Unformatted text preview: Chapter 2 SOME BASIC SEQUENCES Copyright c 1996 by Ali H. Sayed. All rights reserved. These notes are distributed only to the students attending the undergraduate DSP course EE113 in the Electrical Engineering Department at UCLA. The notes cannot be reproduced without written consent from the instructor: Prof. A. H. Sayed, Electrical Engineering Department, UCLA, CA 90095, [email protected] Complex numbers play a critical role in characterizing discretetime signals and discretetime systems. For example, it will be seen in later chapters that complex numbers are needed to describe the frequency content of a sequence and the frequency response of a system. Complex numbers are also needed to describe some basic sequences even in the time domain. For this reason, this chapter provides a brief review of complex numbers and explains how they are useful in describing some important sequences. Complex Numbers Every complex number has the form z = a + jb, where a and b are real numbers and √ j = −1. Every such number can be expressed in a socalled polar form, in terms of a magnitude factor and a phase factor. Speciﬁcally, z can be expressed as a + jb = Aejφ , −π ≤ φ ≤ π where the magnitude A is a positive real number that is computed as follows: A= √ a 2 + b2 while the phase φ is measured in radians and it lies within the interval [−π, π ]. The value of φ can be determined from the scalars {a, b}, with care, as follows. First, we determine the angle ψ ∈ − π , π whose arctan is given by 22 ψ = arctan b a . Then we may need to modify ψ by adding ±π to it depending on which quadrant in the complex plane the number z belongs to — see Fig. 2.1. Speciﬁcally, we obtain φ from ψ as 5 6 follows: ψ Some Basic Sequences Chapter 2 if z belongs to quadrants I or IV (i.e., if a > 0). if z belongs to quadrant III (i.e., if a < 0 and b < 0). if z belongs to quadrant II (i.e., if a < 0 and b > 0). φ= The correction to ψ (i.e., the addition of ±π ) is deﬁned in such a way that the resulting argument φ always lies within the interval [−π, π ]. ψ−π ψ+π Figure 2.1. Division of the complex plane into four quadrants. Consider the following example. Let z2 = −1 − j. √ Both complex numbers have the same magnitude, A = 2, and lead to the same phase angle, π ψ = arctan(1) = . 4 However, z1 and z2 are distinct numbers: one lies in quadrant I while the other lies in π quadrant III. The correct phase angle for z2 is therefore π − π = − 34 , so that we arrive at 4 √ √ 3π π z2 = 2 e−j 4 . z1 = 2 e j 4 , Basic Sequences The following list of sequences comes up time and again in the study of discretetime signals and systems. The reader is well advised to become familiar with the list. Some of the sequences are so common (like the unitsample sequence and the unitstep sequences) that they have their own standard notation. 1. The unitsample sequence is denoted by δ (n) and is deﬁned as δ (n) = 1 0 n=0 n=0 z1 = 1 + j, 7 In other words, the sequence is zero everywhere except at time n = 0 when it is equal to unity. The unitsample sequence is often referred to more simply as the impulse sequence. [As can be seen from this deﬁnition, the impulse signal in discretetime is far simpler to deﬁne than the impulse signal in continuoustime.] 2. The unitstep sequence is denoted by u(n) and is deﬁned as u(n) = 1 0 n≥0 n<0 In other words, the sequence is equal to unity for all nonnegative time instants. Fig. 2.2 plots the terms of the unitsample and unitstep sequences over the interval −10 ≤ n ≤ 10.
