Appendix: Solution to Selected Problems

Solutions, Chapter 2

2.1 Since X(O) = I: x(t)dt,

the transforms in Table 2-2 can be used. The functions II(t),sinc(t),A(t) and jinc(t) all have unit area. Jo(27rt) has an area of 1/7r.

2.2 (a) 6(t) = limA_oo AA(At).

(b) 6(t) = limA_oo Ajinc(At).

(c) 6(t) = limA_oo Asinc2(At).

2.3

In each case, the limit of the transform approaches one. For example

A jinc(At) ~ )1- (u/A)2II(~) 2A

---+ 1 as A -+ 00.

Selected solutions

(a) Derivative

(!)nx(t) - 100 X(u) ( !£) n ej21rutdu -00 dt

- I: (j27rut X(u)ej21rutdu.

280 Solutions

(b) Convolution

x(t) * h(t) +---+ i:[i: x(r)h(t - r)dr]e-j21rutdt.

Reverse integration orders and applying the shift the­orem gives

x(t) * h(t) +---+ H(u) i: x(t)e-j21rutdt.

( c) Inversion

i: X(v)ej21rvtdv +---+ i:[i: X(v)ej21rvtdv]e-j21rutdu.

Reverse integration order and note

Proof results from sifting property.

(d) Duality

2.4 (a)

(b)

X(t) +---+ i: X(t)ej21r(-u)tdt

= x{-u).

S(t) +---+ i: S(t)e-j21rutdt

=1

by the sifting property of the Dirac delta

IT(t) +---+ 11/2 e-j21rutdt -1/2

_ -1 -.!...e - j21rut 11/2 ~ 7rt j2 -1/2

= sinc{u).

281

(c) Use (2.14) and the convolution property in Table 2.

(d) Use (2.20). The result follows from linearity.

(e) Use (2.15) with v = 1 and apply (2.18).

2.5 (a) g(t) ~ a(t - to) around the neighborhood t = to.

(b)

Since b(t) is nonzero only when e = 0, we conclude

and

b(g(t)) - b(a(t - to)) 1

- ~b(t - to).

g(t) = In(tlb)j to = b g'(t) = lit ~ a = lib

b(ln(tlb)) = M(t - b).

(c) The argument has an infinite number of zero cross­mgs:

sin(7rtn) = 0 j tn = 0, ±1, ±2, ....

Since

we have

2.6 (a) From duality

g'(tn) - 7rcos(7rtn )

- (-lt7r

00 1 - 7r nI=oo I an Ib(t - tn)

= comb(t).

sinc(t) ~ IJ(u)

282 Solutions

2.7 (a) (i) Use Parseval's theorem with x(t) = sinc2 (t):

f11 A2(u)du

2 10\1 - u)2du

2/3.

(ii) Same approach, but x(t) = jinc(t):

f: jinc2(t)dt

(iii) Parseval's theorem gives

f11 (1 - u2)du

4/3.

1 11 du 71"2 -1 1 - u 2

00.

(b) Using the power theorem and (2.19) gives

f: sinc(t)jinc(t)dt =

2.8 (a) Use the Poisson Sum Formula:

00

L sinc(t - n)

11/2 VI - u2du

-1/2

J3 71" -+-. 4 6

00 L II( n )ej21rnt

n=-oo n=-oo

1

since II(n) = b[n].

(b) Same result since A(n) = b[n].

(c) Same result since VI - n2II(n/2) = b[n].

283

(d) sinc(at) +---+ IT(u/a)/a. Thus

00

L sinc[a(t-n)] - ! f: IT (~) ej27rnt a n=-oo a n=-oo

where N is the greatest integer not exceeding a/2. Using a geometric series:

~ . [( )] sin[(2N + 1 )7rt] L..J smc a t - n = -::...:-.----:---:-'-~

n=-oo a sm( 7rt)

Part (a) is a special case for a = 1 (N = 0).

00 1 N L sinc[a(t - n)] = -[1 + 2 L cos(27rnt)].

n=-oo a ~1

Part (a) is special case for a = l(N = 0).

(e) Use (2.24) with T = 2

00

L jinc(2n) ! f: VI - (n/2)2IT(n/4) 2 n=-oo n=-oo

(1 + V3)/2.

(f) Use (2.24) with T = 1/4 :

00 00

L jinc2(n/4) = 4 L C{4n) n=-oo n=-oo

where

C(u) - VI - u2 IT{u/2) * VI - u2IT(u/2) C(u)IT(u/4).

Thus: 00

L jinc2(n/4) = 4C(0) n=-oo

284 Solutions

where

C(O) {II (1 - u2 )du

4/3.

(g) The n > 0 and n < 0 terms cancel. The n = 0 term IS zero.

2.9 (a) (ft)sinc(t) = 0 ==> 0 = tan(O);O = 7rt. The nth extrema location, On, can be founded iteratively by On[m] ---+ On as m ---+ 00 with

On[m+l] = n7r+arctan(On[m]) ; n = 0, ±1, ±2, ....

The first few locations and the corresponding ex­trema are listed in Table A.I.

(b) On ---+ n + 1/2 as n ---+ 00 and

( _l)n sinc( t n ) ---+ ( 1 r

7rn+'2

(To justify, compare plots of 0 and tan( 0) vs. 0).

