Digital Signal Processing (Solution Manual) - 3rd Edition by Mitra

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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,

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Chapter 2

2.1 (a) 22.85, 9.1396, ₁ 4.81, 2 1 1 1 = x = x _∞ = x (b) 18.68, 7.1944, ₂ 3.48. 2 2 1 2 = x = x _∞ = x 2.2 Hence, Thus, ⎩ ⎨ ⎧ < ≥ = µ . 0 , 0 , 0 , 1 ] [ n n n ⎩ ⎨ ⎧ ≥ < = − − µ . 0 , 0 , 0 , 1 ] 1 [ n n n x[n]=µ[n]+µ[−n−1].

2.3 (a) Consider the sequence defined by If n < 0, then k = 0 is not included in the sum and hence, x[n] = 0 for n < 0. On the other hand, for k = 0 is included in the sum, and as a result, x[n] =1 for Therefore,

. ] [ ] [ = ∑ δ −∞ = n k k n x , 0 ≥ n . 0 ≥ n ]. [ , 0 , 0 , 0 , 1 ] [ ] [ n n n k n x n k µ = ⎩ ⎨ ⎧ < ≥ = ∑ δ = −∞ =

(b) Since it follows that Hence,

⎩ ⎨ ⎧ < ≥ = µ , 0 , 0 , 0 , 1 ] [ n n n ⎩ ⎨ ⎧ <≥ = − µ . 1 , 0 , 1 , 1 ] 1 [ n n n ]. [ , 0 , 0 , 0 , 1 ] 1 [ ] [ n n n n n =δ ⎩ ⎨ ⎧ ≠ = = − µ − µ 2.4 Recall µ[n]−µ[n−1]=δ[n]. Hence, ] 3 [ 4 ] 2 [ 2 ] 1 [ 3 ] [ ] [n =δ n + δn− − δn− + δ n− x ]) 4 [ ] 3 [ ( 4 ]) 3 [ ] 2 [ ( 2 ]) 2 [ ] 1 [ ( 3 ]) 1 [ ] [ (µ −µ − + µ − −µ − − µ − −µ − + µ − −µ − = n n n n n n n n ]. 4 [ 4 ] 3 [ 6 ] 2 [ 5 ] 1 [ 2 ] [ + µ − − µ − + µ − − µ − µ = n n n n n 2.5 (a) [ ]= [− +2]={2 0 −3 −2 1 5 −4}, ↑ n x n c (b) [ ] [ 3] { 2 7 8 0 1 3 6 0 0}, ↑ − − − = − − = y n n d (c) [ ] [ ] {5 2 0 1 2 2 3 0 0}, ↑ − − = − =w n n e (d) [ ]= [ ]+ [ −2]={−4 5 1 −2 3 −3 1 0 8 7 −2}, ↑ n y n x n u (e) [ ]= [ ]⋅ [ +4]={0 15 2 −4 3 0 −4 0}, ↑ n w n x n v (f) [ ]= [ ]− [ +4]={−3 4 −5 0 0 10 2 −2}, ↑ n w n y n s (g) [ ]=3.5 [ ]={21 −10.5 −3.5 0 2.8 24.5 −7}. ↑ n y n r 2.6 (a) x[n]=−4δ[n+3]+5δ[n+2]+δ[n+1]−2δ[n]−3δ[n−1]+2δ[n−3], ], 5 [ 2 ] 4 [ 7 ] 3 [ 8 ] 1 [ ] [ 3 ] 1 [ 6 ] [n = δn+ − δ n −δ n− + δ n− + δ n− − δn− y ], 8 [ 5 ] 7 [ 2 ] 5 [ ] 4 [ 2 ] 3 [ 2 ] 2 [ 3 ] [n = δ n− + δ n− + δn− −δn− − δ n− + δ n− w (b) Recall δ[n]=µ[n]−µ[n−1]. Hence, ]) [ ] 1 [ ( ]) 1 [ ] 2 [ ( 5 ]) 2 [ ] 3 [ ( 4 ] [n n n n n n n x =− µ + −µ + + µ + −µ + + µ + −µ ]) 4 [ ] 3 [ ( 2 ]) 2 [ ] 1 [ ( 3 ]) 1 [ ] [ ( 2 µ −µ − − µ − −µ − + µ − −µ − − n n n n n n

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], 4 [ 2 ] 3 [ 2 ] 2 [ 3 ] 1 [ ] [ 3 ] 1 [ 4 ] 2 [ 9 ] 3 [ 4µ + + µ + − µ + − µ −µ − + µ − + µ − − µ − − = n n n n n n n n 2.7 (a) z_1 z_1 + + h[0] h[1] h[2] x[n] y[n] x[n-1] x[n-2]

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

(b) h[0] z_1 z_1 + + z_1 z_1 + + 11 β β12 22 β 21 β x[n] y[n] x[n 1]_ x[n 2]_ w[n 1]_ w[n 2]_ w[n]

From the above figure we get w[n]=h[0](x[n]+β₁₁x[n−1]+β₂₁x[n−2]) and ]. 2 [ ] 1 [ ] [ ] [n =w n +β₁₂w n− +β₂₂w n−

y Making use of the first equation in the second

we arrive at ]) 2 [ ] 1 [ ] [ ]( 0 [ ] [n =h x n +β₁₁x n− +β₂₁x n− y ]) 3 [ ] 2 [ ] 1 [ ]( 0 [ ₁₁ ₂₁ 12 − +β − +β − β + h x n x n x n ]) 4 [ ] 3 [ ] 2 [ ]( 0 [ ₁₁ ₂₁ 22 − +β − +β − β + h x n x n x n ] 2 [ ) ( ] 1 [ ) ( ] [ ] 0 [

(

+ β₁₁+β₁₂ − + β₂₁+β₁₂β₁₁+β₂₂ − =h x n x n x n . ] 4 [ ] 3 [ ) (β₁₂β₂₁+β₂₂β₁₁ − +β₂₂β₂₁ −

)

+ x n x n

(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.

z_1 z_1 + + + + 0.6 0.3 0.2 __0.8 0.5 _ y[n] w[n 1]_ w[n 2]_ w[n] z_1 + 0.4 x[n] u[n] y[n+1]

