signals and systems discrete time -...

Dr. Mahmoud M. Al-Husari

Signals and SystemsDiscrete Time

This set of lecture notes are never to be considered as a substituteto the textbook recommended by the lecturer.

Contents

7 Discrete Time Signals and Systems 17.1 Exponential Sequences . . . . . . . . . . . . . . . . . . . . . . . . 17.2 Discrete-Time Sinusoids . . . . . . . . . . . . . . . . . . . . . . . 37.3 Fourier Analysis of Discrete Time Systems . . . . . . . . . . . . . 4

7.3.1 The Discrete Time Fourier Series (DTFS) . . . . . . . . . 47.3.2 The Discrete Time Fourier Transform . . . . . . . . . . . 57.3.3 Fourier Transform of DT Periodic Sequences . . . . . . . 97.3.4 Properties of the DTFT . . . . . . . . . . . . . . . . . . . 9

7.4 Relationship between CTFT and DTFT . . . . . . . . . . . . . . 107.5 The Discrete Fourier Transform . . . . . . . . . . . . . . . . . . . 137.6 Examples of the Use of The DFT . . . . . . . . . . . . . . . . . . 16

iii

iv CONTENTS

Chapter 7Discrete Time Signals andSystems

In the preceding chapters, we discussed mainly techniques for the analysis ofcontinuous time signals and systems. In Chapter 3 we did some system analysisfor discrete time systems, concepts such as impulse response and discrete timeconvolution was introduced. In Chapter 2 we introduced elementary discretetime signals such as the delta function, unit step function, energy and powersignals, we now discuss exponentials, sine and cosine waveforms.

7.1 Exponential Sequences

A discrete time exponential signal is given by

x[n] = Cαn (7.1)

where, in general, C and α are complex numbers. The exponential term αn canalso be expressed as

αn = eβn (α = eβ or β = lnα) (7.2)

For example, 5n = e1.609n because 5 = e1.609. Similarly, e3n = (e3)n =(20.086)n. We now consider different cases of the discerete exponential sig-nal in (7.1).

For C and α real, x[n] increases with increasing n if |α| > 1. Similarly, if |α| < 1,we have a decreasing exponential. Suppose β in (7.2) is purely imaginary, sothat we have

x[n] = CejΩn

The exponential ejΩn is plotted in the complex plane for various values of n, asillustrated in Figure 7.1 The function x[n] = ejΩn takes on values ej0, ejΩ, ej2Ω,ej3Ω, · · · at n = 0, 1, 2, 3, · · · , respectively. Note that

ejΩn = rejθ, r = 1, and θ = nΩ

1

2 CHAPTER 7. DISCRETE TIME SIGNALS AND SYSTEMS

Locus of ejΩn

n = 3

n = 2

n = 0

n = 1

Figure 7.1: Locus of ejΩn.

This fact shows that the magnitude and angle of ejΩn are 1 and nΩ, respectively.Therefore, the points ej0, ejΩ, ej2Ω, ej3Ω, · · · , ejΩn, · · · lie on a circle of unitradius at angles 0, Ω, 2Ω, 3Ω, · · · , nΩ, · · · respectively. For each unit increasein n, the function x[n] = ejΩn moves along the unit circle counterclockwise byan angle Ω. Therefore, the locus of ejΩn may be viewed as a phasor rotatingcounterclockwise at a uniform speed of Ω radians per unit sample interval.

This signal is closely related to the discrete time sinusoids using Euler’s formula,

ejΩn = cos Ωn+ j sin Ωn

If we assume n is dimensionless, then the frequency of ejΩn is Ω and has theunits of radians. We recall that the continuous time signal x(t) = ejωt is periodicwith period T = 2π/ω for any ω. The interseting question here: is a discretetime exponential, ejΩn, always periodic for any Ω. A discrete time signal x[n]is said to be periodic if

x[n] = x[n+N ]

Therefore, for the case x[n] = ejΩn we must have

ejΩn = ejΩ(n+N) = ejΩnejΩN

This result is possible only ifejΩN = 1

For this to hold, ΩN must be an integer multiple of 2π, so that

ΩN = 2πm m = 0,±1,±2, · · ·

orΩ

2π=m

N

Because both m and N are integers, thus, x[n] is periodic only if Ω/2π is arational number. To compute the fundamental period N0, we must choosethe smallest value of m that will make m(2π/Ω) an integer. For example, ifΩ = 4π/17, then the smallest value of m that will make

m2π

Ω= m

17

2

7.2. DISCRETE-TIME SINUSOIDS 3

an integer is 2. Therefore

N0 = m2π

Ω= 2

17

2= 17

Let Example 7.1x[n] = ej(

7π9 )n

Find the fundamental period.

