<br />
⎞<br />
⎜ +<br />
2 2 2 2<br />
1+ T ω<br />
⎟ ⎜<br />
1+ T ω<br />
⎟<br />
⎝ ⎠ ⎝ ⎠<br />
2 2<br />
j
Im<br />
ϕ = ϕB<br />
− ϕN<br />
= −ϕ<br />
N<br />
= −arctg( )<br />
Re<br />
Tω<br />
ϕ = − arctg( ) = −arctg(T ω)<br />
1<br />
N<br />
⎛ −KPωT<br />
⎞<br />
⎜ 2 2<br />
arctg 1 T ⎟<br />
ϕ = + ω<br />
⎜<br />
K<br />
⎟<br />
⎜<br />
P<br />
⎟<br />
2 2<br />
⎝ 1+ T ω ⎠<br />
= arctg( −T ω)<br />
= −arctg(T ω)<br />
ω<br />
G<br />
ϕ<br />
0<br />
K P<br />
0<br />
1<br />
T<br />
K<br />
P<br />
2<br />
π<br />
-<br />
4<br />
∞<br />
0<br />
π<br />
-<br />
2<br />
Im<br />
ω = ∞<br />
- 45°<br />
ω=0<br />
K<br />
P<br />
Re<br />
K P<br />
2<br />
ω =<br />
1<br />
T<br />
1<br />
ω = = ωL<br />
LOMNA FREKVENCIJA<br />
T<br />
- član 1. reda ne može imati kašnjenje veće od<br />
π<br />
-<br />
2
P 2 član<br />
G(s) =<br />
G( jω ) =<br />
G =<br />
ϕ = ϕ − ϕ<br />
ω<br />
G<br />
ϕ<br />
B<br />
K<br />
P<br />
2 2<br />
T2 s + T1s + 1<br />
KP<br />
KP<br />
=<br />
2 2<br />
2 2<br />
T<br />
2<br />
( jω ) + T<br />
1( jω ) + 1 1 − T2 ω + T1<br />
ωj<br />
K<br />
N<br />
P<br />
(1− T ω ) + (T ω)<br />
0<br />
K P<br />
0<br />
2 2 2 2<br />
2 1<br />
Im<br />
Re<br />
= −ϕN<br />
= −arctg( )<br />
N<br />
1<br />
T 2<br />
T ω<br />
= −arctg( )<br />
1−<br />
T<br />
1<br />
2 2<br />
2<br />
ω<br />
2<br />
KP<br />
0 T<br />
1<br />
ω = ∞<br />
ω=0<br />
π<br />
K P<br />
Re<br />
-<br />
T<br />
2<br />
∞<br />
-π<br />
Im<br />
- član 2. reda može imati maksimalno kašnjenje :<br />
ϕ max = -π<br />
ϕ<br />
max<br />
π<br />
= −n<br />
2<br />
K<br />
P<br />
T<br />
T<br />
2<br />
1
D 0 član<br />
G(s) = Tds<br />
G(j ω ) = T d<br />
jω<br />
G = Tdω<br />
ϕ = ϕ − ϕ<br />
B<br />
T<br />
arctg( d<br />
ω<br />
ϕ = ) =<br />
0<br />
Im<br />
Re<br />
N<br />
= ϕB<br />
= arctg( )<br />
B<br />
π<br />
2<br />
ω<br />
G<br />
0<br />
1<br />
T D<br />
∞<br />
0 1 ∞<br />
Im<br />
1<br />
ω =<br />
T<br />
D<br />
ω = ∞<br />
1<br />
ϕ<br />
π<br />
2<br />
π<br />
2<br />
π<br />
2<br />
ω=0 Re
I 0 član<br />
G(s) =<br />
1<br />
Ts<br />
i<br />
G( jω ) =<br />
1<br />
Ti<br />
jω<br />
1<br />
G =<br />
Tω<br />
i<br />
ϕ = ϕB<br />
− ϕN<br />
= −ϕN<br />
= −arctg( Im<br />
)<br />
N<br />
Re<br />
T<br />
arctg( i<br />
ω<br />
ϕ = − ) = −<br />
π<br />