Unit−sample sequence 1 0.8 0.6 0.4 0.2 0 −10 −8 −6 −4 −2 0 n 2 4 6 8 10 Unit−step sequence 1 0.8 0.6 0.4 0.2 0 −10 −8 −6 −4 −2 0 n 2 4 6 8 10 Figure 2.2. The top plot shows the terms of the unitsample sequence δ (n) over the interval −10 ≤ n ≤ 10, while the bottom plot shows the terms of the unitstep sequence u(n) over the same interval of time. 3. The real exponential sequence is deﬁned by: x(n) = Aαn with A and α both realvalued. Fig. 2.3 plots the terms of two real exponential sequences over the interval −10 ≤ n ≤ 10. In one case, α is positive and equal to 0.8, while in the other case α is negative and equal to −0.8. 4. The sinusoidal sequences are deﬁned by x(n) = sin(ωo n + θo ), x(n) = cos(ωo n + θo ) 8
Real exponential sequence: 0.8n 3.5 3 2.5 2 1.5 1 0.5 0 −5 0 n Real exponential sequence: (−0.8)n 3 2 1 0 −1 −2 −3 −4 −5 0 n 5 5 Some Basic Sequences Chapter 2 10 10 Figure 2.3. The top plot shows the terms of the real exponential sequence 0.8n over the interval −10 ≤ n ≤ 10, while the bottom plot shows the terms of the real exponential sequence (−0.8)n over the same interval of time. for some realvalued quantities {ωo , θo }. Fig. 2.4 plots the terms of two sinusoidal sequences over the interval −10 ≤ n ≤ 10. In one case, ωo = π/5 and θo = 0, while in the other case ωo = π/5 and θo = π/3. 5. The complex exponential sequence is deﬁned by x(n) = Aαn where now either A or α is complexvalued. This sequence plays a fundamental role in the study of discretetime signals and systems. Usually, A will be a real positive scalar and α will be of the form α = ejωo , for some ωo . In this case, the expression for x(n) would reduce to x(n) = Aejωo n More generally, if both A and α happen to be complexvalued, say A = Aejθo , α = αejωo , ωo , θo ∈ [−π, π ] then the expression for x(n) = Aαn can be reexpressed in the equivalent form x(n) = A αn ej (ωo n+θo ) An important special case is the sequence x(n) = ejωo n 9
Sinusoidal sequence: sin( n/5) π 1 0.5 0 −0.5 −1 −10 −8 −6 −4 −2 0 n 2 4 6 8 10 Sinusoidal sequence: sin( n/5 + π/3) π 1 0.5 0 −0.5 −1 −10 −8 −6 −4 −2 0 n 2 4 6 8 10 Figure 2.4. The top plot shows the terms of the sinusoidal sequence sin(πn/5) over the interval −10 ≤ n ≤ 10, while the bottom plot shows the terms of the sinusoidal sequence sin( πn + π ) over the same interval of time. 5 3 which corresponds to the choices A = 1 and α = ejωo . Fig. 2.5 shows the magnitude and phase plots of the complex exponential sequence that corresponds to the choices A = ejπ/2 and α = 0.5ejπ/3 , i.e., to the special case x(n) = 0.5n ej ( over the interval −5 ≤ n ≤ 5.
π πn 3 +2) Figure 2.5. The top plot shows the amplitude andπ the phase components of πn the terms of the exponential sequence 0.5n ej ( 3 + 2 ) over the interval −5 ≤ n ≤ 5. It is seen that the amplitude in this example decays exponentially fast. 10 OneSided Sequences Consider the sequence x(n) = 1 2
n Some Basic Sequences Chapter 2 This is a real exponential sequence that is deﬁned for all values of n, −∞ < n < ∞. On the other hand, the sequence n 1 x(n) = u(n) 2 is again a real exponential sequence but one that has nonzero values only for n ≥ 0; it is zero for n < 0. This example illustrates a useful property of the step sequence u(n) — it allows us to deﬁne onesided sequences. Polar Plots A convenient way to plot complex exponential sequences is by means of a polar plot. For example, consider the sequence π x(n) = ej 4 n u(n) The ﬁrst 8 terms of this sequence, corresponding to time instants n = 0, 1, . . . , 7, are given by π 3π 5π 3π 7π π 1, ej 4 , ej 2 , ej 4 , −1, ej 4 , ej 2 , ej 4 All 8 terms are complex numbers with unit magnitude and, hence, they all lie on a circle of unit radius. Moreover, these terms are π radians apart on the unit circle. The arrow in 4 Fig. 2.6 indicates that the above 8 terms cover the circle in a counterclockwise direction. Fig. 2.7 shows the ﬁrst four terms of the alternative sequence x(n) = which are 1 2
