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Please don't modify ! Contact me in the discussion. THX.
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }
A
s
+
B
(
62
,
9
s
+
1
)
+
C
(
62
,
9
s
+
1
)
2
{\displaystyle {\frac {A}{s}}+{\frac {B}{\left(62,9s+1\right)}}+{\frac {C}{\left(62,9s+1\right)^{2}}}}
y
(
t
)
=
K
∗
(
1
+
e
−
t
T
+
t
T
⋅
e
−
t
T
)
{\displaystyle y(t)=K*\left(1+e^{\frac {-t}{T}}+{\frac {t}{T}}\cdot e^{\frac {-t}{T}}\right)}
Y(s)
Y
(
s
)
=
0
,
313
(
62
,
9
s
+
1
)
2
⋅
1
s
{\displaystyle Y(s)={\frac {0,313}{(62,9s+1)^{2}}}\cdot {\frac {1}{s}}}
Přechodova fce
F
(
s
)
=
0
,
313
3956
,
4
s
2
+
62
,
9
s
+
1
{\displaystyle F(s)={\frac {0,313}{3956,4s^{2}+62,9s+1}}}
DR
3959
,
4
y
′
′
+
62
,
9
y
′
+
y
=
0
,
313
u
{\displaystyle 3959,4y^{\prime \prime }+62,9y^{\prime }+y=0,313u}
Y
(
s
)
=
F
(
s
)
⋅
U
(
s
)
{\displaystyle Y(s)=F(s)\cdot U(s)}
K
=
Δ
y
Δ
u
{\displaystyle K={\frac {\Delta y}{\Delta u}}}
Δ
u
=
u
(
∞
)
−
u
(
0
)
{\displaystyle \Delta u=u(\infty )-u(0)}