1

Partial Answers to End-of-Chapter Problems

Chapter 0

[1] A 75 min. recording gives 7.938×108 bytes for maximum frequency of 22.05 KHz, 8 bits/sample,two channels (stereo); (b) vocoder reduces number of bits transmitted; (c) texting is cheaper;(d) DVD audio 192 KHz sampling rate, and 24 bits/sample; (e) 352× 240× 60 pixels/sec.

[2] Analog and discrete signals for Ts = 0.1 are very similar, not for Ts = 1.

[3] y(t) = −8π sin(2πt), finite difference for Ts = 0.01 looks like derivative, not so for Ts = 0.1.

[4] (a) ∆1[x[n+ 1]] = ∆[x[n]]; (b) average gives better approximation to derivative.

[5] (a) diL(t)/dt+ iL(t) = is(t) dual of the difference equation in the chapter.

[6] (a) Σkcak = cΣkak, Σk[ak + bk] = Σkak + Σkbk, a0 + a1 + a2 + a3 = a0 + a2 + a3 + a1 (i.e., inany order); (b) S = 0.5(N(N + 1)); (c) S1 = (α+ βN/2)(N + 1).

[7] The definite integral is 0.5, then [(N − 1)(N − 2) + 2(N − 1)]/(2N2) ≤ 0.5 ≤ [(N − 1)(N − 2) +

2(N − 1)]/(2N2) + 1/N .

[8] S1 = S + 50, S2 = S.

[9] (a) S = N ; (b)(c) S = (1 − αN )/(1 − α), α 6= 1 if α = 1 S = N ; (d) S = 1/(1 − α) |α| < 1; (e)S1 = α/(1− α2).

[10] (c) α = ea; (d) vc(1) = 0.79.

[11] (a) |z| =√

2, ∠z = π/4; (b) |u| =√

2, ∠u = −π/4; (c) z/w = w/v = u/z = −j; (d) y = 10−6z,negligible magnitude, same phase as z.

[12] (a) log(w) = z; (c) w + w∗ = 2e cos(1); (e) | log(w)|2 = 2.

[13] Compare ejαejβ (use Euler’s) to ej(α+β).

[14] Use sin(α) = cos(α− π/2) and sin(β) = cos(β − π/2).

[15] (a)(b) zz∗ = x2+y2, 1/z = z∗/(zz∗), using polar representation is easier; (b) (z+w)∗ = z∗+w∗,(zw)∗ = z∗w∗; (d) zw = rρej(θ+φ).

[16] (a) z = x + jy = |z|ejθ, |x| = |z|| cos(θ)| ≤ |z| since | cos(θ)| ≤ 1; (b) add two vectors graphi-cally.

[17] (a)(b) x(t) = (1− t2)+ j2t, y(t) = −2t+2j; (c)∫ 1

0x(t)dt = (2/3)+ j; (d)

∫ 1

0x∗(t)dt = (2/3)− j.

[18] ejπn = (−1)n; (b)(c) use Euler’s identity∫ T0

0sin(πt) cos(πt)dt = 0.

2

[19]∫ 1

0sin2(2πt)dt = 0.5.

[20] z2 = 1 has roots−1 and 1; z2 = −1 has roots−j and j; z3 = 1 has roots 1 and e±j2π/3; z3 = −1

has roots −1 and e±j2π/3.

[21] log(−2) = log(2)± jπ, log(2ejπ/4) = log(2) + jπ/4.

[22] cosh(θ) = cos(jθ) is even; sinh(θ) = −j sin(jθ) is odd.

[23] (a) sin(Ω0t) = cos(Ω0t− π/2); (b) phasor Ae−jπ/2 generates sine; (c) phasor Aejπ/2.

Chapter 1

[1] (a) E2 = 1; (b) 0 < x2(t) < x(t) so x(t) absolutely integrable; (c) Ey = 0.5; for C = 1mF thenR = 1 KΩ.

[2] (a) Pa = 0; (b) Pa = 0.25; (c) Pa = 0.5; (d) the real part of the complex power, P = 0.5VsI∗ for

phasors Vs and I , in given cases gives same results.

[3] (a) x(t) periodic of period T0 = 2; (b)(c) Px = Px1+ Px2

= 2.5; (d) y(t) not periodic, Py =

Py1+ Py2

.

[4] (a)(b) T1 = 2 period of x1(t), T2 = 2/3 period of x2(t); (b) x(t) periodic of period T0 = T1 =

3T2 = 2; (c) T1/T2 = 1/4 so that the period is T0 = 4T1 = T2.

[5] (a) x(t + 1) is x(t) advanced by 1, i.e., shifted to the left by 1; (b) x(−t + 1) is x(−t) (reversalof x(t)) shifted to the right by 1; (c) y(t) equals Λ(t) except y(0) = 2 while Λ(0) = 1.

[6] (a)-(c) cosh(t) = 0.5(et + e−t), −∞ < t < ∞ is even, sinh(t) = 0.5(et − e−t), −∞ < t < ∞ isodd; (d) x(t) = e−tu(t).

[7] (a) Triangular pulse has width 2∆ and height 1/∆ thus unit area, as ∆ → 0 it gives δ(t); (b)S∆(t) = sin(πt/∆)/(πt), S∆(0) = 1/∆, and

∫∞−∞ S∆(t)dt = 1, so that as ∆→ 0 it gives δ(t).

[8] (a) y(t) = −Ω0 sin(Ω0t); (b) z(t) = −Ω0 sin(Ω0t)u(t) + δ(t); (c)∫ t−∞ z(t′)dt′ = cos(Ω0t)u(t).

[9] (a) Steady state: voltage across capacitor equals voltage of source, so no current throughresistor; (b) vc(t) = 1 in steady state; (c) (d) capacitor discharges through resistor.

[10] (a) r(t) =∫ t

0u(τ)dτ ; (b) dr(t)/dt = u(t) + tδ(t) = u(t) and derivative of integral∫ ∞

0

du(t− τ)/d(t− τ)× d(t− τ)/dt dτ =

∫ t

−∞δ(µ)dµ = u(t)

[11] ssT (t) = Σ∞k=0u(t − kT ), a stairway to the stars; (b) digitized ramp and cosine, using ssT (t)

approximate signal.

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[12] (a) y(t) 6= z(t), operations do not commute.

[13] (a) x(2t) = 2t[u(t)−u(t−0.5)], i.e., x(t) has been compressed; (b) x(t/2) = 0.5t[u(t)−u(t−2)],i.e., x(t) has been expanded; (c) play it faster (contraction) or slower (expansion) than thespeed at which it was recorded.

[14] (a) Even-odd decomposition

xe(t) =

0.5(1− t) 0 < t ≤ 1

0.5(1 + t) −1 < t ≤ 0

1 t = 0

xo(t) =

0.5(1− t) 0 < t ≤ 1

−0.5(1 + t) −1 < t ≤ 0

0 t = 0

(b) Ex = Exe + Exo

[15] (a) g(t) = u(t)−2u(t−1)+u(t−2); (b) x(t) =∑∞k=−∞ g(t+2k); (d) w(t) = 2

∑∞k=−∞ δ(t−2k),

periodic of period T0 = 2.

