Chapter 6
Noise in Modulation Systems 6.1
Problems
Problem 6.1 The signal power at the output of the lowpasss …lter is PT . The noise power is N0BN , where BN is the noise-equivalent bandwidth of the …lter. From (5.116), we know that the noise-equivalent bandwidth of an nth order Butterworth …lter with 3 dB bandwidth W is Bn (n) =
¼W=2n sin (¼=2n)
Thus, the signal-to-noise ratio at the …lter output is SNR =
PT sin (¼=2n) PT = N0BN ¼=2n N0 W
so that f (n) is given by sin (¼=2n) ¼=2n We can see from the form of f (n) that since sin (x) ¼ 1 for x ¿ 1, f (1) = 1. Thus for large n, the SNR at the …lter output is close to PT =N0 W . The plot is shown In Figure 6.1. f (n) =
Problem 6.2 We express n (t) as ·
¸ · ¸ 1 1 n (t) = nc (t) cos !c t § (2¼W) t + µ + ns (t) sin ! ct § (2¼W ) t + µ 2 2
where we use the plus sign for the U SB and the minus sign for the LSB. The received signal is xr (t) = Ac [m (t) cos (!c t + µ) ¨ m b (t) sin (!c t + µ)] 1
2
CHAPTER 6. NOISE IN MODULATION SYSTEMS
1 0.9 0.8 0.7
f(n)
0.6 0.5 0.4 0.3 0.2 0.1 0
1
2
3
4
5
6
7
8
9
10
n
Figure 6.1: Multiplying xr (t) + n (t) by 2 cos (! ct + µ) and lowpass …ltering yields yD (t) = Acm (t) + nc (t) cos (¼W t) § ns (t) sin (¼W t) From this expression, we can show that the postdetection signal and noise powers are given by SD = A2c m2 ST = Ac m2
ND = N0W NT = N0 W
This gives a detection gain of one. The power spectral densities of nc (t) and ns (t) are illustrated in Figure 6.2. Problem 6.3 The received signal and noise are given by xr (t) = Acm (t) cos (! ct + µ) + n (t) At the output of the predetection …lter, the signal and noise powers are given by 1 ST = A2c m2 2
NT = n2 = N0 BT
The predetection SNR is (SN R)T =
A2c m2 2N0 BT
6.1. PROBLEMS
3
S nc ( f ) , Sns ( f ) N0
f
1 − W 2
1 W 2 Figure 6.2:
Snc ( f ) N0
f
1 − BT 2
0
1 B 2 T
Figure 6.3: If the postdetection …lter passes all of the nc (t) component, yD (t) is y D (t) = Ac m (t) + nc (t) The output signal power is A2c m2 and the output noise PSD is shown in Figure 6.3. Case I: BD > 12 BT For this case, all of the noise, nc (t), is passed by the postdetection …lter, and the output noise power is Z 1 BT 2 ND = N0df = 2N0 BT This yields the detection gain
¡ 12 BD
(SNR)D A2 m2=N0BT = c =2 (SN R)T A2c m2 =2N0BT
4
CHAPTER 6. NOISE IN MODULATION SYSTEMS
Case II: BD < 12 BT For this case, the postdetection …lter limits the output noise and the output noise power is Z BD ND = N0df = 2N0BD ¡BD
This case gives the detection gain
(SN R)D A2m2=2N0BD BT = c = (SNR)T BD A2c m2=2N0 BT Problem 6.4 This problem is identical to Problem 6.3 except that the predetection signal power is i 1 h ST = A2c 1 + a2m2n 2 and the postdetection signal power is
SD = A2c a2 m2n The noise powers do not change. Case I:
BD > 12 BT (SNR)D A2 a2 m2n =N0BT 2a2 m2n h c i = = (SN R)T 1 + a2 m2n A2c 1 + a2m2n =2N0BT
Case II:
BD < 12 BT (SNR)D A2a2m2n=2N0 BD a2m2n B hc i = = = T 2 2 (SN R)T BD 2 2 2 1 + a mn Ac 1 + a mn =2N0BT
Problem 6.5 Since the message signal is sinusoidal mn (t) = cos (8¼t)