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The second-order section can be redrawn as shown below without changing its input-output relation. z_1 z_1 + + + + 0.6 0.3 0.2 __0.8 0.5 _ w[n 1]_ w[n 2]_ w[n] x[n] y[n] z_1 + 0.4 u[n] y[n+1] z_1 z_1

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

z_1 z_1 + + + + 0.6 0.3 0.2 __0.8 0.5 _ x[n] y[n] z_1 + 0.4 u[n] y[n+1] z_1 z_1 x[n _1] x[n _2] u[n _1] u[n _2] s[n]

Analyzing the above structure we arrive at

], 2 [ 2 . 0 ] 1 [ 3 . 0 ] [ 6 . 0 ] [n = x n + x n− + x n− s ], 2 [ 5 . 0 ] 1 [ 8 . 0 ] [ ] [n =s n − u n− − u n− u ]. [ 4 . 0 ] [ ] 1 [n u n y n y + = +

From Substituting this in the second equation we get after some

algebra ]. [ 4 . 0 ] 1 [ ] [n y n y n u = + − ]. 2 [ 8 . 0 ] 1 [ 18 . 0 ] [ 4 . 0 ] [ ] 1 [n+ =s n − y n − y n− + y n−

y Making use of the first

equation in this equation we finally arrive at the desired input-output relation

]. 3 [ 2 . 0 ] 2 [ 3 . 0 ] 1 [ 6 . 0 ] 3 [ 2 . 0 ] 2 [ 18 . 0 ] 1 [ 4 . 0 ] [n + y n− + y n− − y n− = x n− + x n− + x n− y

(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: z_1 z_1 + + + 0.3 0.2 __0.8 0.5 _ w[n 1]_ w[n 2]_ w[n] x[n] z_1 z_1 y [n]₂

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Interchanging the order of the two sections we arrive at an equivalent structure shown below: q[n] z_1 z_1 + + __0.8 0.5 _ x[n] + 0.3 0.2 z_1 z_1 y [n]₂ y [n 1]₂ _ _ y [n 2]₂

Analyzing the above structure we get

], 2 [ 2 . 0 ] 1 [ 3 . 0 ] [n = x n− + x n− q ]. 2 [ 5 . 0 ] 1 [ 8 . 0 ] [ ] [ ₂ ₂ 2 n =q n − y n− − y n− y

Substituting the first equation in the second we have

]. 2 [ 2 . 0 ] 1 [ 3 . 0 ] 2 [ 5 . 0 ] 1 [ 8 . 0 ] [ ₂ ₂ 2 n + y n− + y n− = x n− + x n− y (2-1)

Analyzing the first-order section of Figure P2.1(d) given below 0.6 x[n]

+

z_1 0.4 u[n] y [n]₁ u[n 1]_ we get ], 1 [ 4 . 0 ] [ ] [n =x n + u n− u ]. 1 [ 6 . 0 ] [ 1 n = u n− y

Solving the above two equations we have

]. 1 [ 6 . 0 ] 1 [ 4 . 0 ] [ ₁ 1 n − y n− = x n− y (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 0.8y₁[n−1]−0.32y₁[n−2]=0.48x[n−2] and

]. 3 [ 3 . 0 ] 3 [ 2 . 0 ] 2 [ 5 .

0 y₁ n− − y₁ n− = x n− Adding the last two equations to Eq. (2-2) we

arrive at y₁[n]+0.4y₁[n−1]+0.18y₁[n−2]−0.2y₁[n−3] ]. 3 [ 3 . 0 ] 2 [ 48 . 0 ] 1 [ 6 . 0 − + − + − = x n x n x n (2-4)

Similarly, from Eq. (2-1) we get

]. 3 [ 08 . 0 ] 2 [ 12 . 0 ] 3 [ 2 . 0 ] 2 [ 32 . 0 ] 1 [ 4 . 0 ₂ − − ₂ − − ₂ − =− − − − − y n y n y n x n x n Adding this

equation to Eq. (2-1) we arrive at

] 3 [ 2 . 0 ] 2 [ 18 . 0 ] 1 [ 4 . 0 ] [ ₂ ₂ ₂ 2 n + y n− + y n− − y n− y ]. 3 [ 08 . 0 ] 2 [ 08 . 0 ] 1 [ 3 . 0 − + − − − = x n x n x n (2-5)

Adding Eqs. (2-4) and (2-5), and making use of Eq. (2-3) we finally arrive at the input-output relation of Figure P2.1(d) as:

]. 3 [ 22 . 0 ] 2 [ 56 . 0 ] 1 [ 9 . 0 ] 3 [ 2 . 0 ] 2 [ 18 . 0 ] 1 [ 4 . 0 ] [n + y n− + y n− − y n− = x n− + x n− + x n− y

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} 5 . 2 1 0 5 . 0 1 1 5 . 1 0 0 0 5 . 1 1 1 5 . 0 0 1 5 . 2 { − − − − = ↑ and ]) [ ] [ ( ] [ 2 1 n w n w n w_od = − − }. 5 . 2 1 0 5 . 0 1 1 5 . 1 0 0 0 5 . 1 1 1 5 . 0 0 1 5 . 2 {− − − − − − = ↑

2.10 (a) x₁[n]= nµ[ +2]. Hence, x₁[−n]=µ[−n+2]. Therefore, ⎪⎩ ⎪ ⎨ ⎧ ≤ −≤ ≤ − ≥ = + − µ + + µ = , 3 , 2 / 1 , 2 2 , 1 , 3 , 2 / 1 ]) 2 [ ] 2 [ ( ] [ 2 1 , 1 n n n n n n x _ev and ⎪⎩ ⎪ ⎨ ⎧ ≤ − − − ≤ ≤ ≥ = + − µ − + µ = . 3 , 2 / 1 , 2 2 , 0 , 3 , 2 / 1 ]) 2 [ ] 2 [ ( ] [ 2 1 , 1 n n n n n n x _od (b) x₂[n]=αnµ[n−3]. Hence, x₂[−n]=α−nµ[−n−3]. Therefore,

(

)

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

(

)

=−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]

)

Hence, is an even sequence.

