EE 360: Communication Systems I, Fall 2017Homework 9 Due: 12:05PM, Dec. 7, 2017Collaboration policy: You are allowed to discuss homework problems with your classmates. How-ever, you must write up and submit your own solutions, representing your own work.Any violation of the collaboration policy will be handled seriously.1. Reading assignment: 6.1, 6.2, 6.3, 6.4, 6.5.2. (3+3=6 points) Problem 6.9 (a)(d) on page 257 in the textbook.Solution:6(1)Next, using the trigonometric identityand the definition of the sinc functionwe may go on to write(2)Accordingly, using Eq. (2) in (1), we getwhich is the desired result, except for the scaling factor .Problem 6.9The bandwidth B of a raised cosine pulse spectrum is 2B0- f1, where B0= 1/2Tband f1= B0(1 - a).Thus B = B0(1 + α). For a data rate of 56 kilobits per second, B0 = 28 kHz.(a) α = 0.25,B = 28 kHz x 1.25 = 35 kHz(b) α = 0.5,B = 28 kHz x 1.5 = 42 kHz(c) α = 0.75,B = 28 x 1.75 = 49 kHz(d) α = 1.0,B = 28 x 2 = 56 kHzProblem 6.10The raised cosine pulse bandwidth BT=2B0- f1, where B0= 1/2Tb. For this channel, BT= 75 kHz.For the given bit duration, B0 = 50 kHz. Then,f1=2B0 - BT= 25 kHzα = 1 - f1/BT= 0.5pt() c2B0t()2πB0t()cos1 16B02t2–-----------------------------⎝⎠⎜⎟⎛⎞sin=A() A()cossin12---2A()sin=c x()sinπx()sinπx-------------------=c2B0t()2πB0t()cossin2πB0t()2πB0t()cossin2πB0t---------------------------------------------------------=4πB0t()sin4πB0t----------------------------=csin 4B0t()=pt()c4B0t()sin1 16B02t2–---------------------------=E3. (3+2=5 points) Problem 6.10 on page 257 in the textbook.Solution:6(1)Next, using the trigonometric identityand the definition of the sinc functionwe may go on to write(2)Accordingly, using Eq. (2) in (1), we getwhich is the desired result, except for the scaling factor .Problem 6.9The bandwidth B of a raised cosine pulse spectrum is 2B0- f1, where B0= 1/2Tband f1= B0(1 - a).Thus B = B0(1 + α). For a data rate of 56 kilobits per second, B0 = 28 kHz.(a) α = 0.25,B = 28 kHz x 1.25 = 35 kHz(b) α = 0.5,B = 28 kHz x 1.5 = 42 kHz(c) α = 0.75,B = 28 x 1.75 = 49 kHz(d) α = 1.0,B = 28 x 2 = 56 kHzProblem 6.10The raised cosine pulse bandwidth BT=2B0- f1, where B0= 1/2Tb. For this channel, BT= 75 kHz.For the given bit duration, B0 = 50 kHz. Then,f1=2B0 - BT= 25 kHzα = 1 - f1/BT= 0.5pt() c2B0t()2πB0t()cos1 16B02t2–-----------------------------⎝⎠⎜⎟⎛⎞sin=A() A()cossin12---2A()sin=c x()sinπx()sinπx-------------------=c2B0t()2πB0t()cossin2πB0t()2πB0t()cossin2πB0t---------------------------------------------------------=4πB0t()sin4πB0t----------------------------=csin 4B0t()=pt()c4B0t()sin1 16B02t2–---------------------------=E14. (5+3=8 points) Problem 6.12 on page 257 in the textbook.Solution:7The design parameters of the required raised-cosine pulse spectrum are f1 = 25 kHz and α = 0.5.Problem 6.11The transmission bandwidth BTis related to the excess bandwidth fvby the formula (see Eqs.