2.10 The Fourier series is 00

y(t) = a L: sinc(an)ei21rnt/T. n=-oo

Thus 00

y(r/2) = a L: sinc(an) cos(n7ra). n=-oo

Write sinc as sin(7rx)/(7rx) and use a trigonometric iden­tity:

00

y( r /2) = a L: sinc(2an). n=-oo

We evaluate using (2.24) which, for x(t) = sinc(t), can be written

00

a L: sinc(2an) n=-oo

~ f= n(~) 2 n=-oo 2 a

1/2.

285

I n I ()n I sinc( ()n) I 1 4.4934 -0.21723 2 7.7253 0.12837 3 10.9041 -0.09133 4 14.0662 0.07091 5 17.2208 -0.05797 6 20.3713 0.04903 7 23.5194 -0.04248 8 26.6661 0.03747 9 29.8116 -0.03353 10 32.9564 0.03033 11 36.1006 -0.02769 12 39.2444 0.02547

Table A.1: The first few extrema of sinc(t).

2.11 Since the integrand in (2.24) is even

J (27rt) (7r/2)1I 11 1 II = (1 - u2Y-2" cos(27rut) duo (2t)1I v;rr(v +~) -1

Thus we have the transform pair

The entries in Table 2-2 for v = 0 ,1 follow immediately using (2.16).

2.12 With attention to Fig.A.l:

T T t-- t+-ZT(t) = II( T/;) - II( T/;)·

Thus Z ( ) .. (Tu) . (7rut)

T u = - ) SIne 2 sm 2 .

286 Solutions

1

z{t)

(' T/2

-1

T

Figure A.I

t

The Fourier coefficients follow as:

1 en = TZT(n/T)

-j sinc(n/2) sin( 7rn/2).

Thus, since z( t) is real :

00

z(t) = L -jsinc(n/2) sin(7rn/2) ei21fnt/T n=-oo

00 • 27rnt L sinc( n/2) sin( 7rn/2) sm( T)' n=-oo

The truncated series is

N. . . 27rnt ZN(t) = L smc(n/2) sm( 7rn/2) sm( T)'

n=-N

Extrema come from:

o d dtZN(t)

27r N 27rnt - T E sinc(n/2) sin(7rn/2) n cos( T)

n=-N

4 N 27rnt = T L: sin2 ( 7rn/2) cos( T ).

n=-N

2 N n 27rnt o = - L: [1- (-1) ] cos(-) T n=-N T

- ~~ E [1- (_l)n] ei21rnt/T. n=-N

Consider the geometric series:

Thus

N

S = L: an = a-N + a-N +1 + ... + aN. n=-N

Substituting: aN+t - a-(N+t>

S=----,--­a1/ 2 _ a-1/ 2

For a = _e-i21rt /T = _ei6

S _ (_l)N cos(N + ~)O urn - cos(0/2) .

For a = ei6

S _ sin(N + ~)O urn - sin(0/2) .

Thus

0= sin(N + !)O _ (_l)N cos(N + !)O sin(0/2) cos(0/2)

or, for N even:

1 tan(N + "2)0 = tan(0/2).

287

288 Solutions

Therefore

or 8 = 27rt = p7r ==> t = pT

T N 2N' The first extrema is at p = 1 corresponding to t = T/2N. Substituting gives

~ sin2( 7rn/2) . ( /N) - n:::N 7rn/2 sm 7rn

_ 2 ~ sin2 (7rn/2) . ( /N) L.J /2 sm 7rn n=l 7rn

2 ~ 1 - (_l)n . ( /N) - - L.J sm 7rn .

7r n=l n

Let 2m + 1 = n

T 4 N/2 • 1I'(2m+1)

( ) "" sm N ZN 2N = -; ~o 2m + 1 .

Define

h(t) = f sin(2m + l)t m=O 2m + 1

AI = N/2

so that T 4

ZN(2N) = -;h(7r/N).

Note that h(O) = O. Now

Thus:

dh(t) dt

N

- L cos(2m + l)t m=O

sin2(M+l)t -

2 sin(t)

h(t) = r sin2(M + l)T dT Jo 2sin(T)

289

and ZN( I...-) = ! r/N sin(N + l)r dr.

2N 7r Jo 2sin(r) Since the interval of integration gets smaller and smaller, sin r ~ r and as N -+ 00

T ZN(-) ~ .i r1r/N sin(N+2}r dr 2N 1r JO 2r

_ 1 r1r+~ sin(e)dt - 1r JO ~ 1,,'

Since Si(7r) = 1.8519370, we conclude that, as N -+ 00,

~ ~ 1 - ~Si(7r)

= 0.1789797.

2.13 (a) ~2 = R~(O) = f~oo S~(u)du. (b) ~2 = Re[O].

Note that (2.28) is a Fourier series with period 1. Thus

~2 = 1r Se(u)du

where integration is over any period.

2.14 These derivations can be found in most any introductory text on stochastic processes.

2.15 Same as 2.14.

2.16 (a) From (2.26)

(b)

R~(r - t) E[~( r) C(t)]* - R~(t - e).

S;(u) i: R~(t) ej21rut dt

i: R~( -r) e-j21rur dr

i: R~(r) e-j21rur dr

S~(u).