[16] (a) x(t/2) is periodic of period 4; (b) x(2t) is periodic of period 1.

[17] (a) y(t) =∑∞k=−∞[u(t− 2k)− 2u(t− 1− 2k) + u(t− 2− 2k)]; (b) z(t) = 0; (c) neither is final

energy.

[18] (a) T0 = 1; (b) z(t) periodic of period T1 = 2; (c) v(t) = ej3πt has period T3 = 2/3.

[19] (a) Py = 0.5; (b) Py < 1

[20] Fs = 8192 samples/sec, NN = 4096 samples.

[21] (a) 1 + 0.7ejφt = (1 + 0.7 cos(φt)) + j0.7 sin(φt) = A(t)ejθ(t), A(t) =√

1.49 + 1.4 cos(φt),θ(t) = tan−1(0.7 sin(φt)/(1 + 0.7 cos(φ(t)).

Chapter 2[1] (a) v0(t) = −(1 + 0.5 cos(20πt))u(t); (b) v0(t) = −R(t)u(t− 50× 10−3); (c) v0(t) 6= v0(t− 50×

10−3), TV system.

[2] (a) Nonlinear system.

[3] (a) h(t) = (1/T )(u(t+ T/2)− u(t− T/2), non-causal ; (b) y(t) = (1/T )(t+ T/2)[u(t+ T/2)−u(t− T/2)].

[4] (a) Convolution integral, h(t) = e−tu(t); (b) causal; (c) s(t) = (1− e−t)u(t); (d) use superposi-tion and TI.

[5] (a) Nonlinear; (b) large q/kT gives large current for positive voltages; (c) nonlinear.

[6] (a) is1(t) = u(t) − u(t − 1), vc(0) = −1 then vc1 = −1 + r(t) − r(t − 1), if is2(t) = 2is1(t),vc(0) = −1, vc2 6= 2vc1 , i.e., nonlinear; (b) nonlinear.

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[7] (b) i1(t) = −2π sin(2πt)[1 + 2 cos(2πt)]; (c) time varying system.

[8] Not LTI, LTI, not LTI, LTI respectively.

[9] x1(t) = u(t) → y1(t) = cos(πt)u(t), x2(t) = u(t − 1) → y2(t) = cos(πt)u(t − 1) 6= y1(t − 1),TV system; (b) TV; (c) x1(t) = u(t) → y1(t) = 0.5[u(t) + u(t − 1)], x2(t) = u(t − 1) → y2(t) =

0.5[u(t− 1) + u(t− 2)] = y1(t− 1), TI system.

[10] (a) Linear, time-varying, causal, BIBO stable system; (b) modulated signal is periodic of pe-riod 28; (c) x(t) = u(t)→ y(t) = u(t)−u(t−2) and x(t−3) = u(t−3)→ u(t−3)[u(t)−u(t−2)] =

0 6= y(t− 3), TV system.

[11] (a) Nonlinear if y(0) 6= 0; (b) dy(t)/dt + 2y(t) = 2x(t); (c) yss(t) = 1, independent of IC; (d)convolution gives y(t) = (1− e−t)u(t); (e) BIBO stable.

[12] Nonlinear, time invariant system.

[13] (a) TV; (b) s(t) =√

2m(t) cos(Ωct− π/4), one modulator with a phase, linear.

[14] (a) LTI system

x(t) = u(t)→ y(t) =

0 t < 0

t 0 ≤ t < 1

1 t ≥ 1

x1(t) = x(t)− x(t− 1)→ y1(t) =

t 0 ≤ t < 1

2− t 1 ≤ t ≤ 2

0 otherwise

(b)(c) if T = T0, output of the averager is zero for t < 0 and t ≥ T0.

[15] (b) y(t) =∑∞k=0(0.1)kx(t− k), echo system is LTI; (c) z(t) =

∑Mk=0 βkx(t− kτ), LTI.

[16] (a) Non-causal; (b) BIBO stable.

[17] (a) h(t) = (1/T )[u(t)−u(t−T )], BIBO stable; (b) r(t) not bounded, output y(t) = (t−T/20u(t)

also not bounded.

[18] (a)(b) x∗(t) ≈ xs(0)[1− e−t/rC ]u(t) 0 ≤ t ≤ ∆, x∗(t) = x∗(∆)e−t/RCu(t) ∆ ≤ t ≤ Ts.

Chapter 3[1] Use L[δ(t− τ)] = e−sτ .

[2] (a)X(s) = cos(s), (c)X(s) = cos(Ω) cosh(σ)+j sin(Ω) sinh(σ) from which you find magnitudeand phase.

[3] (a) Y (s) = 2π(1 − e−s)/(s2 + 4π2) with ROC the whole plane; (b) X(s) = (1 − e−s)((1/s) −(s/(s2 + 4π2)) with ROC the whole plane except for the origin; (c) 0.5[(1/s) + (s/(s2 + 4)].

[4] No Laplace transform, empty ROC.

5

[5] X(s) = (es−1s+ es−1 − 2)/(s2 − 1), ROC σ > −1.

[6] (a) d2y(t)/dt2 + 4y(t) = x(t), (b) y(0) = 0 and y′(0) = −1.

[7] H(s) = (s+ 2)/(s(s+ 1)2) with ROC σ > 0.

[8] (a) First two terms: −4δ(t) and dδ(t)/dt; (b) x(t) = δ(t) − 2e−tu(t) − e−2tu(t); (c) x(t) =

−2u(t) + 7e−tu(t)− 5e−2tu(t).

[9] (a) 0.1; (b) 0.2; (c) no steady-state response; (d) yt(t) = −0.5e−t cos(t)u(t) + 0.5e−t sin(t)u(t).

[10] (b) x(t) = 0.5(1− e−2t)u(t)− e−2(t−1)u(t− 1) = e−tsinh(t)u(t)− e−2(t−1)u(t− 1)

[11] (a) y(t) = 0.5(1−e−2t)u(t)−0.5(1−e−2(t−3))u(t−3); y(t) = r(t)−2r(t−1)+r(t−2) smootherthan x(t) as output of an averager; (c) 3M.

[12] (a) x(t) = 4u(t)− 4e−tu(t); (b) h(t)=u(t)-u(t-1).

[13] s(t) = (2/√

3)e−0.5t sin(√

3t/2)u(t) then response to x1(t) is y1(t) = s(t) − s(t − 1), responseto x3(t) is

∫ t0s(τ)dτ .

[14] (a) P (s) = (1 − e−s)/s, whole s-plane. (b) pole s = 0, zeros sk = −j2πk for −∞ < k < ∞,pole/zero cancellation.

