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) ~

�

�� � � � � �

� �� �

� � � �

� �� �

� � �� � �

� �� �

� ��

�� � �

��

�� �

��

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