]. [ ] [ ] [n x n v n x_od _od = = v[n]

2.13 (a)Since x[n] is causal, x[n]=0, n<0. Also, x[−n]=0, n>0. Now,

(

(

[ ] [ ]

)

. 2 1 n y n y n y_cs = + ∗ − Hence, [0]

(

[0] [0]

)

[0] 2 1 re cs y y y y = + ∗ = and . 0 ], [ ] [ 2 1 _> = n n n

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

2.14 Since x[n] is causal, x[n]=0, n<0. From the solution of Problem 2.13 we have

]. [ ] [ ) cos( 2 , 0 , 0 , 0 1 , 0 ), cos( 2 , 0 , 0 , 0 ], [ , 0 ], [ 2 ] [ n n n n n n n n n n x n n x n x _o o ev ev δ − µ ω = ⎪⎩ ⎪ ⎨ ⎧ < = > ω = ⎪⎩ ⎪ ⎨ ⎧ <= > =

2.15 (a) {x[n]}={Aαn} where A and α are complex numbers with α <1. Since for n

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(b) where ⎩ ⎨ ⎧ < ≥ α = µ α = , 0 , 0 , 0 , ] [ ] [ n n A n A n

y n n A and α are complex numbers with

. 1 <

α Here, αn ≤1,n≥0. Hence y[n] ≤ A for all values of Hence, is a bounded sequence.

.

n { ny[ ]}

(c) {h[n]}=Cβnµ[n] where C and β are complex numbers with β >1. Since for n

n> ,0 β can become arbitrarily large, { nh[ ]} is not a bounded sequence.

(d) {g[n]}=4cos(ω_on). Since g[n] ≤4 for all values of is a bounded sequence. ]} [ { , g n n (e) ⎪⎩ ⎪ ⎨ ⎧ ≤ ≥ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = . 0 , 0 , 1 , 1 ] [ 2 1 n n n v _n Since 1₂ <1 n for n>1 and 2 1 1 = n for n=1, [ ] <1 n v for

all values of n. Thus { nv[ ]} is a bounded sequence.

2.16 [ ] ( 1) [ 1]. 1 − µ − = + n n n x n Now [ ] ( 1) 1 . 1 1 1 ∞ = ∑ = ∑ − = ∑ ∞ = ∞ = + ∞ −∞ = n n n n n n n x Hence is

not absolutely summable.

]} [ { nx 2.17 (a) x₁[n]=αnµ[n−1]. Now <∞ α − α = ∑ α = ∑ α = ∑ ∞ = ∞ = ∞ −∞ = 2[ ] ₁ _n ₁ 1 n n n n n x , since . 1 <

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

(b) x₂[n]=αnµ[n−1]. Now ∑ = ∑ α = ∑ α∞ = ∞ = ∞ −∞ = 2 1 1 ] [ n n n n n n n n x , ) 1 ( 2 ∞ < α − α = since . 1 2 _<

α Hence, {x₂[n]} is absolutely summable.

(c) x₃[n]=n2αnµ[n−1]. Now n n n n n n n n x = ∑ α = ∑ α ∑ ∞ = ∞ = ∞ −∞ = 1 2 1 2 3[ ] K + α + α + α + α = ₂2 2 ₃2 3 ₄2 4 ) ( 5 ) ( 3 ) (α + α2 + α3 + α4 +_K + α2 + α3 + α4 +_K + α3 + α4 + α5 +_K =

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) ( 7 α4 + α5 + α6 +_K + = +_K α − α + α − α + α − α + α − α 1 7 1 5 1 3 1 4 3 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∑ α − ∑ α α − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∑ − α α − = ∞ = ∞ = ∞ =1 1 1 2 1 1 ) 1 2 ( 1 1 n n n n n n n n _⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ α − α − α − α α − = 1 ) 1 ( 2 1 1 2 . ) 1 ( ) 1 ( 3 <∞ α − α + α

= Hence, {x₃[n]} is absolutely summable.

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

2.19 (a) A sequence x[n] is absolutely summable if ∑∞ [ ] <∞.

−∞ = n n x By Schwartz inequality we have [ ]2 [ ] [ ]_⎟⎟<∞. ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∑ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∑ ≤ ∑ ∞ −∞ = ∞ −∞ = ∞ −∞ = n n n n x n x n

x Hence, an absolutely summable

sequence is square summable and has thus finite energy.

Now consider the sequence x[n]= 1µ[n−1].

n 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 Then the series and the integral

both converge or both diverge. For . 1 ≥ x ∑∞ =1 n n a ∫ ∞ 1 ) ( dxx f . ) ( , 1 1 x n n f x a = = But ∫ = ∞ =∞− =∞ ∞ 0 ) (ln 1 1 1 x dx x . Hence, ∑ = ∑∞ = ∞ −∞ = 1 1 ] [ n n n n

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

n is not

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(b) To show that { nx[ ]} is square-summable, we observe that here 1₂ n n a = , and thus, . ) ( 1₂ x x f = Now, 1 1 1 1. 1 1 1 2 ⎟_⎠ =−∞+ = ⎞ ⎜ ⎝ ⎛− = ∫ ∞ ∞ x dx x Hence, ∑∞ =1 1 2 n n converges, or in other words, x[n]= 1µ[n−1] n is square-summable.

2.20 See Problem 2.19, Part (a) solution.

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

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

not absolutely summable. Now,

] [ 2 n x ∞ ∞ ω ⋅ ω π ω ⋅ π = ∫ _πω 1 1 ) int( cos cos cos 1 cos x x x x dx x x c c c c _where

is the cosine integral function. Since int cos dx x x c ∫ _πω ∞ 1 cos diverges, ∑ π ω ∞ =1 cos n c n n also diverges. Hence, x₂[n] is not absolutely summable.