n ej 4 n u(n) π 1 π 1 π 1 3π 1, ej 4 , ej 2 , ej 4 2 4 8 These values are again π radians apart but they now lie on cocentric circles of diminishing 4 radii. The outer circle has unit radius, the next one has radius equal to 1/2, the third one has radius equal to 1/4, the fourth one has radius equal to 1/8, and so on. Energy Sequences The energy of a sequence x(n) is deﬁned by Ex =
∆ ∞ n=−∞ x(n)2 When Ex < ∞, we say that the sequence is an energy sequence. In other words, energy sequences have ﬁnite energy. For example, the step sequence, x(n) = u(n), is not an energy sequence since
∞ ∞ n=−∞ u(n)2 = n=0 1 → ∞. 9
onesided sequences. Polar Plots
A convenient way to plot complex exponential sequences is by means of a polar plot. For example, consider the sequence π x(n) = ej 4 n u(n) The ﬁrst 8 terms of this sequence, corresponding to time instants n = 0, 1, . . . , 7, are given by 1, ej 4 , ej 2 , ej
π π 3π 4 , −1, ej 5π 4 , ej 3π 2 , ej 7π 4 All 8 terms are complex numbers with unit magnitude and, hence, they all lie on a circle of unit 1 radius. Moreover, these terms are π radians apart on the unit circle. The arrow in Fig. 2.??? 1 4 indicates that the above 8 terms cover the circle in a counterclockwise direction. n=2 n=1 n=3 π 4 n=0 n=4 n=7 n=5 n=6 Figure 2.6. the ﬁrst four the ﬁrst terms of the sequence x( Fig. 2.??????? shows A polar plot of terms of 8the alternative sequencen) = ej 4 n u(n). π x(n) = 1 2 n ej 4 n u(n) π On the are which other hand, the sequence 1 π 1 π 1 3π 1, ej 4 , ej 2 ,n ej 4 2 41 8 These values are again π radians apart but they 2 now lie on cocentric circles of diminishing radii. 4 The outer circle has unit radius, the next one has radius equal to 1/2, the third one has radius is equal to 1/4,sequence since has radius equal to 1/8, and so on. an energy the fourth one x(n) = u(n) Energy Sequences ∞ n=−∞ ∞ x(n)2 = n=0 1 4 n = 1 1− 1 4 = 4 < ∞. 3
1n 4 In this second calculation, we used the fact that the values { a geometric series with ratio 1/4. , n ≥ 0} are the terms of Power Sequences The average power of a sequence x(n) is deﬁned by Px = lim
∆ N 1 x(n)2 2N + 1 n=−N N →∞ That is, we compute the energy of the sequence x(n) over a symmetric interval −N ≤ n ≤ N , normalize the result by the number of terms in this interval, which is 2N + 1, and then evaluate the limit as the size of the interval increases indeﬁnitely. 12
10 Some Basic Sequences
Some Basic Sequences Chapter 2 Chapter 2 n=2 n=3 n=1 n=0 Figure 2.7. A polar plot of the ﬁrst 8 terms of the sequence x(n) = The energy of a sequence x(n) is deﬁned by 1 n jπn e 4 u(n). 2 Ex = ∆ ∞ n=−∞ x(n)2 When Ex < ∞,< ∞, we say that the is an energy sequence. In other words, energy sequences power When Px we say that the sequence sequence is a power sequence. In other words, have ﬁnite energy. For sequences have ﬁniteexample, the step sequence, x(n) = u(n), is not sequence,sequence since), is not power. For example, although the step an energy x(n) = u(n ∞ an energy sequence it is nevertheless a power∞ sequence since, for any N , 2 1 On the other hand, the 2N + 1 sequence n=−∞ N u(n) = n=0 1 → ∞. n=−N u(n)2 =
1 2 x(n) = 1 N +1 1= 2N + 1 n=0 2N + 1 n
u(n) N so that the limit as N → ∞ is Px = 1/2. is an energy sequence since
∞ Signal Transformations−∞ n= x(n)2 = ∞ n=0 1 4 n = 1 1− 1 4 = 4 < ∞. 3 In this second calculation, we in the fact that versions { 4 Sequences will often appearusedtransformed the values and it, n ≥ 0} arethattermsreader become is useful the the of a geometric series with ratio 1/4. acquainted with the following common signal transformations. If x(n) is a given sequence, then 1n Power Sequences The average power of a sequence x(n) is deﬁned by vertical axis. 1. x(−n) corresponds to a sequence that is obtained from x(n) by reﬂecting it about the