Solution First we haveΩ

2π=m

N=

7

18

Thus, the sequence is periodic and the fundamental period, obtained by choosingm = 7, is given by N = 18.

Let Example 7.2x[n] = ej(

79 )n

We haveΩ

2π=

7

18π

which is not rational. Thus the sequence is not periodic.

7.2 Discrete-Time Sinusoids

A continuous-time sinusoid cosωt has a unique waveform for every value of ω inthe range 0 tp∞. Increasing ω results in a sinusoid of ever increasing frequency.Such is not the case for the discrete-time sinusoid cos Ωn because

cos(Ω± 2πk)n = cos(Ωn± 2πkn)

TO BE CONTINUED

Figure 7.2: Highest oscillation rate in a discrete-time sinusoid occurs at Ω = π.


7.3 Fourier Analysis of Discrete Time Systems

In chapters 4, 5, and 6, we studied the ways to representing a continuous timesignal as a sum of sinusoids or exponentials. In this section we shall discusssimilar development for discrete time signals. We first represent a periodic x[n]as a Fourier series formed by a discrete time exponential and its harmonics.

7.3.1 The Discrete Time Fourier Series (DTFS)

A continuous time periodic signal of period T can be represented using theexponential form of the Fourier series consisting of exponentials ej0t, e±jω0t,e±j2ω0t, e±j3ω0t, · · · where ω0 = 2π/T is the fundamental frequency. In analogywith our representation of periodic signals in continuous time, we can expectthat we can obtain such a representation in terms of the exponentials ej0n,e±jΩ0n, e±j2Ω0n, · · · , e±jkΩ0n, · · · , and so on. That is we seek a representationfor x[n], of the form

x[n] =

∞∑k=−∞

akejkΩ0n (7.3)

It would appear that we need infinite number of harmonics, however it turnsout that we have to include only N terms in the summation on the right side in(7.3). The reason being that discrete time exponentials of frequencies seperatedby integer multiples of 2π are identical as discussed next.

Let xk[n] define the set of functions

xk[n] = ejkΩ0n k = 0,±1,±2, · · ·

with Ω0 = 2π/N . Therefore, the set is

· · · , e−jΩ0n, 1, ejΩ0n, ej2Ω0n, · · · , ejkΩ0n, · · · , ej(k+N)Ω0n, · · ·

where ejkΩ0n is the kth harmonic to the fundamental signal ejΩ0n. Consider the(k +N)th harmonic

xk+N [n] = ej(k+N)Ω0n = ejkΩ0nejNΩ0n = ejkΩ0nej2πn = xk[n]

Here we have used the fact that ej2πn = 1. Hence there are only N distinctwaveforms in the set given above. These correspond to the frequencies Ωk = kΩ0

for k = 0, 1, · · · , N − 1. The first harmonic is identical to the N + 1 harmonic,the second hamrmonic is identical to the N + 2 harmonic, and so on. Theconsequence of this result on the Fourier series representation of a discrete timeperiodic signal x[n] in (7.3) is that the summation is only needed over anyN consecutive values of k, for example k = 0, 1, 2, · · · , N − 1. Therefore, therepresentation for x[n] can be written as

x[n] =∑

k=<N>

akejkΩ0n (7.4)

where k =< N > indicates the range of summation of any N succesive integers.

The coefficients ak in the Fourier series (7.4) can be computed using (given herewithout proof)

ak =1

N

∑n=<N>

x[n]e−jkΩ0n (7.5)

7.3. FOURIER ANALYSIS OF DISCRETE TIME SYSTEMS 5

Since xk[n] = xk+N [n], it is clear that

ak+N = ak

and if x[n] is a real periodic signal then

ak = a∗−k+N

Let x[n] be a periodic extenstion of the sequence Example 7.3

2,−1, 1, 2

Determine the Fourier coefficients ak.

