SOLUTIONS MANUAL Digital Signal Processing: A Computer-Based Approach Third Edition PDF

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SOLUTIONS MANUAL to accompany Digital Signal Processing: A Computer-Based Approach Third Edition Sanjit K. Mitra Prepared by Chowdary Adsumilli, John Berger, Marco Carli, Hsin-Han Ho, Rajeev Gandhi, Chin Kaye Koh, Luca Lucchese, and Mylene Queiroz de Farias Not for sale. 1 Chapter 2 2.1 (a) x1 1 = 2...

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SOLUTIONS MANUAL to accompany

Digital Signal Processing: A Computer-Based Approach Third Edition

Sanjit K. Mitra

Prepared by Chowdary Adsumilli, John Berger, Marco Carli, Hsin-Han Ho, Rajeev Gandhi, Chin Kaye Koh, Luca Lucchese, and Mylene Queiroz de Farias

Not for sale.

1

Chapter 2 2.1

(a) x1 1 = 22.85, x1 2 = 9.1396, x1 ∞ = 4.81, (b) x 2 1 = 18.68, x 2 2 = 7.1944, x 2 ∞ = 3.48.

2.2

1, n ≥ 0, 1, n < 0, µ[ n] = ⎧⎨ Hence, µ[ −n − 1] = ⎧⎨ Thus, x[ n] = µ[ n] + µ[ − n − 1]. ⎩0, n < 0. ⎩0, n ≥ 0.

2.3

(a) Consider the sequence defined by x[ n] = ∑ δ[ k ]. If n < 0, then k = 0 is not included

n

k = −∞

in the sum and hence, x[n] = 0 for n < 0. On the other hand, for n ≥ 0, k = 0 is included in the sum, and as a result, x[n] =1 for n ≥ 0. Therefore, n 1, n ≥ 0, x[ n] = ∑ δ[ k ] = ⎧⎨ = µ[ n]. 0 ⎩ , n < 0, k = −∞

2.4

2.5

1, n ≥ 0, 1, n ≥ 1, it follows that µ[ n − 1] = ⎧⎨ Hence, (b) Since µ[ n] = ⎧⎨ 0 , n 0 , < ⎩ ⎩0, n < 1. 1, n = 0, µ[ n] − µ[ n − 1] = ⎧⎨ = δ[ n]. ⎩0, n ≠ 0, Recall µ[ n] − µ[ n − 1] = δ[ n]. Hence, x[ n] = δ[ n] + 3δ[ n − 1] − 2δ[ n − 2] + 4δ[ n − 3] = (µ[ n] − µ[ n − 1]) + 3(µ[ n − 1] − µ[ n − 2]) − 2(µ[ n − 2] − µ[ n − 3]) + 4(µ[ n − 3] − µ[ n − 4]) = µ[ n] + 2µ[ n − 1] − 5µ[ n − 2] + 6µ[ n − 3] − 4µ[ n − 4]. (a) c[ n] = x[ − n + 2] = {2 0 − 3 − 2 1 5 − 4}, ↑

