Solutions of Chapter 3

Part 2/2

Problem 3.5-4 Solve the following equation recursively (first three terms only):

y[n+2]+3y[n+1]+2y[n] = x[n+2]+3x[n+1]+3x[n]

with x[n] = 3nu[n],y[−1] = 3 and y[−2] = 2.

Solution:Write the difference equation as follows.

y[n+2] =−3y[n+1]−2y[n]+ x[n+2]+3x[n+1]+3x[n]

Setting n =−2 and substituting y[−1] = 3,y[−2] = 2,x[−1] = x[−2] = 0,x[0] = 1, yields

y[0] =−3y[−1]−2y[−2]+ x[0]+3x[−1]+3x[−2] =−12

Setting n =−1 and substituting y[0] =−12,y[−1] = 3,x[−1] = 0,x[0] = 1,x[1] = 31 = 3, yields

y[1] =−3y[0]−2y[−1]+ x[1]+3x[0]+3x[−1] = 36

Proceeding along same lines, we obtain

y[2] =−3y[1]−2y[0]+ x[2]+3x[1]+3x[0] =−3(36)−2(−12)+(32)+3(3)+3(1) =−63

Problem 3.6-2 Solvey[n+2]+2y[n+1]+ y[n] = 0

if y[−1] = 1,y[−2] = 1.

Solution: We write the equation as(E2 +2E +1)y[n] = 0

The characteristic equation isγ2 +2γ +1 = (γ +1)2 = 0

from which we have γ1 = γ2 =−1, thus

y[n] = (c1 + c2n)(−1)n

Setting n =−1 and n =−2 and substituting initial conditions yields{y[−1] = (c1 + c2(−1))(−1)(−1) =−c1 + c2 = 1y[−2] = (c1 + c2(−2))(−1)(−2) = c1 −2c2 = 1

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from which we have c1 =−3,c2 =−2 and

y[n] =−(3+2n)(−1)n

Problem 3.6-3 Solvey[n+2]−2y[n+1]+2y[n] = 0

if y[−1] = 1,y[−2] = 0.

Solution: We write the equation as(E2 −2E +2)y[n] = 0

The characteristic equation is

γ2 −2γ +2 = (γ −1− j1)(γ −1+ j1) = 0

from which we have the roots γ1,2 = 1± j1 =√

2e± jπ/4, thus

y[n] = c(√

2)n cos(π4

n+θ)

Setting n =−1 and n =−2 and substituting initial conditions yields{y[−1] = c√

2cos(−π

4 +θ) = c√2( 1√

2cosθ + 1√

2sinθ) = 1

y[−2] = c2 cos(−π

2 +θ) = c2 sinθ = 0

from which we have c = 2,θ = 0 and

y[n] = 2(√

2)n cos(π4

n)

Problem 3.7-3 Find the unit impulse response h[n] of the system

y[n]−6y[n−1]+25y[n−2] = 2x[n]−4x[n−1] (1)

Solution: We write the system in the advanced form as

y[n+2]−6y[n+1]+25y[n] = 2x[n+2]−4x[n+1] (2)

Setting x[n] = δ [n], we have

h[n+2]−6h[n+1]+25h[n] = 2δ [n+2]−4δ [n+1] (3)

The characteristic equation is γ2 −6γ +25 = 0. The characteristic roots are

γ1,2 = 3±4 j = 5e± j φ , φ = 0.927

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The general form of h[n] can be written as

h[n] = A0δ [n]+ c(5)n cos(φn+θ) (4)

Substituting (4) into (3) yields

A0(δ [n+2]−6δ [n+1]+25δ [n]) = 2δ [n+2]−4δ [n+1]

Setting n = 0 yields A0 = 0.To determine c and θ in (4), we need to find two values of h[n] (not initial conditions sice h[n] is defined

in n ≥ 0, however, by using iterative procedure we can derive two auxiliary conditions h[0] and h[1].)Setting n =−2 in equation (3) yields

h[0]−6(0)+25(0) = 2(1)−4(0)⇒ h[0] = 2

Setting n =−1 in equation (3) yields

h[1]−6(2)+25(0) = 2(0)−4(1)⇒ h[1] = 8

Substituting A0,h[0],h[1] into (4) yields{h[0] = ccosθ = 2h[1] = 5ccos(0.927n+θ) = 5c(cos0.927cosθ − sin0.927sinθ) = 8

from which we have c = 2.0616,θ =−0.245rad.Therefore,

h[n] = 2.0616(5)n cos(0.927n−0.245)u[n]

