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Lecture 5 – 6

Z - Transform

ByByDileep 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 the A frequency domain representation can also include information on the

phasephase shift that must be applied to each sinusoid in order to be able to shift that must be applied to each sinusoid in order to be able to recombine the frequency components to recover the original time recombine the frequency components to recover the original time signal. signal.

The frequency domain relates to the Fourier transform or Fourier series The frequency domain relates to the Fourier transform or Fourier series by decomposing a function into an infinite or finite number of 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 infinitely waveform can be expressed as a sum of sinusoids (sometimes infinitely many.)many.)

In using the Laplace, Z-, or Fourier transforms, the frequency spectrum In using the Laplace, Z-, or Fourier transforms, the frequency spectrum is complex and describes the frequency magnitude and phase. In many is complex and describes the frequency magnitude and phase. In many applications, phase information is not important. By discarding the applications, phase information is not important. By discarding the phase information it is possible to simplify the information in a phase information it is possible to simplify the information in a frequency domain representation to generate a frequency spectrum or frequency domain representation to generate a frequency spectrum or spectral density. A spectrum analyser is a device that displays the spectral density. A spectrum analyser is a device that displays the spectrum.spectrum.

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The Direct Z-TransformThe z-transform of a discrete time signal is defined as the power seriesThe z-transform of a discrete time signal is defined as the power series (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 byX(z) = Z{x[n]}X(z) = Z{x[n]}

Since the z-transform is an infinite series, it exists only for those values Since the z-transform is an infinite series, it exists only for those values of z for which this series converges. The Region of Convergence 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 this series (ROC) of X(z) is the set of all values of z for which this series converges.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 > 0Solution: 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 > 0Solution: 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

z211

1

.......z21z

211

z21z

21

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:|ROCaz11

.......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)

thenthenaa11xx11[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:

11nn

1n

z3114

z2113]34)2(3[z

az11]]n[ua[z

Example 4: Determine the z-transform of the signal (cosw0n)u[n]

20

10

1

1jw1jw0

njwnjw0

zwcosz21wcosz1

ze11

21

ze11

21]n[unwcosz

e21e

21]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

z11.......zz1

zz]n[u]]n[u[z

1

N

1N

1

z1z1

z11z

z11]]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 signalaann(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 zwcosz21wcosz1]n[u)nw[cos(z

220

10

1

0 cos21cos1]][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

az1az

az11

dzdz]]n[una[z

az11]]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

mxZmxZmxrxZmxmxZmxrxZ

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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

z8.0z8.11z2)z(X

Solution: 21

111

z8.0z8.11z2z1)z(Xz1

1

1

11

11

z5.01z2

z5.01z1z2z1

The final value theorem yields

102.0

2z8.01

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 z5.0z5.111)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

zz22z4)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

zz32z)z(X

Solution: 5.0z1zz

5.0z5.0z5.1z

5.0zz32

z)z(X 2321

3

5.0z4

1z1

z1

z3

5.0z1zz5.0

z)z(X

22

5.0zz)4(

1zz

z13)z(X

or11

1

z5.0114

z11z3)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 z2.01z2.011)z(Y

Solution:

23

2.0z2.0zz)z(Y

22

2

2.0z1.0

2.0z75.0

2.0z25.0

2.0z2.0zz

z)z(Y

22.0zz1.0

2.0zz75.0

1zz25.0)z(Y

21

1

2.01.0

11 z2.01z2.0

z2.01175.0

z2.01125.0

]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 givesY(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 11

121

6.0z2.0z1z2z02.0z2.01)z(Y 1

1

121

111

21

211

21

z1.01z2.01z1z2.0z6.04.1

z02.0z1.01z1z2.0z6.04.1)z(Y

1.0z2.0z1zz2.0z6.0z4.1 23

1.0z830.0

2.0z567.0

1z136.1

z)z(Y

111 z1.011830.0

z2.011567.0

z11136.1)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