290 Solutions

1 1(0,1)1(1,00)1

I~I ! I ~ I Table A.2

2.17 Using the shift theorem

comb(t - 1/2) +-+ comb ( u) e-j1rU

= ,",00 b(u _ n) e- j1rn i..Jn=-oo = L~=-oo( _1)n b(u - n).

Separating the sum into even and odd n completes the problem.

2.18 (a) See Table A.2.

(b) on the interval (0, 1), x( t) has finite area and energy. On the interval (0, 00), the area is infinite and the energy is finite.

( c) A < 00 ===} F < 00. This follows from the proof that the space of 11 sequences is subsumed in 12 [Lu­enburger: Naylor and Sell]. The converse is not true. Consider

2.19 (a)

x(t) = sinc(Bt).

Clearly, E = 1/ B. However, since

we have the divergent series

n=O odd n otherwise

2 00 1 A = 1 + - L = 00.

7r k=O 2k + 1

I x(t) I = I J~oo X(u) ejhut du I $ J~oo IX(u) I du

=A.

2.20

291

Thus, C = A. Note, then, that x(t) is bounded if X ( u) has finite area.

(b) A counterexample is x( t) = sgn( t).

1 x(t) 12 = 1 J~B X(u) ej27rut du 12

~2BJ~B IX(u)12 du =2BE.

2.21 (a) Yes, because applying Parseval's theorem to the deriva­tive theorem gives

Ep i: 1 x(p)(t) 12 dt

i: 1 (j27ru)P X(u) 12 duo

Since (u/ B)2P < lover the interval 1 u 1 < B,

Ep = (27rB)2P J~B (U/B)2p IX(u) 12 du

~ (27rB)2p J~B 1 X(u) 12 du = (27r B)2p E

where E is the energy of x(t). Also, x(p)(t) is clearly bandlimited.

(c) Yes:

Ix(p)(t) 1 = IJ~B (j27ru)P X(u) ei27rut du 1

~ J~B 127ru IP 1 X(u) 1 du

= (27rB)P J~B lu/BIP IX(u)1 du ~ (27r B)P A

where A is the area of X( u).

292 Solutions

2.22 The series converges absolutely if

S(t) = f: It - ~ 1m I X(m)(T) I < 00.

m=O m.

From the derivative theorem for Fourier transforms,

S(t) = E:=o It:r I I~B (j211"u)m X(u) ei21fut du I :::; E:=o It:r I~B (211"u)2m du]1/2

·[I~B IX(u) 12 du]1/2

where, in the second step, we have used Schwarz's inequal­ity. Since x( t) has finite energy,

E = i: I X(u) 12 du

is finite. Thus

S(t) :::; .j2jj E ,,00 (21fB j-T 12)m L.."m=O m! 2m+l

< . Iij]j E ,,00 (271B 1 t-T Um

V~D L.."m=O m!

= .j2jj E e21fBlt-T I.

This bound is finite for all finite t and T.

2.23 (a) From the derivative theorem

I x(M) (t) 12 I i: (j211"u)M X(u) du 12

< 2(211" )2M E i: u2M du

which, when evaluated, gives (2.39).

(b) Clearly

I x(t + T) - x(t) I - 11t+T X'(~) d~ I

< it+T I X'(~) I d~

< j (211" B)3 E It+T 3 t d~

which, when evaluated, gives (2.40).

293

2.24 Define Ef = i: 1 f(t) 12 dt.

Then, applying Schwarz's inequality to the integral gives

Note that Exercise 2.19 is a special case for the Fourier inversion integral over the interval 1 u 1 $ B.

294 Solutions

Solutions, Chapter 3

3.1 The Fourier dual of the Poisson sum formula is 00 00

2B L X(u - 2nB) = L x(.!:.-) e-j1rn11./B.

n=-oo n=-oo 2B

Since X(u) = X(u) Il(u/2B), we multiply both sides by Il( 2~) and inverse transform. The sampling theorem series results.

3.2 Use the Fourier dual of the Poisson sum formula again: 00 00 L X(u - n) = T L x(nT) e-j21rn11.T.

n=-oo T n=-oo

For B < liT < 2B, the sum on the left will be the over­lapping aliased version in Fig.A.2. Note that no spectra overlap the zeroth order spectrum at u = o. Thus

00 n L X(u - T) 111.=0 = X(O) ; B < liT < 2B

n=-oo

and 00 i: x(t) dt = T L x(nT);T < liB.

n=-oo

3.3 If x(t) is bandlimited, so is x(t + 0:). Thus

00 n x(t + 0:) = nI=oo x(2B + 0:) sinc(2Bt - n).

Substitute t - 0: for t and we're done.

3.4 Since (-j21rt)m i--+ 8(m)(t)

we use (3.4) and ask the question:

t m ? 1 sin(21r Bt) f 1r(2B)m n=-oo

(_l)n nm 2Bt - n .

The result is clearly a divergent series for m > 1.

295

2B X(u-2B)

~ 2B X(u)~

------ ----,-

u

Figure A.2

3.5 00

s(t) = L: y(nT) h(t - nT). n=-oo

Thus 00

8(u) = L: y(nT) e-j21fnuT

n=-oo

8( u) is periodic with Fourier coefficients:

l IlT A

y(nT) = T S(u) ej21fnuT duo -lIT

Thus l IlT A

T S(u) du = y(O). -lIT

From the inversion formula:

y(t) = i: Y(u) ej21fut duo

Thus y(O) = i: Y(u) du

and our exercise is complete.