6.1. PROBLEMS
5
Thus, m2n = 0:5 The e¢ciency is therefore given by Ef f =
(0:5) (0:8)2 = 0:2424 = 24:24% 1 + (0:5) (0:8)2
From (6.29), the detection gain is (SN R)D = 2Eff = 0:4848 = ¡3:14dB (SNR)T and the output SNR is, from (6.33), (SN T)D = 0:2424
PT N0 W
Relative to baseband, the output SNR is (SNR)D = 0:2424 = ¡6:15 dB PT =N0W If the modulation index in increased to 0:9, the e¢ciency becomes Ef f =
(0:5) (0:9)2 = 0:2883 = 28:83% 1 + (0:5) (0:9)2
This gives a detection gain of (SN R)D = 2Eff = 0:5765 = ¡2:39dB (SNR)T The output SNR is (SNR)D = 0:2883
PT N0W
which, relative to baseband, is (SNR)D = 0:2883 = ¡5:40 dB PT =N0W This represents an improvement of 0:75 dB. Problem 6.6
6
CHAPTER 6. NOISE IN MODULATION SYSTEMS
The …rst step is to determine the value of M. First we compute µ ¶ Z 1 1 2=2¾2 M ¡y p P fX > Mg = e dy = Q = 0:005 ¾ 2¼¾ M This gives
M = 2:57 ¾
Thus, M = 2:57¾ and
m (t) 2:57¾
mn (t) = Thus, since m2 = ¾2 m2n =
m2 ¾2 = = 0:151 6:605¾2 (2:57¾)2
Since a = 12 , we have Ef f = and the detection gain is
0:151
¡ 1 ¢2 2
1 + 0:151
¡ 1 ¢2 = 3:64% 2
(SNR)D = 2E = 0:0728 (SN R)T Problem 6.7 The output of the predetection …lter is
e (t) = Ac [1 + amn (t)] cos [!c t + µ] + rn (t) cos [! c t + µ + Án (t)] The noise function rn (t) has a Rayleigh pdf. Thus fRn (rn ) =
r ¡r2 =2N e N
where N is the predetection noise power. This gives 1 1 N = n2c + n2s = N0B T 2 2 From the de…nition of threshold 0:99 =
Z
0
Ac
r ¡r 2=2N e dr N
6.1. PROBLEMS
7
which gives
2
0:99 = 1 ¡ e¡Ac =2N
Thus,
¡ which gives
A2c = 1n (0:01) 2N A2c = 9:21 N
The predetection signal power is i 1 1 h PT = Ac 1 + a2 m2n ¼ A2c [1 + 1] = A2c 2 2
which gives
PT A2 = c = 9:21 ¼ 9:64dB N N
Problem 6.8 Since m (t) is a sinusoid, m2n = 12 , and the e¢ciency is Ef f =
1
1 2 2a + 12 a2
and the output SNR is (SN R)D = Eff = In dB we have (SNR)D;dB = 10log10 For a = 0:4,
µ
=
a2 2 + a2
a2 PT 2 2 + a N0 W
a2 2 + a2
¶
+ 10 log 10
PT N0 W
(SNR)D;dB = 10log10
PT ¡ 11:3033 N0W
(SNR)D;dB = 10 log 10
PT ¡ 9:5424 N0 W
(SNR)D;dB = 10 log 10
PT ¡ 7:0600 N0 W
(SNR)D;dB = 10 log 10
PT ¡ 5:4022 N0 W
For a = 0:5,
For a = 0:7,
For a = 0:9,
8
CHAPTER 6. NOISE IN MODULATION SYSTEMS
The plot of (SNR)dB as a function of PT =N0W in dB is linear having a slope of one. The modulation index only biases (SNR)D;dB down by an amount given by the last term in the preceding four expressions. Problem 6.9 Let the predetection …lter bandwidth be BT and let the postdetection …lter bandwidth be BD. The received signal (with noise) at the predetection …lter output is represented xr (t) = Ac [1 + amn (t)] cos !c t + nc (t) cos ! c t + ns (t) sin ! ct which can be represented xr (t) = fAc [1 + amn (t)] + nc (t)g cos ! c t ¡ ns (t) sin !c t The output of the square-law device is y (t) = fAc [1 + amn (t)] + nc (t)g2 cos2 ! ct
¡fAc [1 + amn (t)] + nc (t)gns (t) cos ! c (t) +n2s (t) sin2 cos !c (t)