Problem 3.7-4 (a) For the general Nth-order difference equation, letting

a0 = a1 = · · ·= aN−1 = 0

results in a general causal Nth-order LTI nonrecursive difference equation

y[n] = b0x[n]+b1x[n−1]+ · · ·+bN−1x[n−N +1]+bNx[n−N]

Find the impulse response h[n] for the system.(b) Find the impulse response of a nonrecursive LTID system described by the equation

y[n] = 3x[n]−5x[n−1]−2x[n−3]

Solution:(a) The characteristic equation is γN = 0. Hence, all the characteristic roots are zero. We use a direct

method to find h[n] by realizing that h[n] is the response to unit impulse input.Setting x[n] = δ [n], and y[n] = h[n] yields

h[n] = b0δ [n]+b1δ [n−1]+ · · ·+bN−1δ [n−N +1]+bNδ [n−N] =N

∑i=0

biδ [n− i]

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(b) From the result in part (a), we can immediately have

h[n] = 3δ [n]−5δ [n−1]−2δ [n−3]

Observe that the impulse response has only a finite (N) number of nonzero elements. For this reason,such systems are called finite-impulse response (FIR) systems. For a general recursive case, the impulseresponse has an infinite number of nonzero elements, and such systems are called infinite-impulse response(IIR) systems.

Problem 3.8-2 Find the (zero-state) response y[n] of an LTID system if the input is x[n] = 3n−1u[n+2] and

h[n] =12[δ [n−2]− (−2)n+1]u[n−3]

Solution:

h[n] =12[δ [n−2]− (−2)n+1]u[n−3] =

12

δ [n−2]u[n−3]− 12(−2)n+1u[n−3] =−1

2(−2)n+1u[n−3]

y[n] = x[n]∗h[n] = h[n]∗ x[n]

Note that

h[m] =−12(−2)m+1u[m−3] =

{(−2)m, m ≥ 30, m < 3

x[n−m] = (3)n−m−1u[n−m+2] ={

3n−13−m, m ≤ n+20, m > n+2

Therefore

y[n] =n+2

∑m=3

(−2)m3n−13−m = 3n−1n+2

∑m=3

(−23)m =

8135

[(−2)n −3n]u[n]

Another solution:

y[n] = x[n]∗h[n] = (−12(−2)n+1u[n−3])∗ (3n−1u[n+2])

Let us advance the first term by 3 units and delay the second term by 2 units, the resulting convolutionyields y[n+1]. Hence

y[n+1] = (−12(−2)n+4u[n])∗ (3n−3u[n]) =− 8

27(−2)n ∗3nu[n]

From convolution Table, we obtain

y[n+1] =− 827

[(−2)n+1 −3n+1

−2−3

]u[n] =

8135

[(−2)n+1 −3n+1]u[n]

Thusy[n] =

8135

[(−2)n −3n]u[n]

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Problem 3.8-10 Find the total response specified by the equation

y[n+1]+2y[n] = x[n+1]

if y[−1] = 10 and the input x[n] = e−nu[n].

Solution: The total response can be written as

total response =N

∑i=1

ciγni︸ ︷︷ ︸

zero−input component

+ x[n]∗h[n]︸ ︷︷ ︸zero−state component

Step 1: solve zero-input component.The characteristic equation of the system is

γ +2 = 0 ⇒ γ =−2

Therefore the zero-input response is y0[n] = c1(−2)n.Setting n =−1 and substituting y[−1] = 10 yields

10 =−c1

2⇒ c1 =−20

Therefore,y0[n] =−20(−2)n,n ≥ 0

Step 2: solve zero-state component.In the system equation, setting x[n] = δ [n] yields

h[n+1]+2h[n] = δ [n+1] (5)