296 Solutions

3.6

3.7 The new signal's spectrum is no longer Hermetian. The signal, therefore, is complex. Although the sampling rate is reduced by a factor of a half, each sample now requires two numbers.

3.8 The resulting low pass kernel is

k(Ujt) = exp(-j21rut) n(2~).

We are thus assured that g( u), the Fourier transform of f(t), is zero outside of the interval I U I :::; B. Since this is the definition of a bandlimited function, the result is of little use. Note that Fourier inversion of (3.23), as we would expect, results in the cardinal series.

297

Solutions, Chapter 4

4.1 (a)

00

y(t; C) - D C2 L x(nT) sinc2 [C(t - nT)] n=-oo

00

y(t; W) - DW2 L x(nT) sinc2 [W(t - nT)] n=-oo

00

x(t) - L x(nT) kr(t - n/2W) n=-oo

where

kr(t) = D [C 2 sinc2 ( Ct) - B2 sinc2(Bt)]

+---+ Kr(u) = D [CA(u/C) - BA(u/B)].

For lu I ~ B,

- D [C{l-~} - B {1- ~}] C B

1 - T ; D=2W(C-B)

and x(t) is recovered.

(b) Here,

and

- D C2 sinc( Ct) u

Kr(u) = K(2B) = D CA(u/C)

Notice, for I u I ~ B,

298 Solutions

4.2 (a) K(2~) = l/(a + j27ru) and

H(2~) = (a + j27ru) II(2~)·

This inverse filter is shown in Fig.A.3.2

(b)

K(~) = 2a 2B a 2 - (j27rU)2

and

U a 1. 2 U H(-) = [- - - (J27ru) ] II(-).

2B 2 2a 2B

The inverse filter is shown in Fig.A.4.

y(t) >

~ __________ ~ ______________ J

H("lB)

Figure A.3

4.3 (b) H(u) = -j sgn(u). Since l/sgn(u) = sgn(u),

J u K(u) = 2B sgn(u) II(2B)

2Note: the LPF should be used first to avoid differentiating discontinuities.

x(t) LPF

and

k(t)

299

y(t)

a '----~2

Figure A.4

. B

JB j sgn( u) [j sin(27rut)] du 2 -B -1 fB B 10 sin(27rut) du

-7r Bt sinc2(Bt).

Thus, if g( t) is the Hilbert transform of f( t) with bandwidth B, then

f(t)=~ f: g(!!....) (-1)ncos(27rBt)-1. 7r n=-oo 2B 2Bt - n

4.5 Doing so gives ~ = S x or

I: {8[n - m] - rsinc(r(n - m))}x(~) ~M 2W

= r I: x( ~) sinc(r(n - m)) nEM 2W

where ~ has elements {gU1) In E M} and

g(nT) = I: {8[n - m] - rsinc(r(n - m))}x( Wn ) ~M 2

can be found from the known data.

300 Solutions

4.6

f(nT) = {~t~)::) , even n o ; odd n, n =f:. ±1

Therefore, for T = 1/2W,

1 4(_1)n/2 . ±1- n g(±T) = -2 L (_ 2) sIne( 2 ).

even n 1r 1 n

Note g(T) = g( -T). Using the + gives

. n - 1 2( -1 )n/2 sIne(-2-) = 1r(1 _ n) ; even n

and

( 2" 1 9 ±T) - (2/1r) L- (1 _ n2)(1 _ n) even n

00 1 (2/1r)2[1 + 2 ];-1 [1 - (2m)2)2

where we have let 2m = n. Numerically,

g(±T) = 1/2.

Now,

[ -1/2 0 1 H = 0 1/2·

Thus f(±T) = 1.

Note: The function these samples were taken from is

f(t) = sine(2Bt - ~) + sine(2Bt + ~).

4.8 The samples are taken from the signal

f( ) - d . () _ eos( 1rt) - sine(t) t - - SIne t - .

dt t

301

Thus 1(0) = O. (Note r = 1/2). Using (4.7) :

x(O) = -2 (_1)n/2 4 ""----'--- - - L ( -1)!!:}!-

even n n 7r odd n

n#O

= 0

since both summands are odd.

4.11 (a) For M = 1 and a = 0, (4.50) becomes

.6.( u) = j27r B.

Equation (4.51) becomes

Thus

_1_ sinc(Bt) sin( 7r Bt) 7rB sin2 (7r Bt)

7r 2 B2t

Equation (4.52) becomes

n2

= 2.. [-( u - B)II( ~ - ~) + (u + B) II( ~ + ~)] B2 B 2 B 2

Thus

1 = B A(u/B).

- sinc2(Bt) sin2(7rBt)

(7rBt)2 .

302 Solutions

(b) The interpolation is

Since kl(pi B) = b[n] and k2(pl B) = 0, the equa­tion reduces to an identity at t = ml B. Differentiate

Clearly:

k~(t) - ~ sinc2(Bt)

2B sinc(Bt) d1(Bt)

and I n

k1( B) = 0

since d1(0) = 0 and sinc(n) is zero everywhere else. Note that

Thus

and

The interpolation therefore also interpolates the derivative samples.

