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Solutions to Chapter 4 Problems Introduction to Communication Systems, by Upamanyu Madhow Problem 4.1(a) While it is easy enough to compute P (f ) directly, it is a bit more convenient to find the Fourier transform for the symmetric version of this pulse, given by q(t) = p(t + 0.5) = cos πt I[− 1 , 1 ](t). Then p(t) = q(t − 0.5) and P (f ) = Q(f )e−jπf . 2 2

R1 Q(f ) = −2 1 cos πte−j2πf t dt 2 R 1 jπt −jπt e−j2πf t dt = −2 1 e +e 2 2   jπt(1−2f ) 1 jπt(1+2f ) 1 = 12 ejπ(1−2f ) −2 1 + ejπ(1+2f ) −2 1 2

2

π

π

=

ej 2 (1−2f ) −e−j 2 (1−2f ) 2jπ(1−2f )

=

je−jπf +jejπf 2jπ(1−2f )

+

π

+

π

ej 2 (1+2f ) −e−j 2 (1+2f ) 2jπ(1+2f )

je−jπf +jejπf 2jπ(1+2f )

π

since e±j 2 = ±j. We can now simplify to obtain Q(f ) =

2 cos πf π(1 − 4f 2 )

from which the desired expression for P (f ) follows. (b) The PSD is proportional to |P (f )|2, hence the 95% power containment bandwidth B satisfies Z

B 2

2

|P (f )| df = 0.95

−B 2

Z

∞

|P (f )|2df

−∞

Since p(t) is time-limited, it is convenient to evaluate the right-hand side using Parseval’s identity: Z ∞ Z ∞ Z 1 1 2 2 |P (f )| df = |p(t)| dt = sin2 πt dt = 2 −∞ −∞ 0 Using the symmetry of |P (f )|2 around the origin, the final equation we must solve for B is 2

Z

B 2

|P (f )|2df = 0.95/2

0

2

cos πf where |P (f )|2 = π24(1−4f 2 )2 . We can now use Matlab or any other convenient software package for numerical integration and to search for B. Since the unit of time is microseconds, the unit of bandwidth is MHz. Problem 4.2(a) The pulse is sketched in Figure 1, which also shows that it can be written as a convolution of two boxes: p = u ∗ v, so that P (f ) = U(f )V (f ). Now, centered versions of u and v have Fourier transforms (1−a)sinc((1−a)f ) and sincaf , respectively, and convolving them would give rise to a centered version of p(t) with Fourier transform P˜ (f ) = (1 − a)sinc((1 − a)f )sincaf . But p(t) is delayed by 21 from its centered version p˜(t), which corresponds to a phase of e−jπf in the frequency domain. Thus, its Fourier transform is given by

P (f ) = P˜ (f ) e−jπf = (1 − a)sinc((1 − a)f )sincaf e−jπf

v(t) 1/a

u(t)

p(t)

1

1

0

a

1−a

1

t

*

= 0

1−a

t

0

a

t

Figure 1: The trapezoidal pulse p(t) in Problem 4.2 can be expressed as a convolution of two boxes. Note that we could have found the phase terms individually for the Fourier transforms of u and v, and then multiplied their Fourier transforms together, but this approach is slicker. 2 2 +12 +32 = 5. (b) For equiprobable 4PAM, the average symbol energy is given by σb2 = (−1) +(−3) 4 The symbol time T = 1. The PSD for the modulated signal u is given by Su (f ) =

σb2 |P (f )|2 = 5(1 − a)2 sinc2 ((1 − a)f )sinc2 af T

(c) Dropping constants from the PSD, since these do not affect bandwidth, the 95% bandwidth B satisfies the equation Z B Z ∞ 2 2 |P (f )| df = 0.95 |P (f )|2 df −B 2

−∞

Since the pulse is timelimited, we can use Parseval’s equality to compute the right-hand side. The pulse energy is given by Z ∞ Z ∞ Z a Z 1−a 4 2 2 2 |P (f )| df = |p(t)| dt = 2 (t/a) dt + 12 dt = 1 − a 3 −∞ −∞ 0 a Using the symmetry of |P (f )|2, we can now write Z