[15] (a) X(s+ α), frequency shift; (b) X(s) = s/(s2 + Ω20); (c) Y (s) = s/(s2 − Ω2

0).

[16] (a) X(s) has infinite number of zeros on jΩ axis; (b) Y (s) =∏∞k=2(s+ 4π2k2).

[17] (a) x1(t) = Ae−3t cos(πt/2 + θ)u(t), x2(t) = Bu(t), x(t) = x1(t) + x2(t); (b) poles 0, −3± 1.57j

with residues 0.087 and −0.044± 0.083j.

[18] (a) Y (s) = X(s)/(s2 + 0.5s+ 0.15); (b) the system is not LTI; (c) Y1(s) = (s2 + 1.5s+ 2)/(s3 +

0.5s2 + 0.15s).

[19] (a) Ignoring IC, (D2 + 2D + 3)y(t) = x(t) where D is derivative operator; (b) y(0) = 1,y′(0) = −1; (c) h(t) = (1/

√2)e−t sin(

√2t)u(t).

[20] (a) The zero-state response is yzs(t) = [1−te−t−e−t]u(t); (b) yzi(t) = e−tu(t); (d) yss(t) = u(t).

[21] (a) System is unstable; (b) limt→∞ y(t)→ 0.

[22] (a)∫∞−∞ |h(t)|dt ≤ |A|

∫∞0e−tdt = |A|; (b) yss(t) = 0.2u(t); (c) steady state response is infinite;

(d) output is not bounded.

[23] (a) h(t) = 1T [u(t)− u(t− T )]; (b) y(t) = r(t)− r(t− 1)− r(t− 2) + r(t− 3).

[24] (a) y(t) =[

16 −

12e−2t + 1

3e−3t]u(t); (b) y(t) = 1

6

[1− e−t cos(

√5t)−

√5

5 e−t sin(

√5t)]u(t).

[25] (a) System is not BIBO stable; (b) x(t) = 2 cos(2t)u(t); (c) No.

6

[26] (a) y(t) = 1T

∫ tt−T x(τ)dτ ; (b) Input is cos(πt)u(t).

[27] (a) d2y1(t)dt2 + 2dy1(t)

dt + 4y1(t) = x(t) + dx(t)dt , d3y3(t)

dt3 + 2d2y3(t)dt2 + 10dy3(t)

dt = −x(t) + dx(t)dt ; (b)

General solution is of the form y1(t) = [A+Be−t cos(√

3t+ θ)]u(t).

[28] (a) X(s) = 1s (e0.5s − e−0.5s), y(t) = r(t+ 1)− 2r(t) + r(t− 1); (b) If ψ(t) = L−1(1/s3) we have

that y3(t) = ψ(t+ 1.5)− 3ψ(t+ 0.5) + 3ψ(t− 0.5)− ψ(t− 1.5).

[29] (a)G(s) = H(s)1+H(s) ; (c) Negative feedback: G(s) = 1/(s+1)

1+1/(s+1) = 1s+2 , positive feedback: G1(s) =

1/(s+1)1−1/(s+1) = 1

s , y(t) = r(t).

[30] If K = 1, g(t) = 2e−tu(t) is absolutely integrable.

[31] (b) Hap(s) = K s−1s+1 ; (c) G(s) = H(s)Hap(s) = s+1

(s−1)(s2+2s+1)−(s−1)s+1 = −1

(s+1)2

[32] (a)(b) Expressions for x1(t) are equivalent; (c) X(s) = 2π(1+e−0.5s)(s2+4π2)(1−e−0.5s) .

[33] (a)(b) Y (s) = X(s)N(s)D(s) = s+2

s2((s+4)2+9) .

[34] (a) E1(s) = 1s(1+1/D(s)) ; (b) Final-value theorem: limt→∞ e1(t) = 0.

Chapter 4[1] No, the eigenfunction property is not valid for non-linear systems as in (a) or time-varying

systems as in (b).

[2] yss(t) = 2|H(j2π)| cos(2πt+ ∠H(j2π)

[3] (a) H(jΩ0) = (1− e−jΩ0T )/(jΩ0T ); (b) H(s) = (1/sT )[1− e−sT ]

[4] (a) T1/T2 = 2π so x(t) not periodic; (b) yss(t) = |H(j)| cos(t + ∠H(j)) + |H(j2π)| cos(2πt +

∠H(j2π)), where H(j) = 15+j = 5−j

26 = 0.196∠−11.31o and H(j2π) = 1/(5 + j2π) =

0.1245∠−51.49o

[5] y(t) = 0.5 + 0.447 cos(t+ 18.4o)

[6] (a) yss(t) = 4|H(j0)| cos(0t+ ∠H(j0)) = 4× 0.5 = 2; (b)

Y (s) =4

s((s+ 1.5)2 + (2− 1.52))=A

s+ · · ·

and yss(t) = A = Y (s)s |s=0= 2.

[7] (a) Xk = j/(2πk); (b) x1(t) = r(t)− r(t− 1)− u(t− 1),

Xk =1

T0X1(s)|s=jkΩ0 =

e−jπk

2jπ2k2sin(πk)− 1

j2πk=

j

2πk, k 6= 0,

X0 =∫ 1

0tdt = 0.5.

7

[8] (a) x1(t) = r(t)− r(t− 1)− u(t− 1), T0 = 2,Ω0 = π

Xk =

j/(2πk) k 6= 0 even−(2 + jπk)/(2π2k2) k odd

X0 = 12

∫ 1

0tdt = 0.25; (b) Yk = X−k; (c)

Zk =

0 k 6= 0 even0.5 k = 0

−2/(π2k2) k odd

[9] (a) x(t) train of pulses; (b)

X1(s) =e−s/2(es/2 − e−s/2)

s;

(c)

Xk =1

2X1(s) |s=jπk= 0.5

sin(πk/2)

πk/2e−jkπ/2

[10] x1(t) = sin(πt)u(t) + sin(π(t− 1))u(t− 1), X1(s) = πe−0.5s

s2+π2 (e0.5s + e−0.5s),Full-wave: T0 = 1, Ω0 = 2π and Xk = 2

(1−4k2)π , X0 = 2π ;

Half-wave: T0 = 2, Ω0 = π and Xk = 1(1−k2)π e

−jπk/2 cos(πk/2), X0 = 1π ; (b) xh(t) =

0.5[xf (t) + sin(πt)].

[11] Xk = 12

sin(πk/2)πk/2 e−jπk/2, Yk = 1

2

(sin(πk/2)πk/2

)2

(−1)k using 1 − cos(πk) = 2 sin2(πk/2). |Yk| =

2|Xk|2 and X0 = 12 = Y0.

[12] Periods: x1(t) = u(t) − u(t − 1), 0 ≤ t ≤ T0 = 2, y1(t) = x1(2t) = u(2t) − u(2t − 1) =

u(t) − u(t − 0.5), 0 ≤ t ≤ T1 = 1, Xk = sin(πk/2)πk e−jπk/2 for kΩ0 = πk and Yk = Xk for

kΩ1 = 2πk.