N 1 ∆ 2. x(n − 1) corresponds to a sequence that is obtained from x(n) by shifting it by one Px = lim x(n)2 N →∞ 2N + 1 n=−N unit of time to the right. 3. x(n + 1) corresponds to a sequence that is obtained from x(n) by shifting it by one unit of time to the left. 13 4. x(−n + 1) corresponds to a sequence that is obtained from x(n) by ﬁrst reﬂecting x(n) about the vertical axis and then shifting it to the right by one unit of time. 5. x(−n − 1) corresponds to a sequence that is obtained from x(n) by ﬁrst reﬂecting it about the vertical axis and then shifting it to the left by one unit of time. x(n) 1 0.5 1 0.5 x(−n) −5 1 0.5 x(n) 5 −5 1 0.5 x(n−1) 5 −5 1 0.5 x(n) 5 −5 1 0.5 x(n+1) 5 −5 1 0.5 x(n) 5 −5 1 0.5 x(−n+1) 5 −5 1 0.5 x(n) 5 −5 1 0.5 x(−n−1) 5 −5 5 −6 −4 −2 2 4 Figure 2.8. Plots of various signal transformations of the sequence x(n) = 0.6n u(n). The plot for x(n) shows only the samples of x(n) over the interval 0 ≤ n ≤ 5. Plotting Sequences Consider a sequence x(n) and deﬁne a new sequence y (n) that is obtained by modifying the argument of x(n) as follows: y (n) = x(an + b) for some integer values of a and b. The signal transformations of the previous section correspond to particular choices of a and b. More generally, for arbitrary integers a and b, this is how we can obtain the plot of y (n) from the plot of x(n): 1. The value of y (n) at n = 0 is simply the value of x at time b, x(b). That is, we set n = 0 in an + b and arrive at the argument b. 2. The value of y (n) at n = 1 is simply the value of x at time a + b, x(a + b). That is, we set n = 1 in an + b and arrive at the argument a + b. 14 Some Basic Sequences Chapter 2 3. The value of y (n) at n = −11 is simply the value of x at time −a + b, x(−a + b). That is, we set n − 1 in an + b and arrive at the argument −a + b. 4. And so forth. In more formal terms, we can describe the procedure of generating the plot of y (n) from the plot of x(n) as follows: 1. We draw the n axis. 2. We deﬁne a new variable m = an + b and draw a new axis mapping the values of n to values of m. We also determine the orientation of this m axis. 3. We plot x(m) versus m. This is the same plot as x(n) versus n except that it is done on the m axis. 4. We ﬁnally replace the horizontal and vertical axes in the m domain by the horizontal and vertical axes in the n axis. We only keep the samples that are deﬁned for the n time instants. 13 PROBLEMS
PROBLEMS 15 Problem 2.1: Plot the sequence x(n) = δ (n + 1) +
Problem 2.1: Plot the sequence x(n) = δ (n + 1) + 1n 2 Problem 2.2: Express the complex exponential sequence √ Problem 2.2: Express the complex exponential sequence 1 √ 3 + jn x(n) = (1 − j ) 3 2 j1 2 + x(n) = (1 − j )
2 2 1n 2 u(n − 3).
n u(n − 3). in polar form and plot its terms at the time instants n = −1, 0, 1.
in polar form and plot its terms at the time instants n = −1, 0, 1. Problem 2.3: 2.3: The samples of asequence xxnn) are zero except attime instantsinstants shown in Problem The samples of a sequence ( ( ) are zero except at the the time shown in Fig. 2.xx???. The amplitudes of of the nonzero samples are ,either 1, 2, the 3. Plot the )sequence h(n) Fig. 2.9. The amplitudes the nonzero samples are either 1 2, or 3. Plot or sequence h(n that that is is deﬁned by deﬁned 33 1 1 h(h() ) = 2x(n + 2) −− δ (n) n)u+ u(n .− 3). n n = x( n + 2) 2 δ ( + (n − 3) 2 2 Plot also the sequences x(3n − 2), x(−2n + 3), and x(−2n − 1). Plot also the sequences x(3n − 2), x(−2n + 3), and x(−2n − 1). x(n) 3 2 1 −2 −1 1 2 3 4 5 6 n Figure 2.9. The sequence x(n) deﬁned in Prob. 2.3. Problem 2.4: Plot in polar coordinates the terms of the sequence the energy and average power of this sequence? √
Problem 2.4: Plot in polar coordinates the terms of the sequence Problem energy Consider the sequencesequence? the 2.5: and average power of this
2 4 √ 2 4
2 4 −j
n √ 2 4 n u(n). What is −j √ u(n). What is 1 Problem 2.5: Consider the sequence x(n) = 3 (a) Find its energy.