B 2

2

|P (f )| df = 2

0

Z

B 2

0

4 (1 − a)2 sinc2 ((1 − a)f )sinc2 af df = 0.95(1 − a) 3

That is, we must solve Z

B 2

0

0.95(1 − 43 a) sinc ((1 − a)f )sinc af df = 2(1 − a)2 2

2

We solve this numerically using Matlab and get the plot in Figure 2. Since the unit of time is 100 ps, the unit of bandwidth is 10 MHz, so we are plotting 10B (MHz) versus a. The plot shows that the 95% power containment bandwidth is minimized for a ≈ 0.25. Problem 4.3(a) As shown in Figure 3, the spectrum S(f ) is trapezoidal, being a convolution of two boxes in the frequency domain. We also show S(f − T1 ) for minimum bandwidth Nyquist signaling using such a trapezoidal pulse. We see from the plot that this requires a = T1 . This also makes sense when thinking in the time domain: the term sinc(at) = sinc(t/T ) provides zeroes at t = kT for nonzero integers k. (b) The physical bandwidth in baseband is the one-sided pulse bandwidth (a + b)/2 = 400 Hz. For 4PAM signaling at 1200 bps, the symbol rate is T1 = 1200 bps/(log2 4 bits/symbol) = 600 symbols/sec. Thus, a = 600 Hz and b = 200 Hz. (c) We now have a passband channel of physical bandwidth 20 MHz. This is now the two-sided bandwidth in complex baseband, so that a+b = 20 MHz. For 60 Mbps using 64QAM, the symbol

2

36

34

95% Bandwidth (MHz)

32

30

28

26

24

22

20

18 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

a

Figure 2: 95% bandwidth for the trapezoidal pulse p(t) in Problem 4.2, when the unit of time is 100 ps. S(f) 1/b 1/a −a/2

* a/2

1/a

= −b/2

S(f−1/T)

(a−b)/2 (a+b)/2

b/2

f

1/(2T) = a/2

Figure 3: The spectrum S(f ) for the pulse in Problem 4.3, which is a convolution of two boxes in the frequency domain.

rate is T1 = 60 Mbps/(log2 64 bits/symbol) = 10 Msymbols/sec. We therefore have a = T1 = 10 MHz and b = 10 MHz. (d) This is already provided in the text, where we discuss the role of excess bandwidth. The product of two time domain sincs decays as 1/t2 , so that the worst-case ISI, and hence the worst-case amplitude, converges. 1/T

P(f− 1/T)

P(f)

A

B 1/(2T)=(A+B)/2

Figure 4: Graphical computation of maximum Nyquist rate for pulse in Problem 4.4. Problem 4.4 From the frequency domain plot in Figure 4, we see that the pulse is Nyquist at maximum rate T1 = A + B = 1.5 Msymbols/sec. But if we compute the time domain pulse (a product of two sincs), we see that there are actually two sets of rates at which the pulse is Nyquist. Up to constant factors, p(t) = sinc(at)sinc(bt) and P (f ) = I[−a/2,a/2] (f ) ∗ I[−b/2,b/2] (f ), so that (a + b)/2 = B and (a − b)/2 = A, which means that a = A + B = 1.5 MHz and b = B − A = 1 MHz. The sinc terms provide zeros spaced by 1/a and 1/b respectively, so that the pulse is Nyquist at symbol rates a/K or b/K, where K is any positive integer. We are now ready to answer parts (a) and (b). (a) True: 3 Mbps using 8PSK (log2 8 = 3 bits/symbol) corresponds to a symbol rate of 1 Msymbols/sec. (b) True: 4.5 Mbps using 8PSK (3 bits/symbol) corresponds to a symbol rate of 1.5 Msym-

3

bols/sec. Problem 4.5(a) Yes, the pulse is Nyquist at rate 1/T , since p(nT R ) = δn0 . (b) No, the pulse is not square root Nyquist at rate 1/T , since p(t)p∗ (t − T )dt 6= 0. 1/T

P(f)

P(f− 1/T)

b=1/(2T) (1+e)

a

1/(2T)=(a+b)/2

Figure 5: Relating pulse parameters to symbol rate and excess bandwidth in Problem 4.6. Problem 4.6 (a) For signaling at 80 Mbps using 16QAM (log2 16 = 4 bits/symbol), the symbol rate 1/T = 20 Msymbols/sec. The fractional excess bandwidth e = 0.5. From Figure 5, we see 1 1 that 2T = 10MHz = (a + b)/2 and 2T (1 + e) = 15MHz = b. Solving, we get a = 5MHz and b = 15MHz. (b) True. Signaling at 20 Mbps using QPSK (log2 4 = 2 bits/symbol) corresponds to a symbol rate of 10 Msymbols/sec. This equals the maximum Nyquist rate 20 Msymbols/sec divided by an integer, hence the pulse is Ny quist at this rate as well. 1/a

b

1/a

1/a

b/2

P(f− 1/T)