2.22 = ∑ = ∑∞

(

+

)

−∞ = + ∞ → − = + ∞ → _K K _K ev od K K n K K x x n ₂ ₁ x n x n 2 1 2 1 2 1 ] [ ] [ lim ] [ lim P

(

)

∑ + + = − = + ∞ → K K n ev od ev od K K n x n x n x n x [ ] [ ] 2 [ ] [ ] lim 2 2 1 2 1

(

[ ] [ ]

)(

[ ] [ ] lim 1 2 1 2 1 n x n x n x n x K K n K K x xev + od + ∑ + − − − = − = + ∞ → P P

)

od ev n K K n od ev x _K _K x n x n x x x P P P P + + = + = _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∑ − − ∑ ⋅ + ∞ → ∞ −∞ = − = [ ] [ ] 2 1 1 2 1 ₂ ₂ lim

as ∑ = ∑ − Now for the given sequence,

− = − = K K n K K n n x n x2[ ] 2[ ]. ∑ + ∞ → = + ∞ → − = + ∞ → ⎟⎠ = ⎞ ⎜ ⎝ ⎛ = ∑ ⎜_⎝⎛ ⎟_⎠⎞ = ∑ = K n od _K _K K n K K K K n od K K x x n 0 1 1 2 1 6 3 1 0 6 3 1 1 2 1 2 1 2 1 lim lim ] [ lim P . lim 6 3 1 2 1 1 2 1 6 3 1 _⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = + + ∞ → K K K Hence, . 10 6 3 1 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − = od ev x x x P P P

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2.23 x[n]=sin(2πkn/N), 0≤n≤ N−1. Now [ ] 1sin (2 / ) 0 2 1 0 2 N kn n x N n N n x = ∑ = ∑ π − = − = E

(

1 cos(4 / )

)

1cos(4 / ). 0 2 1 2 1 0 2 1 ∑ π − = ∑ − π = − = − = N n N N n N kn N kn Let = ∑ π and − = 1 0 ) / 4 cos( N n N kn C Then . ) / 4 sin( 1 0 ∑ π = − = N n N kn S 0. 1 1 / 4 4 1 0 / 4 ₌ − − = ∑ = + − ₋− _ππ = π − N k j kn j N n N kn j e e e jS C This implies Hence . 0 = C . 2 N x = E 2.24 (a) x[n]= Aαµ[n]. Then . 1 ] [ 2 2 0 2 2 2 α − = ∑ α = ∑ = ∞ = ∞ −∞ = A A n x n n n a x_a E (b) x [n]= 1₂µ[n−1]. n b Then ∑ = ∑ = ∑ ∞ = ∞ = ∞ −∞ = 1 1 1 1 2 4 2 ] [ n n n n n b x_b x n E . 90 4 π =

2.25 (a) x₁[n]=(−1)n. Then average power

, 1 ) 1 2 ( 1 2 1 lim ] [ lim ₁ 2 1 2 1 1 = ∑ = _→_∞ + + = − = + ∞ → x n K K K K K n K K x P and energy . 1 ] [ 2 1 1 = ∑ = ∑ =∞ ∞ −∞ = ∞ −∞ = n n x x n E

(b) x₂[n]=µ[n]. Then average power

, 2 1 1 2 1 lim 1 1 2 1 lim ] [ 1 2 1 lim 0 2 2 2 + = + = ∑ + = ∑ + = ∞ → = ∞ → − = ∞ → K K K n x K K K n K K K n K x P and energy . 1 ] [ 0 2 2 2 = ∑ = ∑ =∞ ∞ = ∞ −∞ = n n x x n E

(c) x₃[n]=nµ[n]. Then average power

, 6 ) 1 2 )( 1 ( lim 1 2 1 lim ] [ 1 2 1 lim 1 2 2 3 3 =∞ + + = ∑ + = ∑ + = ∞ → = ∞ → − = ∞ → K K K n K n x K K K n K K K n K x P and energy [ ] . 0 2 2 3 3 = ∑ = ∑ =∞ ∞ = ∞ −∞ = n n x x n n E

(d) [ ] 0 . Then average power

0 4 n j e A n x = ω ∑ + = − = ∞ → K K n K x x n K P ₄[ ]2 1 2 1 lim 4

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, ) 1 2 ( 1 2 1 lim 1 2 1 lim 1 2 1 lim ₀ 0 2 ₀2 _A₀2 _K _A₀2 K A K e A K K K K n K K K n n j K = + ⋅ + = ∑ + = ∑ + = ∞ → − = ∞ → − = ω ∞ → and energy ₃[ ]2 ₀ 0 2 ₀2 . 3 = ∑ = ∑ = ∑ =∞ ∞ −∞ = ∞ −∞ = ω ∞ −∞ = n n n j n x x n A e A E (e) ₅[ ] cos 2 ⎟. ⎠ ⎞ ⎜ ⎝ ⎛ ₊_φ = π M n A n

x Note x₅[n] is a periodic sequence.Then average power

. 1 cos 2 1 cos 1 ] [ 1 1 0 2 4 2 1 0 2 2 1 0 2 5 5 ∑ ⎟⎠ ⎞ ⎜ ⎝ ⎛ _⎟₊ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = ∑ ⎜_⎝⎛ +φ⎟_⎠⎞ = ∑ = − = + φ π − = π − = M n M n M n M n M n x A M A M n x M P Let ∑ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ _φ = − = 1 0 4 2 cos M n M πn C and 1cos 2 . 0 4 ∑ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ _φ = − = M n M πn S Then . 0 1 1 / 4 4 2 1 0 / 4 2 1 0 2 4 = − − ⋅ = ∑ = ∑ = + − φ _ππ = π φ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π ₊ _φ M j j j M n M n j j M n j e e e e e e jS C M n Hence C =0. Therefore . 2 1 2 1 1 2 0 2 5 A A M P M n x = ⋅ ∑ = − =

Since x₅[n] is a periodic sequence, it has infinite energy.

2.26 In each of the following parts, Ndenotes the fundamental period and r is a positive integer.

(a) ~x₁[n]=4cos(2πn/5). Here N and r must satisfy the relation 2 .

5 2 r N = π ⋅ π

Among all positive solutions for N and r, the smallest values are N =5 and r =1. Hence the average power is given by

. 8 cos 4 5 1 ] [ ~ 1 4 2 0 5 2 1 0 2 1 1 ∑ ⎟⎠ = ⎞ ⎜ ⎝ ⎛ = ∑ = = π − = n n N n x x n N P

(b) ~x₂[n]=3cos(3πn/5). Here N and r must satisfy the relation 2 .

5 3 r N = π ⋅ π

Among all positive solutions for N and r, the smallest values are N =10 and r=3. Hence the average power is given by

. 5 . 4 cos 3 10 1 ] [ ~ 1 9 2 0 5 3 1 0 2 2 2 ∑ ⎟⎠ = ⎞ ⎜ ⎝ ⎛ = ∑ = = π − = n n N n x x n N P

(c) ~x₃[n]=2cos(3πn/7). Here N and r must satisfy the relation 2 .