(6.21) and (6.22))where B0= 1/(2Tb). We may therefore express the bit rate 1/Tbas a function of the excessbandwidth fv as follows:(1)From Eq. (1), we see that the bit rate 1/Tbdecreases linearly with the excess bandwidth fvfor afixed channel bandwidth BT. Specifically, with BT= 3 kHz, the bit rate versus excess bandwidthgraph takes the form shown in Fig. 1. Note that the excess bandwidth fvattains its largest valuewhen the roll-off factor α equals unity, in which case fv = 3 kHz.Problem 6.12We are given the following specifications:(a) The transmission bandwidth is related to the roll-off factor by the formula (see Eq. (6.21))(1)whereTherefore, with (1/Tb) = 4.5 kilobits/s, we haveHence, solving Eq. (1) for the roll-off factor, we getBTB0fv+=1Tb------2 BTfv–()=..6420 1 2 3 Excess bandwidth, fv (kHz)Bit rate1/Tb(kilobits/s)Figure 1BT3 kHz=1Tb------4.5 kilobits/s=BTB01 α+()=B012Tb()⁄=B02.25 kHz=αBTB0------1–=32.25----------1–=8(b) The excess bandwidth is defined (see Eq. (6.22))Problem 6.13According to Eq. (6.30), the pulse-shaping criterion for zero-intersymbol interference isembodied in the relation(1)where P(f) is pulse-shaping spectrum and 1/T is the signaling rate.(a) The pulse-shaping spectrum of Fig. 6.13(a) is defined by(2)Substituting Eq. (2) into (1) leads to the following condition on the signaling rateor, equivalently,(3)(b) The pulse-shaping spectrum of Fig. 6.12(b) is defined by(4)Substituting Eq. (3) into (1) leads to the following condition on the signaling rate13---=fvαB0=13---2.25×=0.75 kHz=PfmT----–⎝⎠⎛⎞m=-∞∞∑constant=Pf()E 2B0()⁄ for f 0=E2B0---------1fB0------–⎝⎠⎛⎞for 0 fB0<<0 for fB0=⎩⎪⎪⎪⎨⎪⎪⎪⎧=1T---B02------=B02 T⁄=Pf()E 2B0()⁄ for 0 ff1<≤E2B0---------1ff1–B0f1–------------------–⎝⎠⎛⎞for f1fB0<<0 for fB0>⎩⎪⎪⎪⎨⎪⎪⎪⎧=1T---12---f1B0+()=Page 2 of 45. (5+5=10 points) Problem 6.13 on page 257 in the textbook.Solution:8(b) The excess bandwidth is defined (see Eq. (6.22))Problem 6.13According to Eq. (6.30), the pulse-shaping criterion for zero-intersymbol interference isembodied in the relation(1)where P(f) is pulse-shaping spectrum and 1/T is the signaling rate.(a) The pulse-shaping spectrum of Fig. 6.13(a) is defined by(2)Substituting Eq. (2) into (1) leads to the following condition on the signaling rateor, equivalently,(3)(b) The pulse-shaping spectrum of Fig. 6.12(b) is defined by(4)Substituting Eq. (3) into (1) leads to the following condition on the signaling rate13---=fvαB0=13---2.25×=0.75 kHz=PfmT----–⎝⎠⎛⎞m=-∞∞∑constant=Pf()E 2B0()⁄ for f 0=E2B0---------1fB0------–⎝⎠⎛⎞for 0 fB0<<0 for fB0=⎩⎪⎪⎪⎨⎪⎪⎪⎧=1T---B02------=B02 T⁄=Pf()E 2B0()⁄ for 0 ff1<≤E2B0---------1ff1–B0f1–------------------–⎝⎠⎛⎞for f1fB0<<0 for fB0>⎩⎪⎪⎪⎨⎪⎪⎪⎧=1T---12---f1B0+()=Note: The solution is not unique. For (a), let T =mB0, m = 1, 2, . . ., or equivalently,1T=B0m, m = 1, 2, . . .we can achieve zero intersymbol interference.Similarly, for (b), let1T=f1+ B0m, m = 1, 2, . . .we can achieve zero intersymbol interference.Page 3 of 46. (3+3=6 points) Problem 6.14 on page 258 in the textbook.Solution:(a) With given bit rate 4.5 Kbps, if 8PAM is used, the corresponding symbol rate1Twould be4.5log28= 1.5 KbpsThus, the minimum bandwidth is B0=12T= 0.75 Kbps.(b) For raised cosine pulse spectrum, the roll-off factor α ∈ [0, 1]. When α = 1,BT= (1 + α)B0= 1.5KHz < 3KHzThus, the maximum
View Full Document