Solution The period is N = 4, so that Ω0 = π/2, from (7.5) we have

ak =1

4

3∑n=0

x[n]e−jk(π/2)n

Hence,

a0 =1

4

3∑n=0

x[n] =1

4(2− 1 + 1 + 2) = 1

a1 =1

4

3∑n=0

x[n]e−j(π/2)n =1

4

3∑n=0

x[n](−j)n

a2 =1

4

3∑n=0

x[n]e−j2(π/2)n =1

4

3∑n=0

x[n](−1)n

a3 =1

4

3∑n=0

x[n]e−j3(π/2)n

TO BE CONTINUED

7.3.2 The Discrete Time Fourier Transform

We now consider the frequency domain representation of discrete-time signalsthat are not necessarily periodic. For continuous-time signals, we obtained sucha representation by defining the Fourier transform of a signal x(t) as

X(ω) = F [x(t)] =

∫ ∞−∞

x(t)e−jωtdt

For discrete-time signals, we consider an analogous definition of the Fouriertransform as

X(Ω) = F [x[n]] =

∞∑n=−∞

x[n]e−jΩn (7.6)

It is important to note here that X(Ω) is periodic with period 2π. Accordingto (7.6) it follows that

X(Ω + 2π) =

∞∑n=−∞

x[n]e−j(Ω+2π)n

=

∞∑n=−∞

e−jΩne−j2πn = X(Ω)


As a consequence, we have to consider values of Ω only over the range [0, 2π].Furthermore, Ω is continuous, this fact makes the spectrum of X(Ω) continuousand periodic function of Ω with period 2π.

To find the inverse relation between X(Ω) and x[n], we replace the variable nin (7.6) by m to get

X(Ω) =

∞∑m=−∞

x[m]e−jΩm

Now multiply both sides by ejΩn and integrate over the range [0, 2π] to get∫2π

X(Ω)ejΩndΩ =

∫2π

∞∑m=−∞

x[m]ejΩ(n−m)dΩ

Interchanging the orders of summation and integration then gives∫2π

X(Ω)ejΩndΩ =

∞∑m=−∞

x[m]

∫2π

ejΩ(n−m)dΩ

It can be verified that ∫2π

ejΩ(n−m)dΩ =

2π, n = m

0, n 6= m

We can, therefore, write

x[n] =1

2π

∫2π

X(Ω)ejΩndΩ

We consider a few examples.

Find the DTFT ofExample 7.4x[n] = αnu[n], |α| < 1

Solution For this example

X(Ω) =

∞∑n=0

αne−jΩn

=

∞∑n=0

(αe−jΩ

)nThis is a geometric progression with a common ration αe−jΩ. Therefore,

X(Ω) =1

1− αe−jΩ=

1

1− α cos Ω + jα sin Ω

provided that |αe−jΩ| < 1 which is the case since |α| < 1. The magnitude isgiven by

|X(Ω)| = 1√(1− α cos Ω)2 + (α sin Ω)

2

=1√

1 + α2 − 2α cos Ω


and the phase by

∠X(Ω) = − tan−1

[α sin Ω

1− α cos Ω

]Figure 7.3 shows x[n] = αnu[n] and its spectra for α = 0.8. Observe that the

n

|X(Ω)|

∠X(Ω)

x[n]

Figure 7.3: Fourier spectra of Emaple 5.4.

frequency spectra are continuous and periodic functions of Ω with the period2π. The magnitude spectrum is an even function and the phase specturm is anodd function of Ω.

Consider the function Example 7.5

x[n] = δ[n]

Solution Its Fourier transform is

X(Ω) =

∞∑n=−∞

δ[n]e−jΩn = 1

and the Fourier transform of δ[n− k] is e−jkΩ.

Let Example 7.6

x[n] = α|n|, |α| < 1


Solution We obtain its Fourier transform as

X(Ω) =

∞∑n=−∞

α|n|e−jΩn

=

−1∑n=−∞

α−ne−jΩn +

∞∑n=0

αne−jΩn

= −1 +

∞∑n=0

(αejΩ

)n+

∞∑n=0

(αe−jΩ

)n=

1− α2

1− 2α cos Ω + α2

In this case, X(Ω) is real, so that the phase is zero.

Find the inverse Fourier transform of the rectangular pulse spectrum shown inExample 7.7Figure 7.4.

n

|X(Ω)|

x[n]

Figure 7.4: Inverse DTFT of a periodic rect spectrum 5.7.