=

* −b/2

P(f)

f −a/2

a/2

(a+b)/2 (a−b)/2

a/2

Figure 6: The pulse P (f ) in Problem 4.7. We have assumed without loss of generality that a ≥ b. Problem 4.7 (a) Since sinc(at) ↔ a1 I[−a/2,a/2] (f ) and sinc(bt) ↔ 1b I[−b/2,b/2] (f ), P (f ) is a convolution of two boxes, which is the trapezoid shown in Figure 6. For b ≥ a, for minimum bandwidth 1 = a2 , so that a = T1 , and Nyquist signaling at rate 1/T with excess bandwidth α, we set 2T 1 (1 + α) = a+b , so that b = Tα . For 40 Mbps using 16QAM (4 bits/symbol), the symbol rate is 2T 2 1 = 10 Msymbols/sec. Thus, a = 10 MHz, and for 50% excess bandwidth, b = 5 MHz. T (b) We would like the pulse to be Nyquist for two signaling rates: 40 Mbps using 16QAM (4 bits/symbol) corresponds to T11 = 10 Msymbols/sec, and 18Mbps using 8PSK (3 bits/symbol) corresponds to T12 = 6 Msymbols/sec. Note that neither is an integer multiple of the other. However, since p(t) = sinc(at)sinc(bt) has zeros at integer multiples of 1/a and 1/b, we can enforce both p(nT1 ) = 0 and p(nT2 ) = 0 for n 6= 0 by setting a = T11 = 10 MHz and b = T12 = 6 MHz. Problem 4.8 The symbol rate is given by T1 = 16 Mbps/(log2 256 bits/symbol) = 2 Msymbols/sec. (a) For the timelimited sine pulse to be square root Nyquist at rate 1/T , the maximum pulse duration (i.e., the minimum bandwidth) equals T . Thus, the unit of time is T = 0.5 microseconds. (b) The pulse is trapezoidal in the frequency domain, and the minimum bandwidth Nyquist corresponds to the picture shown in Figure 7. This corresponds to 1/T = 2 units of frequency. Since the symbol rate is 2 Msymbols/sec, the unit of frequency is 1 MHz and hence the unit of time is 1 microsecond. Another way of seeing this is to realize that the minimum bandwidth is when the sinc(2t) term provides the zeros at t = kT , so that 1/T = 2 units.

4

P(f) P(f−1/T)

=

* −1

1

−1/2

1/2

1/2

3/2

f

1/(2T) = 1

Figure 7: The minimum bandwidth configuration for the pulse in Problem 4.8(b) to be Nyquist. P(f)

=

* 1.5

f (MHz)

f (MHz)

f (MHz) −1.5

−1

1

−2.5

−0.5

0.5

2.5

Figure 8: The frequency domain pulse for Problem 4.9. Problem 4.9 (a) We have sinc3t ↔ 13 I[−1.5,1.5] (f ) , sinc2t ↔ 21 I[−1,1] (f ), so that p(t) = sinc3tsinc2t ↔ 1 I (f ) ∗ I[−1,1] (f ). We sketch P (f ) in Figure 8, ignoring scale factors on the y axis, but 6 [−1.5,1.5] noting that the units of frequency are MHz, since the units of time are microseconds. (a) The two sincs provide zeros at rate 3MHz and 2MHz, so the maximum Nyquist rate is 3 Msymbols/sec. The maximum bit rate using 16QAM (log2 16 = 4 bits/symbol) is therefore 3 Msymbols/sec × 4 bits/symbol = 12 Mbps. From Figure 8, the pulse bandwidth is 5 MHz, whereas the minimum bandwidth for signaling at 3 Msymbols/sec is 3 MHz, so that the fractional excess bandwidth is (5 − 3)/3 = 2/3. (b) True. 4 Mbps using QPSK (log2 4 = 2 bits/symbol) corresponds to a symbol rate of 2 Msymbols/sec. By virtue of the sinc2t term, the pulse is Nyquist at this rate. Problem 4.10 True. Since p(t) has duration at most T , there is no overlap between p(t) and ) for k a nonzero integer. We therefore have p(t)p∗ (t − kT ) ≡ 0, which implies that Rp(t − kT ∗ p(t)p (t − kT )dt = 0. Problem 4.11 (a) The spectra are sketched in Figure 9. C(f) π /2a

R(f) 1

−1/2

1/2

−a/2

f

a/2

f

Figure 9: The spectra R(f ) and C(f ) for Problem 4.11.

