Solution Manual F undamentals of Communication Systems John G. Proakis Masoud Salehi Second Edition 1 / 4
Chapter 2 Problem 2.1 1.Π(2t+5)=Π ζ 2 ζ t+ 5 2 ηη . This indicates first we have to plotΠ(2t)and then shift it to left by 5 2
. A plot is shown below:
✻ − 11 4 − 9 4 ✲ t Π(2t+5) 1 2.P ∞ n=0 Λ(t−n)is a sum of shifted triangular pulses. Note that the sum of the left and right side of triangular pulses that are displaced by one unit of time is equal to 1, The plot is given below ✲ ✻ t x2(t) −1 1
- It is obvious from the definition of sgn(t)that sgn(2t)=sgn(t). Thereforex3(t)=0.
4.x4(t)is sinc(t)contracted by a factor of 10.
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Problem 2.2 1.x[n]=sinc(3n/9)=sinc(n/3).
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0.2 0.4 0.6 0.8 1 2.x[n]=Π θ n 4 −1 3 ff . If− 1 2 ≤ n 4 −1 3 ≤ 1 2 , i.e.,−2≤n≤10, we havex[n]=1.
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0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 3.x[n]= n 4 u−1(n/4)−( n 4 −1)u−1(n/4−1). Forn <0,x[n]=0, for 0≤n≤3,x[n]= n 4 and forn≥4,x[n]= n 4 − n 4 +1=1.
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0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Problem 2.3 x1[n]=1 andx2[n]=cos(2πn)=1, for alln. This shows that two signals can be different but their sampled versions be the same.Problem 2.4 Letx1[n]andx2[n]be two periodic signals with periodsN1andN2, respectively, and letN= LCM(N1, N2), and definex[n]=x1[n]+x2[n]. Then obviouslyx1[n+N]=x1[n]andx2[n+N]= x2[n], and hencex[n]=x[n+N], i.e.,x[n]is periodic with periodN.For continuous-time signalsx1(t)andx2(t)with periodsT1andT2respectively, in general we cannot find aTsuch thatT=k1T1=k2T2for integersk1andk2. This is obvious for instance if T1=1 andT2=π. The necessary and sufficient condition for the sum to be periodic is that T1 T2 be a rational number.Problem 2.5
Using the result of problem 2.4 we have:
- The frequencies are 2000 and 5500, their ratio (and therefore the ratio of the periods) is
- The frequencies are 2000 and
- The sum of two periodic discrete-time signal is periodic.
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rational, hence the sum is periodic.
5500 π . Their ratio is not rational, hence the sum is not periodic.