Z-Transform

Z-Transform

Linear time-invariante systems:

input: x(n); output: y(n); impulse response: h(n)

y(n) =∑k

h(k)x(n − k)

For example, if x(n) = zn:

y(n) =∑k

h(k)x(n − k) =∑k

h(k)zn−k = zn∑k

h(k)z−k

So:

y(n) = H(z)zn

Then, zn are ”eigenvectors” with H(z) as ”eigenvalue”.

Z-Transform

Because of the linearity it is convenient to look at linearcombinations:

x(n) =∑k

akznk

y(n) =∑k

akH(zk)znk

H(z) =∑k

h(k)z−k

is called the z-transform of h.

It is closely related to the DFT:

X (k) =∑n

x(n)e−j 2πN

kn

Z-Transform

DTFT:

X (ω) =∞∑

n=−∞x(n)e−iωn

so, it is the Z-transfor for z = e iω.

Recall: e iω = cos(ω) + isin(ω).

So, if we restrict the z-transform to complex numbers z so that,|z | = 1 then they are the same thing.

Z-Transform

Z-transform. Example

Let u(n) be the unit step. It is 1 if n ≥ 0 and 0 otherwise.

Consider now x(n) = anu(n).

X (z) =∞∑

n=−∞anu(n)z−n =

∞∑n=0

(az−1)n =1

1− az−1=

z

z − a

This can be done if |z | > |a|.

Z-Transform

Z-transform. Properties

- It is linear.

- Convolutions go to products (like with the Fourier transform).

x(n) = x1(n) ∗ x2(n) =⇒ X (z) = X1(z)X2(z)

Because of this, if we recall that h is the impulse response:

y(n) = h(n) ∗ x(n) =⇒ Y (z) = H(z)X (z)

and then

H(z) =Y (z)

X (z)

Z-Transform

Some system properties.

Causality:

The output depends only on present and past inputs.

It is necessary and sufficient that h(n) = 0, n < 0.

Stability:

If the input has bounded infinity norm then the output does also.It is necessary and sufficient that the 1−norm of h is bounded.

In frequency domain the region of convergence of H(z) mustcontain the unit circle.

Z-Transform

Causal linear time-invariant systems.

An LTI is called causal if:

a0y(n) + a1y(n − 1) + ...+ aMy(n −M) =

b0x(n) + b1x(n − 1) + ...+ bNx(n − N)

By taking the Z-transform on both sides, and using the fact thatZ (x(n − k)) = z−kZ (x(n)) we get:

a0Y (z) + a1z−1Y (z) + ...+ aMz−MY (z) =

b0X (z) + b1z−1X (z) + ...+ bNz−NX (z)

Z-Transform

Causal linear time-invariant systems.

And then:

Y (z)

X (z)=

b0 + b1z−1 + ...+ bNz−N

a0 + a1z−1 + ...+ aMz−M

By convention a0 = 1 and, also, we can take b0 out:

Y (z)

X (z)=

1

b0

1 + c1z−1 + ...+ cNz−N

1 + a1z−1 + ...+ aMz−M

But, as we know, this H(z).

Then we can study the zeros and poles of that expression.

For example, a filter that contains no other y(n − 1), y(n − 2)...will yield an H(z) with no poles.

An a filter that contains no other x(n − 1), x(n − 2)... will yield anH(z) with no zeros.

Z-Transform

Causal linear time-invariant systems.

So, the idea is that, by designing rational function with the zerosand poles that we want, we can, by using Y (z) = H(z)X (z) eitherkill or amplify certain frequencies.

Z-Transform

Frequency Response

If we remember the definition of H(z) and the fact that we canreplace z = e iω.

We define:

Frequency response of a system: H(e iω).

Magitud response of a system as : |H(e iω)|.

Phase response of a system as : ∠H(e iω).

We know that we can write: H(e iω) = |H(e iω)|e i∠H(e iω).

Z-Transform

Example: ideal delay system

Suppose that y(n) = x(n − nd).

Take as input x(n) = e iωn. Therefore y(n) = e iωne−iωnd .

By the definition of H(e iω) we see that in this case:

H(e iω) = e−iωnd

Page 41.

Z-Transform

Example: moving average

y(n) =1

M1 + M2 + 1

M2∑k=−M1

x(n − k)

In this case:

h(n) =1

M1 + M2 + 1if −M1 ≤ n ≤ M2

Therefore, the frequency response is given by:

H(e iω) =1

M1 + M2 + 1

M2∑k=−M1

e−iωk

The absolute value of the frequency response decays at highfrequencies.

So, moving averages attenuate fast variations. Page 45, graph.

Z-Transform

Example: lowpass filter

Suppose that we want to create an h that does not modify lowfrequencies and kills high frequencies.

We can define:

Hlp =

{1 if 0 < |ω| < ωc

0 if ωc < |ω| < π

How do we find h so that H has that form?

hlp(n) =1

2π

∫ ωc

−ωc

e iωndω =1

2πin(e iωcn − e−iωcn)

=1

πnsin(ωcn)

Z-Transform

IIR & FIR

Page 251

Z-Transform

Relationship between phase and magnitud (page 270)

|H(e iω|2 = H(e iω)(H(e iω))∗

But z−1 = z∗ if z = e iω. Then:

|H(e iω|2 = H(z)H(1

z∗)

which is:

C (z) =b0

a0

ΠMk=1(1− ckz−1)(1− c∗k z)

ΠNk=1(1− dkz−1)(1− d∗k z)

So, if we know that H(z) is causal we can, from the poles of C (z)identify the poles of H.

However, the same can’t be said for the zeros (since there is norestriction for the zeros of a stable system).

Z-Transform

Relationship between phase and magnitud (page 270)

The poles and zeros of C (z) are conjugate reciprocals. Recall:

1

c + di=

c

c2 + d2− i

d

c2 + d2

So:

(1

c + di)∗ =

c

c2 + d2+ i

d

c2 + d2

So, conjugate reciprocals lie in the same direction but withreciprocal absolute values.

See example 5.11 (page 271).

Z-Transform

All-Pass Systems

Example 5.12: how many stable, causal systems with three polesand three zeros?

Only 4.

If we do not restrict the number of poles and zeros the number ofpossible H(z)s is unlimited.

The frequency response magnitude of

Hap(z)z−1 − a∗

1− az−1

is independent of ω, |Hap(e iω)| = 1.

A system that does this is called all-pass system.

Z-Transform

Minimum Phase Systems

So, in order for the system to be stable and causal the poles of thetransfer function have to be inside the unit circle.

However, there are no restriction on the zeros.

In some cases it is useful to impose the condition that the inversesystem (the one with transfer function 1

H(z)) is also stable andcasual.

These type of systems are called minimum-phase systems.

Z-Transform

Minimum Phase Systems

If we now have the function C (z) and we know that the associatedH(z) is a minimum-phase system then H(z) is uniquelydetermined.

So, from the square of the magnitude of the frequency response wecan’t uniquely determine H(z) since we could take H and multiplyit by all-pass factors and those can’t be seen from the C (z).

Stable, causal systems can be decomposed as:

H(z) = Hmin(z)Hap(z)

Z-Transform

Minimum Phase Systems

Page 280

Z-Transform