(b) d[ n] = y[ − n − 3] = {− 2 7 8 0 − 1 − 3 6 0 0}, ↑

(c) e[ n] = w[ − n] = {5 − 2 0 − 1 2 2 3 0 0}, ↑

(d) u[ n] = x[ n] + y[ n − 2] = {− 4 5 1 − 2 3 − 3 1 0 8 7 − 2}, ↑

(e) v[ n] = x[ n] ⋅ w[ n + 4] = {0 15 2 − 4 3 0 − 4 0}, ↑

(f) s[ n] = y[ n] − w[ n + 4] = {− 3 4 − 5 0 0 10 2 − 2}, ↑

(g) r[ n] = 3.5 y[ n] = {21 − 10.5 − 3.5 0 2.8 24.5 − 7}. ↑

2.6

(a) x[ n] = −4δ[ n + 3] + 5δ[ n + 2] + δ[ n + 1] − 2δ[ n] − 3δ[ n − 1] + 2δ[ n − 3], y[ n] = 6δ[ n + 1] − 3δ[ n] − δ[ n − 1] + 8δ[ n − 3] + 7δ[ n − 4] − 2δ[ n − 5], w[ n] = 3δ[ n − 2] + 2δ[ n − 3] + 2δ[ n − 4] − δ[ n − 5] − 2δ[ n − 7] + 5δ[ n − 8], (b) Recall δ[ n] = µ[ n] − µ[ n − 1]. Hence, x[ n] = −4(µ[ n + 3] − µ[ n + 2]) + 5(µ[ n + 2] − µ[ n + 1]) + (µ[ n + 1] − µ[ n]) − 2(µ[ n] − µ[ n − 1]) − 3(µ[ n − 1] − µ[ n − 2]) + 2(µ[ n − 3] − µ[ n − 4])

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2

= −4µ[ n + 3] + 9µ[ n + 2] − 4µ[ n + 1] − 3µ[ n] − µ[ n − 1] + 3µ[ n − 2] + 2µ[ n − 3] − 2µ[ n − 4],

2.7

(a) x[n-1]

_ z 1

x[n]

x[n-2]

_ z 1

h[2]

h[1]

h[0]

y[n]

+

+

From the above figure it follows that y[ n] = h[ 0] x[ n] + h[1] x[ n − 1] + h[2] x[ n − 2]. (b) x[n]

h[0]

w[n]

+ _1

+

y[n]

_1

z

z

β11

x[n _ 1]

+

_ z 1

+

_ z 1

β 21

x[n _ 2]

β12

w[n _ 1]

w[n _ 2]

β 22

From the above figure we get w[ n] = h[0]( x[ n] + β11 x[ n − 1] + β 21 x[ n − 2]) and y[ n] = w[ n] + β12 w[ n − 1] + β 22 w[ n − 2]. Making use of the first equation in the second we arrive at y[ n] = h[0]( x[ n] + β11 x[ n − 1] + β 21 x[ n − 2]) + β12 h[0]( x[ n − 1] + β11 x[ n − 2] + β 21 x[ n − 3]) + β 22 h[0]( x[ n − 2] + β11 x[ n − 3] + β 21 x[ n − 4]) = h[0]( x[ n] + (β11 + β12 ) x[ n − 1] + (β 21 + β12 β11 + β 22 ) x[ n − 2] + (β12 β 21 + β 22 β11 ) x[ n − 3] + β 22 β 21 x[ n − 4]). (c) Figure P2.1(c) is a cascade of a first-order section and a second-order section. The input-output relation remains unchanged if the ordering of the two sections is interchanged as shown below. x[n]

0.6

w[n]

+ _

+ _ z 1

0.8

u[n]

+

y[n+1] _ z 1

0.3

+

+

0.4

y[n]

w[n _ 1] _ z 1 _ 0.5

0.2

w[n _ 2]

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3

The second-order section can be redrawn as shown below without changing its inputoutput relation. x[n]

0.6

w[n]

+ _

+

+

+

y[n+1] _ z 1

_ _ z 1 z 1 0.8 w[n _ 1]

0.3

0.4

y[n]

+

_ z 1

_ z 1 _ 0.5

u[n]

0.2

w[n _2]

The second-order section can be seen to be cascade of two sections. Interchanging their ordering we finally arrive at the structure shown below: 0.6

+

x[n]

s[n]

_ z 1

_

0.3 x[n _ 1]

u[n]

+ 0.8

_ z 1

x[n _ 2]

y[n+1] _ z 1

_ z 1

+

+

+

0.4

u[n _ 1]

y[n]

_ z 1 _ 0.5

0.2

u[n _ 2]