[13] (a) Derivative of period of x(t) in −0.5 ≤ t ≤ 0.5 gives

y1(t) =dx1(t)

dt= 2δ(t+ 0.25)− 2δ(t− 0.25)

and y(t) =∑k 4j sin(πk/2)ej2kπt. (b) x1(t) = 2u(t+ 0.25)− 2u(t− 0.25), Yk = jk2πXk.

[14] (a) δTs(t) =∑∞k=−∞ δ(t − kTs) =

∑∞k=−∞

1TsejkΩst; (b) line spectrum of δTs(t) is periodic of

period Ωs and magnitude Ts.

[15] (a) x(t) periodic of period T0, x(t+T0) =∑k Ak cos(Ωk(t+T0)+θk) = x(t) then ΩkT0 = 2πk or

Ωk = 2πT0k multiples of Ω0, the fundamental frequency; (b) x(t) = 2 cos(0Ω0 +0)+1 cos(1Ω0t+

0)− 3 cos(3Ω0t+ π/4) then A0 = 2, θ0 = 0; A1 = 1, θ1 = 0 and A3 = −3 and θ3 = π/4; (c) notperiodic.

[16] (a) Xk = 2(−1)k

(1−4k2)π , X0 = 2/π; (b) y(t) = X0H(j0) +∑∞k=−∞,k 6=0XkH(j2πk)ej2πkt, if y(t) =

X(0) = 2/π then H(j0) = 1 and H(j2πk) = 0 for k 6= 0.

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[17] (a) Zk = Xk + Yk; (b)

Wk =

Xk k oddXk + Yk k even

[18] y(t) periodic of period T0 and

Yk =

2X0 − 3 k = 0

2Xk k 6= 0

z(t) is periodic of period T0 and Zk = Xk(1 + e−2jΩ0k); w(t) is periodic of period T0/2 and

Wk =

Xk/2 k even0 otherwise

[19] (a) y1(t) = dx1(t)/dt = u(t)− 2u(t− 1) + u(t− 2), Yk = 2/(jπk) for k 6= 0 odd;

(b) z(t) = 1 + 4∑

k>0, odd

sin(πkt)

πk;

(d) x(t) = 0.5 +

∞∑k=1, odd

4 cos(πkt− π)

(πk)2.

[20] N = 5.

[21] (a) Zn =∑kXkYn−k; (b) Zn =

∑kXkY(n−k)/2

[22] (a) s(t) =∫ t

0x1(τ)dτ − t, 0 ≤ t ≤ 1; (b) p(t) =

∑k,odd(4/(kπ))ej2πkt; (c)(d)

∫ 0.5

−0.5p(τ)dτ =∑

k, odd4kπ

sin(πk)πk = 0

[23] X0 = 1/2, X2 = X−2 = −1/4; (b) Yk = 2/(π(1 − 4k2)); (c) x(t) 6= y(t); (d) low-pass filter ofmagnitude 2 and cutoff frequency 0 < Ωx < 4π gives a steady state z(t) = 1.

[24] (a) Wk = 2e−jkπ(cos(πk) − cos(πk/2))/(−Ω20k

2) Ω0 = 352π; (b) s(t) has harmonics of w(t)

shifted to central frequency 2πfA.

[25] π = 4∑∞k=1 sin(πk/2)/k

[26]

ε =

∫T0

x2(t)dt− 2∑k

∫T0

x(t)Xkφk(t)dt+∑k

∑`

XkX`

∫T0

φk(t)φ∗` (t)dt

Chapter 5[1] (a) X0 = 1

T0

∫ 0.5

−0.5dt = 1

T0; (b) Xk = 1

T0

sin(πk/T0)kπk/T0

; (c) spectrum gets smaller and denser as T0

increases (so multiply Xk by T0)

[2] (b)X1(Ω) = sin(Ω/2)/(Ω/2),X2(Ω) = 4πe−j(Ω/4) cos(Ω/4)/(4π2−Ω2),X3(Ω) = 2(1−cos(Ω))/Ω2

9

[3] (b) Absolutely integrable,∫∞−∞ |x(t)|dt = 2

∫∞0

2e−2tdt = 2; (c) X(Ω) = 84+Ω2 ; (d) X(s) when

s = jΩ is FT.

[4] (a) x(t) absolutely integrable and finite energy, (b)

X(Ω) = X(s) |s=jΩ= 2jΩ + 2

(jΩ + 2)2 + 4π2

[5] (a)(c)X1(Ω) = 12+jΩ andX3(Ω) = 1

(2+jΩ)2 from their Laplace transforms, x2(t) = r(t) ↔ X1(s) =

1/s2, σ > 0 so we can’t find its FT (jΩ-axis not included)

[6] (a) IFT x(t) = Aπt sin Ω0t so A = π and Ω0 = 1; (b) X1(t) = sin(t/2)/(t/2) ↔ 2π[u(Ω + 0.5) −

u(Ω− 0.5)], (b) X1(2t) = sin(t)/t ↔ π[u(Ω + 1)− u(Ω− 1)]

[7] (a) x(t) even, y(t) odd; (b) X(Ω) = 2∫∞

0e−|t| cos(Ωt)dt since cos(.) is even then X(−Ω) =

X(Ω); (c) X(Ω) = 21+Ω2 (d) Y (Ω) = −j

∫∞−∞ y(t) sin(Ωt)dt, Y (−Ω) = −Y (Ω); (e) Y (Ω) = −2jΩ

1+Ω2

[8] (a) T0 = 1, X(Ω) = sin(Ω/2)Ω/2 e−jΩ/2 and |X(2πk)| = 0. For T0 = 10, X(Ω) = sin(5Ω)

5Ω e−j5Ω and|X(kπ/5)| = 0, i.e., expansion in time corresponds to contraction in frequency

[9] (b) X(Ω) = sin(Ω/2)Ω/2 ; (c) Y (Ω) = X(Ω) + 0.5X(Ω− π) + 0.5X(Ω + π); (d) y(t) is smoother than

x(t).

[10] (a) x(t) = r(t+1)−2r(t)+r(t−1), triangular pulse, y(t) = dx(t)/dt = u(t+1)−2u(t)+u(t−1),two pulses; (b) X(Ω) = sin2(Ω/2)

(Ω/2)2 ; (c) Y (Ω) = jΩ sin2(Ω/2)(Ω/2)2 , |Y (Ω)| = |X(Ω)||Ω|

[11] (b) X(Ω) = j (sin(Ω/2))2

Ω/4 , imaginary as x(t) is odd; (c) Y (Ω) =(

sin(Ω/2)Ω/2

)2

, real as y(t) is even;(d) y(t) is smoother.

