HW Solutions

EE 321B Spring 2000

Last updated April 4, 2000

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HW 1 Solutions

1.3-1 f2(t) = f(t-1) + f1(t-1)

1.3-2 f3(t) = f(t-1) + f1(t-1)

1.4-5 (h) -e^2

1.4-10 -phi dot (0)

1.7-1 (e) nonlinear

1.7-1 (f) linear

1.7-1 (h) nonlinear

1.7-2 (d) TV

1.7-2 (e) TV

1.7-2 (f) TI

1.7-7 (c) noncausal

1.7-7 (d) noncausal

1.7-8 (c) not invertible for even values of n, invertible for odd values of n

1.7-8 (d) not invertible

extra question: see class notes

HW 2 Solutions

2.2-3 y 0 (t) = 2 - e^(-t)

2.2-6 y 0 (t) = 5 + 2t - e^(-t)

2.3-4 h(t) = ( 2 + 3t)(e^(-3))[u(t)]

2.4-2 (1/a)c(at), a > 0; -(1/a)c(at), a < 0

2.4-6 (i) (1-cost)[u(t)] (ii) sint[u(t)]

2.4-8 (a) [(1/6) - (2/3)e^-3t + 0.5e^(-2t)]u(t)

(b) (e^(-2t) - e^(-3t))[u(t)]

2.4-16 (a, d, e, h) see instructor for answers

2.4-19 see class notes

2.6-1 (a) A.S.

(b) M.S.

(d) U

2.6-3 (a) 0

(b) M.S

(d) Ideal integrator

HW 3 Solutions

3.1-3: c = 1.5; E sub e = 0.25.

3.1-4 (a) c = - (1/pi);

(b) e(t) = t + (1/pi)*sin(2*pi*t) on the interval [0,1], and zero outside this interval.

E sub e = (1/3) - (1/2*pi^2)

3.2-1 (1) c sub n = 0

(2) c sub n = -1

(3) c sub n = 0

(4) c sub n = 1.414/pi

3.4-1 for sketches see instructor

f(t) = 1/3 + [4/(pi^2)]*[sum from 1 to inf]{[(-1)^n/n^2]*cos(n*pi*t)}, on the interval [-1,1]

3.4-3 (a) f(t) = (4/pi)*{cos(pi*t/2) - (1/3)cos(3*pi*t/2) + (1/5)cos(5*pi*t/2) - (1/7)cos(7*pi*t/2) + …)

(b) f(t) = (1/5) + [sum from 1 to inf]{a sub n cos(n*t/5)}, where an = [2/(pi*n)]sin(n*pi/5)

(d) f(t) = (4/pi^2)cos(2t - pi/2) + (1/pi)cos(4t - pi/2) + (4/9*pi^2)cos(6t + pi/2) + (1/pi)cos(8t + pi/2) + …

(e) C sub 0 = 1/6;

C sub n = (3/2*pi^2*n^2){sqrt[2 + (4*pi^2*n^2)/9 - 2cos(2*pi*n/3) - [(4*pi*n)/3]sin(2*pi*n/3)]}

Theta sub n = arctan{[(2*pi*n/3)cos(2*pi*n/3) - sin(2*pi*n/3)] / [cos(2*pi*n/3) + [(2*pi*n)/3]sin(2*pi*n/3) - 1] }

(f) f(t) = 0.5 + (6/pi^2){cos(pi*t/3) - (2/9)cos(pi*t) + (1/25)cos(5*pi*t/3) + (1/49)cos(7*pi*t/3) + …}

3.4-5 (a) C sub 0 = 0.504

C sub n = 0.504*{2/[sqrt(1 + 16*n^2)]}, theta sub n = -arctan(4n)

(b) identical to the F.S. in Eq. 3.56a, with t replaced by 2t

(c) scaling by a factor k scales the fundamental frequency by the same factor k. everything else remains the same. Time compressing a periodic signal by a factor k increases the original fundamental frequency by a factor of k.

3.4-11 Typo in problem statement: use f(t) in Fig. P3.4-10 (not P3.4-11).

c sub 1 = 0.2; c sub 2 = -0.25; c sub 3 = c sub 5 = c sub 6 = c sub 7 = 0; c sub 4 = -0.125; c sub 8 = - (1/16).

Let E sub e (N) = energy of error signal in the approximation using first N terms. Then

E sub e ( 1) = 0.0833; E sub e ( 2 ) = 0.0204; E sub e ( 3) = 0.0052; E sub e ( 4) = 0.001302

The Walsh Fourier series gives smaller error than the trig F.S. in Prob. 3.4-10 for same number of terms in the approximation.

3.5-1 (a) D sub 0 = 0

D sub n = (2/pi*n)sin(n*pi/2); |n| > =1

(b) D sub n = (1/pi*n)sin(n*pi/5)

(d) D sub 0 = 0; for n != 0, D sub n = (-j/(pi*n))*{(2/pi*n)sin(pi*n/2) - cos(pi*n/2)}

(e) |D sub n | = (3/4*pi^2*n^2){sqrt[2 + (4*pi^2*n^2)/9 - 2cos(2*pi*n/3) - [(4*pi*n)/3]sin(2*pi*n/3)]}

angle(D sub n)= arctan{[(2*pi*n/3)cos(2*pi*n/3) - sin(2*pi*n/3)] / [cos(2*pi*n/3) + [(2*pi*n)/3]sin(2*pi*n/3) - 1] }

(f) D sub 0 = 0.5; D sub n = (3/pi^2*n^2)*{cos(n*pi/3 - cos(2*pi*n/3)}

3.5-2 f(t) = 3 + 2cos(2t - pi/6) + cos(3t - pi/2) + 0.5cos(5t - 2*pi/3)

f(t) = 3 + exp(j(2t - pi/6)) + exp(-j(2t - pi/6)) + 0.5exp(j(3t - pi/2)) + 0.5exp(-j(3t - pi/3) + 0.25exp(-j(5t - 2*pi/3)) + 0.25exp(-j(5t - 2*pi/3))

3.5-3 (a) f(t) = (2*sqrt(2)exp(j*pi/4))*exp(-j3t) + 2*exp(j*pi/2)exp(-jt) + 3 + 2exp(-j*pi/2)exp(jt) + (2*sqrt(2)exp((-j*pi/4))exp(j3t)

(b) f(t) = 3 + 4cos(t - pi/2) + 4sqrt(2)cos(3t - pi/4)