Analyzing the above structure we arrive at s[ n] = 0.6 x[ n] + 0.3 x[ n − 1] + 0.2 x[ n − 2], u[ n] = s[ n] − 0.8u[ n − 1] − 0.5u[ n − 2], y[ n + 1] = u[ n] + 0.4 y[ n]. From u[ n] = y[ n + 1] − 0.4 y[ n]. Substituting this in the second equation we get after some algebra y[ n + 1] = s[ n] − 0.4 y[ n] − 0.18 y[ n − 1] + 0.8 y[ n − 2]. Making use of the first equation in this equation we finally arrive at the desired input-output relation y[ n] + 0.4 y[ n − 1] + 0.18 y[ n − 2] − 0.2 y[ n − 3] = 0.6 x[ n − 1] + 0.3 x[ n − 2] + 0.2 x[ n − 3]. (d) Figure P2.19(d) is a parallel connection of a first-order section and a second-order section. The second-order section can be redrawn as a cascade of two sections as indicated below: x[n]

w[n]

+

+

_ _ z 1 z 1 _ 0.8 w[n _ 1]

_ z 1 _ 0.5

+

y 2[n]

_ z 1

w[n _2]

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0.3

0.2

4

Interchanging the order of the two sections we arrive at an equivalent structure shown below: 0.3

x[n]

_1

+

z

q[n]

y 2[n]

+

_ z 1

_

0.2

_ z 1

0.8

+

y 2 [n _ 1] _ z 1

_ 0.5

y 2[n _ 2]

Analyzing the above structure we get q[ n] = 0.3 x[ n − 1] + 0.2 x[ n − 2], y2 [ n] = q[ n] − 0.8 y2 [ n − 1] − 0.5 y2 [ n − 2]. Substituting the first equation in the second we have y2 [ n] + 0.8 y2 [ n − 1] + 0.5 y2 [ n − 2] = 0.3 x[ n − 1] + 0.2 x[ n − 2].

(2-1)

Analyzing the first-order section of Figure P2.1(d) given below u[n] _ 1 u[n _ 1] 0.6 y 1 [n] z x[n] + 0.4

we get

u[ n] = x[ n] + 0.4u[ n − 1], y1[ n] = 0.6u[ n − 1]. Solving the above two equations we have y1[ n] − 0.4 y1[ n − 1] = 0.6 x[ n − 1]. (2-2) The output y[n] of the structure of Figure P2.19(d) is given by y[ n] = y1[ n] + y2 [ n]. (2-3) From Eq. (2-2) we get 0.8 y1[ n − 1] − 0.32 y1[ n − 2] = 0.48 x[ n − 2] and 0.5 y1[ n − 2] − 0.2 y1[ n − 3] = 0.3 x[ n − 3]. Adding the last two equations to Eq. (2-2) we arrive at y1[ n] + 0.4 y1[ n − 1] + 0.18 y1[ n − 2] − 0.2 y1[ n − 3] (2-4) = 0.6 x[ n − 1] + 0.48 x[ n − 2] + 0.3 x[ n − 3]. Similarly, from Eq. (2-1) we get − 0.4 y2 [ n − 1] − 0.32 y2 [ n − 2] − 0.2 y2 [ n − 3] = −0.12 x[ n − 2] − 0.08 x[ n − 3]. Adding this equation to Eq. (2-1) we arrive at y2 [ n] + 0.4 y2 [ n − 1] + 0.18 y2 [ n − 2] − 0.2 y2 [ n − 3] (2-5) = 0.3 x[ n − 1] + 0.08 x[ n − 2] − 0.08 x[ n − 3]. Adding Eqs. (2-4) and (2-5), and making use of Eq. (2-3) we finally arrive at the inputoutput relation of Figure P2.1(d) as: y[ n] + 0.4 y[ n − 1] + 0.18 y[ n − 2] − 0.2 y[ n − 3] = 0.9 x[ n − 1] + 0.56 x[ n − 2] + 0.22 x[ n − 3]. Not for sale.

5

2.8

(a) x1* [ n] = {1 − j 4 − 2 − j 5 3 + j 2 − 7 − j 3 − 1 − j}, ↑

x1* [ −n] = {− 1 − j − 7 − j 3 3 + j 2 − 2 − j 5 1 − j 4}. Therefore ↑

x1, cs [n] =

1 2

x1, ca [n] =

1 2

(x [n] + x [−n]) = { j1.5

− 4.5 + j 3 − 4.5 − j − j1.5},

* 1

1

↑

(x1[n] − x1*[−n]) = {1 + j2.5

(b) x 2 [ n] = e

jπn / 3

. Hence, x 2* [ n] = e

(

2.5 + j 4 − j 2 − 2.5 + j 4 − 1 + j 2.5}. ↑ − jπn / 3

)

and thus, x 2* [ −n] = e jπn / 3 = x 2 [ n].