[12] (a) X(Ω) = 2 cos(Ω); (b) dX(Ω)dΩ = Y (Ω) so that y(t) = x(t)(−jt) = −jδ(t − 1) + jδ(t + 1),

Y (Ω) = −2 sin(Ω)

[13] (a) X(Ω) = 2 cos(ΩT1); (b) Y (Ω) = 1 + 2∑∞k=1 cos(kT0Ω), both y(t) and Y (Ω) are periodic of

period 2π/T0

[14] (b) ∆k = 1Ts

, δTs(t) =∑∞k=−∞

1TsejkΩst, Ωs = 2π

Tswith FT ∆(Ω) = 2π

Ts

∑k δ(Ω− kΩs); (c) δTs(t)

and ∆Ts(Ω) periodic, of periods Ts and 2π/Ts

[15] (b) X(Ω) =(

sin(Ω/2)Ω/2

)2

e−jΩ

[16] (a)(b)X(Ω) = X1(Ω)∑k e−jkT0Ω = 2πX1(Ω)

T0

∑k δ(Ω−kΩ0) where Ω0 = 2π/T0; (c)

∑k e−jkT0Ω =

2πT0

∑m δ(Ω−mΩ0)

[17] (b) P (Ω) = 2 sin(Ω)Ω ; (c) X(Ω) = P (Ω) + 1

2 [P (Ω− 2π) + P (Ω + 2π)]

[18] (a) jΩY (Ω) = 0.5[X(Ω)−X(Ω)e−j2Ω] so that Y (Ω) = X(Ω) sin(Ω)e−jΩ

Ω ,X(Ω) = −4 sin2(Ω/2)jΩ e−jΩ;

(b) Z(Ω) = X(Ω)/(jΩ) = (sin(Ω/2)/(Ω/2))2e−jΩ so that z(t) = r(t)− 2r(t− 1) + r(t− 2)

10

[19] (a) s(t) =∑6k=4 2x(t) cos(kΩ0t); (b) If X(Ω) = πδ(Ω− Ω0) + πδ(Ω + Ω0) then

S(Ω) = π

6∑k=4

δ(Ω− (k + 1)Ω0) + δ(Ω− (k − 1)Ω0) + δ(Ω + (k − 1)Ω0) + δ(Ω + (k + 1)Ω0)

(c) Same message available at different carrier frequencies

[20] (a) Fourier coefficients: Xk = cos(πk/2)e−jπk/2/(π(1− k2)), X0 = 1/π, |X1| = 1/4 and |X2| =1/(3π); (b) Low-pass filter: amplitude π and cut-off frequency 0 < Ωc < π

[21] Capacitor: VC(s)/Vs(s) = 1/(s2+s+1), low-pass filter. Inductor: VL(s)/Vs(s) = s2/(s2+s+1)

high-pass filter. Resistor: VR(s)/Vs(s) = s/(s2 + s+ 1), band-pass filter

[22] (a) X(Ω) = 4π[δ(Ω − 1000) + δ(Ω + 1000)], Y (Ω) = 0.5[X(Ω − 10000) + X(Ω + 10000)]

FT of the transmitted signal y(t). Filter output is 10y(t), the received signal; (b) z(t) =

10y(t) cos(10000t), Z(Ω) = 5X(Ω) + 2.5[X(Ω − 20000) + X(Ω + 20000)]. Pass z(t) throughlow-pass filter G(Ω) of magnitude 1/5 and Ωc > 1000 to get x(t)

[23] (a) h(t) = sin(πt)/(πt), non-causal; (b) g(t) = h(t)2 cos(5πt) with magnitude responseG(jΩ) =

H(j(Ω− 5π)) +H(j(Ω + 5π))

[24] H1(s) = (s − 1)((s − 1)2 + π2)/((s + 1)((s + 1)2 + π2)), all-pass filter. H2(s) = (s + jπ)(s −jπ)/((s+1)(s+1+jπ)(s+1−jπ)), notch-filter. H3(s) = (s−1)/((s+1)(s+1+jπ)(s+1−jπ)),low-pass filter

[25] m(t) = p(t) sin(2πt), for p(t) = u(t)−u(t−1), M(Ω) = (1/2j)[P (Ω + 2π)−P (Ω−2π)]; y1(t) issuppressed-carrier AM, Y1(Ω) = 0.5(M(Ω + 20π) +M(Ω− 20π)); y2(t) is AM with envelopesimilar to message

Chapter 6

[1] (a) V3(s)/V1(s) = 1/(s2 + 3s + 1); (b) V3(s)/V1(s) = V3(s)/V2(s) × V2(s)/V1(s) = 1/(s + 1)2;(c) cascading three filters with transfer functions H1(s) = 1000/(s + 100), H2(s) = 1/(s + 1),H3(s) = 1/1000 gives G(s)

[2] (a) Ω0 = 40π × 103 rad/sec; (b) passing s(t) with spectrum S(Ω) = (α/4)(M(Ω) + M(Ω +

2Ωc) + M(Ω − 2Ωc) through a unit gain ideal lowpass filter and bandwidth 2πBW we get(α/4)m(t), to get m(t) let α = 4

[3] −(R2/R1)Vi(s) = V0(s) + (V0(s)/A)(1 + R2/R1) can be written as negative feedback. AsA→∞ V0(s)/Vi(s) = −R2/R1

[4] (a) RC circuit transfer function V0(s)/Vi(s) = (1/RC)/(s + 1/RC) from which to obtain thefeedback system; (b) steady-state error e(t) = vi(t)− v0(t) = 0

11

[5] Differential equation corresponding to transfer function is

d2vo(t)

dt2+dvo(t)

dt+ vo(t) = vi(t)

can be implemented using two integrators and an adder; (b) Vo(s)/Vi(s) = (1/s)/(1 + (s +

1/s)) = G(s)/(1 +G(s)H(s)) so that G(s) = 1/s and H(s) = s2 + 1 for the negative feedbacksystem; limt→∞ e(t) = 0

[6] (a) Y (s)/X(s) = G(s)/(1 +G(s)H(s)) = 1/(1 + s/K); (b)If G(s) = s, H(s) = s/K the transferfunction of the feedback system corresponds to a band-pass filter

[7] If T (s) = N(s)/N(−s) all-pass filter; negative feedback: G(s) = N(s)/(N(−s)−N(s)H(s)) =

((s−α)2+β2)/4αs,H(s) = 1; positive feedback: H(s) = −1 andG(s) = ((s−α)2+β2)/(2(s2+

α2 + β2)). Replace α = β =√

2/2 to get feedback implementations for the given T (s)

[8] (a) H(s) = 1/(s2 + (K − 1)), for K > 0 it is not possible to stabilize system; (b) H(s) =

G(s)Ha(s) = 1/(s+ 1)2

[9] (a) E(s) = (X(s)D(s))/(D(s) + N(s)), D(s) = sD1(s) and roots of D(s) + N(s) in left-hands-plane for steady-state error to be zero; (b) e(0) = 1 and ess(t) = 2/3

