Transcript Charlie_Files-CorrectedMM3 L6 - The Z

MM3FC Mathematical Modeling 3
LECTURE 6
Times
Weeks 7,8 & 9.
Lectures : Mon,Tues,Wed 10-11am, Rm.1439
Tutorials : Thurs, 10am, Rm. ULT.
Clinics : Fri, 8am, Rm.4.503
Dr. Charles Unsworth,
Department of Engineering Science, Rm. 4.611
Tel : 373-7599 ext. 82461
Email : [email protected]
This Lecture
What are we going to cover & Why ?
• The Z-Transform.
(Makes convolution easy)
• Cascading systems with the z-transform.
(simplifies the whole process)
The Z-Transform
• We will now introduce the z-transform for FIR filters and
finite length sequences in general.
• We will use the FIR case to introduce the important
concept of ‘domains of representation’.
• We will learn that the z-transform brings polynomials and
rational functions into the analysis of discrete time
systems.
• We will show that convolution is equivalent to polynomial
multiplication and that common algebraic operations, such as
multiplying, dividing and factoring polynomials can be
interpreted as combining or decomposing LTI systems.
Definition of the Z-Transform
• A finite length signal x[n] can be represented as :
N
x[n] =
∑x[k] δ[n - k ]
k =0
• And the conventional z-transform of such a signal is given by :
N
X [ z] =
∑x[k] z
-k
k =0
… (5.1)
Where, z represents any complex number.
•
All we are doing is ‘transforming’ the data x[n] into another
representation X[z] to make the system easier to solve.
•
We say we are performing the ‘z-transform of x[n]’.
• It is instructive to note, though, that X[z] can be re-written as :
N
X [z] =
∑x[k] (z
-1 k
)
k =0
… (5.2)
This emphasizes that the variable (z-1) is a polynomial of degree k=N.
• There is nothing complicated about performing a transform.
• All we do to obtain X(z) is to construct a polynomial whose coefficients
are the values of the sequence x[n]
Example 1 : Let’s determine the z-transform of our finite-sequence .
n
n<-1
-1
0
1
2
3
4
5
n>5
x[n]
0
0
2
4
6
4
2
0
0
The z-transform is :
X (z) = 2 + 4z-1 + 6z-2 + 4z-3 + 2z-4
Example 2 :
Determine the sequence x[n] from the z-transform X(z).
X(z) = 1 –2z-1 + 3z-3 – z-5
n
n<-1
-1
0
1
2
3
4
5
n>5
x[n]
• You have just performed what is known as the ‘Inverse z-transform’.
• Similarly, we can recover x[n] from X(z) by extracting the co-efficient
values of X(z) and placing them in their corresponding kth position in the
sequence x[n].
Domains
• In general, a ‘z-transform pair’ is a sequence and it’s corresponding ztransform.
N
x[n] = ∑x[k] δ[n - k]
N
⇔
k =0
n – Domain (or Time-Domain)
X [ z] =
∑x[k] z
-k
k =0

z – Domain
… (5.3)
• Notice how (n) is the independent variable of the sequence x[n].
• Here, we say the signal exists in the ‘n-Domain’.
• Since, (n) is often an index that counts time, we also say this is the
‘time-Domain’ of the signal.
• Similarly, (z) is the independent variable of the z-transform X(z).
• Thus, we the signal exists in the ‘z-Domain’.
• During a ‘z-transform’ we move from the ‘time-Domain’  ‘z-Domain’.
• In an ‘Inverse transform’ we move from the ‘z-Domain’  ‘time-Domain’.
Z-Transform of an Impulse
• A simple but very important example is the z-transform pair of an
impulse of a sequence x[n].
• As we know this can be defined as :
x[n] = [n- n0]
The co-efficient of the polynomial z-1 = magnitude of the impulse = 1.
The kth power of the polynomial z-1 = the offset of the impulse = n0.
x[n] = δ[n - n 0 ] ⇔ X(z) = 1.(z -1 ) n0 = z -n 0
n – Domain