Assuming that the postdetection …lter removes the double frequency terms, yD (t) can be written 1 1 yD (t) = fAc [1 + amn (t)] + nc (t)g2 + n2s (t) 2 2 Except for a factor of 2, this is (6.50). We now let mn (t) = cos ! mt and expand. This gives yD (t) =
1 2 A [1 + a cos !m t]2 + Ac nc (t) 2 c 1 1 +Acanc (t) cos !m t + n2c (t) + n2s (t) 2 2
We represent yD (t) by y D (t) = z1 (t) + z2 (t) + z3 (t) + z4 (t) + z5 (t) where zi (t) is the ith term in the expression for yD (t). We now examine the power spectral density of each term. The …rst term is µ ¶ 1 2 1 2 1 z1 (t) = Ac 1 + a + A2c a cos !m t + A2c a2 cos 2! mt 2 2 4 z2 (t) = Acnc (t) z3(t) = Acanc (t) cos !m t
6.1. PROBLEMS
9 1 z4(t) = n2c (t) 2
and
1 z5(t) = n2s (t) 2 A plot of the PSD of each of these terms is illiustrated in Figure 6.4. Adding these …ve terms together gives the PSD of yD (t). Problem 6.10 Assume sinusoidal modulation since sinusoidal modulation was assumed in the development of the square-law detector. For a = 1 and mn (t) = cos (2¼fm t) so that m2n = 0:5, we have, for linear envelope detection (since PT =N0 W À 1, we use the coherent result), (SNR)D;` = Ef f
1 PT a2 m2n PT PT 1 PT = = 2 1 = 2 2 N0 W 3 N0W 1 + 2 N0 W 1 + a mn N0W
For square-law detection we have, from (6.61) with a = 1 µ
a (SNR)D;S` = 2 2 + a2
¶2
2 PT 9 N0 W 1 PT 3 N0 W
=
PT 2 PT = N0W 9 N0W
Taking the ratio (SN R)D;S` = (SNR)D;`
2 = ¡1:76 dB 3
This is approximately ¡1:8 dB. Problem 6.11 For the circuit given H (f ) = and
R 1 ³ ´ = R + j2¼f L 1 + j 2¼f L R
jH (f)j2 = The output noise power is Z 1 N0 N= ¡1 2
1+
df ³
2¼fL R
³
1+
´2 = N0
Z
0
1 2¼f L R
1
´2
1 1 + x2
µ
¶ R NR dx = 0 2¼L 4L
10
CHAPTER 6. NOISE IN MODULATION SYSTEMS
Sz1 ( f
) 2
1 4 AC 64
1 4 1 2 AC 1 + a 4 2 1 4 2 AC a 4
1 4 2 AC a 4
−2 f m
− fm −
1 4 4 AC a 64 f
0
Sz 2 ( f
)
fm −
2 fm AC2 N0 f
1 − BT 2
0
Sz3 ( f
)
1 BT 2
f
1 1 − BT − f m − BT + f m 0 2 2
1 BT − f m 2 Sz4 ( f ) , Sz5 ( f
1 BT + f m 2
)
1 2 N0 BT 4 f 0
Figure 6.4:
6.1. PROBLEMS
11
The output signal power is S=
1 A2R2 2 R2 + (2¼fc L)2
This gives the signal-to-noise ratio S 2A2 RL h i = N N0 R 2 + (2¼fc L)2 Problem 6.12 For the RC …lter
S 2A2 RL h i = N N0 1 + (2¼fc RC)2
Problem 6.13 The transfer function of the RC highpass …lter is H (f ) =
R j2¼f RC = 1 1 + j2¼f RC R + j2¼C
so that H (f) =
(j2¼f RC)2 1 + (j2¼fRC)2
Thus, The PSD at the output of the ideal lowpass …lter is ( 2 N0 (2¼fRC) jf j < W 2, 2 1+(2¼fRC) Sn (f ) = jf j > W 0, The noise power at the output of the ideal lowpass …lter is Z W Z W (2¼f RC)2 N= Sn (f ) df = N0 df 1 + (2¼fRC)2 ¡W 0 with x = 2¼f RC, the preceding expression becomes Z 2¼RCW N0 x2 N= dx 2¼RC 0 1 + x2 Since
x2 1 =1¡ 1 + x2 1 + x2
12
CHAPTER 6. NOISE IN MODULATION SYSTEMS
we can write N0 N= 2¼RC or N= which is
½Z
2¼RCW
0
df ¡
Z
2¼RCW
0
dx 1 + x2
¾
¢ N0 ¡ 2¼RCW ¡ tan¡1 (2¼RCW ) 2¼RC
N = N0 W ¡
N0 tan¡1 (2¼RCW ) 2¼RC
The output signal power is 1 A2 (2¼fc RC)2 S = A2 jH (fc )j2 = 2 2 1 + (2¼fcRC)2 Thus, the signal-to-noise ratio is S A2 (2¼fcRC)2 2¼RC = N 2N0 1 + (2¼fc RC)2 2¼RCW ¡ tan¡1 (2¼RCW ) Note that as W ! 1;
S N
! 0.