[10] (a) Z(s) = P (s)Q(s) = 2 + 5s+ 6s2 + 5s3 + 2s4 + s5

[11] (a) E(s) = X(s)(s(s + 1)(s + 2))/(1 + s(s + 1)(s + 2)); (b) for x(t) = u(t), e1ss = 0; (c) forx(t) = r(t), e2ss = 2

[12] (a) L1 = 48.5× 103; (b) r(t) = 1.9 cos(2π × 1950t− 2π × 0.065)

[13] (a) C =√

2, L = 1/√

2; (b) VR(s)/Vi(s) = (√

2s)/(s2 +√

2s+ 1)

[14] (a) Nb = 6, Nc = 4; (b)(c) Butterworth: Ωhp = 1787.4, α(Ωp) = 0.5 dB and α(Ωs) = 32.023 dB;Chebyshev: ε = 0.3493, Ωhp = 1639.7, α(Ωp) = 0.5 dB and α(Ωs) = 36.65 dB

[15] Ωhp = 1280.9 rad/sec. Ωp = 999.82 rad/sec

[16] Butterworth: K = 10 dc gain, N = 6 minimum order, Ωhp = 1.79×103 half-power frequency;Chebyshev: K = 10.6, N = 4, ε = 0.349, Ωhp = 1.64× 103; minimum order depends on ratioof the two frequencies

Chapter 7

[1] (a) fs ≥ 10 KHz, bits/hour = 288× 106; (b) fs ≥ 44KHz

[2] (a) X(Ω) = π[u(Ω + 1)− u(Ω− 1)]; (b) Ts ≤ π sec/sample; (c) Ts ≤ π/2; (d) Ts = π

12

[3] (a) Y (Ω) = e−jΩ(sin(Ω/2)/(Ω/2))2; (b)(c) y(t) smoother than x(t), since Ωx = 5.6 > Ωy = 2.6

rad/sec, Tsy > Tsx, aliasing if x(t) is sampled using Tsy ; (d) Tsx appropriate for sampling x(t)

and y(t)

[4] (a)(b) Ωmax ≈ 22 rad/sec; computing ∆(t) and ∆(Ω) the uncertainty bound is satisfied.

[5] (a)X(Ω) = 2π[u(Ω+0.5)−u(Ω−0.5)]; (b)(c) bandlimited, Ts ≤ 2π; (d)Xs(Ω) = (1/(2π))∑kX(Ω0k) =

1, passing x(t) through a low-pass filter with amplitude 2π and cutoff frequency 1/2 we ob-tain the original signal

[6] (b) Maximum frequency is 3.3 rad/sec; (d) antialiasing filter: ideal low-pass of magnitude 1and cut-off frequency 3.3 rad/sec

[7] (a)X(Ω) = 0.5[M(Ω−Ωc)+M(Ω+Ωc)]; (b) Ts ≤ (1/64)×10−3 sec/sample; (c) use bandwidthof signal

[8] (a) Ωmax y = 4000π; (b) maximum frequency of Y (Ω) remains the same, Ts ≤ 0.25× 10−3; (c)No.

[9] (a) Ts = 1, Ωs = 2π ≥ 2Ωmax, low-pass filter of magnitude 1 and cutoff frequency Ωmax <

Ωc < 2π − Ωmax; (b) maximum frequency should be less than π

[10] CD player is reproducing a signal that has been processed by an antialiasing filter beforebeing sampled, quantized and coded. High frequencies might have been filtered out and alsoquantization error is present possibly making the reproduced signal of less quality than theoriginal

[11] (a)-(c) Sampled signal values x(nTs) = 0.8 cos(2πn0.025) + 0.15, 0 ≤ n ≤ 40

Chapter 8[1] x[3] = 2, x[4] = 3, x[5] = 5

[2] (b) x[n] has finite energy, 4/3

[3] (a) A = 1, ω0 = 3πTs; (b) Ts = (2m)/(3N) for integers m and N

[4] (a) x[−n] = (1 + 0.25n)[u[−n]− u[−n− 5]]; (b) even component

xe[n] =

0.5(1− 0.25n) 1 ≤ n ≤ 4

1 n = 0

0.5(1 + 0.25n) −4 ≤ n ≤ −1

(e) ε2x = ε2

xe + ε2xo

[5] (a) r[n] = u[n− 1] + u[n− 2] + · · ·

13

[6] (a) Period for n ≥ 0 x[n] = r[n]− 2r[n− 1] + r[n− 2]; (b) period 3

[7] (a) z[n] = x[2n] = cos(4πn/7); (b) y[n] = x[n/2] periodic of period 14; (c) upsampled signalcannot be obtained by decreasing the sampling period.

[8] (a) α = e−2; (b) x[n] is absolutely summable; (c) x[n] is finite energy.

[9] x[n] is periodic of period N0 = 2; z[n] is not periodic; v[n] is periodic.

[10] (a) z[n] is periodic of period 12; w[n] is periodic.

[11] (a) For non-divisible integers K and M , x[n] + y[n] is periodic of period MN1 = KN2; (c)when x[n] has period N1 = 3 and y[n] is periodic of period N2 = 2, the gdc[2, 3] = 1, soN0 = 6.

[12] (a) Ti = 2Ri/343

[13] (a) T0 = 3Ts; (b) sampled envelope is

e(nTs) = (1/3) [r(nTs)− r((n− 30)Ts)− 0.1r((n− 200)Ts)− r((n− 300)Ts)]

[14] (a) Sampler is linear; (b) sampler is time-varying

[15] (a)(b) Quantizer is non-linear but time-invariant

[16] (a)(b) Windowing is linear but time-varying, consider x[n] = nu[n], w[n] = u[n] − u[n − 6]

then output y[n] = nw[n], for x[n− 6] then x[n− 6]w[n] = 0 6= y[n− 6]

[17] (a) h[n] = 0.15(n/2) for n ≥ 0 and even, 0 otherwise

[18] (a) h[n] =∑5k=1 kδ[n− k]; (b) causal; (c) s[n] =

∑5k=1 ku[n− k]; (d) bounded ouput

[19] (a)(b) By superposition and by convolution sum: y[n] = h[n] + h[n− 1] + h[n− 2]

[20] (a) a = 0.5; (b) h[n] = 0.5nu[n]

[21] (a) A; (b) either

[22] (a) h1[n] = (−0.5)nu[n]; (b)(c) both IIR and FIR systems are BIBO stable; (d) FIRs always stable

[23] (a) If s[n] is unit-step response then h[n] = s[n]− s[n− 1]; (b) the ramp response is

ρ[n] =

∞∑k=0

s[n− k]

[24] (b) y[0] = 1, y[1] = 1.5, y[2] = 7/4, y[n] = 15× 0.5n for n ≥ 3; (b) use superposition

14

Chapter 9[1] (a) Poles s = −1± j1 are mapped into z = e−1e±j1; (b) z = eσejΩ for Ts = 1

[2] (a) s = log(z)/Ts; (b) z = 1 is mapped into s = 0, z = −1 into s = ±jπ/Ts

[3] (a) S1(z) = 1/(1− z−1), ROC |z| > 1; (b) No z-transform

[4] (a) X(z) = −1/(1− z/α), ROC |z| < |α|; (b) −α/(z − α)2 ⇔ (n− 1)α(n−1)u[−n]

[5] (a) X2(z) = 1/(1− z−1), |z| < 1, (b) without ROCs it would not be possible to distinguish theZ-transforms of x1[n] 6= x2[n].