4.12 We have implicitly assumed that C±N = o. If P = 2N + 1 then, in the spectral replication, the Dirac delta at u = NIT would be aliased by the shifted Dirac delta originally at u = -NIT.

303

4.13 We evaluate the Fourier coefficients in (4.70) with the familiar formula

1jT/2 . en = - x( r) eJ27rnr/T dr. T -T/2

Substituting (4.71 ) followed by manipulation com­pletes the problem.

4.14 Since v(t) is real, V(u) is Hermitian. Thus, as illus­trated in Fig. A.5, X ( u) has a four fold symmetry. We can, therefore, hetrodyne the center frequency fa (rather than the lower frequency fL to the origin). Let

y(t) x(t) cos(27r fat) v(t) cos2 (27r fat) 1 1 2v(t) + 2v(t) cos(7rfot).

A lowpass filter gives

1 z(t) = 2v(t)

which can be sampled at the Nyquist rate of B.

~ B/2

.A IX(U)

A -f a fa

Figure A.S

u

u

304 Solutions

4.15 Since y(t) > 0 and approaches zero for large 1 t I, no matter how small ~, there exists some value of 1 t 1

above which there will be no more samples. Thus, the number of samples is finite and the signal is not uniquely determined.

4.16 Clearly

k(n) = sinc(n/2) cos[1l"(2N + 1)n/2].

Except for n = 0, sinc( n/2) is zero for even n. The cosine term is always zero for odd n. Therefore, (4.23) is satisfied.

4.17 Real Nth order polynomials.

4.18 We wish to compute the interpolation function corre­sponding to a p • The Lagrangian kernel in (4.82) can be partitioned as

where t - (mTN + a p )

ap(t) = II m#O a p - (mTN + a p )

is due to sample locations located distances mT from a p and the contribution due to the remaining terms IS

Using the product formula for sin(z) (Section 4.8),

ap(t) = sinc[2BN(t - a p)].

After factoring out the zero term, the m product in bp ( t) can be written as

305

Expressing both products as sincs and simplifying re­veals the resulting interpolation function to be iden­tical to that in (4.47).

4.19 No. Lagrangian interpolation would result in the con­ventional cardinal series. Recall that Lagrangian in­terpolation results in an interpolation where only the sample value contributes to the interpolation at that point. Equation(4.2), on the other hand, usually has every sample value contributing to the sample at t = m/2W.

4.20 We write (4.2) as

x(t) = rx(0)sinc(2Bt)+r L x(n/2B) sinc(2Bt-rn). n=ji:O

Substitute (4.9) and simplify. The same result can be obtained by filtering (4.10).

4.22 The matrix to solve is (for ~ < u < B )

S2(U) = - j27ru j27r(u _ 2B) [Sl(U)] 2B[ 1 1

S3( u) 3 (j27ru)2 [j27r( U - 21)F

x [ F(:~u~B) ]. F(u _ 4~)

Using the third order Vandermonde determinant

and Cramer's rule, we have

F(u) = 1 3 3 2B 4B - (-) [(u - -)(u - -)51(u) 2 2B 3 3

+j(2.-)( 3B )2(u-B)52(u) 27r 2

1 3)3 () B -(-)(- 53 U • - < u < B 87r2 2B '3

306 Solutions

2B F(u- -)

3

4B F(u- -)

3

To construct F( u), we evaluate the equation

Solving gives

and

With a bit of work we inverse Fourier transform to the interpolation functions in (4.57).

307

Solutions, Chapter 5

5.1 The explanation is unreasonable because the noise level on a bandlimited signal could, in the limit, be reduced to zero. The resolution is that any physi­cal continuous noise has a finite correlation length. Thus, as the samples are taken closer and closer, the noise eventually must become correlated and the white noise assumption is violated. For continuous white noise, all noise samples are uncorrelated. But the noise level is infinite.

5.3 '112 = Co.

5.4 Consider a power spectral density that is nonzero only over the interval B < I u I < W. Interpolating and filtering would yield a zero NINV.

5.6 Clearly

From (5.73) and Parseval's theorem,

100 1 1B -00 k2(t) dt = (2W)2 -B I <T>o(u) 1-2du.

The case ofy(t) can be obtained by k(t) = r sinc(2Bt). Since

r2 i: sinc2(2Bt) dt - (2~)2 i: du

We conclude that Ez ~ Ey •

5.7 Clearly

2~ :5 i: k2(t) dt,

7J(t) = x(t) - x(t) i- o.

308 Solutions

5.8 The power spectral density of the noise follows from the transform of (5.6);

_ 22" 1 ~ cos(7rnu/W) ~ Sll(u) - 2W, e [(2W,)2 +2 ~ n2 + (2W,)2] I1(2W)'

Since [Gradsteyn & Ryzhik #1.445.2]

~ cos( kx) = !!.- cosh a( 7r - x) __ 1_ L...J ; 0 < x ::; 27r, k=l k2 + a2 2a sinh( a7r) 2a2

we conclude that

Substituting into (5.9) and evaluating gives

sinh(27r, B) coth(27r, W)

sinh(27rr, W) coth(27r, W).