… (5.4)
z – Domain
• Now we’ve learned how to transform a signal from the n-Domain to the
z-Domain …… but why ? ……
• Next, we are going to answer this very question ……
The z-transform & Linear Systems
• The z-transform is indispensable in the design of LTI systems.
• The reason is the way an LTI system responds to a particular input
signal x[n] = zn for -  n  .
Recall, the general difference equation of an FIR filter :
M
y[n] =
∑b
k
x[n - k ]
k
zn -k
k =0
M
y[n] =
∑b
k=0
=( ∑b k z - k) z n
M
k=0
The ‘system function’ of the FIR
… (5.5)
• We can see that the ‘system function’ of the FIR is a polynomial whose
form depends on the coefficients of the FIR.
• Thus, we define the ‘system function’ of an FIR filter to be :
M
H[ z ] =
∑b
M
k
z-k =
k =0
∑h[k] z
-k
… (5.6)
k =0
• The ‘system function’ is also the z-transform of the ‘impulse response’.
M
h[n] = ∑b k δ[n - k]
M
⇔
H[z] =
k =0
-k
b
z
∑k
k =0
… (5.7)
• Thus, FIR filters with an input x[n] = zn for -  n   give an output
y[n].
y[n] = h[n] * zn = H(z)zn
… (5.8)
Example 3 : Find the system function H(z) of the FIR filter with impulse
response. h[n] = [n-1] – 7 [n-2] – 3 [n-3]
Simply, replace the (n-k)’s with the corresponding z-k.
Gives :
H(z) = z-1 –7z-2 –3z-3
Example 4 : Find the impulse response h[n] of the FIR filter whose system
response is. H(z) = 4(1 – z-1)(1 + z-1)(1 + 8z-1).
Properties of the z-transform
• LINEARITY : The z-transform obeys the ‘principle of superposition’ or
‘linearity’
ax1[n] + bx2[n]  aX1(z) + bX2(z)
… (6.9)
Example 5 : Calculate the linear superposition of the z-transforms
of the two signals below. For a=1, b=1
x1[n] = [n] + [n-1] + 2 [n-2] + 3 [n-5]
n
x2
-2
-2
-1
3
0
-4
1
2
2
1
3
0
4
0
5
-2
• Time Delay Property : The z-transform of a signal x[n] delayed by (n0)
samples is equivalent to z-transform X(z) of x[n] mutiplied by z-n.
A delay of 1 sample multiplies the z-transform by z-1
x[n-1]

z-1X(z)
A delay of n0 samples multiplies the z-transform by z-n0
x[n – n0]

z-n0X(z)
… (6.10)
Example 6 : The signal x1[n] = [n] + [n-1] + 2 [n-2] + 3 [n-5] is delayed
by 6 time samples. Calculate its z-transform
The General z-Transform Formula
• So far we have defined the z-transform H(z) only for finite input
sequences x[n].
N
H(z) = ∑x[n] z - n
n =0
• The z-transform extends for infinite-length signals too.
∞
H(z) =
∑x[n] z
n = -∞
-n
… (6.11)
The z-Transform in Block Diagrams
• As we have seen a ‘unit delay’ becomes a z-1 operator in the z-Domain.
• Similarly in a block diagram, all delays can be replaced by a z-n0
operator.
x[n]
x[n]
Unit
delay
z-1
x[n-1]
x[n]H(z) = x[n-1]
=
Equivalent
Systems
Example 7 : Draw a block diagram for the system: y[n] = (1 – z-1)x[n]
Convolution & the z-Transform
• Convolution in the ‘n-domain’ is multiplication in the ‘z-Domain’.
M
y[n] = x[n] * h[n] =
∑h[k] x[n
- k]
k =0
M
Y(z) =
-k
(
h[k]
z
X(z) )
∑
k=0
M
(∑h[k] z
=
k=0
-k
) X(z)
= H(z)X(z)
= system function = H(z)
Convolution in the ‘n-Domain is multiplication in the z-Domain.
y[n] = h[n] * x[n]
y(z) = H(z)X(z)

… (4.12)
Example 8 : The signal x[n] = [n-1] –[n-2] + [n-3] – [n-4] is passed
through the filter y[n] = x[n] – x[n-1]. Determine the output from the
filter using convolution time-domain and the z-domain.
In the Time-Domain
n
x[n]
h[n]
n< 1
0
1
1
1
-1
2
-1
3
1
4
-1
5
0
n>4
0
h[0]
h[1]
0
1
0
-1
-1
1
1
-1
-1
0
1
0
0
y[n]
0
1
-2
2
-2
1
0
 y[n] = [n-1] –2[n-2] + 2[n-3] – 2[n-4] + 5[n-5]
For the z-Domain, express the signal & filter in terms of z’s and multiply.
Then inverse transform to get the signal. (Just replace the z’s with ’s).
Y[N] = X[N]*H[N] = (z-1 – z-2 + z-3 – z-4 )(1 – z-1)
= z-1 – 2z-2 + 2z-3 – 2z-4 + z-5
 y[n] = [n-1] –2[n-2] + 2[n-3] – 2[n-4] + 5[n-5]
1.
Transforming to the z-Domain is a simpler way to solve the system.
2. Convolution in the z-Domain is just algebraic multiplication.
Cascading Systems with the z-Transform
• One of the main applications of the z-transform is in system design.
• Remember, the impulse response of a 2 cascaded LTI systems.
• We can now use the z-transform to determine the same thing.
[n]
LTI 1
h1[n]
h1[n]
LTI 2
h2[n]
H2(z)
H1(z)
h[n] = h1[n] * h2[n]
H(z) = H1(z) H2(z)
The system function for a cascade of 2 LTI systems
is the product of the individual system functions.
h[n] = h1[n] * h2[n]