(b) By the symmetry of R and C, we have that S = R ∗ C is also symmetric. We therefore only evaluate the convolution for f > 0. Z Z S(f ) = R(f − ν)C(ν)dν = R(ν − f )C(ν)dν

using the symmetry of R. The range of integration is determined as shown in Figure 10. For f ≤ (1 − a)/2, we have R(ν − f ) = 1 wherever C(ν) > 0, so that Z a/2 S(f ) = C(ν)dν = 1 −a/2

5

C(ν )

C(ν )

C(ν ) R(ν −f)

f−1/2

−a/2 a/2

ν

f+1/2

R(ν −f)

R(ν −f) −a/2f−1/2 a/2

0 < f < (1−a)/2

f+1/2

ν

−a/2

(1−a)/2 < f < (1+a)/2

a/2

f−1/2

ν

f+1/2

f> (1+a)/2

Figure 10: Convolution of R(f ) and C(f ) in Problem 4.11(b).

For (1 − a)/2 < f < (1 + a)/2, we have S(f ) =

Z

a/2 f − 21

π C(ν)dν = 2a

a/2

Z

f − 21

   1 π 1 cos(πν/a)dν = 1 − sin (f − ) 2 a 2

To see that this is indeed a raised cosine shape, set f˜ = πa (f − (1 − a)/2), so that πa (f − 12 ) = f˜− π2 . For (1 − a)/2 < f < (1 + a)/2, we have 0 < f˜ < π2 , and i π i 1h 1h ˜ ˜ ˜ 1 − sin(f − ) = 1 + cos f S(f ) = 2 2 2

For f > (1 + a)/2, there is no overlap between R(ν − f ) and C(ν), so that S(f ) = 0. (c) The time domain pulse s(t) = r(t)c(t), where r(t) = sinc(t) and c(t) =

R

C(f )ej2πf t df = π

π 2a π

R a/2

=

π 4a

n

ej( a +2πt)f j( π +2πt) a

=

π 4a

n

ej( 2 +πat) −ej(− 2 −πat) j( π +2πt) a

=

π 4a

=

cos πat 1−4a2 t2

n

+

+


ejπf /a + e−jπf /a ej2πf t df

of =a/2

ej(− a +2πt)f j(− π +2πt) a π

π

2j cos πat j( π +2πt) a

1 −a/2 2

f =−a/2 π

+

−2j cos πat j(− π +2πt) a

π

ej(− 2 +πat) −ej( 2 −πat) j(− π +2πt) a

o

o

Thus, the time domain pulse is given by s(t) = sinc(t)

cos πat 1 − 4a2 t2

(d) The raised cosine pulse s(t) decays as t13 . Setting g(t) = s(t/T ), consider a linearly modulated waveform X X X x(t) = b[n]g(t − nT ) = b[n]s((t − nT )/T ) = b[n]s(t/T − n) n

n

n

Fixing t, we realize that s(t/T − n) decays roughly as 1/n3 . The convergence of that the sum in the preceding equation converges to a finite value for any t. Problem 4.12-4.14: Software exercises. Solutions omitted.

6

P

n

1/n3 implies

Problem 4.15 (a) Clearly dmin = 2 for the chosen scaling. (b) For computing Es , note that there are four inner points with I/Q coordinates ±1, four outer points with I/Q coordinates ±3, and eight intermediate points with one coordinate from ±1 and one from ±3, yielding average energy Es = (c) We have




4 2 4 2 8 2 1 + 12 + 3 + 32 + 1 + 32 = 10 16 16 16 Eb =

10 5 Es = = log2 16 4 2

ηP =

d2min 22 8 = = Eb 5/2 5

and power efficiency

Problem 4.16 (a) 16QAM (4 bits/symbol) at 50 Mbps corresponds to a symbol rate of 50/4 = 12.5 Msymbols/second. With an excess bandwidth of 50%, the bandwidth is 12.5 × (1 + 0.5) = 18.75 MHz. For a carrier frequency of 5.2 GHz, this 18.75 MHz occupies the frequency interval [5.190625, 5.209375] GHz. (b) For the same excess bandwidth and the same band, the symbol rate becomes fixed at 12.5 Msymbols/second. Since QPSK corresponds to 2 bits/symbol, the bit rate is 25 Mbps. (c) The power efficiency of QPSK is ηP = 4, compared to a power efficiency of ηP = 59 for 16QAM, so that it requires less power for a given bit rate. Since the bit rate is also going down by a factor of 2, the required power, relative to the 16QAM system, is scaled by a factor ηP (16QAM) 1 9 = ηP (QP SK) 2 40 Since the power in the 16QAM system is 100 mW, the power required for the QPSK system is 22.5 mW. T