7 3 r N = π ⋅ π

Among all positive solutions for N and r, the smallest values are N =14 and r=3. Hence the average power is given by

. 2 cos 2 14 1 ] [ ~ 1 13 2 0 7 3 1 0 2 3 3 ∑ ⎟⎠ = ⎞ ⎜ ⎝ ⎛ = ∑ = = π − = n n N n x x n N P

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(d) ~x₄[n]=4cos(5πn/3). Here N and r must satisfy the relation 2 . 3 5 r N = π ⋅ π

Among all positive solutions for N and r, the smallest values are N =6 and r =5. Hence the average power is given by

. 8 cos 4 6 1 ] [ ~ 1 5 2 0 3 5 1 0 2 4 4 ∑ ⎟⎠ = ⎞ ⎜ ⎝ ⎛ = ∑ = = π − = n n N n x x n N P

(e) x~₅[n]=4cos(2πn/5)+3cos(3πn/5). We first determine the fundamental period

of Here and

1

N cos(2πn/5). N₁ r must satisfy the relation ₁ 2 .

5 2 r N = π ⋅ π Among all positive solutions for N₁ and , the smallest values are r N₁ =5 and We next determine the fundamental period of

. 1 =

r

2

N cos(3πn/5). Here and r must satisfy the relation 2 N . 2 2 5 3 r N = π ⋅ π

Among all positive solutions for N₂ and r, the smallest values are N₂ =10 and r=3. The fundamental period of ~x₅[n] is then given by

. 10 ) 10 , 5 ( ) , (N₁ N₂ = LCM = LCM

Hence the average power is given by

2 4 0 5 3 5 2 1 0 2 5 4cos 3cos 10 1 ] [ ~ 1 5 ∑ ⎟⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∑ = = π π − = n n n N n x x n N P . 5 . 12 0 5 . 4 8 cos cos 24 cos 9 cos 16 10 1 11 0 5 3 5 2 5 3 11 0 2 5 2 11 0 2 _≅ ₊ ₊ ₌ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∑ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ = = π π π = π = n n n n n n n

(f) ~x₆[n]=4cos(5πn/3)+3cos(3πn/5). We first determine the fundamental period of Here and r must satisfy the relation

1 N cos(5πn/3). N₁ ₁ 2 . 3 5 r N = π ⋅ π Among all positive solutions for N₁ and , the smallest values are r N₁ =6 and We next determine the fundamental period of

. 5 =

r

2

N cos(3πn/5). Here and r must satisfy the relation 2 N . 2 2 5 3 r N = π ⋅ π

Among all positive solutions for and r, the smallest values are and The fundamental period of is then given by

2 N 10 2 = N r=3. ~x₆[n] . 30 ) 10 , 6 ( ) , (N₁ N₂ = LCM = LCM

Hence the average power is given by

2 29 0 5 3 3 5 1 0 2 6 4cos 3cos 30 1 ] [ ~ 1 6 ∑ ⎟⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∑ = = π π − = n n n N n x x n N P . 5 . 12 0 5 . 4 8 cos cos 24 cos 9 cos 16 30 1 29 0 5 3 3 5 5 3 30 0 2 3 5 29 0 2 _≅ ₊ ₊ ₌ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∑ ⎜_⎝⎛ ⎟_⎠⎞ ⎜_⎝⎛ ⎟_⎠⎞ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ = = π π π = π = n n n n n n n

2.27 Now , from Eq. (2.38) we have~[ ]= ∑∞ [ + ]. Therefore

−∞ = k kN n x n y

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. ] [ ] [ ~ ₊ ₌ _∑∞ ₊ ₊ −∞ = k N kN n x N n y Substituting r= k+1 we get ]. [ ~ ] [ ] [ ~_y _n _N _x _n _rN _y _n r = ∑ + = + ∞ −∞

= Hence ~ ny[ ] is a periodic sequence with a period N.

2.28 (a)N =5. Now ~ [ ]= ∑∞ [ + 5]. The portion of in the range

−∞ = n p n x n k x ~x_p[n] 0≤ n≤4 is given by x[n−5]+x[n]+x[n+5]={0 0 −4 5 1} . 4 0 }, 1 7 4 3 2 { } 0 0 0 0 0 { } 0 2 0 3 2 {− − + = − − − ≤ ≤ + n

Hence, one period of ~x_p[n] is given by {−2 −3 −4 7 1},0≤n≤4.

Now ~ [ ]= ∑∞ [ + 5]. The portion of in the range is given by

−∞ = n p n y n k y ~ ny_p[ ] 0≤ n≤4 } 6 0 0 0 0 { ] 5 [ ] [ ] 5 [n− +y n +y n+ = y . 4 0 }, 13 8 0 1 5 { } 0 0 0 0 2 { } 7 8 0 1 3 {− − + − = − − ≤ ≤ + n

Hence, one period of ~ ny_p[ ] is given by {−5 −1 0 8 13},0≤n≤4.

−∞ = n p n wn k w w~_p[n] 0≤ n≤4 } 0 0 0 0 0 { ] 5 [ ] [ ] 5 [n− +w n +w n+ = w . 4 0 }, 2 7 1 0 1 { } 0 5 2 0 1 { } 2 2 3 0 0 { + − − = − ≤ ≤ + n

Hence, one period of w~_p[n] is given by {−1 0 1 7 2},0≤n≤4. (b) N =7. Now ~ [ ]= ∑∞ [ + 7]. The portion of in the range

−∞ = n p n x n k x ~x_p[n] 0≤ n≤6 is given by x[n−7]+x[n]+x[n+7]={0 0 0 0 −4 5 1} } 0 0 0 0 0 0 0 { } 0 0 0 2 0 3 2 {− − + + . 6 0 }, 1 5 4 2 0 3 2 {− − − ≤ ≤

= n Hence, one period of ~x_p[n] is given by

. 6 0 }, 1 5 4 2 0 3 2 {− − − ≤n≤

−∞ = n p n y n k y ~ ny_p[ ] 0≤ n≤6 } 6 0 0 0 0 0 0 { ] 7 [ ] [ ] 7 [n− +x n +x n+ = x } 0 0 0 0 0 0 0 { } 0 2 7 8 0 1 3 {− − − + + . 6 0 }, 6 2 7 8 0 1 3 {− − − ≤ ≤