The characteristic root is γ =−2, so

h[n] = A0δ [n]+ c0(−2)nu[n] (6)

Substituting (6) into the unit impulse response equation yields

A0δ [n+1]+2A0δ [n] = δ [n+1]

Setting n = 0 yields A0 = 0Setting n =−1 in (5) yields

h[0]+2h[−1] = δ [0]⇒ h[0] = 1

Substituting h[0] = 1 into (6) yields

h[0] = c0(−2)0 = 1 ⇒ c0 = 1

Thereforeh[n] = (−2)nu[n]

The zero-state response is

y[n] = x[n]∗h[n] = (e−nu[n])∗ ((−2)nu[n])

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The above convolution can be found using the convolution Table to be

y[n] =e

2e+1[e−(n+1)− (−2)n+1]u[n] =

[1

2e+1(e)−n +

2e2e+1

(−2)n]

u[n]

Another solution for the convolution:

y[n] =n

∑m=0

e−m(−2)n−m = (−2)n 1− [− 12e ]

n+1

1− (− 12e)

=

[1

2e+1(e)−n +

2e2e+1

(−2)n]

u[n]

Step 3: the total response

total response = y0[n]+ y[n]

=

[−20(−2)n +

12e+1

(e)−n +2e

2e+1(−2)n

]u[n]

=1

2e+1[−(38e+20)(−2)n +(e)−n] u[n]

Problem 3.8-19 Using the slide-tape algorithm, find x[n]∗g[n] for the signals shown in Fig. P3.8-19

Figure 1: Fig3.8-19

Solution: From Fig. P3.8-19, we obtain the x tape and g tape. Then we rotate the g tape about the verticalaxis, as shown in the figure. We observe that:

n = 0, y[n] = 0+1+2+3+4+5 = 15

n = 1, y[n] = 1+2+3+4+5 = 15

n = 2, y[n] = 2+3+4+5 = 14

n = 3, y[n] = 3+4+5 = 12

n = 4, y[n] = 4+5 = 9

n = 5, y[n] = 5

n = 6, y[n] = 0

n ≥ 6, y[n] = 0

n < 0, y[n] = 15

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Figure 2: Solution of Fig3.8-19

Problem 3.9-3 (a) Use the classical method to solve

y[n+2]+3y[n+1]+2y[n] = x[n+2]+3x[n+1]+3x[n]

with the input x[n] = (3)n and the auxiliary conditions y[0] = 1,y[1] = 3.(b) Repeat part (a) for auxiliary conditions y[−1] = y[−2] = 1.

Solution:(E2 +3E +2)︸ ︷︷ ︸

Q[E]

y[n] = (E2 +3E +3)︸ ︷︷ ︸P[E]

x[n]

The characteristic equation is γ2 +3γ +2 = 0, the characteristic roots are γ1 =−1,γ2 =−2.Therefore, the natural response is

yn[n] =C1(−1)n +C2(−2)n

For input x[n] = (3)n, the form of the forced response is

yϕ [n] =C(3)n

Substituting yϕ [n] into the system equation yields

C(3)n+2 +3C(3)n+1 +2C(3)n = (3)n+2 +3(3)n+1 +3(3)n ⇒C =2120

Thus, the total response is

y[n] =C1(−1)n +C2(−2)n +2120

(3)n,n ≥ 0

(a) Setting n = 0,1 and substituting y[0] = 1,y[1] = 3, yields{y[0] =C1(−1)0 +C2(−2)0 + 21

20(3)0 = 1

y[1] =C1(−1)1 +C2(−2)1 + 2120(3)

1 = 3

from which we have C1 =−14 ,C2 =

15

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Thereforey[n] =−1

4(−1)n +

15(−2)n +

2120

(3)n,n ≥ 0

(b) We solve system equation iteratively to find y[0] and y[1] using the given auxiliary conditions y[−1] =y[−2] = 1.

