Z transform.pdf

EE480.3 Digital Control Systems Part 2. z-transform Kunio Takaya Electrical and Computer Engineering University of Saskatchewan January 14, 2008

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Contents 1 z-Transform

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2 z-transform of simple functions

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3 Properties of the z-transform

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4 Continue to discrete time systems

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z-Transform

The definition of the z-Transform is X(z) =

+∞ X

x(k)z −k .

k=−∞

where, x(k) is a discrete time sequence (sampled data). When x(k) is defined for k ≥ 0, i.e. causal, one sided z-transform is given by X(z) =

+∞ X

x(k)z −k .

k=0

The variable z is complex, so is X(z).

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2

z-transform of simple functions

δ function   1 δ(k) =  0 Z[δ(k)] =

if k = 0 otherwise

+∞ X

δ(k)z −k = 1

k=0

unit step function   1 u(k) =  0 Z[u(k)] =

+∞ X

if k ≥ 0 if k < 0

u(k)z −k = z 0 + z −1 + z −2 + +z −3 · · ·

k=0

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Since the sum of n terms of a geometric progression is given by the formula, n−1 X 1 − rn k ar = a 1−r k=0

1 1 − z −n Z[u(k)] = lim = . −1 n→∞ 1 − z −1 1−z This z-trnasform is valid only for jzj < 1, in the region of convergence in the z-plane. Decaying exponential function f (t) = e−at t ≥ 0, sampled with a sampling period T ⇒ f (k) = e−aT k , k ≥ 0 5

Z[f (k)]

=

∞ X

e−aT k z −k =

k=0

=

∞ X

(e−aT z −1 )k

k=0

1 1 − e−aT z −1

The region of convergence is e−aT < jzj. If a > 0, the system is stable and the pole at z = e−aT < 1 is inside the unit circle. If a < 0, the system is unstable and the pole at z = e−aT > 1 is outside the unit circle. Damped cosine wave f (k) = e−aT k cos ωT k, k ≥ 0

Z[f (k)]

=

∞ X

e−aT k cos ωT k z −k

k=0

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= = = =

∞

∞

k=0

k=0

1 X −aT +jωT −1 k 1 X −aT −jωT −1 k (e z ) + (e z ) 2 2 1 1 1 1 + 2 1 − e−aT +jωT z −1 2 1 − e−aT −jωT z −1 1 − e−aT cos ωT z −1 1 − 2e−aT cos ωT z −1 + e−2aT z −2 z 2 − e−aT cos ωT z z 2 − 2e−aT cos ωT z + e−2aT

The region of convergence is e−aT < jzj. • Exercise: Find the z-transform of a discrete sequence f (k) = 2−k

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for k ≥ 0

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Properties of the z-transform

Shift Operations - One step delay with u(k) Z[x(k − 1)u(k)]

=

=

∞ X

x(k − 1)z −k

k=0 ∞ X

x(k)z −(k+1)

k=−1

=

∞ X

x(k)z −k−1 + x(−1)z 0

z −1

∞ X

k=0

=

x(k)z −k + x(−1)

k=0

=

z −1 Z[x(k)u(k)] + x(−1)

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Shift Operations - One step delay with u(k − 1) Z[x(k − 1)u(k − 1)]

=

=

∞ X

k=1 ∞ X

x(k − 1)z −k x(k)z −(k+1)

k=0

=

z −1

∞ X

x(k)z −k

k=0

=

z −1 Z[x(k)u(k)]

Shift Operations - m step delay with u(k) Z[x(k − m)u(k)]

=

∞ X

x(k − m)z −k

k=0

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=

∞ X

x(k)z −(k+m)

k=−m

=

∞ X

x(k)z −k−m +

k=0

=

z −m

∞ X

x(k)z −k−m

k=−m m X

x(k)z −k +

k=0

=

−1 X

z −m Z[x(k)u(k)] +

x(−k)z k−m

k=1 m X

x(−k)z k−m

k=1

Shift Operations - m step delay with u(k − m) Z[x(k − m)u(k − m)]

=

=

∞ X

k=m ∞ X

x(k)z −(k+m)

k=0

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x(k − m)z −k

=

z −m Z[x(k)u(k)]

Shift Operations - m step time advance with u(k + m) Z[x(k + m)u(k + m)]

=

= =

∞ X

x(k + m)z −k

k=−m ∞ X

x(k)z −(k−m)

k=0 +m

z

Z[x(k)u(k)]

Initial Value theorem lim X(z) = lim

z→∞

z→∞

+∞ X

x(k)z −k = x(0)

k=0

Final Value theorem Z[x(k + 1)u(k)] − Z[x(k)u(k)] 11

= =

zX(z) − zx(0) − X(z) ∞ ∞ X X x(k + 1)z −k − x(k)z −k

k=0

zX(z) − X(z) lim (z − 1)X(z)

z→1

=

=

k=0

zx(0) +

x(0) +

∞ X

x(k + 1)z −k −

k=0 ∞ X

x(k + 1) −

k=0

=

k=0 ∞ X

k=0

x(∞)

Derivative dX(z) dz

=

∞ d X x(k)z −k dz k=0

=

∞ X

k=0

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d −k x(k) z dz

∞ X

x(k)

x(k)z −k

= = Z[kx(k)] =

∞ X

−x(k)kz −k−1

k=0 −1

−z

Z[kx(k)] dX(z) −z dz

Example: Z[ku(k)]