Therefore, x2, cs [n] = x2 [n] + x2* [−n] = e j 2 πn / 3 = x2 [n], and 2 1

x2, ca [n] =

1 2

(x2 [n] − x2* [−n]) = 0.

(c) x 3 [ n] = j e − jπn / 5 . Hence, x 3* [ n] = − j e jπn / 5 and thus,

(

)

x 3* [ − n] = − j e − jπn / 5 = − x 3 [ n]. Therefore, x3, cs [n] = x3 [n] + x3*[− n] = 0, and 2 1

x3, ca [n] = 2.9

1 2

(x3[n] − x3*[−n]) = x3[n] = j e − jπn / 5 .

(a) x[ n] = {− 4 5 1 − 2 − 3 0 2}. Hence, x[ − n] = {2 0 − 3 − 2 1 5 − 4}. Therefore, x ev [ n] =

↑ 1 ( x[ n] + 2

↑

x[ − n]) =

1 {− 2 2

5 − 2 − 4 − 2 5 − 2} ↑

= {− 1 2.5 − 1 − 2 − 1 2.5 − 1}

and x od [ n] =

↑ 1 1 ( x[ n] − x[ −n]) = {− 6 2 2

5 4 0 − 4 − 5 6} ↑

= {− 3 2.5 2 0 − 2 − 2.5 3}. ↑

(b) y[ n] = {0 0 0 0 6 − 3 − 1 0 8 7 − 2}. Hence, ↑

y[ − n] = {− 2 7 8 0 − 1 − 3 6 0 0 0 0}. ↑

Therefore, yev [ n] = and y od [ n] =

1 ( y[ n] + 2

y[ − n]) = {− 1 3.5 4 0 2.5 − 3 2.5 0 4 3.5 − 1} ↑

1 ( y[ n] − y[ −n]) = {1 2

− 3.5 − 4 0 3.5 0 − 3.5 0 4 3.5 − 1}. ↑

(c) w[ n] = {0 0 0 0 0 0 0 0 0 0 3 2 2 − 1 0 − 2 5}. Hence, ↑

w[ − n] = {5 − 2 0 − 1 2 2 3 0 0 0 0 0 0 0 0 0 0}. Therefore ↑

wev [ n] =

1 (w[ n] + w[ − n]) 2

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6

= {2.5 − 1 0 − 0.5 1 1 1.5 0 0 0 1.5 1 1 − 0.5 0 − 1 2.5} and ↑

wod [ n] =

1 (w[ n] − w[ − n]) 2

= {− 2.5 1 0 0.5 − 1 − 1 − 1.5 0 0 0 1.5 1 1 − 0.5 0 − 1 2.5}. ↑

2.10

(a) x1[ n] = µ[ n + 2]. Hence, x1[ −n] = µ[ −n + 2]. Therefore, n ≥ 3, ⎧⎪1 / 2, 1 x1, ev [ n] = (µ[ n + 2] + µ[ − n + 2]) = ⎨ 1, − 2 ≤ n ≤ 2, and 2 ⎪⎩1 / 2, − 3 ≤ n, n ≥ 3, ⎧⎪ 1 / 2, 1 − 2 ≤ n ≤ 2, x1, od [ n] = (µ[ n + 2] − µ[ − n + 2]) = ⎨ 0, 2 ⎪⎩− 1 / 2, − 3 ≤ n. (b) x 2 [ n] = α n µ[ n − 3]. Hence, x 2 [ −n] = α − n µ[ −n − 3]. Therefore,