= n Hence, one period of ~ ny_p[ ] is given by

. 6 0 }, 6 2 7 8 0 1 3 {− − − ≤n≤

−∞ = n p n wn k w w~_p[n] 0≤ n≤6 } 0 0 0 0 0 0 0 { ] 7 [ ] [ ] 7 [n− +w n +w n+ = w } 0 0 0 0 0 5 2 { } 0 1 2 2 3 0 0 { − + − +

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. 6 0 }, 0 1 2 2 3 5 2 {− − ≤ ≤

= n Hence, one period of w~_p[n] is given by

. 6 0 }, 0 1 2 2 3 5 2 {− − ≤n≤ 2.29 ~x[n]= Acos(ω_on+φ). (a) ~x[n]={1 −1 −1 1 1 −1 −1 1}. Hence A= 2,ω_o =π/2,φ=π/4. (b) x~[n]={0 − 3 0 3 0 − 3 0 3}. Hence A= 3, ω_o =π/2, . 2 / π = φ (c) x~[n]={1 −0.366 −1.366 −1 0.366 1.366}. Hence A= 2, ω_o =π/3, . 4 / π = φ (d) ~x[n]={2 0 −2 0 2 0 −2 0}. Hence A=2,ω_o =π/2,φ=0.

2.30 The fundamental period Nof a periodic sequence with an angular frequency ω _o satisfies Eq. (2.47a) with the smallest value of N and . r

(a) ω_o =0.5π. Here Eq. (2.47a) reduces to 0.5πN =2πr which is satisfied with . 1 , 4 = = r N

(b) ω_o =0.8π. Here Eq. (2.47a) reduces to 0.8πN =2πr which is satisfied with . 2 , 5 = = r N

(c) We first determine the fundamental period of In this case, Eq. (2.47a) reduces to

1 N Re{ejπn/5}=cos(0.2πn). 1 1 2 2 .

0 πN = πr which is satisfied with N₁=10,r₁=1. We next determine the fundamental period of In this case, Eq. (2.47a) reduces to

2 N Im{ejπn/10 = jsin(0.1πn). 2 2 2 1 .

0 πN = πr which is satisfied with N₂ =20,r₂ =1. Hence the fundamental period N of ~ nx_c[ ] is given by

. 20 ) 20 , 10 ( ) , (N₁ N₂ = LCM = LCM

(d) We first determine the fundamental period of In this case, Eq. (2.47a) reduces to 1 N 3cos(1.3πn). 1 1 2 3 .

1 πN = πr which is satisfied with N₁ =20,r₁ =13. We next determine the fundamental period N₂ of 4sin(0.5πn+0.5π). In this case, Eq. (2.47a) reduces to 0.5πN₂=2πr₂ which is satisfied with N₂ = r4, ₂ =1. Hence the

fundamental period N of ~x₄[n] is given by LCM(N₁,N₂)= LCM(20,4)=20. (e) We first determine the fundamental period of In this case, Eq. (2.47a) reduces to

1 N 5cos(1.5πn+0.75π). 1 1 2 5 .

1 πN = πr which is satisfied with N₁ = r4, ₁ =3. We next determine the fundamental period N₂ of 4cos(0.6πn). In this case, Eq. (2.47a) reduces to 0.6πN₂=2πr₂ which is satisfied with N₂ =10,r₂ =3. We finally

determine the fundamental period N₃ of sin(0.5πn). In this case, Eq. (2.47a) reduces to 0.5πN₃=2πr₃ which is satisfied with N₃ = r4, ₃ =1. Hence the fundamental period

N of ~x₅[n] is given by LCM(N₁,N₂,N₃)= LCM(4,10,4)=20.

2.31 The fundamental period Nof a periodic sequence with an angular frequency ω _o satisfies Eq. (2.47a) with the smallest value of N and . r

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(a) ω_o =0.6π. Here Eq. (2.47a) reduces to 0.6πN =2πr which is satisfied with . 3 , 10 = = r N

(b) ω_o =0.28π. Here Eq. (2.47a) reduces to 0.28πN =2πr which is satisfied with . 7 , 50 = = r N

(c) ω_o =0.45π. Here Eq. (2.47a) reduces to 0.45πN =2πr which is satisfied with . 9 , 40 = = r N

(d) ω_o =0.55π. Here Eq. (2.47a) reduces to 0.55πN =2πr which is satisfied with . 11 , 40 = = r N

(e) ω_o =0.65π. Here Eq. (2.47a) reduces to 0.65πN =2πr which is satisfied with . 13 , 40 = = r N

2.32 ω_o =0.08π. Here Eq. (2.47a) reduces to 0.08πN =2πr which is satisfied with For a sequence . 1 , 25 = = r

N ~x₂[n]=sin(ω₂n) with a fundamental period of N =25, Eq. (2.47a) reduces to 25ω₂ =2πr. For example, for r=2 we have

Another sequence with the same fundamental period is obtained by setting which leads to

. 16 . 0 25 / 4 2 = π = π ω 3 =

r ω₃ =6π/25=0.24π. The corresponding periodic sequences are therefore ~x₂[n]=sin(0.16πn) and ~x₃[n]=sin(0.24πn).

2.33 The three parameters A,Ω_o,and φ of the continuous-time signal can be determined from ) (t x_a ) cos( ) ( ] [n =x nT =A Ω nT +φ

x _a _o by setting 3 distinct values of

For example . n , cos ] 0 [ = A φ=α x , sin ) sin( cos ) cos( ) cos( ] 1 [− =A −Ω T +φ = A Ω T φ+A Ω T φ=β x _o _o _o , . sin ) sin( cos ) cos( ) cos( ] 1 [ = A Ω T+φ =A Ω T φ−A Ω T φ=γ x _o _o _o

Substituting the first equation into the last two equations and then adding them we get α γ + β = Ω 2 )

cos( _oT which can be solved to determine Ω . Next, from the second _o equation we have Asinφ=β−Acos(Ω_oT)cosφ=β−αcos(Ω_oT). Dividing this equation by the last equation on the previous page we arrive at

T T o o Ω α Ω α − β = φ sin( ) cos( tan

which can be solved to determine φ. Finally, the parameter is determined from the first equation of the last page.

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Now consider the case _T 2 2 _o.