Solution Using

x[n] =1

2π

∫2π

X(Ω)ejΩndΩ

=1

2π

∫ π/4

−π/4ejΩndΩ

=1

j2πnejΩn

∣∣∣∣π/4−π/4

=sin(π4n)

πn=

1

4sinc

(n4

)


7.3.3 Fourier Transform of DT Periodic Sequences

For continuous-time periodic signals, we obtained the Fourier transform of aperiodic signal x(t) as

X(ω) =

∞∑n=−∞

2πcnδ(ω − nω0)

For discrete-time periodic signals, we consider an analogus definition of theFourier transform as

X(Ω) =

∞∑k=−∞

2πakδ(Ω− kΩ0)

Find the discrete-time Fourier transform for the periodic signal Example 7.8

x[n] = 2 cos

(3π

8n+

π

3

)+ 4 sin

(π2n)

Solution First we need to determine the DTFS. The fundamental frequencyis Ω0 = π/8. Therefore, only the third and fourth harmonics exist in the DTFSspectrum. Using Euler identities we deduce

ak =

−2/j, k = −4

e−jπ/3, k = −3

ejπ/3, k = 3

2/j, k = 4

0, otherwise for −7 ≤ k ≤ 8

Hence the DTFT over one period is

X(Ω) = −4π

jδ(

Ω +π

2

)+2πe−jπ/3δ

(Ω +

3π

8

)+2πejπ/3δ

(Ω− 3π

8

)+

4π

jδ(

Ω− π

2

)for π < Ω ≤ π.

7.3.4 Properties of the DTFT

The properties of the discrete-time Fourier transform closely parallel those ofthe continuous-time transform. In this section, we consider some of the moreuseful properties.

Periodicity

We saw that the discrete-time Fourier transform is periodic in Ω with period2π, so that

X(Ω + 2π) = X(Ω)

Linearity

Let x1[n] and x2[n] be two sequences with Fourier transforms X1(Ω) and X2(Ω),respectively. Then

F [α1x1[n] + α2x2[n]] = α1X1(Ω) + α2X2(Ω)


Time and Frequency Shifting

It can be easily ahown that If

x[n]F←→ X(Ω)

thenx[n− n0]

F←→ e−jΩn0X(Ω)

andejΩ0nx[n]

F←→ X(Ω− Ω0)

Differentiation in Frequency

nx[n]F←→ j

dX(Ω)

dΩ

Find the discrete-time Fourier transform for the signal x[n] = nαnu[n], withExample 7.9|α| < 1.

Solution Then, by using the results in Example 5.4, we can write

X(Ω) = jd

dΩF [αnu[n]] = j

d

dΩ

1

1− αe−jΩ

=αe−jΩ

(1− αe−jΩ)2

Convolution

x[n] ∗ h[n]F←→ X(Ω)H(Ω)

Multiplication

x[n].h[n]F←→ 1

2πX(Ω) ∗H(Ω)

Parseval’s Theorem

∞∑n=−∞

[x[n]]2 =1

2π

∫2π

|X(Ω)|2dΩ

7.4 Relationship between CTFT and DTFT

The sampling theorem was originally presented in Chapter 6. We showed thatsampling is equivalent to amplitude modulation of a periodic impulse train. Thesignal being sampled, x(t), multiplies the impulse train δT (t) to produce a newsignal xs(t),

xs(t) = x(t)δT (t)

The periodic impulse train samples the signal x(t) at the times at which theimpulses occur. The resulting continuous-time signal xs(t) is an impulse trainwhere the strengths of the impulses are the values of x(t) at the times equal

7.4. RELATIONSHIP BETWEEN CTFT AND DTFT 11

xs(t) = x(t)δT (t)

Figure 7.5: (a) Signal x(t) (solid line) and modulated impulse train xs(t), and (b)sequence of samples x[n] = x(nT ).

to nT . This is shown in Figure 7.5(a). If we extracted the sample values fromthe impulses and arranged them in a sequence, we would have the discrete-timesequence of samples x[n] = x(nT ), shown in Figure 7.5(b).