⎧ 1 αn, n ≥ 3, ⎪⎪ 2 n −n α µ[ n − 3] + α µ[ −n − 3] = ⎨ 0, − 2 ≤ n ≤ 2, and 1 − n ⎪ α , − 3 ≤ n, ⎪⎩ 2 ⎧ 1 αn, n ≥ 3, ⎪⎪ 2 n −n α µ[ n − 3] − α µ[ −n − 3] = ⎨ 0, − 2 ≤ n ≤ 2, 1 ⎪− α − n , − 3 ≤ n. ⎪⎩ 2

x 2, ev [ n] =

1 2

(

)

x 2, od [ n] =

1 2

(

)

(c) x 3 [ n] = n α n µ[ n]. Hence, x 3 [ − n] = − n α − n µ[ − n]. Therefore,

( (

) )

1 n 1 n α n µ[ n] + ( − n) α − n µ[ − n] = n α 2 2 1 n 1 x 3, od [ n] = n α n µ[ n] − ( − n) α − n µ[ − n] = n α . 2 2

x 3, ev [ n] =

n

(d) x 4 [ n] = α . Hence, x 4 [ −n] = α 1

1

1

1

−n

=α

n

and

= x 4 [ n]. Therefore, n

x 4, ev [ n] = ( x 4 [ n] + x 4 [ − n]) = ( x 4 [ n] + x 4 [ n]) = x 4 [ n] = α 2 2

and

x 4, od [ n] = ( x 4 [ n] − x 4 [ −n]) = ( x 4 [ n] − x 4 [ n]) = 0. 2 2 2.11 x ev [ n] = (x[ n] + x[ − n]). Thus, x ev [ −n] = (x[ − n] + x[ n]) = x ev [ n]. Hence, x ev [n] is 2 2 1

1

an even sequence. Likewise, x od [ n] = (x[ n] − x[ − n]). Thus, 2 1

x od [ − n] =

1 2

(x[ −n] − x[ n]) = − xod [ n].

Hence, x od [n] is an odd sequence.

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7

2.12 (a) g[ n] = x ev [ n]x ev [ n]. Thus, g[ −n] = x ev [ − n]x ev [ − n] = x ev [ n]x ev [ n] = g[ n]. Hence, g[n] is an even sequence. (b) u[ n] = x ev [ n]x od [ n]. Thus, u[ − n] = x ev [ − n]x od [ − n] = x ev [ n](− x od [ n]) = −u[ n]. Hence, u[n] is an odd sequence. (c) v[ n] = x od [ n]x od [ n]. Thus, v[ −n] = x od [ − n]x od [ − n] = (− x od [ n])(− x od [ n]) = x od [ n]x od [ n] = v[ n]. Hence, v[n] is an even sequence. 2.13 (a) Since x[n] is causal, x[ n] = 0, n < 0. Also, x[ − n] = 0, n > 0. Now, Hence, x ev [0] = (x[0] + x[0]) = x[0] and 2 ⎧2 x ev [ n], n > 0, ⎪ 1 x ev [ n] = x[ n], n > 0. Combining the two equations we get x[ n] = ⎨ x ev [ n], n = 0, 2 ⎪⎩ 0, n < 0.

x ev [ n] =

1 2

(x[ n] + x[ −n]).

Likewise, x od [ n] =

x od [ n] =

1 2

1 2

1

(x[ n] − x[−n]).

Hence, x od [0] =

1 2

(x[0] − x[0]) = 0

and

2 x [ n], n > 0, x[ n], n > 0. Combining the two equations we get x[ n] = ⎧⎨ ev n ≤ 0. ⎩ 0,

(b) Since y[n] is causal, y[ n] = 0, n < 0. Also, y[ − n] = 0, n > 0. Let y[ n] = yre [ n] + jyim [ n], where yre [n] and yim [n] are real causal sequences. 1

(

)

1

(

)

Now, yca [ n] = y[ n] − y ∗ [ − n] . Hence, yca [0] = y[0] − y ∗ [0] = jyim [0] and 2 2

yca [ n] =

1 y[ n], n 2

> 0. Since yre [0] is not known, y[n] cannot be fully recovered from

yca [n] . 1

(

)