Setting n =−2, we obtain

y[0]+3y[−1]+2y[−2] = (3)0 +3(0)+3(0)⇒ y[0] =−4

Setting n =−1, we obtain

y[1]+3y[0]+2y[−1] = (3)1 +3(3)0 +3(0)⇒ y[1] = 16

In the total response

y[n] =C1(−1)n +C2(−2)n +2120

(3)n,n ≥ 0

Setting n = 0,1, and substituting y[0] =−4,y[1] = 16, yields{y[0] =C1(−1)0 +C2(−2)0 + 21

20(3)0 =−4

y[1] =C1(−1)1 +C2(−2)1 + 2120(3)

1 = 16

from which we have C1 =114 ,C2 =−39

5Therefore

y[n] =114(−1)n − 39

5(−2)n +

2120

(3)n,n ≥ 0

Problem 3.10-2 Each of the following equations specifies an LTID system, Determine weather each ofthese systems is BIBO stable or unstable. Determine weather each of these systems is asymptotically stable,unstable, or marginally stable.

(a) y[n+2]+0.6y[n+1]−0.16y[n] = x[n+1]−2x[n](b) y[n]+3y[n−1]+2y[n−2] = x[n−1]+2x[n−2](c) (E −1)2(E + 1

2)y[n] = x[n](d) y[n]+2y[n−1]+0.96y[n−2] = x[n](e) y[n]+ y[n−1]−2y[n−2] = x[n]+2x[n−1](f) (E2 −1)(E2 +1)y[n] = x[n]

Solution:(a) The characteristic polynomial is

γ2 +0.6γ −1.6 = (γ −0.2)(γ +0.8)

The characteristic roots are 0.2 and -0.8. Both are inside the unit circle. The system is BIBO stable andasymptotically stable.

(b) The characteristic polynomial is

γ2 +3γ +2 = (γ +2)(γ +1)

The characteristic roots are -1 and -2. One root lies outside the unit circle. Thus, the system is BIBOunstable and asymptotically unstable.

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(c) The characteristic polynomial is

(γ −1)2(γ +12)

The characteristic roots are 1 (repeated twice) and −12 . Repeated roots on the unit circle. Thus, the

system is BIBO unstable and asymptotically unstable.(d) The characteristic polynomial is

γ2 +2γ +0.96 = (γ +0.8)(γ +1.2)

The characteristic roots are -0.8 and -1.2. One root lies outside the unit circle. Thus, the system is BIBOunstable and asymptotically unstable.

(e) The characteristic polynomial is

γ2 + γ −2 = (γ +0.5+ j 1.5)(γ +0.5− j 1.5)

The characteristic roots are 0.5± j 1.5. Both roots lie outside the unit circle. Thus, the system is BIBOunstable and asymptotically unstable.

(f) The characteristic polynomial is

(γ2 −1)(γ2 +1) = (γ +1)(γ −1)(γ + j 1)(γ − j 1)

The characteristic roots are ±1 and ± j 1. All roots are simple and on the unit circle. Thus, the system isBIBO unstable and marginally stable.

Problem 3.10-3 Consider two LTIC systems in cascade, as illustrated in Fig.3.23. The impulse response ofthe system S1 is h1[n] = 2nu[n] and the impulse response of the system S2 is h2[n] = δ [n]−2δ [n−1]. Is thecascade system asymptotically stable or unstabl? Determine the BIBO stability of the composite system.

Figure 3: Fig3.23

Solution:For system S1, since

∞

∑n=−∞

|h1[n]| → ∞

Thus, S1 is asymptotically unstable and BIBO unstable.For system S2, since

∞

∑n=−∞

|h2[n]|= 3

Thus, S2 is BIBO stable. However, the asymptotical stability is inconclusive. Please note that we cannotdraw conclusion about internal (asymptotical) stability of a system from just its impulse response or fromexternal (BIBO) stability, due to possible pole-zero cancellation.

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If the two systems are cascaded, the impulse response of the composite system is

h[n] = h1[n]∗h2[n] = 2nu[n]∗ (δ [n]−2δ [n−1]) = 2nu[n]−2(2)n−1u[n−1] = δ [n]

Therefore, the composite system is BIBO stable. However, the system S1 will eventually burn out (orsaturate) because its output contains the signal of the form 2n which has unbounded characteristic responsegenerated by intended or intended initial conditions.

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