I component

Q component T/2

Figure 11: The I and Q components of a typical realization of an MSK waveform. Problem 4.17: (a) As shown in Figure 11, the I and Q components of an MSK waveform are offset from each other by T2 . (b) Over any interval [mT /2, (m + 1)T /2], the I and Q components are sinusoids offset in phase by ± π2 . Since sin2 (x) + cos2 (x) = 1, the envelope is constant. (c) The I and Q components have identical PSDs (the time offset does not affect the PSD). The I and Q components are independent, since the bits sent over the I and Q channels are independent. The PSDs therefore add up, and we obtain that Ss (f ) = E[|bc [n]|2 ]

|GT X (f )|2 |GT X (f )|2 |GT X (f )|2 + E[|bs [n]|2 ] =2 T T T

7

The PSD is not changed by considering the symmetric version of the transmitted pulse: gT X (t) = cos πt/T I[−T /2,T /2] . Setting T = 1 for convenience, we obtain GT X (f ) =

R 1/2

−1/2

ejπt +e−jπt −j2πf t e dt 2

=

1 2

h

ejπ(1−2f )t jπ(1−2f )

=

1 2

h

ejπ/2 e−jπf −e−jπ/2 ejπf jπ(1−2f )

=

1 2π

=

2 cos πf π 1−4f 2

h

+

e−jπf +ejπf 1−2f

e−jπ(1+2f )t −jπ(1+2f )

+

it=1/2

+

t=−1/2 e−jπ/2 e−jπf −ejπ/2 ejπf −jπ(1+2f )

e−jπf +ejπf 1+2f

i

i

When we timescale by T , we have GT X (f ) = The PSD is given by Ss (f ) =

2T cos πf T π 1 − 4f 2 T 2

8T 2 cos2 πf T π 2 (1 − 4f 2 T 2 )2

PSD of MSK

(d) We can set T = 1 once more to compute 99% energy containment bandwidth: Z B1 /2 Z ∞ Z ∞ 2 2 |GT X (f )| df = 0.99 |GT X (f )| df = 0.99 |gT X (t)|2 dt = 0.99/2 −B1 /2

−∞

−∞

Using symmetry, we arrive at the following equation (which must be numerically solved): Z

B1 /2 0



2 cos πf π 1 − 4f 2

2

df = 0.99/4

This yields B1 = 1.2. This is much better than the corresponding result B1 = 10.2 for a rectangular pulse. Of course, for signaling at rate T1 , the bandwidth simply scales to BT1 . (e) In order to normalize the area under the PSD curve, we should fix the energy of the transmit pulse to be the same in each case. Setting T = 1, for MSK as considered above, the energy is 21 . A rectangular pulse of the same energy is given by gT X (t) = √12 I[−1/2,1/2] ↔ √12 sinc(f ). In each case, Ss (f ) = 2|GT X (f )|2 , so that ( 8 cos2 πf MSK 2 (1−4f 2 )2 π Ss (f ) = 2 sinc (f ) OQPSK Plotting these yields Figure 4.6. Problem 4.18: We have si (t) = cos(2πfi t + φi )I[0,T ] (t), i = 0, 1. The sinusoids are passband signals, and the time windowing by multiplying by I[0,T ] corresponds to convolution with sinc(f T ) (upto constant) in the frequency domain. The latter is narrow for large T , hence the signals si (t), while not strictly bandlimited, are passband for fi ≫ T1 . Letting s˜i (t) denote the complex envelope of si with respect to reference frequency f0 (any other reference near f0 and f1 would

8

also work), we can express the passband inner product in terms of the complex baseband inner product as follows: 1 hs1 , s0 i = Re (h˜ s1 , s˜0 i) 2 We can read off the complex envelopes as s˜1 (t) = ej(2π(f1 −f0 )t+φ1 ) I[0,T ] (t),

s˜0 (t) = ejφ0 I[0,T ] (t)

Let us now evaluate the complex inner product: h˜ s1 , s˜0 i =

R

s˜1 (t)˜ s∗0 (t)dt =

= ej(φ1 −φ0 ) e

j2π(f1 −f0 )T −1

RT 0

ej(2π(f1 −f0 )t+φ1 ) e−jφ0 dt

j2π(f1 −f0 )