1

(

)

Likewise, ycs [ n] = y[ n] + y ∗ [ − n] . Hence, ycs [0] = y[0] + y ∗ [0] = yre [0] and 2 2

ycs [ n] =

1 y[ n], n 2

> 0. Since yim [ 0] is not known, y[n] cannot be fully recovered from

ycs [n] . 2.14 Since x[n] is causal, x[ n] = 0, n < 0. From the solution of Problem 2.13 we have ⎧2 x ev [ n], n > 0, ⎧2 cos(ωo n), n > 0, ⎪ ⎪ 1 n = 0, = 2 cos(ωo n)µ[ n] − δ[ n]. x[ n] = ⎨ x ev [ n], n = 0, = ⎨ 0 , n < 0, ⎪⎩ 0, ⎪ n < 0, ⎩ 2.15 (a) {x[ n]} = {Aα n } where A and α are complex numbers with α < 1. Since for

n < 0, α

n

can become arbitrarily large, {x[ n]} is not a bounded sequence.

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⎧ n (b) y[ n] = Aα n µ[ n] = ⎨ Aα , n ≥ 0, where A and α are complex numbers with n < 0, ⎩ 0, n

α < 1. Here, α ≤ 1, n ≥ 0. Hence y[ n] ≤ A for all values of n. Hence, {y[ n]} is a bounded sequence. (c) {h[ n]} = Cβ n µ[ n] where C and β are complex numbers with β > 1. Since for

n > 0, β

n

can become arbitrarily large, {h[ n]} is not a bounded sequence.

(d) {g[ n]} = 4 cos(ωo n). Since g[ n] ≤ 4 for all values of n,{g[ n]} is a bounded sequence. 1 ⎧⎛ 1 1 ⎪⎜1 − 2 ⎞⎟, n ≥ 1, (e) v[ n] = ⎨⎝ n ⎠ Since 2 < 1 for n > 1 and 2 = 1 for n = 1, v[ n] < 1 for n n ⎪⎩ 0, n ≤ 0. all values of n. Thus {v[ n]} is a bounded sequence. ∞ ∞ ( −1) n +1 ∞ 1 (−1) n +1 µ[ n − 1]. Now ∑ x[ n] = ∑ = ∑ = ∞. Hence {x[ n]} is 2.16 x[ n] = n n n = −∞ n =1 n =1 n not absolutely summable.

∞

2.17 (a) x1[ n] = α n µ[ n − 1]. Now

∑

n = −∞

∞

∞

n =1

n =1

x 2 [ n] = ∑ α n = ∑ α

n

=

α 1− α

< ∞ , since

α < 1. Hence, {x1[ n]} is absolutely summable. ∞

(b) x 2 [ n] = α n µ[ n − 1]. Now

α

2

∑

n = −∞

∞

∞

n =1

n =1

x 2 [ n] = ∑ nα n = ∑ n α

n

α

=

(1 − α ) 2

< ∞, since

< 1. Hence, {x 2 [ n]} is absolutely summable.

(c) x 3 [ n] = n 2 α n µ[ n − 1]. Now 2

3

∞

∑

n = −∞

∞

∞

n =1

n =1

x 3 [ n] = ∑ n 2 α n = ∑ n 2 α

n

4

= α + 2 2 α + 32 α + 4 2 α + K 2

3

4

2

3

4

3

4

5

= ( α + α + α + α + K) + 3( α + α + α + K) + 5( α + α + α + K)

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9

4

5

6

+ 7( α + α + α + K) =

= =

α 1− α

⎛ ∞ 1 n⎞ ⎜⎜ ∑ (2n − 1) α ⎟⎟ = ⎠ 1− α ⎝ n =1

1 1− α

α (1 + α ) (1 − α ) 3

+

3α

2

1− α

+

5α

3

7α

+

1− α

4

1− α

+K

∞ ⎛ ∞ 1 n n⎞ ⎜⎜ 2 ∑ n α − ∑ α ⎟⎟ = n =1 ⎠ 1− α ⎝ n =1

⎛ 2α α ⎜ − ⎜ (1 − α ) 2 1 − α ⎝

⎞ ⎟ ⎟ ⎠

< ∞. Hence, {x 3 [ n]} is absolutely summable.