Problem 6.14 For the case in which the noise in the passband of the postdetection …lter is negligible we have ²2Q = ¾2Á ; SSB and QDSB and
3 ²2D = ¾4Á ; 4
DSB
Note that for reasonable values of phase error variance, namely ¾2Á ¿ 1, DSB is much less sensitive to phase errors in the demodulation carrier than SSB or QDSB. Another way of viewing this is to recognize that, for a given acceptable level of ²2 , the phase error variance for can be much greater for DSB than for SSB or QDSB. The plots follow by simply plotting the two preceding expressions. Problem 6.15 From the series expansions for sin Á and cos Á we can write 1 1 cos Á = 1 ¡ Á2 + Á4 2 24 1 3 sin Á = Á ¡ Á 6
6.1. PROBLEMS
13
Squaring these, and discarding all terms Ák for k > 4, yields 1 cos2 Á = 1 ¡ Á2 + Á4 3 1 sin2 Á = Á2 ¡ Á4 3 Using (6.70) and recognizing that m1 (t), m2 (t), and Á (t) are independent, yields "2 = ¾2m1 ¡ 2¾2m1 cos Á + ¾2m1 cos2 Á + ¾2m2 sin2 Á + ¾2n Assuming Á (t) to be a zero-mean process and recalling that Á4 = 3¾4Á gives 1 1 cos Á = 1 + ¾2Á + ¾4Á 2 8 2 4 2 cos Á = 1 ¡ ¾Á + ¾Á sin2 Á = ¾2Á ¡ ¾4Á
This yields "2 = ¾2m1 ¡ 2¾2m1
µ ¶ ¡ ¢ ¡ ¢ 1 1 1 + ¾2Á + ¾4Á + ¾2m1 1 ¡ ¾2Á + ¾4Á + ¾2m2 ¾2Á ¡ ¾4Á + ¾2n 2 8
which can be expressed
3 "2 = ¾2m1 ¾4Á + ¾2m2 ¾2Á ¡ ¾2m2 ¾4Á + ¾2n 4 For QDSB we let ¾2m1 = ¾2m2 = ¾2m. This gives µ ¶ 1 4 2 2 2 " = ¾m2 ¾Á ¡ ¾Á + ¾2n 4 For ¾2Á >> ¾4Á, we have "2 = ¾2m ¾2Á + ¾2n which yields (6.73). For DSB, we let ¾2m1 = ¾2m and ¾2m2 = 0. This gives (6.79) 3 "2 = ¾2m ¾4Á + ¾2n 4
14
CHAPTER 6. NOISE IN MODULATION SYSTEMS
Problem 6.16 From (6.73) and (6.79), we have ¾2n ; SSB ¾2m 3 ¡ 2 ¢2 ¾2n ¾ + 2 ; DSB 4 Á ¾m
0:05 = ¾2Á + 0:05 =
Thus we have the following signal-to-noise ratios ¾2m = ¾2n ¾2m = ¾2n
1 ; 0:05 ¡ ¾2Á
SSB
1
³ ´2 ; 0:05 ¡ ¾2Á 3 4
DSB
The SSB characteristic has an asymptote de…ned by ¾2Á = 0:05 and the DSB result has an asymptote de…ned by
or
3 ¡ 2 ¢2 ¾ = 0:05 4 Á ¾2Á = 0:258
The curves are shown in Figure 6.5. The appropriate operating regions are above and to the left of the curves. It is clear that DSB has the larger operating region. Problem 6.17 Since we have a DSB system
3 ¾2 "2N = ¾4Á + 2n 4 ¾m
Let the bandwidth of the predetection …lter be BT and let the bandwidth of the pilot …lter be B Bp = T ® This gives ¾2n NB B NB ® = 0 2 T = T 02 p = 2 ¾m ¾m Bp ¾m ½
6.1. PROBLEMS
15
1000 900 800
SSB
700
DSB 600 500 SNR
400 300 200 100 0 0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Phase error variance
Figure 6.5: From (6.85) and (6.86), we have ¾2Á = so that "2N =
1 1 k2 2½
3 1 ® + 4 2 16 k ½ ½
For an SNR of 15 dB ½ = 101:5 = 31:623 Using this value for ½ and with k = 4, we have ¡ ¢ "2N = 7:32 10¡7 + 0:032®