0 2<br />
Im<br />
ω<br />
G<br />
ϕ<br />
0<br />
∞<br />
π<br />
-<br />
2<br />
1<br />
T i<br />
∞<br />
1 0<br />
π π<br />
- -<br />
2 2<br />
ω =<br />
1<br />
T<br />
i<br />
ω = ∞<br />
1<br />
ω=0<br />
Re
ZADATAK 2.<br />
Za regulacijsku stazu prema slici potrebno je izračunati frekvencijske karakteristike i dati prikaz<br />
amplitudno – frekvencijske karakteristike (AFK) i fazno-frekvencijske karakteristike (FFK) u<br />
<strong>Nyquistov</strong>om <strong>dijagram</strong>u.<br />
x u<br />
K<br />
1<br />
1<br />
T1 s + 1 T2 s + 1<br />
x i<br />
K = 3<br />
T = T = 1<br />
1 2<br />
G(s) =<br />
G( jω ) =<br />
K =<br />
3<br />
2<br />
(s + 1)(s + 1) s + 2s + 1<br />
3<br />
2<br />
( jω ) + 2( jω ) + 1<br />
⎡<br />
2 2<br />
G Re + Im ⎤<br />
= =<br />
⎣ ⎦<br />
⎡ ⎤<br />
⎣ ⎦<br />
B<br />
G G 2 2<br />
N Re + Im<br />
=<br />
B<br />
N<br />
3<br />
− ω + jω<br />
2<br />
(1 ) 2<br />
G =<br />
3<br />
(1 − ω ) + 4ω<br />
2 2 2<br />
=<br />
3<br />
=<br />
3 ... AFK<br />
ω<br />
4<br />
+<br />
2<br />
2<br />
2 ω + 1 ω + 1
ϕ = ϕ − ϕ<br />
B<br />
N<br />
arctg(<br />
2ω<br />
)<br />
1− ω<br />
= −ϕN<br />
= −<br />
2<br />
... FFK<br />
ω<br />
0<br />
1<br />
2<br />
∞<br />
Im<br />
G<br />
3<br />
1,5<br />
0,6<br />
0<br />
ϕ<br />
0<br />
-<br />
π<br />
2<br />
-126,87°<br />
-π<br />
ω = ∞<br />
ω=2<br />
0,6<br />
ω=1<br />
1,5<br />
ω=0<br />
3<br />
Re<br />
G( jω ) =<br />
3<br />
− ω + jω<br />
2<br />
(1 ) 2<br />
⋅<br />
(1<br />
2<br />
) 2<br />
(1<br />
2<br />
) 2<br />
− ω − jω<br />
− ω − jω<br />
=<br />
2<br />
3(1 − ω )<br />
(1 − ω ) + (2 ω)<br />
2 2 2<br />
+<br />
−6ω<br />
(1 ) (2 )<br />
2 2 2<br />
− ω + ω j
ZADATAK 3.<br />
Regulacijska staza se sastoji od integralnog i proporcionalnog člana, a zadana je slijedećom<br />
prijenosnom funkcijom K<br />
G(s) =<br />
s(s + 3)<br />
Odrediti pojačanje K staze, tako da za frekvenciju ω=1 rad/s pojačanje amplitude izlaznog signala<br />
iznosi 10 . Izračunati AFK i FFK i dati prikaz u <strong>Nyquistov</strong>om <strong>dijagram</strong>u.<br />
G(s)<br />
=<br />
G( ω ) =<br />
K<br />
s(s + 3)<br />
=<br />
K<br />
s<br />
K<br />
jω( jω + 3)<br />
1<br />
s + 3<br />
j<br />
2<br />
G =<br />
ϕ = ϕ − ϕ<br />
B<br />
K<br />
( ) (3 )<br />
K<br />
= −ω + 3 jω<br />
K<br />
= ... AFK<br />