In Section 6.3 we derived a frequency-domain representation of the sampledsignal xs(t) using the multiplication property as

Xs(ω) =1

T

∞∑n=−∞

X(ω − nωs) (7.7)

where ωs = 2π/T . An alternative form to X(ω) in (7.7) could be obtained usingthe shifting property of Fourier transforms. The sampled signal xs(t) can bewritten as

xs(t) = x(t)δT (t)

= x(t)

∞∑n=−∞

δ(t− nT )

=

∞∑n=−∞

x(t)δ(t− nT )

=

∞∑n=−∞

x(nT )δ(t− nT )


Using the shifting property the Fourier transform of the signal xs(t) is

xs(t) =

∞∑n=−∞

x(nT )δ(t− nT )F←→ Xs(ω) =

∞∑n=−∞

x(nT )e−jωnT

which can be written as

Xs(ω) =

∞∑n=−∞

x[n]e−jωnT (7.8)

Clearly the frequency-domain represention Xs(ω) in (7.7) should be equivalentto the one in (7.8), that is

∞∑n=−∞

x[n]e−jωnT =1

T

∞∑n=−∞

X(ω − nωs)

Using the definition of the DTFT in (7.6), we can state the following fundamen-tal relation between the DTFT of a sequence of samples x[n] = x(nT ) and thecontinuous-time Fourier transform of x(t) in sampling:

X(Ω) = Xs(ω)

∣∣∣∣ω=

ΩT

Thus, X(Ω) can be obtained from Xs(ω) by replacing ω with Ω/T . Therefore,X(Ω) is identical to Xs(ω) frequency scaled by a factor T , as shown in Figure7.6.

x(t)

xs(t)

x[n]

X(ω)

Xs(ω)

X(Ω)

n

Figure 7.6: Connection between DTFT and CTFT

7.5. THE DISCRETE FOURIER TRANSFORM 13

7.5 The Discrete Fourier Transform

In order to motivate the study of the DFT, let us assume we are interestedin finding the Fourier transform of a continuous-time signal x(t) using a digitalcomputer. Since a digital computer can only store and manipulate a finite set ofnumbers, it is necessary to represent x(t) by a finite set of values. The first stepin doing so is to sample the signal to obtain a discrete sequence xs[n]. Since thesignal may not be time-limited, the next step is to obtain a finite set of samplesof the discrete sequence by a process of truncation. Without loss of generality,we can assume that these samples are defined for n in the range [0, N − 1]. Letus denote this finite sequence by x[n].

Since we now have a discrete sequence, we can take the discrete-time Fouriertransform of the sequence as

X(Ω) =

N−1∑n=0

x[n]e−jΩn

This in still not in a form suitable for digital computation, sinc Ω is a continuousvariable taking values in the range [0, 2π]. Therefore, we need to evaluate X(Ω)at only a finite number of values, Ωk, by a process of sampling uniformly in therange [0, 2π] as

X(Ωk) =

N−1∑n=0

x[n]e−jΩkn, k = 0, 1, . . . ,M − 1 (7.9)

where

Ωk =2π

Mk

The number of frequency samples, M , can be of any value. However, we chooseit to be the same as the number of time samples, N . With this modification,and writing X(Ωk) as X[k], we finally have

X[k] =

N−1∑n=0

x[n]e−j2πN nk (7.10)

Equation (7.10) is known as the Discrete Fourier Transform. We saw thatX(Ω) is periodic in Ω with period 2π, so that X(Ωk) = X(Ωk + 2π). This canbe written as

X[k] = X(Ωk) = X(Ωk + 2π) = X

(2π

N(k +N)

)= X[k +N ]

That is, X[k] is periodic with period N . Often, the length-N DFT sequence isreferred to as the N -point DFT.

The inverse discrete Fourier transform (IDFT) relation is given by

x[n] =1

N

N−1∑n=0

X[k]ej2πN nk (7.11)


It can be shown that x[n] as determined from this equation is also periodic withperiod N ,

x[n+N ] =1

N

N−1∑n=0

X[k]ej2πN (n+N)k

=1

N

N−1∑n=0

X[k]ej2πN nk

= x[n]

That is, the IDFT operation yields a periodic sequence, of which only the firstN values, corresponding to one period, are evaluated. Thus, in all operationsinvolving DFT and the IDFT, we are effectively replacing finite sequence x[n]by its periodic extension, xp[n], as shown in Figure 7.7. One would expect thereis a connection between the Fourier series expansion of periodic discrete-timesequences discussed earlier and the DFT. A connection explored next.

DFT and DTFS, Any Connection?