[6] (a) F (z) = z−1/(1− z−1 − z−2); (b) poles φi, i = 1, 2 are such that φ1 + φ2 = 1

[7] (a) x(0.1n) = [u(t)− u(t− 1)]t=0.1n; (b) N = 11; (d) X(z) = Xs(s)|z=e0.1s

[8] (a) (b) X(z) = (1− z−10)/(1− z−1); (c) X(z) has not poles on the unit circle

[9] (b) X(z) = 2/(1− αz−1), ROC |z| > |α|

[10] (b)(c) X(z) = 1/(1− z−2) |z| > 1

[11] (a) Y (z) = X(z)/(1 + 0.5z−1)− 0.5y[−1]/(1 + 0.5z−1); (c) H(z) = 1/(1 + 0.5z−1)

[12] (a)(b) H(z) = z2/[(z − rejω0)(z − re−jω0); (c) For ω0 = π/2 or 3π/2 we get the same transferfunction

[13] (a) y[n] = 2 cos(ω0)y[n− 1]− y[n− 2] +Ax[n]−A cos(ω0)x[n− 1], n ≥ 0, zero ICs, x[n] = δ[n];(b) Let y[n− 1] = y1[n] where y[n] is above solution

[14] (b) X(z) =∑9k=0 z

−k, 9th-order polynomial; (c) pole-zero cancellation

[15] (a) y[−2] = 4, y[−1] = 2; (b) steady state 4/3

[16] (a) x[n] = δ[n] + 0.5δ[n− 1]; (b) h[n] = 0.5[(−0.5)n + 0.5n]u[n]

[17] (b) y[n] = δ[n] + 2δ[n− 1] + 3δ[n− 2] + 2δ[n− 3] + δ[n− 4]

[18] x[n] = [−3(−0.25)n + 4(−0.5)n]u[n], zero steady state

[19] (b) f [n] = u[n− 2] + u[n− 3]

[20] (c) y[n] = [4 + 3.16(0.707)n cos(πn/4− 161.5o)]u[n] steady state 4

[21] (a) Equation for denominator coefficients:h[M − 1] h[M − 2] · · · h[M −N + 1]

h[M ] h[M − 1] · · · h[M −N + 2]...

... · · ·...

h[M +N − 3] h[M +N − 4] · · · h[M − 1]

a1

a2

...aN−1

=

−h[M ]

−h[M + 1]...

−h[M +N − 2]

15

After denominator coefficients are found, replace them in the following equation to get thebk:

h[0] 0 · · · 0

h[1] h[0] · · · 0...

... · · ·...

h[M − 1] h[M − 2] · · · h[M −N ]

1

a1

...aN−1

=

b0

b1...

bM−1

(b) H(z) = 1/(1− 0.5z−1)

[22] (a)(b) Hd(z) = 1/(1− 0.9−1); (c) compute convolution (h1 ∗h2)[n] and the find Prony of order(2,2)

[23] (a) (n + 1)an+1u[n] ⇔ a/(1 − az−1)2; (b) X(z) = 0.5z−1/(1 − 0.5z−1)2 with an inversex[n] = 0.5nnu[n] according to Z-transform table and also using result in (a) in the expansion

[24] A = 2, C = 2, B = −4, and D = 2, F = 4, E = −2 giving same x[n] = 2(1− 2n(1− n))u[n].

[25] (a) H(z) = 2z/(1 − 0.5z−1) use Prony of order (1, 1) assuming h[n] is causal and adjust forshift; (b) separate the signal into causal and noncausal

Chapter 10

[1] (a) H(ejω) = (1− 2ejω)/(1− 0.5ejω); (b) |H(ejω)| = 2

[2] (a) H1(ej0) = 4, H1(ejπ) = 0; (b) h2[n] = (−1)nh1[n]

[3] (a) X(ejω) = 0.75/(5/4− cos(ω)); (b) X(1) = 3; (c) x[0] = 1; (c) zero phase

[4] (a) H(ejω) = e−jω(1 + 2 cos(ω))/3; (b) H1(ejω) = H(ej(ω−π))

[5] (a) x[n] is not periodic; (b)x[n] = cos(0.5πn+ π/4)/π − 2 cos(0.71n− π/8)

[6] (a) IF (n) = πn/L; (b) chirp would jam all possible discrete frequencies

[7] (a) h[n] symmetric with respect to n = N ; (b) the low-pass filter is shifted to ±π/2

[8] (a) P (ejω) = (1− e−jωN )/(1− e−jω); (c) y[n] = 0 in steady state

[9] (a) 0.5X1(ejω/2) 6= X(ejω)

[10] (a) Equivalent to cascade of upsampler of rate M , the lowpass filter of the decimator withgain M , and downsampler of rate M ; (c) interpolating with L = 3 the input signal with 100

samples gives a signal with 300 samples, which when decimated with M = 2 gives a signalwith 150 samples

16

[11] (a) Angle of complex numbers in second and third quadrants is obtained like in the first andfourth quadrant and subtracted and added to π; (c) X(z) = z − 1 then X(ej0) = 0 and anyvalue can be chosen for phase, when X(z) = 1/(1− z) then X(ej0)→∞ and the phase is notdefined

[12] (a) |X(ejω)| = 2, ∠X(ejω) = −4ω

[13] (a) X(ejω) = (A+ 1) +∑9n=1 2 cos(nω); (b) phase corresponding to x1[n] is linear

[14] (a)Ak = 3−|k| for−2 ≤ k ≤ 2, zero otherwise; (b) T (ejω) has coefficientsB0 = 3,Bk = 2(3−k)

for k = 1, 2 and zero otherwise

[15] (c) X(ejω) = e−jω + 2e−j2ω , Y (ejω) = ejω + 2ej2ω , Z(ejω) = 2 cos(ω) + 4 cos(2ω), |Y (ejω)| =

|X(ejω)|

[16] (a) X(ej0) = 9, X(ejπ) = 1, (c) phase of X(ejω) is −2ω when X(ejω)ej2ω > 0 and −2ω ± πwhen that term is negative

[17] (c) Xe(ejω) = 0.5(2− cos(ω))/(1.25− cos(ω)), Xo(e

jω) = −0.5j sin(ω)/(1.25− cos(ω)); (d) useParseval’s result to show that Ex = Exe + Exo

[18] (b) There is a zero at zero and poles at z = 2 and z = 0.5; (c) C(ej0) = 3, zero phase

[19] (a) The convolution sum gives z[n] = δ[n − 2] + 6δ[n − 3] + 11δ[n − 4] + 12δ[n − 5]; (b)V (z) = 1 + z−1 + z−2, W (z) = (3z2 + 2z + 2z−1 + 3z−2)V (z)

[20] (a) y[n] = w[n] sin(0.1πn), Y (ejω) = (1/2j)(W (ejω−0.1π)−W (ejω+0.1π)

[21] (a) Using the z-transform Xk = (1/N)X1(z)|z=ej2πk/N where X1(z) is the z-transform ofx1[n] = x[n](u[n]− u[n− 10])

[22] (a) X = Sx; (c) forN = 4,Xk = 0.25x[0]+[x[2](−1)k+(−j)kx[1]+x[3](−1)k, k = 0, 1, 2, 3.