(a) For φ1 = φ0 = 0, we obtain that h˜ s1 , s˜0 i =

ej(2π(f1 −f0 )T − 1 cos(2π(f1 − f0 )T ) + j sin(2π(f1 − f0 )T ) − 1 = j(2π(f1 − f0 ) j2π(f1 − f0 )

from which we see that 1 sin(2π(f1 − f0 )T 1 s1 , s˜0 i) = hs1 , s0 i = Re (h˜ 2 2 2π(f1 − f0 ) For f1 6= f0 , the smallest nonzero value of the difference that makes this inner product vanish 1 is when |2π(f1 − f0 )T | = π, or |f1 − f0 | = 2T . This is the minimum separation required for orthogonality in coherent FSK. (b) Now, suppose that φ1 , φ0 are arbitrary. We want to set

where


1 1 s1 , s˜0 i) = Re ej(φ1 −φ0 ) Z = 0 hs1 , s0 i = Re (h˜ 2 2 Z=

ej2π(f1 −f0 )T − 1 j2π(f1 − f0 )

regardless of the values of φ1 , φ0 . The only way this can happen is if Z = 0. To see why, note that φ1 = φ0 implies that we must have Re(Z) = 0, and that φ1 − φ0 = π2 implies that we must have Im(Z) = 0 (check!). For f1 − f0 6= 0, the smallest separation for which Z = 0 is given by |2π(f1 − f0 )T | = 2π, which simplifies to |f1 − f0 | = T1 . This is the minimum separation required for orthogonality in noncoherent FSK. Problem 4.19 (a) The Walsh-Hadamard codes for 8-ary orthogonal signaling are given by   1 1 1 1 1 1 1 1  1 −1 1 −1 1 −1 1 −1      1 1 −1 −1 1 1 −1 −1      1 −1 −1 1 1 −1 −1 1  H2 H2   H3 = =  1 1 1 1 −1 −1 −1 −1 H2 −H2    1 −1 1 −1 −1 1 −1 1     1 1 −1 −1 −1 −1 1 1  1 −1 −1 1 −1 1 1 −1 9

(b) Omitted. (c) We need 8 complex dimensions for 8-ary orthogonal signaling, and the spectral efficiency is log2 8 = 83 bps/Hz. For biorthogonal signaling, we can fit 16 signals into the same number of 8 dimensions, so that the spectral efficiency becomes log82 16 = 84 bps/Hz. The fractional increase in spectral efficiency is 33%, as computed below: 4 8

−

3 8

3 8

=

1 3

Problem 4.20 The complex envelopes for the signals are given by u = a + jb, v = b + ja, w = b − ja, x = a − jb, dropping the t dependence from the notation for convenience. Note that ha, bi = 0 and ha, ai = hb, bi. (a) True. For coherent orthogonal signaling with signals s1 and s2 , we only need Rehs1 , s2 i = 0, or hs1c , s2c i + hs1s , s2s i = 0. We can now check that this condition holds for each pair of the 4 signals given: (u, v) : ha, bi + hb, ai = 0 (u, w) : ha, bi − hb, ai = 0 (u, x) : ha, ai − hb, bi = 0 (v, w) : hb, bi − ha, ai = 0 (v, x) : hb, ai − ha, bi = 0 (w, x) : hb, ai + ha, bi = 0 (b) False. The complex inner products must be zero for noncoherent orthogonal signaling, but this condition does not hold: −ju = −j(a + jb) = b − ja = w, so that u and w are complex multiples of each other. Problem 4.21 For QPSK (2 bits/symbol), the symbol rate is 5 Msymbols/sec. For an excess bandwidth of 50%, the bandwidth is 5(1 + 0.5) = 7.5 MHz. For 64-QAM (6 bits/symbol), the symbol rate is 10/6 Msymbols/sec, and the bandwidth is 10/6 (1 + 0.5) = 2.5 MHz. For 64-ary noncoherent orthogonal modulation, we have log642 64 = 6/64 bits/chip, so that the bandwidth is 10 × 64/6 × (1 + 0.5) = 160 MHz. Problem 4.22 For 64-ary noncoherent orthogonal modulation using Walsh-Hadamard codes, the bandwidth efficiency is log642 64 = 6/64 bits/chip. Thus, the chip rate is 20 × 64/6 KHz. With an excess bandwidth of 25%, the bandwidth required is 20 × 64/6 × (1 + 0.25) = 266.67 KHz. For coherent orthogonal modulation, the bandwidth efficiency is doubled, so that the bandwidth required is halved to 133.33 KHz.

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