Download - Lecture 5 – 6 Z - Transform
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Lecture 5 – 6
Z - Transform
ByBy
Dileep KumarDileep Kumar
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Frequency domain vs Time domain
Frequency domain is a term used to describe the analysis of Frequency domain is a term used to describe the analysis of mathematical functions or signals with respect to frequency.mathematical functions or signals with respect to frequency.
((communications point of viewcommunications point of view) A plane on which signal strength can ) A plane on which signal strength can be represented graphically as a function of frequency, instead of a be represented graphically as a function of frequency, instead of a function of time. function of time.
control systemscontrol systems) Pertaining to a method of analysis, particularly useful ) Pertaining to a method of analysis, particularly useful for fixed linear systems in which one does not deal with functions of for fixed linear systems in which one does not deal with functions of time explicitly, but with their Laplace or Fourier transforms, which are time explicitly, but with their Laplace or Fourier transforms, which are functions of frequency. functions of frequency.
Speaking non-technically, a Speaking non-technically, a time domaintime domain graph shows how a signal graph shows how a signal changes over time, whereas a frequency domain graph shows how changes over time, whereas a frequency domain graph shows how much of the signal lies within each given frequency band over a range much of the signal lies within each given frequency band over a range of frequencies. of frequencies.
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Cont: A frequency domain representation can also include information on A frequency domain representation can also include information on
the the phasephase shift that must be applied to each sinusoid in order to be able shift that must be applied to each sinusoid in order to be able to recombine the frequency components to recover the original time to recombine the frequency components to recover the original time signal. signal.
The frequency domain relates to the Fourier transform or Fourier The frequency domain relates to the Fourier transform or Fourier series by decomposing a function into an infinite or finite number of series by decomposing a function into an infinite or finite number of frequencies. This is based on the concept of Fourier series that any frequencies. This is based on the concept of Fourier series that any waveform can be expressed as a sum of sinusoids (sometimes waveform can be expressed as a sum of sinusoids (sometimes infinitely many.)infinitely many.)
In using the Laplace, Z-, or Fourier transforms, the frequency In using the Laplace, Z-, or Fourier transforms, the frequency spectrum is complex and describes the frequency magnitude and spectrum is complex and describes the frequency magnitude and phase. In many applications, phase information is not important. By phase. In many applications, phase information is not important. By discarding the phase information it is possible to simplify the discarding the phase information it is possible to simplify the information in a frequency domain representation to generate a information in a frequency domain representation to generate a frequency spectrum or spectral density. A spectrum analyser is a frequency spectrum or spectral density. A spectrum analyser is a device that displays the spectrum.device that displays the spectrum.
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The Direct Z-Transform
The z-transform of a discrete time signal is defined as the power The z-transform of a discrete time signal is defined as the power seriesseries
(1)(1)
Where z is a complex variable. For convenience, the z-transform of a Where z is a complex variable. For convenience, the z-transform of a signal x[n] is denoted bysignal x[n] is denoted by
X(z) = Z{x[n]}X(z) = Z{x[n]}
Since the z-transform is an infinite series, it exists only for those Since the z-transform is an infinite series, it exists only for those values of z for which this series converges. The Region of values of z for which this series converges. The Region of Convergence (ROC) of X(z) is the set of all values of z for which Convergence (ROC) of X(z) is the set of all values of z for which this series converges.this series converges.
We illustrate the concepts by some simple examples.We illustrate the concepts by some simple examples.
n
nz]n[x)z(X
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Example 1: Determine the z-transform of the following signals
(a)(a) x[n] = [1, 2, 5, 7, 0, 1]x[n] = [1, 2, 5, 7, 0, 1]
Solution: X(z) = 1 + 2zSolution: X(z) = 1 + 2z-1-1+ 5z+ 5z-2-2 + 7z + 7z-3-3 + z + z-5,-5,
ROC: entire z plane except z = 0ROC: entire z plane except z = 0
(b) y[n] = [1, 2, 5, 7, 0, 1](b) y[n] = [1, 2, 5, 7, 0, 1]
Solution: Y(z) = zSolution: Y(z) = z22 + 2z + 5 + 7z + 2z + 5 + 7z-1-1 + z + z-3-3
ROC: entire z-plane except z = 0 and z = ROC: entire z-plane except z = 0 and z = ..
(c)(c) z[n] = [0, 0, 1, 2, 5, 7, 0, 1]z[n] = [0, 0, 1, 2, 5, 7, 0, 1]
Solution: zSolution: z-2-2 + 2z + 2z-3-3 + 5z + 5z-4-4 + 7z + 7z-5-5 + z + z-7, -7, ROC: all z except z=0ROC: all z except z=0
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(d) p[n] = (d) p[n] = [n][n]
Solution: P(z) = 1, ROC: entire z-plane.Solution: P(z) = 1, ROC: entire z-plane.
