Download - Helmut Büch, Gifhorn

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π
i = √ −1
π = 3.1415927 . . .
e = 2.718281 . . .
2.20D − 16

π
i = √ −1
π = 3.1415927 . . .
e = 2.718281 . . .
2.20D − 16

π
i = √ −1
π = 3.1415927 . . .
e = 2.718281 . . .
2.20D − 16

π
i = √ −1
π = 3.1415927 . . .
e = 2.718281 . . .
2.20D − 16

3 × 3
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A
a 1 × 6

1 − 3i
n
m

1
1
A 3 × 3

A
A
b l × 2
:
A 3 × 2
1

1
2 × 2

∗.

∧
A
e A =
∞
i=0
A i
i!

exp(A)
A \ B
AB −1

Ax = b
A
A
xb = A
A I x
|
A
B aij bij

cij = aij ·bij
.∗ ./ .
A n A
A
A n
c A
c. A c aij
D
D

%i
T
′

[]
l × 1

−1 2
s

s

∗ /

=

=

=

=

F (s) =
3(s + 2)
(s + 3) ∗ (s + 1)
F (s) = 1.5 1.5
+
s + 1 s + 3

1 + s
F (s) =
5 + s
z − 1
s =
z + 1
F (z) = F (s)| s= z−1
z+1
= z
2 + 3z

1 + s
F (s) =
5 + s
z − 1
s =
z + 1
F (z) = F (s)| s= z−1
z+1
= z
2 + 3z

1 + s
F (s) =
5 + s
z − 1
s =
z + 1
F (z) = F (s)| s= z−1
z+1
= z
2 + 3z

1 + s
F (s) =
5 + s
z − 1
s =
z + 1
F (z) = F (s)| s= z−1
z+1
= z
2 + 3z

1 + s
F (s) =
5 + s
z − 1
s =
z + 1
F (z) = F (s)| s= z−1
z+1
= z
2 + 3z

z

A

A
A
Ax = b A
Ā = [Ab]
Ā1 A
b
Ā1 = [Aib1]
A1 b1
AX = B
B

A
A
Ax = b A
Ā = [Ab]
Ā1 A
b
Ā1 = [Aib1]
A1 b1
AX = B
B

A
A
Ax = b A
Ā = [Ab]
Ā1 A
b
Ā1 = [Aib1]
A1 b1
AX = B
B

A
A
Ax = b A
Ā = [Ab]
Ā1 A
b
Ā1 = [Aib1]
A1 b1
AX = B
B

∼
==
<
>
≤
≥
∼=

∼
==
<
>
≤
≥
∼=

∼
==
<
>
≤
≥
∼=

∼
==
<
>
≤
≥
∼=

∼
==
<
>
≤
≥
∼=

T ∈ 33(R)

T −1 T = I
I
T
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[−l, 3]
0.1

[−l, 3]
0.1

[0, 1]
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· + × ⊕ ♦ △ ▽ ♣ ○

· + × ⊕ ♦ △ ▽ ♣ ○

· + × ⊕ ♦ △ ▽ ♣ ○

· + × ⊕ ♦ △ ▽ ♣ ○

· + × ⊕ ♦ △ ▽ ♣ ○

· + × ⊕ ♦ △ ▽ ♣ ○

x = f(x, y)(3.1)
f(x, y) = cos(x)cos(y) (x, y) ∈ [0, 2π] × [0, 2π](3.2)
z(i, j) =
f(x(i), y(j))

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. . .
640 × 480

. . .
640 × 480

. . .
640 × 480

. . .
640 × 480

. . .
640 × 480

. . .
640 × 480

. . .
640 × 480

. . .
640 × 480

. . .
640 × 480

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˙x = Ax + Bu
y = Cx + Du
−5 −1
A =
6 0
−1
B =
1
C = −1 0
D = 0
x(0) = [0 0] T

˙x = Ax + Bu
y = Cx + Du
−5 −1
A =
6 0
−1
B =
1
C = −1 0
D = 0
x(0) = [0 0] T

10 −16
F 1 s = 2

A, B.C.D

A A

k
k = 3

k
k = 3

x = 0.5
1 + kF (s) = 0
F (s)
F (s) =
s + 10
(s + 2)(s + 3)

x = 0.5
1 + kF (s) = 0
F (s)
F (s) =
s + 10
(s + 2)(s + 3)