The plot is obviously linear in ® with a slope of 0:032. The bias, 7:32 £ 10 ¡7 , is negligible. Note that for k > 1 and reasonable values of the pilot signal-to-noise ratio, ½, the …rst term (the bias), which arises from pilot phase jitter, is negligible. The performance is limited by the additive noise. Problem 6.18 The mean-square error
© ª "2 (A; ¿) = E [y (t) ¡ Ax (t ¡ ¿)]2
16
CHAPTER 6. NOISE IN MODULATION SYSTEMS
can be written © ª "2 (A; ¿) = E y 2 (t) ¡ 2Ax(t ¡ ¿)y (t) + A2 x2 (t ¡ ¿)
In terms of autocorrelation and cross correlation functions, the mean-square error is "2 (A; ¿) = Ry (0) ¡ 2ARxy (¿) + A2Rx (0) Letting Py = R y (0) and Px = Rx (0) gives "2 (A; ¿) = Py ¡ 2ARxy (¿) + A2Px In order to minimize "2 we choose ¿ = ¿ m such that Rxy (¿) is maximized. This follows since the crosscorrelation term is negative in the expression for "2 . Therefore, "2 (A; ¿ m) = Py ¡ 2ARxy (¿ m ) + A2 Px The gain Am is determined from d"2 = ¡2ARxy (¿ m ) + A2 Px = 0 dA which yields Am =
Rxy (¿ m ) Px
This gives the mean-square error "2 (Am; ¿ m) = Py ¡ 2
2 (¿ ) R2xy (¿ m ) Rxy m + Px Px
which is "2 (Am; ¿ m ) = Py ¡
2 (¿ ) Pxy m Px
The output signal power is
which is
© ª SD = E [Amx (t ¡ ¿ m)]2 = A2m Px SD =
R2xy (¿ m) R2 (¿ m) = xy Px Rx (0)
Since ND is the error "2(Am; ¿ m ) we have R2xy (¿ m ) SD = ND Rx (0) Ry (0) ¡ R2xy (¿ m )
6.1. PROBLEMS
17
Note: The gain and delay of a linear system is often de…ned as the magnitude of the transfer and the slope of the phase characteristic, respectively. The de…nition of gain and delay suggested in this problem is useful when the magnitude response is not linear over the frequency range of interest. Problem 6.19 The single-sided spectrum of a stereophonic FM signal and the noise spectrum is shown in Figure 6.6. The two-sided noise spectrum is given by SnF (f ) =
2 KD N0f , A2C
¡1< f < 1
The predetection noise powers are easily computed. For the L + R channel Z 15;000 2 2 ¡ 12¢ KD KD 2 Pn;L+R = 2 N f df = 2:25 10 N0 0 A2C A2C 0 For the L ¡ R channel
Pn;L¡R = 2
Z
53;000
23;000
2 2 ¡ 12¢ KD KD 2 N f df = 91:14 10 N0 0 A2C A2C
Thus, the noise power on the L ¡ R channel is over 40 times the noise power in the L + R channel. After demodulation, the di¤erence will be a factor of 20 because of 3 dB detection gain of coherent demodulation. Thus, the main source of noise in a stereophonic system is the L ¡ R channel. Therefore, in high noise environments, monophonic broadcasting is preferred over stereophonic broadcasting. Problem 6.20 The received FDM spectrum is shown in Figure 6.7. The kth channel signal is given by xk (t) = Ak mk (t) cos 2¼kf1t Since the guardbands have spectral width 4W , f k = 6kW and the k th channel occupies the frequency band (6k ¡ 1) W · f · (6k + 1) W Since the noise spectrum is given by
