mm3fc mathematical modeling 3 lecture 6 times weeks 7,8 & 9. lectures : mon,tues,wed 10-11am,...
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MM3FC Mathematical Modeling 3LECTURE 6
Times
Weeks 7,8 & 9.Lectures : Mon,Tues,Wed 10-11am,
Rm.1439Tutorials : Thurs, 10am, Rm. ULT.
Clinics : Fri, 8am, Rm.4.503Dr. Charles Unsworth,Department of Engineering Science, Rm. 4.611
Tel : 373-7599 ext. 82461Email : [email protected]
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This LectureWhat are we going to cover &
Why ?• The Z-Transform. (Makes convolution easy)
• Cascading systems with the z-transform. (simplifies the whole process)
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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.
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Definition of the Z-Transform• A finite length signal x[n] can be represented as :
• And the conventional z-transform of such a signal is given by :
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]’.
∑N
0=k
]-[ x[k]=][x knn δ
∑N
0=k
k- x[k]=][ zzX… (5.1)
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• It is instructive to note, though, that X[z] can be re-written as :
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
∑N
0=k
k1- )(z x[k]=[z]X… (5.2)
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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 5n>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].
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Domains• In general, a ‘z-transform pair’ is a sequence and it’s
corresponding z-transform.
n – Domain (or Time-Domain) z – Domain
• 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’.
∑∑N
0=k
k-N
0=k
x[k]=][ ⇔ k]-[n x[k] = x[n] zzXδ
… (5.3)
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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.
n – Domain 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 ……
00 -n-10 z=)1.(z=X(z)⇔]n-δ[n=x[n] n
… (5.4)
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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
0=kk ]-[ b=][y knxn
∑
∑M
0=k
nk-k
M
0=k
k-nk
z z b=
z b=][y n
( )
The ‘system function’ of the FIR
… (5.5)
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• 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 :
• The ‘system function’ is also the z-transform of the ‘impulse response’.
• Thus, FIR filters with an input x[n] = zn for - n give an output y[n].
y[n] = h[n] * zn = H(z)zn
∑∑M
0=k
k-M
0=k
k-k z h[k]=z b=][H z
∑∑M
0=k
k-k
M
0=kk z b=H[z] ⇔ k]-δ[n b = h[n]
… (5.6)
… (5.7)
… (5.8)
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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).
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Properties of the z-transform• LINEARITY : The z-transform obeys the ‘principle of superposition’
or ‘linearity’ ax1[n] + bx2[n] aX1(z) + bX2(z)
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 -2 -1 0 1 2 3 4 5x2 -2 3 -4 2 1 0 0 -2
… (6.9)
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• 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)
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
… (6.10)
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• So far we have defined the z-transform H(z) only for finite input sequences x[n].
• The z-transform extends for infinite-length signals too.
The General z-Transform Formula
∑N
0=n
n-z x[n] = H(z)
∑∞
∞-=n
n-z x[n] = H(z)… (6.11)
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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-1]Unit
delay
x[n] x[n]H(z) = x[n-1]z-1
= EquivalentSystems
Example 7 : Draw a block diagram for the system: y[n] = (1 – z-1)x[n]
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Convolution & the z-Transform• Convolution in the ‘n-domain’ is multiplication in the ‘z-Domain’.
∑M
0=k
k]- x[nh[k]=h[n]*x[n]=y[n]
( )
H(z)X(z)=
X(z) z h[k]=
X(z)z h[k]=Y(z)
∑
∑M
0=k
k-
M
0=k
k-
( )
= system function = H(z)
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Convolution in the ‘n-Domain is multiplication in the z-Domain.
y[n] = h[n] * x[n] y(z) = H(z)X(z)
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 n< 1 1 2 3 4 5 n>4x[n] 0 1 -1 1 -1 0 0h[n] 1 -1
h[0] 0 1 -1 1 -1 0 0h[1] 0 -1 1 -1 1 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]
… (4.12)
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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.
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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.
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)
• 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.
[n] h1[n]LTI 1h1[n]
LTI 2h2[n]
h[n] = h1[n] * h2[n]
H1(z) H2(z) H(z) = H1(z) H2(z)
… (4.13)
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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].)
3
x[n] x[n-1]z-1
-1
y[n]
w[n-1]z-1
-12
w[n]
2 cascaded 1st order systems.Order = power of highest z.
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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.
• The equivalent circuit is faster. 3 multipliers & 2 adders compared to the cascaded systems 4 multipliers and 2 adders.
6
a[n]
x[n] x[n-1]z-1
x[n-2]z-1
-5 1
A 2nd order system
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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]
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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] w[n]LTI 1 LTI 2
y[n]
H1(z) = z-1 – 1 H2(z) = 1 - z-1 + z-2
• Thus, the 3rd order LTI can be factorised into a cascaded 1st & 2nd order system.
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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
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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.