(e) q[n] = (e) q[n] = [n – k], k > 0[n – k], k > 0
Solution: Q(z) = zSolution: Q(z) = z-k-k, entire z-plane except , entire z-plane except z=0.z=0.
(f) r[n] = (f) r[n] = [n+k], k > 0[n+k], k > 0
Solution: R(z) = zSolution: R(z) = zkk, ,
ROC: entire z-plane except z = ROC: entire z-plane except z = ..
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Example 2: Determine the z-transform of x[n] = (1/2)nu[n]
Solution:Solution:
ROC: |1/2 zROC: |1/2 z-1-1| < 1, or equivalently |z| > 1/2| < 1, or equivalently |z| > 1/21
21
n
0n
1nn
0n
n
n
z21
1
1
.......z2
1z
2
11
z2
1z
2
1
z]n[x)z(X
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Example 3: Determine the z-transform of the
signal x[n] = anu[n]Solution: Solution:
|a||z:|ROCaz1
1
.......azaz1
azza)z(X
1
211
n
0n
1n
0n
n
|a||z:|ROCaz1
1
.......azaz1
azza)z(X
1
211
n
0n
1n
0n
n
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Properties of z-transform LinearityLinearity
IfIf x x11[n] [n] X X11(z)(z)
and xand x22[[n] [[n] X X22(z)(z)
thenthen
aa11xx11[n] + a[n] + a22xx22[n] [n] a a11XX11(z) + a(z) + a22XX22(z)(z)
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Example: Determine the z-transform of the signal x[n] = [3(2n) – 4(3n)]u[n]
Solution: Solution:
11
nn
1n
z31
14
z21
13]34)2(3[z
az1
1]]n[ua[z
Example 4: Determine the z-transform of the signal (cosw0n)u[n]
20
10
1
1jw1jw0
njwnjw0
zwcosz21
wcosz1
ze1
1
2
1
ze1
1
2
1]n[unwcosz
e2
1e
2
1]n[unwcos
00
00
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Time Shifting Property:If x[n] X(z) then x[n-k] z-kX(z)
Proof: since
then the change of variable m = n-k produces
n
nzknxknxz ][]][[
)z(Xzz]m[xz
z]m[x]]kn[x[z
k
m
mk
m
)km(
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Example: Find the z-transform of a unit step function. Use time shifting property to find z-transform of u[n] – u[n-N].
The z-transform of u[n] can be found asThe z-transform of u[n] can be found as
Now the z-transform of u[n]-u[n-N] may be Now the z-transform of u[n]-u[n-N] may be found as follows:found as follows:
121
0n
n
n
n
z1
1.......zz1
zz]n[u]]n[u[z
1
N
1N
1
z1
z1
z1
1z
z1
1]]Nn[u]n[u[z
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Scaling in the z-domain If x[n] If x[n] X(z) X(z)Then aThen annx[n] x[n] X(a X(a-1-1z)z)For any constant a, real or complex.For any constant a, real or complex.Proof: Proof:
Example 5: Determine the z-transform of the signalExample 5: Determine the z-transform of the signal
aann(cosw(cosw00n)u[n].n)u[n].
Solution: since Solution: since
zaXza]n[xz]n[xa]n[xaz 1
n
n1n
n
nn
20
10
1
0 zwcosz21
wcosz1]n[u)nw[cos(z
22
01
01
0 cos21
cos1]][cos[
zawaz
waznunwaz n
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Time reversal If x[n] If x[n] X(z) then x[-n] X(z) then x[-n] X(z X(z-1-1))Proof:Proof:
Example 6: Determine the z-transform of Example 6: Determine the z-transform of u[-n].u[-n].Solution: since z[u[n]] = 1/(1 – zSolution: since z[u[n]] = 1/(1 – z -1-1))Therefore,Therefore,
Z[u[-n]] = 1/(1-z)Z[u[-n]] = 1/(1-z)
m m
1m1m
n
n )z(Xz]m[xz]m[xz]n[x]]n[x[z
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Differentiation in the z - Domain
x[n] x[n] X(z) then nx[n] = -z(dX(z)/dz) X(z) then nx[n] = -z(dX(z)/dz)
Tutorial 4: Q1: Prove the differentiation Tutorial 4: Q1: Prove the differentiation property of z – transform.property of z – transform.
Example 7: Determine the z-transform of the Example 7: Determine the z-transform of the signal x[n] = na signal x[n] = nannu[n].u[n].