A semilog plot of the NINV is shown in Figure A.6 as a function of ,W for various values of r.

5.9 By deleting every other sample, the sampling rate parameter becomes 2r. The NINV for a single sample is r/(l - r). Since

2r < r/(l - r) ; r < 1/2,

we conclude that (4.9) yields a smaller NINV.

5.10 Since x(t) is bandlimited, it is analytic. Thus, it's Taylor series expanded about any given point con­verges everywhere. We choose an interval for which x(t) is identically zero. The resulting Taylor series, however, converges to x(t) = 0 everywhere.

> z Z

309

-yW 10.1 '----'-----'----'------'------''----'-----'---"'----'------'

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure A.6

310 Solutions

Figure A.7

Solutions, Chapter 6

6.15 (a) The minimum raster sampling rate occurs when sampling with parallel lines with a slope of 9/5 (or -9/5 ). The resulting minimum sampling rate is equal to the shortest distance between two parallel sides of the parallelogram which can be shown, after a bit of work, to be l/T = 2V56.

(b) Use horizontal lines at a sampling interval T = 1/2.

(c) Sample with lines of unit slope separated by T = 1 intervals. Corresponding spectral replications are shown in Figure A. 7 .

Solutions, Chapter 7

7.1

9c(t) = f(t)[l - Tl-a(tIT)]

00

1 - Tl-a(tIT) = L en e-j27rnt/T

n=-oo

c,. = { ~C; ~ a) .ine[(! - a)n]

The bn coefficients are solutions of

M

n=O n#O

L bm Cn-m = 6[n] ; Inl ~ M. m=-M

And:

m=-M

7.2 (a)

f(t) = 2~ [100 F(u) e-j27rut dU] .

311

If f(t) is bandlimited, the upper integration limit is B.

(b) From the Figure:

F(u) = Co G1(u) 2- Cl G(u - liT). CO - C-l Cl

(c)

312 Solutions

Co cos(rrBt) - C1 cos[rr(B--t)t] cos(rrBt)

g(t)

Co sin (rrBt) - c1 sin[rr(B--t)t] sin (rrBt)

Figure A.8: Restoration of first order aliased data from the shift method illustrated in Fig. 7.27, using Hermetian symmetry. The filters are unity for I u I < B /2 and zero elsewhere.

7.3

-Cl cos 7r(B - ~) tn * BsincBt] cos 7r Bt

+ (2 2 ) [{g(t) [co sin7rBt Co - C-ICI

-Cl sin 7r(B - ~) tn * BsincBt] sin 7r Bt.

(d) See Figure A.S

00

2B sinc[2B(t - r)] = L an tPn(t) n=O

where

an 2B i: sinc[2B(t - r)] tPn(r) dr

- tPn(t).

Thus: 00

2B sinc[2B(t - r)] = L tPn(t)tPn(r). n=O

313

7.4 (a) From the power theorem:

i: sinc(2Bt - n) sinc(2Bt - m) dt

= J [~e-i1rnu/B II(..!:..)] [~e-i1rmu/B II(..!:..)] I

2B 2B 2B 2B 1

= 2B 6[n - m].

(b) Clearly:

f(t) 00 n - E f( -B) sinc(2Bt - n)

-00 2

Multiply both sides by tPn(t) and integrate. Using (7.30) and (4.47) gives:

(c) Here, multiply by sinc(2Bt - m) and integrate

Both are well-posed.

7.5 (a)

i: tPn(t) tPm(t) dt T

- 2B i: tPn(t) J tPm(r) sinc[2B(t - r)] dr dt 2

2B i: tPm(r) i~ tPn(t) sinc[2B(t - r)] dt dr 2

- An i: tPn(r) tPm(r) dr

- An h[n - m].

314 Solutions

(b) Let

Where in the second step we have substituted t = Tu/2B. From which (7.33) follows.

7.6 The ambiguity is removed by (7.30) which requires each PSWF to have unit energy.

7.7 (a)

i: 1 f(t) 12 dt _ f: an f: am 100 1 ~n(t) 12 dt n=O m=O -00

00

- La;. n=O

(b)

7.8

x(t) t

- x(t) II(T) 00

- L bn ~n(t) n=O

y(t) - x(t) * 2B sinc(2Bt)

- i: x( T) sinc[2B( t - T)] dT

- f: bn J ~n(T) sinc[(2B(t - T)] dT n=O

00

- L bn An ~n(t). n=O

315

From (4.20), the energies of x and yare

00

Ex = L An I bn 12 n=O

and

n=O n=O We wish to maximize Ey subject to Ex = 1. Since Ao is the largest eigenvalue, we choose bn = A8[n] and

x(t) = tPo(t) II(~)/Fo. The output has energy Ey = Ao.

7.9 The results are the same as for extrapolation except that, An becomes (1- An) and (1- D T ) replaces D T •

Thus, instead of (7.52), we have:

7.10

00

fN(t) = 9i(t) + DT L an(1- A~)tPN(t) n=O

as with the extrapolation case, the validity of this equation can be proven by induction. Convergence again follows due to (7.53).

00

ge(t) L 9n ei21rnt/T n=O 00

L an tPn(t) ; It 1< T/2. n=O

(a) Multiplying both sides by tPm(t) and integrate over I t I < T /2 :

316 Solutions

or, using (7.36):

The restoration is ill-posed due to the 1/.;r: coefficient.