ω + 9 ω<br />
2 2 2 4 2<br />
−ω + ω<br />
N<br />
= −ϕN<br />
3ω<br />
= −arctg( )<br />
2<br />
−ω<br />
... FFK<br />
K = ? za ω = 1 rad/ s G( jω ) = 10<br />
10<br />
K<br />
= , uz ω = 1, K = 10<br />
ω<br />
4 +<br />
2<br />
9 ω
- za lakše crtanje:<br />
G( )<br />
10<br />
jω = −ω<br />
2 + 3 j ω<br />
⋅<br />
2<br />
−ω −<br />
2<br />
−ω −<br />
3<br />
3<br />
jω<br />
jω<br />
2<br />
−10ω − 30 ω<br />
j<br />
= ω<br />
4 −<br />
2 2<br />
9 j ω<br />
G( jω ) =<br />
2<br />
−10ω<br />
ω + 9ω<br />
4 2<br />
−30ω<br />
+ ω + 9 ω<br />
j<br />
4 2 2<br />
−10<br />
30<br />
= ω +<br />
3<br />
9<br />
+ −<br />
j ω + 9 ω<br />
ω<br />
Re<br />
Im<br />
G<br />
ϕ<br />
0 1 3 ∞<br />
10<br />
−<br />
9<br />
−1 −0,556 0<br />
−3 −0,556 0<br />
−∞<br />
∞<br />
-<br />
π<br />
2<br />
10<br />
0,785<br />
−108,43 −135°<br />
0<br />
−π<br />
10<br />
−<br />
9<br />
ω=1<br />
Im<br />
ω = ∞<br />
ω=3 Re<br />
ω=0
ZADATAK 4.<br />
Za sustav prikazan blok-<strong>dijagram</strong>om odrediti AFK i FFK i dati prikaz u <strong>Nyquistov</strong>om <strong>dijagram</strong>u.<br />
x u<br />
1<br />
K<br />
1<br />
2s + 1 3s + 2 s + 1<br />
x i<br />
K=10<br />
G(s) =<br />
10<br />
(2s + 1)(3s + 2)(s + 1)<br />
=<br />
10<br />
3 2<br />
6s + 13s + 9s + 2<br />
G( jω ) =<br />
10<br />
3 2<br />
6( jω ) + 13( jω ) + 9 jω + 2<br />
10<br />
= − ω + + ω − ω<br />
2 3<br />
( 13 2) j (9 6 )<br />
G =<br />
10<br />
( −13ω + 2) + (9ω − 6 ω )<br />
2 2 3 2<br />
=<br />
10<br />
6 4 2<br />
36ω + 61ω + 29ω + 4<br />
... AFK<br />
ϕ = ϕ − ϕ<br />
B<br />
N<br />
= −ϕ N<br />
3<br />
9ω − 6ω<br />
= −arctg( )<br />
2<br />
−13ω + 2<br />
... FFK
10<br />
G( jω ) = − ω + + ω − ω<br />
2 3<br />
( 13 2) j(9 6 )<br />
⋅<br />
2 3<br />
( −13ω + 2) − j(9ω − 6 ω )<br />
2 3<br />
( −13ω + 2) − j(9ω − 6 ω )<br />
G( jω ) =<br />
2<br />
−130ω + 20<br />
6 4 2<br />
36ω + 61ω + 29ω + 4<br />
+<br />
−90ω + 60ω<br />
6 4 2<br />
36ω + 61ω + 29ω + 4<br />
3<br />
j<br />
ω<br />
Re<br />
0 ∞ 1,22<br />
0,39<br />
5<br />
0<br />
−0,58<br />
0<br />
Im<br />
Im<br />
G<br />
ϕ<br />
0<br />
5<br />
0<br />
−3,17<br />
3,17<br />
π<br />
-<br />
2<br />
0<br />
0,58<br />
−π<br />
−<br />
0<br />
0<br />
3π<br />
2<br />
ω=1,22<br />
-0,58<br />
ω = ∞<br />
ω=0<br />
5 Re<br />
ω=0,39<br />
-3,17