T = Ω

π =

Ω In this case x[n]= Acos(nπ+φ)=β and

(

+ π+φ

)

= π+φ =β

=

+1] cos( 1) cos( )

[n A n A n

x . Since all sample values are equal, the

three parameters cannot be determined uniquely.

Finally consider the case _T 2 2 _o.

T < Ω

π =

Ω In this case x[n]= Acos(Ω_onT+φ) implying

) cos(ω +φ

=A _on ω_o =Ω_oT >π. As explained in Section 2.2.1, a digital sinusoidal sequence with an angular frequency ω greater than assumes the identity _o of a sinusoidal sequence with an angular frequency in the range . Hence,

cannot be uniquely determined from

π

. 0≤ω<π o

Ω x[n]= Acos(Ω_onT +φ).

2.34 x[n]=cos(Ω_onT). If x[n] is periodic with a period N, then

(

)

[ ] cos( ).

cos ]

[n N ₀nT ₀NT x n ₀nT

x + = Ω +Ω = = Ω This implies Ω_oNT =2πrwith

r any nonzero positive integer. Hence the sampling rate must satisfy the relation If i.e.,

. /

2 r N

T = π Ω_o Ω_o =20, T =π/8, then we must have N⋅π =2πr

8

20 . The

smallest value of N and r satisfying this relation are N =4 and The fundamental period is thus

. 5 = r 4 = N .

2.35 (a) For an input x_i[n],i=1,2, the output is

. 2 , 1 ], 2 [ ] 1 [ ] 2 [ ] 1 [ ] [ ] [n =b₀x n +b₁x n− +b₂x n− +a₁y n− +a₂y n− i = y_i _i _i _i _i _i Then, for

an input x[n]= Ax₁[n]+Bx₂[n], the output is y[n]=b₀(Ax₁[n]+Bx₂[n])

]) 1 [ ] 1 [ ( ]) 2 [ ] 2 [ ( ]) 1 [ ] 1 [ ( ₁ ₂ ₂ ₁ ₂ ₁ ₁ ₂ 1 − + − + − + − + − + − +b Ax n Bx n b Ax n Bx n a Ay n By n ]) 2 [ ] 2 [ ( ₁ ₂ 1 − + − +a Ay n By n = A(b₀x₁[n]+b₁x₁[n−1]+b₂x₁[n−2]+a₁y₁[n−1] ]) 2 [ ] 1 [ ] 2 [ ] 1 [ ] [ ( ]) 1 [ ₁ ₂ ₂ ₂ ₃ ₂ ₁ ₂ ₂ ₂ 2 2 − + + − + − + − + − +a y n B b x n b x n b x n a x n a x n ]. [ ] [ ₂ 1 n By n Ay +

)

5 2 1[ ] [ ] ] [n Ax n Bx n y = + . ]) [ ( ]) [ (x₁ n 5 B x₂ n 5 A + ≠

For an input the output is the impulse response As for and the system is causal.

], [ ] [n n x =δ h[n]=(δ[n])5. 0 ] [n = h n<0,

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For a bounded input x[n] ≤ B<∞, the magnitude of the output samples are . ] [ ]) [ ( ] [n = x n 5 = x n 5 ≤B5 <∞

y As the output is also a bounded sequence, the

system is BIBO stable.

Finally, let y[n] and y₁[n] be the outputs for inputs x[n] and x₁[n], respectively. If ]

[ ] [

1 n x n no

x = − then y₁[n]=(x₁[n])5 =

(

x[n−n_o]

)

5 = y[n−n_o]. Hence, the system is time-invariant. (c) =β+ ∑ − with = 3 0 ] [ ] [ l l n x n

y β a nonzero constant. For an input the

output is Then, for an input

the output is , 2 , 1 ], [n i= x_i . 2 , 1 , ] [ ] [ 3 0 = ∑ − + β = = x n i n y_i _i l l ], [ ] [ ] [n Ax₁ n Bx₂ n x = +

(

)

=β+ ∑ − + ∑ − ∑ − + − + β = = = = 3 0 2 3 0 1 3 0 1 2 ] [ ] [ ] [ ] [ ] [ l l l l l l l n Bx n Ax n Bx n Ax n y ]. [ ] [ ₂ 1 n By n Ay +

≠ Hence the system is nonlinear.

For an input the output is the impulse response As

for the system is noncausal. ], [ ] [n n x =δ [ ] [ ]. 0 ∑δ − + β = ∞ = l l n n h 0 ] [n ≠ h n<0,

For a bounded input x[n] ≤ B<∞, the magnitude of the output samples are .

4 ]

[n ≤β+ B<∞

y As the output is also a bounded sequence, the system is BIBO stable.

Finally, let y[n] and y₁[n] be the outputs for inputs x[n] and x₁[n], respectively. If ]

[ ] [

1 n x n no

x = − then Hence, the system is

time-invariant. ]. [ ] [ ] [ 3 0 1 n x n no y n no y =β+ ∑ − − = − = l l

(d) y[n]=ln

(

2+ x[n]

)

. For an input x_i[n],i=1,2, the output is y_i[n]=ln

(

2+ x_i[n]

)

, .

2 , 1 =

i Then, for an input x[n]= Ax₁[n]+Bx₂[n], the output is

(

2 [ ] [ ]

)

[ ] [ ].

ln ]

[n Ax₁ n Bx₂ n Ay₁ n By₂ n

y = + + ≠ + Hence the system is nonlinear.

For an input x[n]=δ[n], the output is the impulse response h[n]+ln

(

2+ δ[n]

)

. For n<0,h[n]=ln(2)≠0. Hence, the system is noncausal.

For a bounded input x[n] ≤ B<∞, the magnitude of the output samples are

(

2

)

.

ln ]

[n ≤ +B <∞

y As the output is also a bounded sequence, the system is BIBO stable.

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] [

1 n x n no

x = − then y₁[n]=ln

(

2+ x[n−n_o]

)

=y[n−n_o]. Hence, the system is time-invariant.

(e) with a nonzero constant. For an input the output

is Then, for an input

], 2 [ ] [n =αx −n+ y x_i[n],i=1,2, ], 2 [ ] [n =αx −n+ y_i _i i =1,2. x[n]= Ax₁[n]+Bx₂[n], the output is ]. [ ] [ ] 2 [ ] 2 [ ] [n A x₁ n B x₂ n Ay₁ n By₂ n

y = α − + + α − + = + Hence the system is linear.