(b) Using the top equation, multiply both sides by ej21rnt/T and integrate over I t I < T /2:

00 1:.

Tgm = L an i~ tPm(t) ej21rnt/T dt. n=O 2

Again, using (7.36)

The result is well-posed.

7.12 (a) All bandlimited signals are analytic. The only function that can be identically zero over any interval is zero everywhere (Taylor series).

(b) The first two iterations are shown in Figure A.9.

7.13 (a) Instead of (7.42), we have

where the nonlinear half wave rectifier operator is defined by

R h(t) = h(t) Jl[h(t)].

1 - D T

Figure A.9

(b) Here we know Se(u) 2:: O. Since

BB = F-1 DB F

our altered algorithm could be:

317

This is the so-called minimum negativity con­straint proposed by Howard. Both are special cases of alternated projections onto convex sets [Stark], [Youla and WebbJ,[Cheung, Marks, and AtlasJ.

7.14 (a) Let h(t) have finite energy. Then BB h(t) = f(t) is bandlimited and can be written as

00

f(t) = L an 1/Jn(t). n=O

(Note that maximum energy occurs when h(t) is chosen to be bandlimited.) Thus,

II H II = sup II Hf(t) II IIf(t)lI=l

318 Solutions

7.15

IIHII = sup II (l-DT)f(t) II IIf(t)ll=l

where f(t) is bandlimited. Now

00

(1 - DT)f(t) = L: an (1- DT) 1/Jn(t) n=O

and

n=O

where we have used (5.41). Since, from Exercise 7.7a,

n=O we choose aoo = 1 (since >'00 = 0) and II H 11= 1.

(b) Same as above, except

II H 11= sup II DT f(t) II . IIf(t)lI=l

Since 00

II DT f(t) 112= L: >'n I an 12 ,

n=O we choose I an I = 1 since >'0 is max and II H II = 0\0 < 1.

XN(O) - I: X N( u)du

- T I: i: {XN_l(0)8(t) 00

+ L: x(nT) 8(t - 'Tn e-j27rut dt du n=O

319

T Joo [XN-l(O) f: x(nT) e-i27rnut]du -00 n=O

00

rXN_l(O) + r L x(nT) sinc(rn) n=O

where r = 2BT. Letting N ~ 00 gives

00

xoo(O) = rxoo(O) + r L x(nT) sinc(rn). n=O

Solving for xoo(O) therefore results in (4.9). Note also that XN(t) ~ x(t).

Index

Abel transform 42 absolutely convergent series

30,88 aliasing 37,71, 72, 139, 187,

196,216,220,231,236, 237,240,243,245,271

analytic functions 9, 50, 88, 308, 316

autocorrelation 23-25,52,112-114,116,138-141,143, 148,157,164,242-244, 274

bandlimited signals 41, 107, 136,153,225,255,268: 316

bandpass signals 92, 93, 96-98, 107, 108, 127

bandwidth 1, 8, 13, 33, 35, 37,42,51,55,58,64, 81, 84, 92, 97, 104, 110,111,117,118,157, 217,229,257,262,271, 273, 299

Bessel functions 16, 83, 177 Black, H.S. 2, 5 Borel, E. 2 bounded signals 8, 29 Bracewell, R. 9, 17, 18, 31,

42,56,113,165,177, 223

bunched samples 68, 76

cardinal series 1-3, 5, 13, 33, 34,38,40,41,44,47, 49-52, 54, 57-59, 61, 63-66,82,89,100,103, 108,111-113,119,128, 145,146,153,155,157, 158,164,271,296,305

characteristic function 147, 150 circular 176, 177, 188,200,

201,204,209,215,216, 220

comb function 21 complete 12, 108, 255, 256,

260 condition number 127 continuous sampling 225, 260,

271 convolution 10, 35, 37, 42,

49, 83, 84, 229, 238, 281

correlation 42, 116 cydostationary processes 163

decimation 203, 204, 210-213, 215, 222

derivative kernel 82, 84, 87, 88, 136

DeSantis, P. 260, 266, 276 Dirac delta 15, 16, 35, 37,

50,51,181,237,240, 280, 302

doublet 51

entire functions 3, 227 Euler 20, 92 expected value 22, 112, 146,

149, 198 extrapolation 225-228, 244,

245,254,257,259-262, 265,266,268,270,274, 315

finite energy signals 30, 107 Fourier series 3, 11-13, 21,

24, 27, 28, 35, 38-41, 130, 176-178, 181,182~ 184,220,229,243,256, 273, 284, 289

Fourier transform 9-11, 13, 14,26,229,240,242, 256,262,276,292,296, 306

Gamma function 16, 242 Gaussian 20, 151 geometric series 92,115,143,

264, 274, 283, 287 Gibb's phenonema 27,28,41 Gori, F. 260, 260, 276

Hamming, R.W. 76, 136, 165 Hankel 42, 177, 219 heterodyne 93, 107, 303 hermetian 24, 55, 92, 296 Higgins, J .R. 2, 5 Hilbert 42, 44, 48, 49, 68, 95,