Comparision of Equations (7.4) and (7.5) with Equations (7.10) and (7.11) showsthat DFT X[k] of finite sequence x[n] can be interpreted as the coefficients akin the Fourier series representation of its periodic extension multiplied by theperiod N . In fact the two can be made identical by including the factor 1/Nwith the DFT rather than with the IDFT. Many authors and textbooks considerboth to be the same.

The different forms of the Fourier representation of discrete-time functions areclosely related to sampling. We saw earlier as shown in Figure 7.6, time-samplinga function produces a periodic specturm. On the other hand, periodic repeti-tions of a signal amounts to sampling its spectrum as depicted in Figure 7.7.The dual of the time-sampling is the frequency-sampling.

Suppose that x[n] = 1, 2, 2, 1, and x[n] = 0 for all other n. Compute the DFTExample 7.10of x[n].

x[n]

xp[n]

X(Ω)

Ω

Ω

n

n

X(Ω)

NN

Ωk

Figure 7.7: Periodic repetition of a signal results in sampling and periodic repetitonof its spectrum.

7.5. THE DISCRETE FOURIER TRANSFORM 15

Solution With N = 4 (i.e., 4-point DFT), from (7.10)

X[k] =

3∑n=0

= x[n]e−jπ2 nk, k = 0, 1, 2, 3

= x[0] + x[1]e−jπk/2 + x[2]e−jπk + x[3]e−jπ3k/2, k = 0, 1, 2, 3

= 1 + 2e−jπk/2 + 2e−jπk + e−jπ3k/2, k = 0, 1, 2, 3

Hence,

X[0] = 1 + 2 + 2 + 1 = 6

X[1] = 1 + 2e−jπ/2 + 2e−jπ + e−jπ3/2

= 1− j2− 2 + j = −1− j

X[2] = 1 + 2e−jπ + 2e−j2π + e−j3π

= 1− 2 + 2− 1 = 0

X[3] = 1 + 2e−jπ3/2 + 2e−j3π + e−jπ9/2

= 1 + j2− 2− j = −1 + j

Again consider the signal in Example 7.10 with the 4-point DFT given by Example 7.11

X[k] =

6, k = 0

−1− j, k = 1

0, k = 2

−1 + j, k = 3

Compute the Inverse DFT to determine x[n].

Solution From (7.11)

x[n] =1

4

3∑n=0

X[k]ejπ2 nk

=1

4

[X[0] +X[1]ejπn/2 +X[2]ejπn +X[3]ejπ3n/2

]Thus,

x[0] =1

4[X[0] +X[1] +X[2] +X[3]] = 1

x[1] =1

4[X[0] + jX[1]−X[2]− jX[3]] =

1

4[8] = 2

x[2] =1

4[X[0]−X[1] +X[2]−X[3]] = 2

x[3] =1

4[X[0]− jX[1]−X[2] + jX[3]] =

1

4[4] = 1


7.6 Examples of the Use of The DFT

The primary application of the DFT is to approximate the Fourier transform ofsignals. To demonstate the use of DFT, consider the function

x(t) = rect

(t− 1

2

)We choose a sampling frequency of ωs = 10π rad/sec or 5 Hz to obtain a discrete-time sequence of samples x[n] a shown in Figure 7.8(a). We choose to use 16samples for the computation. The 16-point DFT is shown in Figure 7.8(b).Because the DFT is calculated from discrete-time samples, we must multiply∣∣X[k]

∣∣ by the sampling period T to cancel ot the factor of 1/T inherent to theFourier transform of sampled signals.

x[n]

∣∣X[k]∣∣

Figure 7.8: A sampled rectangular pulse and its 16-point DFT.

7.6. EXAMPLES OF THE USE OF THE DFT 17

Figure 7.9 shows a comparision of the Fourier transform approximation foundby computing the DFT and the actual Fourier transform.

Figure 7.9: Comparision of the DFT with the Fourier transform.

Bibliography

[1] Roberts, M. J., Signals and Systems: Analysis using Transform Meth-ods and MATLABR©. International edition, McGraw-Hill 2003.

[2] Haykin, S., and Van Veen, B., Signals and Systems, New York, JohnWiley & Sons, 1999.

[3] Philips, C., Parr, J., and Riskin, E., Signals, Systems, and Transforms,Upper Saddle River, NJ, Prentice-Hall, 2008.

[4] Stremler, F., Introduction to Communication Systems, Reading, MAAddison-Wesley 1990.

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