∞ ∞ 1 ∞ 1 1 1 2.18 (a) x a [ n] = = ∑ = = 2 < ∞. Hence, µ[ n]. Now ∑ x a [ n] = ∑ n n 1 n = −∞ n=0 2 n=0 2 2n 1− 2

{x a [ n]} is absolutely summable. ∞ ∞ 1 1 µ[ n]. Now ∑ x b [ n] = ∑ (n + 1)(n + 2) n = −∞ n = 0 (n + 1)(n + 2) ∞ ⎛ 1 1 ⎞ ⎛ 1⎞ ⎛1 1⎞ ⎛1 1⎞ ⎛1 1⎞ = ∑ ⎜ − ⎟ = ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + K = 1 < ∞. Hence, 2⎠ ⎝2 3⎠ ⎝3 4⎠ ⎝ 4 5⎠ n=0 ⎝ n + 1 n + 2 ⎠ ⎝ {x b [ n]} is absolutely summable.

(b) x b [ n] =

2.19 (a) A sequence x[n] is absolutely summable if

∞

∑

n = −∞

x[ n] < ∞. By Schwartz inequality

∞

⎞⎛ ∞ ⎞ 2 ⎛ ∞ x[ n] ≤ ⎜⎜ ∑ x[ n] ⎟⎟⎜⎜ ∑ x[ n] ⎟⎟ < ∞. Hence, an absolutely summable n = −∞ ⎝ n = −∞ ⎠⎝ n = −∞ ⎠ sequence is square summable and has thus finite energy.

we have

∑

1 n

Now consider the sequence x[ n] = µ[ n − 1]. The convergence of an infinite series can be shown via the integral test. Let a n = f (x ), where a continuous, positive and ∞

∞

decreasing function is for all x ≥ 1. Then the series ∑ a n and the integral ∫ f ( x )dx n =1

1

1

1 ∞

∞

1

both converge or both diverge. For a n = , f ( x ) = . But ∫ dx = (ln x ) 1 = ∞ − 0 = ∞ . n x x 1 ∞

Hence,

∑

n = −∞

∞ 1 n =1 n

x[ n] = ∑

1 n

does not converge. As a result, x[ n] = µ[ n − 1] is not

absolutely summable.

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10

(b) To show that {x[ n]} is square-summable, we observe that here a n = ∞

1 n2

, and thus,

∞

∞ 1 1 ⎛ 1⎞ f ( x ) = 2 . Now, ∫ 2 dx = ⎜ − ⎟ = − + 1 = 1. Hence, ∑ 2 converges, or in other ∞ x ⎝ x ⎠1 n =1 n 1x

1

1

1 n

words, x[ n] = µ[ n − 1] is square-summable. 2.20 See Problem 2.19, Part (a) solution. 2

∞ ∞ ⎛ cos ω n ⎞ ∞ cos ωc n 1 2 c µ[ n − 1]. Now, ∑ x 2 [ n] = ∑ ⎜ 2.21 x 2 [ n] = . Since, ⎟ ≤ ∑ 2 2 πn n = −∞ n =1 ⎝ πn ⎠ n =1 π n 2 π 2 ∞ ⎛ cos ωc n ⎞ 1 , ∑ ⎜ ⎟ ≤ . Therefore x2 [ n] is square-summable. ∑ 2 = 6 n =1 ⎝ πn ⎠ 6 n =1 n

∞

1

Using the integral test (See Problem 2.19, Part (a) solution) we now show that x2 [ n] is

cos ωc x ∞ cos ω x πx 1 c ⋅ cos int(ωc x ) dx = ⋅ not absolutely summable. Now, ∫ πx π cos ωc x 1

∞

where 1

∞ cos ω x c

cos int is the cosine integral function. Since ∫

πx diverges. Hence, x2 [ n] is not absolutely summable.

∞

cos ωc n ...