x(n) = 1 3 n
n u(n − 1) +
u(n − 1) + 1 2 1 2−1 n n−1 u(n − 2) u(n − 2) (a) n) = x energy. (b) Let y (Find its (−2n − 3). For what values of n is y (n) zero? Problem 2.6: Consider the sequence (b) Let y (n) = x(−2n − 3). For what values of n is y (n) zero?
n−1 Problem 2.6: Consider the sequence x(n) = 1 + 1 (a) Find its power. 1 n−1 u(n − 2) u(n − 2) x(n) = 1 + 2 2 (b) Let y (n) = x(−2n + 3). For what values of n is y (n) zero? 16
(a) Find its power. (b) Let y (n) = x(−2n + 3). For what values of n is y (n) zero? Some Basic Sequences Chapter 2 Problem 2.7: The samples of ejπn x(1 − n) are {−1, 2, 1 , 0, −2, 3, 1}, where the box denotes the origin of times (i.e., time n = 0). Samples to the right of the box occur at positive time instants while samples to the left of the box occur at negative time instants. Samples outside the speciﬁed interval are all zero. Which terms of x(n) can you determine from this information? Problem 2.8: Given a sequence x(n), we perform the following three operations: (a) We plot the sequence y (n) = x(n − 1) and then scale the time axis and plot y (2n). (b) We plot the sequence z (n) = x(n − 2) and then scale the time axis and plot z (2n). Which procedure results in the right plot for x(2n − 2)? How would you modify the wrong procedure(s)? Problem 2.9: Consider the sequence x(n) =
5 2 (c) We plot the sequence w(n) = x(2n) and then shift it and plot w(n − 2). 1 7 15 31 63 127 ,− , , , , , ,... 4 8 16 32 64 128 where the box denotes the origin of times (i.e., time n = 0). Express x(n) in terms of the sequences n δ (n), u(n), and 1 . 2 Problem 2.10: Answer True or False. In each case, either prove your answer or give a counterexample. (a) A power sequence is necessarily an energy sequence. (b) Every energy sequence has zero average power. (d) If x(n) is an energy sequence then x(n) → 0 as n → ∞. (c) The sequence x(n) = 1/(n + 1), n ≥ 0, is an energy sequence. (e) There does not exist a sequence with inﬁnite average power. (f) The sum of two energy sequences, {x(n) = x1 (n) + x2 (n)}, is an energy sequence. Problem 2.11: Give examples of nonzero sequences x(n) such that (b) x(n)x(−n + 1) = 0 for n = 0, 1, 2. (c) x(n) + x∗ (n) = cos
π (n 3 (a) x(n) and x(n − 2) are identical. − 1) . Here the symbol ∗ denotes complex conjugation. Problem 2.12: Which of the following statements is False? (a) All energy signals are power signals. (b) Some energy signals are power signals. (c) All power signals are energy signals. (d) Some power signals are energy signals. Problem 2.13: How can you construct x(−n + 3) from x(n)? (a) First reﬂect x(n) about the vertical axis and then shift it three units of time to the left. (b) First shift x(n) three units of time to the left and then reﬂect the resulting signal about the vertical axis. (c) First shift x(n) three units of time to the right and then reﬂect the resulting signal about the vertical axis. 17
(d) Both parts (a) and (b) are correct. Problem 2.14: In order to reconstruct x(n), it is enough to know which of the following signals? (a) x(2n) and x(n2 ). (b) x(−2n + 3) and x(2n + 4). (d) Both parts (a) and (b) are correct. Problem 2.15: Which of the following identities is incorrect? (b) δ (n) = δ (5n). (a) δ (3n − 6) = δ (−3n + 6). (c) δ (5n − 1) = δ (4n + 3). (c) x(2n + 3) and x(2n − 1). (d) δ (5n − 1) = δ (5(n − 1)). ...
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 Spring '08
 Walker
 Electrical Engineering, Digital Signal Processing, Signal Processing, Complex number, ej

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