SnF =
2 KD N0 f 2 , 2 AC
jfj < BT
The noise power in the kth channel is Z (6k+1)W i BW 3 h Nk = B f 2df = (6k + 1)3 ¡ (6k ¡ 1)3 3 (6k¡1)W
18
CHAPTER 6. NOISE IN MODULATION SYSTEMS
Pilot
L ( f ) + R( f )
0
15
Noise Spectrum
S nF ( f )
L( f ) + R( f )
23
53
f ( kHz )
Figure 6.6: where B is a constant. This gives Nk =
¢ BW 3 ¡ 216k2 + 2 » = 72BW 3 k2 3
The signal power is proportional to A2k . Thus, the signal-to-noise ratio can be expressed as µ ¶2 ¸A2k Ak (SNR)D = 2 = ¸ k k where ¸ is a constant of proportionality and is a function of KD,W ,AC ,N0. If all channels are to have equal (SNR)D, Ak =k must be a constant, which means that Ak is a linear function of k. Thus, if A1 is known, Ak = k A1; k = 1; 2; :::; 7. A1 is …xed by setting the SNR of Channel 1 equal to the SNR of the unmodulated channel. The noise power of the unmodulated channel is Z W B N0 = B f 2df = W 3 3 0 yielding the signal-to-noise ratio
(SNR)D = where P0 is the signal power. Problem 6.21
P0 B W3 3
6.1. PROBLEMS
19
Noise PSD
f 0
f1
f3
f2
f5
f4
f6
f7
Figure 6.7: From (6.132) and (6.119), the required ratio is ³ ´ K2 3 W ¡1 W 2 AD N f ¡ tan 2 0 3 f3 f3 C R= 2 K 2 DN W3 0 3 A2 C
or R=3
µ
f3 W
¶3 µ
W W ¡ tan¡1 f3 f3
¶
This is shown in Figure 6.8. For f3 = 2:1kHz and W = 15kHz, the value of R is µ ¶2 µ ¶ 2:1 15 ¡1 15 R=3 ¡ tan = 0:047 15 2:1 2:1 Expressed in dB this is R = 10log10 (0:047) = ¡13:3 dB
The improvement resulting from the use of preemphasis and deemphasis is therefore 13:3 dB. Neglecting the tan ¡1 (W=f3) gives an improvement of 21 ¡ 8:75 = 12:25 dB. The di¤erence is approximately 1 dB. Problem 6.22 From the plot of the signal spectrum it is clear that k=
A W2
20
CHAPTER 6. NOISE IN MODULATION SYSTEMS
1 0.9 0.8 0.7
R
0.6 0.5 0.4 0.3 0.2 0.1 0 0
1
2
3
4
5 W/f3
6
7
8
Figure 6.8: Thus the signal power is
Z
W
A 2 2 f df = AW 2 W 3 0 The noise power is N0B. This yields the signal-to-noise ratio S =2
(SNR)1 =
2 AW 3 N0B
If B is reduced to W, the SNR becomes (SNR)2 =
2 A 3 N0
This increases the signal-to-noise ratio by a factor of B=W . Problem 6.23 From the de…nition of the signal we have x(t) = A cos 2¼fc t dx = ¡2¼fc A sin 2¼f ct dt d2x = ¡(2¼fc )2 A cos 2¼fct dt2
9
10
6.1. PROBLEMS
21
The signal component of y (t) therefore has power 1 SD = A2 (2¼fc )4 = 8A2¼4 fc4 2 The noise power spectral density at y (t) is Sn (f ) =
N0 (2¼f )4 2
so that the noise power is N ND = 0 (2¼)4 2
Z
W
f 4df =
¡W
16 N ¼ 4W 5 5 0
Thus, the signal-to-noise ratio at y (t) is 5 A2 (SNR)D = 2 N0W
µ
fc W
¶4
Problem 6.24 Since the signal is integrated twice and then di¤erentiated twice, the output signal is equal to the input signal. The output signal is ys (t) = A cos 2¼f ct and the output signal power is