H(z) = H1(z)H2(z)
… (4.13)
• The z-transform, like convolution, is commutative so the LTI systems
could be cascaded in either order.
• As we can see, the formula would extend for (N) cascaded LTI systems.
Example 9 : Use the z-transform to cascade the two systems below and
determine their combined difference equation. Draw the block diagram of
the 2 cascaded system and its equivalent system. Which is the faster
system ?
w[n] = 3x[n] - x[n-1] ;
y[n] = 2w[n] – w[n-1]
Block diagram (here we see x[n] is the input to w[n] which inputs to y[n].)
x[n]
3
z-1
x[n-1]
2 cascaded 1st order systems.
Order = power of highest z.
-1
w[n]
2
w[n-1]
z-1
-1
y[n]
Just change to z’s and multiply out. The system functions are :
W(z) = 3 - z-1
and
Y(z) = 2 –z-1
Combined system function A(z) = W(z)Y(z)
A(z) = (3 - z-1)(2 –z-1) = 6 – 3z-1 – 2z-1 + z-2
= 6 – 5z-1 + z-2
Thus, a[n] = 6x[n] – 5x[n-1] + x[n-2]
Equivalent circuit.
x[n]
z-1
x[n-1]
x[n-2]
z-1
A 2nd order system
6
-5
1
a[n]
• The equivalent circuit is faster. 3 multipliers & 2 adders compared to the
cascaded systems 4 multipliers and 2 adders.
Example 10 : Use z-transform to combine the following cascaded systems.
Again draw the block diagram of the 2 cascaded system and its equivalent
system. What are the orders of the systems ? Which is the faster system
?
w[n] = x[n] + x[n-1]
and
y[n] = w[n] – w[n-1] + w[n-2]
Factoring z-Polynomials
• If we can multiply z-transforms to get higher-order systems we can also
factorise z-polynomials to break down a large system into smaller
modules.
Example : Consider H(z) = 1 –2z-1 + 2z-2 – z-3
First we visually inspect for roots that may exist.
We can see one root at z = 1. Therefore, (z-1 – 1) is a factor.
Now, H(z) = H1(z)H2(z)
Therefore, H2(z) = H(z)/H1(z)
= (1 –2z-1 + 2z-2 – z-3)/ (z-1 – 1)
= 1 - z-1 + z-2
x[n]
LTI 1
H1(z) = z-1 – 1
w[n]
LTI 2
y[n]
H2(z) = 1 - z-1 + z-2
• Thus, the 3rd order LTI can be factorised into a cascaded 1st & 2nd order
system.
Deconvolution (Inverse filtering)
• The cascading property leads to the question ……
• “ Can we use the 2nd filter in cascade to undo the effect of the 1st
filter” ?
• If H1(z) is known then we can construct a filter H2(z) to undo the effect
of the first by setting H(z) = 1.
H(z) = H1(z)H2(z) = 1
Thus, H2(z) = 1/H1(z)
• The 2nd filter tries to ‘undo the convolution’ that exists, so the process
is referred to as ‘deconvolution’.
• The process is also called ‘Inverse Filtering’.
Example : Determine the inverse filter for H1(z) = 1 – z-1 + ½z-2.
Now, H1(z)H2(z) = 1
Thus, H2(z) = 1/ 1 – z-1 + ½z-2
• What are we to make of this example ?
• It seems that the inverse filter for an FIR filter must have a system
function that is not a polynomial.
• Instead, the inverse filter is a rational function (ratio of 2 polynomials).
• Thus, the inverse filter cannot be an FIR filter.
• And Deconvolution is not as simple as it appeared !
• Later, we will learn about other LTI systems that do have rational system
functions and will return to the problem of deconvolution.