P 1 =
s + 1
s(s + 0.1)
P 2 = s + 10
s(s + 1)
P 3 = s + 100
s(s + 10)
P 4 = s + 1000
s(s + 100)

z s
k
1 + kF (s) = 0 k
z s

F (jω)
(−90 ◦ , 0)
F (jω) =
10 + jω
jω(jω + 3)
XdB = 20 log 10(X)

10 −3 , 10 3
z = e jω
θ ∈ ]−π, π] θ ∈ [0, π]
z = e 2πjωT
ω
Hz ω ∈ 10 −3 , 0.5
T = 1
θ = π ω = 0.5

F (jω) =
20(10 + jω)
(jω + 3)(jω + 5)
rad/sec 4
4 × 2π ≈ 25rad/s
10rad/s 10/2π ≈ 1.6Hz

F (jω) = 106 + jω
10 5 + jω

F (jω) = 106 + jω
10 5 + jω

F (jω) =
1
(jω + 3)
ω = 3

s = 2
T
· z − 1
z + 1
T
Sc T fp
u

sl u
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sl u
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sl u
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sl u
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sl u
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sl u
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F (s) =
1
s 2 + s + 1
F (s)

¡

• l
• m
• µ
• ϑ
•
• τg
µ ¨ ϑ = τg
µ = ml 2
τg = mgl sin(ϑ)
ϑ
l
m
mg

ϑ g
¨ϑ + g
sin(ϑ) = 0.
l
˙x = f(x, u)
(5.1)
y = g(x, u)
u
g
f(x, u)
x0
˙x x
g(x, u)
˙ϑ
− g
l sin(ϑ)
˙ϑ
¨ϑ
˙ϑ0
y = ϑ u = g
ϑ
x = ˙ϑ
⎧
⎨ ˙ϑ
¨ϑ
⎩
y
=
=
ϑ
˙ϑ
− g
l sin(ϑ)
ϑ0
ϑ
˙ϑ
ϑ
y
y
(5.2)
x1 = ϑ x2 = ˙ ϑ
˙x1 = x2

g
f
˙ϑ
− g
l sin(ϑ)
˙ϑ
¨ϑ
ϑ0
˙ϑ0
m = 1 l = 1
g = ˆg = 9.81 2 .
g
− g
l sin(ϑ)
¨ϑ
˙ϑ0
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•
ϑ
˙ϑ
˙ϑ
ϑ0
ϑ
ϑ
y

•
−(g/l) sin(ϑ)
g ϑ
g ϑ
− g
sin(ϑ) → −u1/l ∗ sin(u2),
l
l
45 ◦ 45 ◦
45 ◦

•
−(g/l) sin(ϑ)
g ϑ
g ϑ
− g
sin(ϑ) → −u1/l ∗ sin(u2),
l
l
45 ◦ 45 ◦
45 ◦

˙ ϑ = 0 ◦ /s ϑ = π/4
= 0
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•

−1
+1

f(ˆx, û) = 0
k ∈ Nk
ˆx1 = ˆ ϑ = kπ ˆx2 =ˆ˙x2 = 0 û = ˆg
δϑ = ϑ − ˆ ϑ δ ˙ ϑ = ˙ ϑ −ˆ˙ϑ = ˙ ϑ
⎧
⎨ ˙ϑ
¨ϑ
⎩
0
=
−
1
ˆg
y
l cos(ϑ)
= δϑ
δϑ
0 ˙ϑ
=
˙ϑ
ˆϑ = 0 ˆ˙ϑ = 0 ˆg = 9.81 2
− ˆg
l cos( ˆ ϑ)δϑ
⎧
⎨ ˙ϑ ˙ϑ
¨ϑ
=
−
⎩
ˆg
l ϑ
y = ϑ
(5.3)
g
− ˆg
l
¨ϑ
˙ϑ0
˙ϑ
ϑ0
ϑ
(5.4

− g
sin(ϑ) → −u1/l ∗ sin(u2)
l
− ˆg
ϑ
l
→ −u1/l ∗ u2
•
•
±45 ◦
5 ◦
5 ◦

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A > B
A B
A B
1 × 1

A > B
A B
A B
1 × 1

A > B
A B
A B
1 × 1

A > B
A B
A B
1 × 1

A > B
A B
A B
1 × 1

A > B
A B
A B
1 × 1

A > B
A B
A B
1 × 1

A > B
A B
A B
1 × 1

A > B
A B
A B
1 × 1