Solution: Solution:
21
1
1n
1n
az1
az
az1
1
dz
dz]]n[una[z
az1
1]]n[ua[z
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Convolution and Correlation To study the LTI systems, convolution plays important To study the LTI systems, convolution plays important
role. Shifting multiplications and summation are role. Shifting multiplications and summation are operations in computation of convolution.operations in computation of convolution.
Correlation which is very much similar to convolution Correlation which is very much similar to convolution provides information about the similarity between the two provides information about the similarity between the two sequences.sequences.
It is used in Radars, digital communication and mobile It is used in Radars, digital communication and mobile communication etc.communication etc.
The main application of correlation is that the The main application of correlation is that the incoming/received signal is correlated with standard incoming/received signal is correlated with standard signals and signal of this set which has maximum signals and signal of this set which has maximum correlation with the incoming/received signal is detected.correlation with the incoming/received signal is detected.
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Convolution of two sequencesIf xIf x11[n] [n] X X11(z) and x(z) and x22[n] [n] X X22(z) then (z) then
xx11[n]*x[n]*x22[n] = X[n] = X11(z)X(z)X22(z)(z)
Proof: Proof:
The convolution of xThe convolution of x11[n] and x[n] and x22[n] is defined as[n] is defined as
k
knxkxnxnxnx ][][][*][][ 2121
The z-transform of x[n] is
n
n
n21
n
n zknx]k[xz]n[x)z(X
Upon interchanging the order of the summationand applying the time shifting property, we obtain
zXzXz]k[xzXzknxkx)z(X 1k
2k
12n
n2
k1
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Example 8: Compute the convolutionof the signals x1[n] = [1, -2, 1] and
Solution:Solution:
XX11(z) = 1 – 2z(z) = 1 – 2z-1-1 + z + z-2-2
XX22(z) = 1 + z(z) = 1 + z-1-1 + z + z-2-2 + z + z-3-3 + z + z-4 -4 + z + z-5-5
Now X(z) = XNow X(z) = X11(z)X(z)X22(z) = 1 – z(z) = 1 – z-1-1 – z – z-6-6 + z + z-7-7
Hence x[n] = [1, -1, 0, 0, 0, 0, -1, 1] Hence x[n] = [1, -1, 0, 0, 0, 0, -1, 1]
Note: You should verify this result from the Note: You should verify this result from the definition of the convolution sum.definition of the convolution sum.
elsewhere,0
5n0,1]n[x2
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Exercise
Find the convolution Find the convolution of sequences?of sequences?
}1 ,2 ,1{ }2 ,3 ,1{ 21 xandx
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Correlation of two sequencesIf x1[n] X1(z) and x2[n] X2(z)
Proof:Proof:
)(z(z)XXm)(n(n)xxmr z
nxx
121 21)(
21
(2) )]([)()(
(1) )][()()(
:
2121
22
2121
21
nmxnxmxrx
ettion, we gabove equan)] in (m[m) as x(nthe term xArranging
mnxnxmxrx
(n)(n)x xsequences on of two correlatiing is theThe follow
n
n
)()()(
2
2121
21mxmxmxrx
written asm) can be ((n) and xxvolution ts the con) represenf Eq. ( the RHS oTherefore,
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Continue:
)()()]([
)()]([ )()]([
)]([)]([)]([
)]()([)]([
12121
12211
2121
2121
zXzXmxrxZ
zXmxZandzXmxZ
mxZmxZmxrxZ
mxmxZmxrxZ
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Correlation of two sequencesIf x1[n] X1(z) and x2[n] X2(z) then rx1x2[k] = X1(z)X2(z-1) Tutorial 4 Q2: Prove this property.Tutorial 4 Q2: Prove this property.
The Initial Value Theorem:If x[n] is causal then )z(Xlim]0[x
z
Proof:....z]2[xz]1[x]0[xz]n[x)z(X 21
0n
n
Obviously, as z , z-n 0 since n >0, this proves the theorem.