1 SD = A2 2 The noise power is the same as in the previous problem, thus ND =
16 N0¼4W 5 5
This gives the signal-to-noise ratio (SNR)D =
5 A2 32¼4 N0W 5
Problem 6.25 The signal-to-noise ratio at y (t) is 2A (SNR)D = N0
µ
f3 W
¶
tan¡1
W f3
22
CHAPTER 6. NOISE IN MODULATION SYSTEMS
1 0.9 0.8
Normalized SNR
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
1
2
3
4
5 W/f3
6
7
8
9
10
Figure 6.9: The normalized (SNR)D, de…ned by (SNR)D =(2A?N0); is illustrated in Figure 6.9. Problem 6.26 The instantaneous frequency deviation in Hz is ±f = fdm (t) = x (t) where x (t) is a zero-mean Gaussian process with variance ¾2x = fd2¾2m. Thus, j±f j = jxj =
Z
1
jxj ¡x2 =2¾2x p e dx ¡1 2¼¾ x
Recognizing that the integrand is even gives j±fj =
r
2 1 ¼ ¾x
Z
1
xe
¡x2 =2¾ 2x
0
Therefore, j±fj =
r
2 f ¾m ¼ d
dx =
r
2 ¾x ¼
6.1. PROBLEMS
23
Substitution into (6.148) gives
(SN R)D =
p 1 + 2 3 BWT Q
³
fd W
´2
m2n NP0TW ¸ q h i 2 A2C 2 fd ¾ m exp ¡AC + 6 N0BT ¼ W 2N0 BT
3 ·q
The preceding expression can be placed in terms of the deviation ratio, D, by letting D=
fd W
and BT = 2 (D + 1) W
Problem 6.27 (Note: This problem was changed after the …rst printing of the book. The new problem, along with the solution follow.) Assume a PCM system in which the the bit error probability Pb is su¢ciently small to justify the approximation that the word error pobability is Pw ¼ nPb. Also assume that the threshold value of the signal-to-noise ratio, de…ned by (6.175), occurs when the two denominator terms are equal, i.e., the e¤ect of quantizating errors and word errors are equivalent. Using this assumption derive the threshold value of PT =N0Bp in dB for n = 4; 8; and 12. Compare the results with Figure 6.22 derived in Example 6.5. With Pw = nPb, equating the two terms in the denominator of (6.175) gives 2¡2n = nPb(1 ¡ 2¡2n) Solving for Pb we have Pn = From (6.178)
1 2¡2n 1 ¼ 2¡2n n 1 ¡ 2¡2n n
· ¸ P 1 exp ¡ T = 2Pb ¼ 2(1¡2n) 2N0 Bp n
Solving for PT =N0Bp gives, at threshold, · ¸ PT 1 ¼ ¡2 ln 2(1¡2n) = 2(2n ¡ 1) ln(2) + 2 ln(2) N0Bp n The values of PT =N0Bp for n = 4; 8; and 12 are given in Table 6.1. Comparison with Figure 6.22 shows close agreement.
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CHAPTER 6. NOISE IN MODULATION SYSTEMS
n 4 8 12
Table 6.1: Threshold value of PT =N0Bp 10:96 dB 13:97 dB 15:66 dB
Problem 6.28 The peak value of the signal is 7:5 so that the signal power is SD =
1 (7:5)2 = 28:125 2
If the A/D converter has a wordlength n, there are 2n quantizing levels. Since these span the peak-to-peak signal range, the width of each quantization level is SD = This gives
¡ ¢ 15 = 15 2¡n n 2
¡ ¢ ¡ ¢ 1 2 1 SD = (15)2 2 ¡2n = 18:75 2¡2n 12 12 This results in the signal-to-noise ratio "2 =
SNR =
¡ ¢ SD 28:125 ¡ 2n ¢ = 2 = 1:5 22n 18:75 "2