[23] (a) x1[n] is periodic of periodN and FS coefficientsXke−j2πN0k/N ; (c)N should be even; (d)(e)

X0 = X2 = 0 and X1 = X3 = 0.5(1 + j)

[24] (b) Independent of the period N , for a cosine of period N X1 = X−1 = 0.5 and for a sineX1 = −X−1

[25] (a)-(c) To recover x[n] we demodulate y[n] by multiplying it by (−1)n and then lowpass filterit

[26] (a) Since the length of x[n] is 70 then 2γ ≥ 70 gives γ ≥ 6.13 so we choose γ = 7

[27] (a) When the signal is aperiodic padding it with zeros increases the frequency resolution,amplitudes remaining the same; (b) the fundamental frequencies of the periodic signal remainthe same when we increase the number of periods to improve the frequency resolution

17

[28] (a) The approximation is good because the impulse response decreases significantly after n =

30; (b) no truncation is done and results are better than before

[29] (a)(b) Z(z) = X(z)Y (z) and letting z = ejω it becomes the DTFT of z[n]; (c) N = 4 < length ofx[n] +length of y[n]−1 = 6 so that the linear and the circular convolutions do not coincide; (d)X(k)Y (k) = e−j(2πk/N)+3e−j2(2πk/N)+6e−j3(2πk/N)+5e−j4(2πk/N)+3e−j5(2πk/N) and forN =

4 e−j4(2πk/4) = e−j0(2πk/4) and e−j5(2πk/4) = e−j1(2πk/4) which when replaced the coefficientsof X(k)Y (k) coincide with the values obtained in the circular convolution of length N = 4.

Chapter 11

[1] (a) H(ejω) = β(2 + 2α cos(ω)); (b) for α = β = 0.5 H(z) has pole z = 0 and double zero atz = −1; (c) G(z) has linear phase

[2] (b) H(z) = (1/9)(z2 + 4z + 6 + 4z−1 + z−2), non-causal filter; (b) h1[n] = h[n − 2] is causalwith linear phase

[3] (a) h1[n] is even with respect to n = 2, h2[n] is odd with respect to n = 2; (b) H1(z) has linearphase; (c) letting F (z) = z2H2(z) it can be shown F (ejω) is purely imaginary and used todetermine the phase of H2(ejω)

[4] (a) |H2(ejω)| = 1; (c) when phase is linear the delay it causes is known and can be reversed,not so for nonlinear phase

[5] Nb = 10 , the Chebyshev filter with the same order has a narrower transition band, [1000 1035],than that of the Butterworth filter [1000 1200] Hz

[6] H(z) = (z + 1)2/((2 +√

2)(z2 + (2−√

2)/(2 +√

2)), ωhp = π/2

[7] Increase in order changes results, warping makes poles cluster around π

[8] (a)(c) G(s) is all-pass with linear phase, H(z) is also all-pass with non-linear phase.

[9] Minimum order N = 7,

H1(z) =1

s+ 1|s=BT

Hi(z) =1

s2 + (1/QI)s+ 1|s=BT

[10] (a)(b) Given filter is all-pass,K = 1.4 gives the desired response, discrete filter is all-pass withapproximately linear phase

[11] Desired filter is a low-pass with fp = 1 KHz, αmax = 3 dB, fst = 2 KHz, αmin = 10 dB

[12] Generate noise using MATLAB expression noise=0.1*randn(1,256)

18

[13] The minimum orders of the filters is Nb = 20, Nc = 10 and Ne = 6

[14] (a) H(z) = K(1− 4z−2)/(1− 0.25z−2); (b) H(z) = K(1− z−2)/(1− 0.25z−2)

[15] (a)(b) H(z) = K(1 + z4)/(1/16 + z4), magnitude response looks like a comb, phase cannot beunwrapped because of zeros on unit circle; (c) notch filters connected in parallel give sharperresponse around notches

[16] Desired bank of filters is obtained by connecting in parallel low-pass, band-pass and high-pass filters

[17] The cutoff frequency for FIR is ωc = 2π(fc/fs), designed filter using rectangular window

H(z) = 0.25 +

20∑n=0,n6=10

sin(π(n− 10)/4

π(n− 10)z−n

[18] Desired impulse response is

hd[n] =

sin(πn/3)/(πn) n 6= 0

0.33 n = 0

[19] (a) Modulating h[n] of IIR filter gives

H(z) = H(ejω0z) +H(−ejω0z)

=B(ejω0z)

A(ejω0z)+B(e−jω0z)

A(e−jω0z)

(b) If a = 0.5 and ω0 = π/2, the band-pass filter is

H2(z) =2

1 + 0.5z−2

[20] (a) The downsampled impulse response is g[n] = 0.25nu[n]; (b) G(ejω) = 1/(1− 0.25e−jω)

[21] (a) h[n] = (1/3)(δ[n] + δ[n− 1] + δ[n− 2]), G(ejω) = H(−ejω)

[22] Use the convolution property, so that if D(z) has d coefficients, then D2(z) has coefficientsd ∗ d, i.e., the convolution sum of the coefficients d with itself.

[23] (a) h1[n] = h[n/2] for n = 0, 2, 4, · · · and zero otherwise, H1(z) = H(z2)

[24] (a) Cascade

H(z) = H1(z)H2(z) =2(1− z1)

1 + 0.5z−1

1 +√

2z−1 + z−2

1− 0.9z−1 + 0.81z−2

(b) Parallel

H(z) = −4.94 +2.16

1 + 0.5z−1+

4.78− 1.6z−1

1− 0.9z−1 + 0.81z−2

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[25] Use the z-transform of

g[n] = 2v[n] + 1.8v[n− 1] + 0.4v[n− 2]

v[n] = x[n] + 1.3v[n− 1]− 0.8v[n− 2]

to get the transfer functionG(z)/X(z). For the second cascade Y (z)/G(z) = (3+4.5z−1)/(1−0.3z−1)