= = =

d 1 dz 1 − z −1 −z −2 −z (1 − z −1 )2

−z

z −1 z = (1 − z −1 )2 (z − 1)2

Convolution

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Consider multiplying G(z) and X(z). G(z)X(z) ∞ ∞ X X = g(k)z −k × x(k)z −k k=0

k=0 −1

+ g(2)z −2 + g(3)z −3 + · · ·]

=

[g(0) + g(1)z

×

[x(0) + x(1)z −1 + x(2)z −2 + x(3)z −3 + · · ·]

=

g(0)x(0) + [g(0)x(1) + g(1)x(0)]z −1

+

[g(0)x(2) + g(1)x(1) + g(2)x(0)]z −2

[g(0)x(3) + g(1)x(2) + g(2)x(1) + g(32)x(0)]z −3 n X + g(k)x(n − k) z −n + · · ·

+

=

k=0 ∞ X n X

g(k)x(n − k) z −n

n=0 k=0

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=

Z[

n X

g(k)x(n − k)] = Z[

k=0

n X

x(k)g(n − k)]

k=0

Input Sequence

Impulse Response

X(z)

G(z)

Output Sequence Y(z)=G(z)X(z)

Figure: Convolution and Transfer Function When an input sequence x(k) is applied to a system having an impulse response g(k), the response y(k) of the system is given the convolution sum. k X y(k) = x(n)g(k − n) n=0

The equivalent expression in the z-transform is Y (z) = G(z)X(z). 15

Inverse z-Transform Partial Fraction Method The z-transform of an exponential sequence x(k) = ak is given by X(z) =

z 1 = . −1 1 − az z−a

Any sequence that starts with a non-zero value at k = 0 usually has the same order in the numerator and denominator of the z-transform. To find the inverse z-transform, one must take partial X(z) instead of X(z) itself. fraction expansion on z X(z) X (z) = z 0

= =

B(z) (z − a1 )k (z − a2 )(z − a3 ) · · · c12 c1k c11 + + ··· + (z − a1 )k (z − a1 )k−1 (z − a1 ) 16

+

c2 c3 + + ··· (z − a2 ) (z − a3 )

Where, unknown coefficients are given by ci = (z − ai ) X 0 (z)jz=ai n−1

c1n

d 1 k 0 (z − a1 ) X (z) = n−1 (n − 1)! dz z=a1

After partial fraction expansion, find the expanded form of X(z) by multiplying X 0 (z) by z, then find the inverse z-transform term-by-term by table look-up. Example E(z)

=

E(z) z

=

1 (z − 1)(z − 2) 11 1 1 1 − + 2z z−1 2z−2 17

E(z) e(k)

1 z 1 z − + 2 z−1 2z−2 1 1 = ( δ(k) − 1k + 2k )u(k) 2 2 =

• Exercise: Find the inverse z-transform of Y (z) =

1 (1 − z −1 )(1 − 2z −1 )

Use of MATLAB for partial fraction expansion >> [r,p,k]=residue([1],[1, -3, 2, 0]) r = 0.5000 -1.0000 0.5000 p = 2 1 18

0 k = []

Difference Equation y(k) + a1 y(k − 1) + a2 y(k − 2) + · · · + ak y(0) =

b0 u(k) + b1 u(k − 1) + · · · + bk u(0)

Where, y(k) is output sequence, and u(k) is input sequence. In control systems, b0 is often 0, as input u(k) does not immediately affect output y(k). Take the z-transform of this difference equation considering u(k) = 0 and y(k) = 0 for k < 0. Z f[y(k) + a1 y(k − 1) + a2 y(k − 2) + · · · + ak y(0)]u(k)g = Z f[b0 u(k) + b1 u(k − 1) + · · · + bk u(0))]u(k)g 19

Since Z[y(k − i)] = z −i Y (z) and Z[u(k − i)] = z −i U (z), Y (z) + a1 z −1 Y (z) + a2 z −2 Y (z) + · · · + ak z −k Y (z) = b0 U (z) + b1 z −1 U (z) + b2 z −2 U (z) + · · · + bk z −k U (z) This leads to a transfer function of the difference equation. b0 + b1 z −1 + b2 z −2 + · · · + bk z −k Y (z) = G(z) = U (z) 1 + a1 z −1 + a2 z −2 + · · · + ak z −k The output Y (z) for a given input U (z) is given by b0 + b1 z −1 + b2 z −2 + · · · + bk z −k Y (z) = G(z)U (z) = U (z) −1 −2 −k 1 + a1 z + a2 z + · · · + ak z Example A difference equation, m(k) = e(k) − e(k − 1) − m(k − 1) 20

is driven by an input sequence   1, k ≥ 0 even e(k) =  0, k ≥ 0 odd Using the z-Transform, M (z)

=

E(z)

=

z−1 E(z) z+1 z2 1 = 2 1 − z −2 z −1

Thus, the solution in the z-Transform is M (z)

=

z2 z2 = z 2 + 2z + 1 (z + 1)2

= 1 − 2z −1 + 3z −2 − 4z −3 + · · ·

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Assignment No. 2 Solve following problems related to the z-transform fundamentals in the textbook pp. 78-79. 1. 2-1 2. 2-2 (a)(b) 3. 2-3 (c) 4. 2-4 5. 2-8 (a)(e) Do only the partial fraction expansion method.

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Continue to discrete time systems

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