100, 104, 105, 299

Index 321

Hilbert space 268, 269 Howard, S.J. 276, 317

ill posed 65, 68,129-132,135, 163,228,245,260,270

implicit sampling 107 incomplete cosine 82 incomplete sine 82 induction 267, 315 interlaced sampling 68, 76,

130, 132, 135 interpolation function 41, 57,

63,64,66,69,76-78, 80,81,89,90,92,100, 105-107,119,123,131, 132, 195, 211, 306

interpolation noise level 57, 87,113,115,116,136, 141,143,163,220,243, 245

interval interpolation 225, 245, 254,260,262,264,273, 274,277

jinc function 17, 26, 279 jitter 111,145-149,151,152,

163

Kotel'nikov, V.A. 1,3, 5 Kramer, H.P. 1, 5, 57, 102,

108 Kramer's generalization 13,

57, 102, 108 Kronecker delta 12, 15, 197

322 Index

Lagrangian interpolation 57, 100, 107, 304, 305

Laplace transform 18, 44, 45, 47

Linden, D.A. 68, 78, 80, 109 lost sample restoration 58-63,

105,109,118-121,123-129,164,165,195-201, 203, 221, 274

low passed kernels 43, 44,47, 48

LPK - see low passed kernels

mean square 11, 12, 39, 40, 52,157,158,160-162, 256

Mellin transform 42 Middleton, D. 5, 167, 192,

223 mInImUm mean square error

156-161 minimum negativity constraint

317 modulation 5, 168, 182 mth-Iet 54

NINV 118,120,121,126-129, 136,138,141,142,150, 163,164,221,241,244-249, 307, 308

noise level 22, 24, 25,82,114, 115,120,121,124-127, 130,135,137,138-140, 163,198-201,203,239, 240,243-245,270,307

Nyquist 3, 5, 37, 42, 43, 50, 54,67-69,76, 77, 80, 93,105,114,118,125, 129,130,135,163,187, 188,190,191-193,195, 199,200,202-204,208, 211-213,216,222,244, 258, 303

optical images 200, 211 orthogonal 12, 13, 102, 197,

254, 256, 257, 268 orthonormal 12, 13, 103, 108,

255 oversampling 59, 60, 63, 65,

81,82,105,107,113, 117,135,199,274,275

Papoulis, A. 5, 6, 9, 22, 57, 68, 69, 70, 106, 129, 145,153,158,163,215, 227, 260, 266, 277

Papoulis-Gerchberg algorithm 260

Parseval's theorem 21,49,131, 153,273,282,291,307

Pask, C. 227, 277 period cell 179 periodic signals 7, 9, 16 periodic continuous sampling

228 periodicity 178-180,204-207,

210, 211, 221, 222 Peterson, D.P. 167, 192 PGA - see Papoulis-Gerchberg

algorithm

piecewise 41-43, 67 Poisson sum formula 21, 54,

65, 71, 92, 130, 137, 149, 282, 294

Pollak, R.O. 254, 277 power spectral density 23-25,

114,115,137,160,243, 307, 308

prediction 226, 228, 262 probability 146,147,149,153 prolate spheroidal wave func-

tions 254-260, 266, 272, 273

PSWF - see prolate spheroidal wave functions

raster sampling 216, 217,222 rectangle function 13, 155, 188,

201, 209, 220 recurrent nonuniform sampling

57, 77, 101, 107 regularize 270 rotation 171, 174, 182 rotation matrix 175

sample and hold 66, 67 sampling density 55,186,191-

193,200,203,204,211-213, 215, 216

sampling matrix 191, 196,204, 220

sampling rate 1, 37, 51, 96, 97,107,111,114,118, 138,139,141,161,163, 211,216,217,230,296, 308

Index 323

sampling rate parameter 58, 101, 127

Schwarz's inequality 29, 30, 39, 156, 292, 293

separable functions 169, 170 separability theorem 169, 182 Shannon, C.E. 1, 3, 68, 78,

110, 129 Shannon number 50, 258 Shannon sampling theorem

1,2, 167, 184 signal level 23 signal-to-noise ratio 160, 162,

258 sinc function 13, 14, 227 sine integral 135 Slepian, D. 2, 6, 254, 277 stationary processes 23, 117,

130, 200, 220, 239 subcell 205-210, 213-216 super resolution 225

Taylor series 8, 30, 80, 81, 88,144,227,228,308, 316

time-bandwidth product 50, 258

Toeplitz 232 trapezoidal integration 41-45,

47, 48 triangle function 26, 139-141,

220, 221, 240 trigonometric polynomial 88,

90,163,232,236-238 truncation error 50, 61, 64,

153, 156, 164

324 Index

uniform 3, 13, 40, 41, 145, 151,159,162,163,185, 200

unit doublet (see doublet) unit step function 17, 45

Vandermonde determinant 76~ 105, 305

Von Neumann's alternating projection theorem 270

Walsh functions 13, 108 well-posed 135, 272-273 Whittaker, E.M. and J.M. 3,

6 Whittaker-Shannon Sampling

Theorem 1 Whittaker-Shannon-Kotel'nikov­

Kramer sampling the­orem 1

wide sense stationarity 23, 24, 52,111,137,148,157, 163

Youla, D.C. 266,268-270,278, 317

Zemanian, A.H. 51, 56