digital control formula sheet
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58
u(k) T
Discrete-Time Systems and the z-Transform
X 0 (k) 1----+-----~ T
(b)
bo
T
xz(k)
Figure 2-9 Equivalent representations of equation (2-51): (a) signal flow graph representation; (b) simulation diagram.
Chap. 2
y(k) ~
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Difference Equations
State Equations (Matrix Form)
State Equations
�⃑�(𝑘 + 1) = 𝐴�⃑�(𝑘) + 𝐵�⃑⃑�(𝑘)
�⃑�(𝑘) = 𝐶�⃑�(𝑘) + 𝐷�⃑⃑�(𝑘)
�⃑⃑�(𝑧) = [ 1𝐶[𝑧𝐼 − 𝐴]− 𝐵 + 𝐷]�⃑⃑⃑�(𝑧)
�⃑⃑⃑�(𝒌) ≠ 𝟎, �⃑⃑⃑�(𝟎) ≠ 𝟎:
�⃑�(𝑧) = 𝑧[𝑧𝐼 − 𝐴]−1�⃑�(0) + [𝑧𝐼 − 𝐴]−1𝐵�⃑⃑⃑�(𝑧)
�⃑⃑⃑�(𝟎) = 𝟎:
�⃑�(𝑧) = [𝑧𝐼 − 𝐴]−1𝐵�⃑⃑⃑�(𝑧)
�⃑⃑⃑�(𝒌) = 𝟎:
�⃑�(𝑧) = 𝑧[𝑧𝐼 − 𝐴]−1�⃑�(0) 𝑧 0
Where 𝑧𝐼 = [ ] 0 𝑧
Matrix Functions
𝑎 𝑏Let 𝐴 = [
𝑐 𝑑] , 𝐵 = [
𝑒 𝑓𝑔 ℎ
] , 𝐶 = [𝑖𝑗]
𝐴 × 𝐵 = [𝑎 𝑏𝑐 𝑑
] × [𝑒 𝑓𝑔 ℎ
𝑎𝑒 + 𝑏𝑔 𝑎𝑓 + 𝑏ℎ] = [ ]
𝐴 × 𝐶 = [𝑎 𝑏𝑐 𝑑
] × [𝑖𝑗] = [
𝑐𝑒 + 𝑑𝑔 𝑐𝑓 + 𝑑ℎ𝑎𝑖 𝑏𝑗
] 𝑐𝑖 𝑑𝑗
𝑑𝑒𝑡(𝐴) = |𝑎 𝑏𝑐 𝑑
| = 𝑎𝑑 − 𝑐𝑏
𝑑𝑒𝑡(𝐴) = [𝑑 −𝑏
−𝑐 𝑎]
𝐴−1 =1
𝑑𝑒𝑡(𝐴)⋅ 𝑎𝑑𝑗(𝐴) =
[𝑑 −𝑏
−𝑐 𝑎]
|𝑎 𝑏𝑐 𝑑
|=
[𝑑 −𝑏
−𝑐 𝑎]
𝑎𝑑 − 𝑐𝑏
wHe(k)}] = E(z) = e(O) + e(l)z-1 + e(2)z-2 + · · · (4-1)
In addition, the starred transform for the time function e(t) was defined in equation (3-7) as
E*(s) = e(O) + e(T)e-Ts + e(2T)e-lTs + · · ·
E(z) E* (s )lesr = z
(4-2)
(4-3) function. We denote the product of the plant transfer function and the zero-order hold transfer function as G(s ), as shown in the figure; that is,
1 - e-n G(s) = G,(s)
s
The derivation above is completely general. Thus given any function that can be expressed as
Hence, from (4-3),
A(s) = B(s)F*(s)
A*(s) = B*(s)F*(s)
A(z) B(z)F(z)
(4-10)
(4-11)
(4-12)
G(s)
where B(s) is a function of sand F*(s) is a function of en; that is, in F*(s), s appears only in the form e78
• Then, in (4-12),
B(z) = ;,[B(s )J, F(z) = F* (s )!er' = z (4-13)
Plant C(s)
G(z) = J[ 1 -se-n c.(s)] = z ~ \[ c.s(s)]
C(z) = D(z)G(z)E(z) ~ e(t) T
E(z)
e(kT)
G(s)
Plant c(t) E(s)
e(t)
Figure 4-4 Open-loop system with a digital filter.
C(s)
c(t)
~ E*(s) A(s) A*(s) C(s) C(s) = G2(s)A*(s) A(s) = Gt(s)E*(s) C(z) = G2(z)A(z) A(z) = Gt(z)E(z) Gt(S) Gz(s)
T T
~ E*(s) ·8 ·8 T
E(s) 8 A(s) / A*(s) Gz(s) • Gt(s) T
R(s) E*(s) C(s)
+
-----t H(s) ,_.. ___ --..~~
E*(s) R*(s)- GH*(s)E*(s)
Solving for E* (s ), we obtain
* _ R*(s) E (s) - 1 + GH*(s)
and from (5-5),
_ R*(s) C(s)- G(s) 1 + GH*(s)
C(s) ..
C(s)
C(z) = GI(.i)G2(z)E(z)
C(s) = G1(s )G2(s )E* (s) C(z) = Gt G2(z )E(z)
G1 Gz(z) = ~[G1(s)Gz(s)] G1 G2(z) =I= Gt(z)Gz(z)
(5-8)
(5-9)
C(s) = G2(s)A *(s) = G2(s)Gt E*(s) C(z) = Gz(z)GtE(z) A(s) = G1(s)E(s)
(5-10)
which yields an expression for the continuous output. The sampled output is, then,
C*(s) = G*(s)E*(s) G*(s)R*(s) 1 + GH*(s)'
C(z) G(z)R(z)
1 + GH(z) (5-11)
Star Transforms
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