For an input x[n]=δ[n], the output is the impulse response h[n]=αδ[−n+2]. For ,

0 <

n h[n]=0. Hence, the system is causal.

]

[n = Bα <∞

y As the output is also a bounded sequence, the system is BIBO stable. Finally, let y[n] and y₁[n] be the outputs for inputs x[n] and x₁[n], respectively. If

] [

1 n x n no

x = − then y₁[n]=αx₁[−n+2]=αx[−(n−n_o)+2]=y[n−n_o]. Hence, the system is time-invariant.

(f) y[n]= nx[ −4]. For an input x_i[n],i=1,2, the output is y_i[n]=x_i[n−4],i=1,2. Then, for an input x[n]= Ax₁[n]+Bx₂[n], the output is y[n]= Ax₁[n−4]+Bx₂[n−4]

]. [ ] [ ₂ 1 n By n Ay +

= Hence the system is linear.

For an input x[n]=δ[n], the output is the impulse response h[n]= nδ[ −4]. For n<0, .

0 ]

[n =

h Hence, the system is causal.

]

[n = B<∞

y As the output is also a bounded sequence, the system is BIBO stable. Finally, let y[n] and y₁[n] be the outputs for inputs x[n] and x₁[n], respectively. If

] [

1 n x n no

x = − then y₁[n]=x[n−n_o−4]=y[n−n_o]. Hence, the system is time-invariant.

2.37 Let and be the outputs of a median filter of length for inputs and , respectively. If ] [n y y₁[n] 2K+1 x[n] ] [ 1 n x x₁[n]=x[n−n_o], then } ] [ , ], 1 [ ], [ ], 1 [ , ], [ { med ] [ ₁ ₁ ₁ ₁ ₁ 1 n x n K x n x n x n x n K y = − _K − + _K + } ] [ , ], 1 [ ], [ ], 1 [ , ], [ { med x n−n_o −K x n−n_o − x n−n_o x n−n_o + x n−n_o +K = _K _K ]. [n n_o y −

= Hence, the system is time-invariant.

2.38 y[n]=x[n+1]−2x[n]+x[n−1]. For an input x_i[n],i=1,2, the output is . 2 , 1 ], 1 [ ] [ 2 ] 1 [ ] [n =x n+ − x n +x n− i=

y_i _i _i _i Then, for an input x[n]= Ax₁[n]+Bx₂[n], the output is ] 1 [ ] 1 [ ] [ 2 ] [ 2 ] 1 [ ] 1 [ ] [n =Ax₁ n+ +Bx₂ n+ − Ax₁ n − Bx₂ n +Ax₁ n− +Bx₂ n− y ]. [ ] [ ₂ 1 n By n Ay +

= Hence the system is linear.

If x₁[n]=x[n−n_o], then y₁[n]= x[n−n_o +1]−2x[n−n_o]+x[−n_on−1]=y[n−n_o]. Hence, the system is time-invariant.

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The impulse response of the system is h[n]=δ[n+1]−2δ[n]+δ[n−1]. Now . 1 ] 0 [ ] 1 [− =δ =

h Since h[n]≠0 for all values of n<0, the system is noncausal.

2.39 y[n]=x2[n]−x[n−1]x[n+1]. For an input x_i[n],i=1,2, the output is . 2 , 1 ], 1 [ ] 1 [ ] [ ] [n =x2 n −x n− x n+ i =

y_i _i _i _i Then, for an input the

output is ], [ ] [ ] [n Ax₁ n Bx₂ n x = +

(

[ ] [ ]

) (

[ 1] [ 1]

)(

[ 1] [ 1]

)

] [n = Ax₁ n +Bx₂ n 2 − Ax₁ n− +Bx₂ n− Ax₁ n+ +Bx₂ n+ y

Hence the system is nonlinear. ]. [ ] [ ₂ 1 n By n Ay + ≠ If x₁[n]=x[n−n_o], then y₁[n]= x₁2[n]−x₁[n−1]x₁[n+1] ] 1 [ ] 1 [ ] [ 2 ₋ ₋ ₋ ₋ ₋ ₊

=x n n_o x n n_o x n n_o =y[n−n_o]. Hence, the system is time-invariant.

The impulse response of the system is Since

for all values of

]. [ ] 1 [ ] 1 [ ] [ ] [n 2 n n n n h =δ −δ − δ + =δ 0 ] [n =

h n<0, the system is causal.

2.40 . ] 1 [ ] [ ] 1 [ 2 1 ] [ _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + − = n y n x n y n

y Now for an input x[n]=αµ[n], the output converges to some constant

] [n

y

K as n→∞. The input-output relation of the system as reduces to ∞ → n ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ α + = K K K 2 1

from which we get K2 =α or in other words .

α =

K

For an input x_i[n],i=1,2, the output is , 1,2. ] 1 [ ] [ ] 1 [ 2 1 ] [ _⎟⎟ = ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + − = i n y n x n y n y i i i i Then,

for an input x[n]= Ax₁[n]+Bx₂[n], the output is . ] 1 [ ] [ ] [ ] 1 [ 2 1 ] [ 1 2 _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + + − = n y n Bx n Ax n y n y

On the other hand,

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + − = + ] 1 [ ] [ ] 1 [ 2 1 ] [ ] [ 1 1 1 2 1 n y n Ax n Ay n By n Ay [ ]. ] 1 [ ] [ ] 1 [ 2 1 2 2 2 y n n y n Bx n By _⎟⎟≠ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + − +

Hence the system is nonlinear.

If x₁[n]=x[n−n_o], then [ ]. ] 1 [ ] [ ] 1 [ 2 1 ] [ 1 1 1 o y n no n y n n x n y n y _⎟⎟= − ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − + − = Hence, the system is time-invariant. 2.41 y[n]=x[n]−y2[n−1]+y[n−1].

For an input the output is Then, for an input

, 2 , 1 ], [n i= x_i y_i[n]= x_i[n]−y_i2[n−1]+y_i[n−1],i=1,2. ], [ ] [ ] [n Ax₁ n Bx₂ n x = + the output is ]. 1 [ ] 1 [ ] [ ] [ ] [n =Ax₁ n +Bx₂ n −y2 n− +y n−