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Final Value TheoremIf x[n] X(z), then )z(Xz1lim][x 1
1z
Tutorial 4 Q3: Prove the Final Value Tutorial 4 Q3: Prove the Final Value TheoremTheorem
Example 9: Find the final value of
21
1
z8.0z8.11
z2)z(X
Solution: 21
111
z8.0z8.11
z2z1)z(Xz1
1
1
11
11
z5.01
z2
z5.01z1
z2z1
The final value theorem yields
102.0
2
z8.01
z2lim][y
1
1
1z
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Inverse z-transformIn general, the inverse z-transform may be In general, the inverse z-transform may be
found by using any of the following found by using any of the following methods:methods:
Power series methodPower series method Partial fraction methodPartial fraction method
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Power Series MethodExample 2: Determine the z-transform of Example 2: Determine the z-transform of
21 z5.0z5.11
1)z(X
By dividing the numerator of X(z) by its denominator, we obtain the power series
...zzzz1zz1
1 416313
8152
471
23
2211
23
x[n] = [1, 3/2, 7/2, 15/8, 31/16,…. ]
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Power Series MethodExample 2:Determine the z-transform of Example 2:Determine the z-transform of
21
1
zz22
z4)z(X
By dividing the numerator of X(z) by its denominator, we obtain the power series
x[n] = [2, 1.5, 0.5, 0.25, …..]
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Partial Fraction Method:Example 1: Find the signal corresponding to Example 1: Find the signal corresponding to
the z-transformthe z-transform
21
3
zz32
z)z(X
Solution: 5.0z1zz
5.0
z5.0z5.1z
5.0
zz32
z)z(X
2321
3
5.0z
4
1z
1
z
1
z
3
5.0z1zz
5.0
z
)z(X22
5.0z
z)4(
1z
z
z
13)z(X
or11
1
z5.01
14
z1
1z3)z(X
]n[u5.04]n[u]1n[]n[3]n[x n
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Partial Fraction Method:Example 2: Find the signal corresponding to the z-transformExample 2: Find the signal corresponding to the z-transform
211 z2.01z2.01
1)z(Y
Solution:
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2.0z2.0z
z)z(Y
22
2
2.0z
1.0
2.0z
75.0
2.0z
25.0
2.0z2.0z
z
z
)z(Y
22.0z
z1.0
2.0z
z75.0
1z
z25.0)z(Y
21
1
2.01.0
11z2.01
z2.0
z2.01
175.0
z2.01
125.0
]n[u2.0n5.0]n[u2.075.0]n[u2.025.0]n[y nnn
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Z-Transform Solution of Linear Difference Equations We can use z-transform to solve the difference We can use z-transform to solve the difference
equation that characterizes a causal, linear, time equation that characterizes a causal, linear, time invariant system. The following expressions are invariant system. The following expressions are especially useful to solve the difference especially useful to solve the difference equations:equations:
z[y[(n-1)T] = zz[y[(n-1)T] = z-1-1Y(z) +y[-T]Y(z) +y[-T] Z[y(n-2)T] = zZ[y(n-2)T] = z-2-2Y(z) + zY(z) + z-1-1y[-T] + y[-2T]y[-T] + y[-2T] Z[y(n-3)T] = zZ[y(n-3)T] = z-3-3Y(z) + zY(z) + z-2-2y[-T] + zy[-T] + z-1-1y[-2T] +y[-2T] +
y[-3T]y[-3T]
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Example: Consider the following difference equation:y[nT] –0.1y[(n-1)T] – 0.02y[(n-2)T] = 2x[nT] – x[(n-1)T]where the initial conditions are y[-T] = -10 and y[-2T] = 20. Y[nT] is the output and x[nT] is the unit step input.Solution:Solution:
Computing the z-transform of the difference Computing the z-transform of the difference equation givesequation gives
Y(z) – 0.1[zY(z) – 0.1[z-1-1Y(z) + y[-T]] – 0.02[zY(z) + y[-T]] – 0.02[z-2-2Y(z) + zY(z) + z-1-1y[-T] y[-T] + y[-2T]] = 2X(z) – z+ y[-2T]] = 2X(z) – z-1-1X(z) X(z)
Substituting the initial conditions we getSubstituting the initial conditions we getY(z) – 0.1zY(z) – 0.1z-1-1Y(z) +1 – 0.02zY(z) +1 – 0.02z-2-2Y(z) – 0.2zY(z) – 0.2z-1-1 –0.4 = –0.4 =
(2 – z(2 – z-1-1)X(z) )X(z)
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6.0z2.0z1
1z2)z(Yz02.0z1.01 1
1121
6.0z2.0z1
z2z02.0z2.01)z(Y 1
1
121
111
21
211
21
z1.01z2.01z1
z2.0z6.04.1
z02.0z1.01z1
z2.0z6.04.1)z(Y
1.0z2.0z1z
z2.0z6.0z4.1 23
1.0z
830.0
2.0z
567.0
1z
136.1
z
)z(Y
111 z1.01
1830.0
z2.01
1567.0
z1
1136.1)z(Y
and the output signal y[nT] is
]nT[u)1.0(830.0]nT[u)2.0(567.0]nT[u136.1]nT[y nn