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### hw10_problems

Course: EE 501, Spring 2012
School: NYU Poly
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of Principles Digital Communications (EL 601) Polytechnic University, Fall 2005 Instructor: Dr. Elza Erkip November 29, 2005 Due: December 6, 2005 (in class) Homework #10 1. Problem 5 from Homework 9. The constraint length in that problem was stated incorrectly. Consider a rate 1/4, constraint length 3 convolutional code with the octal generators (5,7,7,7). (a) Draw and label the state transition diagram. (b)...

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of Principles Digital Communications (EL 601) Polytechnic University, Fall 2005 Instructor: Dr. Elza Erkip November 29, 2005 Due: December 6, 2005 (in class) Homework #10 1. Problem 5 from Homework 9. The constraint length in that problem was stated incorrectly. Consider a rate 1/4, constraint length 3 convolutional code with the octal generators (5,7,7,7). (a) Draw and label the state transition diagram. (b) Find the transfer function T (D). (c) Find df ree , the minimum free distance of the code. (d) Show the path corresponding to df ree on the trellis diagram. How many information bits does this path correspond to? 2. Problem 8.26 from Proakis, Digital Communications. Assume that the system starts from all zero state and is padded by zero information bits. 3. Problem 8.27 from Proakis, Digital Communications. Assume that the system starts from all zero state and is padded by zero information bits. 4. Problem 8.36 from Proakis. 5. Proakis Table 8.2-2 lists rate 1/3 maximum free distance codes. We will consider constraint length 3 code whose generators in octal are (5,7,7). (a) Draw the state diagram and trellis diagram for this code. (b) Find the transfer function T (D, N, J ). (c) Find minimum free distance df ree . Identify the corresponding path with weight df ree on the trellis diagram. What is the length of this path? How many bit errors does this correspond to? 6. Use the transfer function calculated in the previous problem to nd an upper bound for the probability of bit error for an AWGN channel with (a) Hard decision decoding (you can use the Cherno bound). 1 (b) Soft decision decoding. Now: (c) Find the coding gain in dB. (d) Compare the performances of soft and hard decision decoding by plotting the results on the same graph. You can use Matlab or your favorite program. Your horizontal axis should be SNR per (b bit = Eb /N0 ) in dB, your vertical axis should be bit error probability in log scale. On the same plot show the bit error probability versus SNR/bit performance of uncoded BPSK. (e) Which one is better hard or soft, by how much? Compare your plot with Proakis Figure 8.2-15. (Note the mislabeling in that gure. Solid line corresponds to soft decision decoding, dashed line corresponds to hard.) From your plot, how much is the soft decision decoding better than uncoded transmission? How does this relate to the coding gain? 7. Consider the constraint length K = 3, rate 1/4 convolutional code C with generator polynomials (5,7,7,7) in octal. Note that according Proakis Table 8-2.3, this code has maximum free distance. The path corresponding to df ree is represented by the state sequence S0 S2 S3 S1 S0 where index of the state represents the contents of the shift registers in decimal. (a) Find the coded bit stream corresponding to the information bit stream 101100. Assume initial contents of the shift registers are zero. (b) Consider the puncturing matrix P1 = 1 1 0 0 1 1 0 0 . What is the rate of the punctured code C1 ? Find the punctured stream corresponding to the information bit stream 101100. (c) Note that the above punctured code C1 corresponds to using only two generators out of four. The resulting code can be found in Proakis Table 8-2.1. What is the free distance df ree,1 ? (d) Consider an alternative puncturing matrix P2 = 0 1 1 1 0 0 0 1 . What is the rate of the punctured code C2 ? Find the punctured stream corresponding to the information bit stream 101100. 2 (e) For C2 , nd the Hamming weight of the punctured codeword represented by the path S0 S2 S3 S1 S0 S0 . . .. Compare with df ree,1 of part (c). Based on this comparison state whether you would prefer C1 or C2 . 3
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NYU Poly - EE - 501
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NYU Poly - EE - 501
NYU Poly - EE - 501
NYU Poly - EE - 501
CHAPTER 2Problem 2.1 :3P (Ai ) =Hence :j =1P (Ai, Bj ), i = 1, 2, 3, 43P (A1 ) =j =1P (A1 , Bj ) = 0.1 + 0.08 + 0.13 = 0.313P (A2 ) =j =1P (A2 , Bj ) = 0.05 + 0.03 + 0.09 = 0.173P (A3 ) =j =1P (A3 , Bj ) = 0.05 + 0.12 + 0.14 = 0.313P (
NYU Poly - EE - 501
CHAPTER 3Problem 3.1 :I (Bj ; Ai ) = log 2P (Bj |Ai )P (Bj , Ai )= log 2P (Bj )P (Bj )P (Ai )Also :0.31, j = 1P (Bj , Ai ) = 0.27, j = 2P (Bj ) =i=10.42, j = 340.31,30.17,P (Ai) =P (Bj , Ai ) = 0.31,j =10.21,Hence :i=1i=2i=3i=4
NYU Poly - EE - 501
CHAPTER 4Problem 4.1 :(a)x(t) =Hence :1x(a)data1 xax(t) = t()a da)1= x(t+bb (db)1 xb= t(+)b db()1= xbb db = x(t)twhere we have made the change of variables : b = a and used the relationship : x(b) = x(b).(b) In exactly the same way as
NYU Poly - EE - 501
CHAPTER 5Problem 5.1 :(a) Taking the inverse Fourier transform of H (f ), we obtain :1ej 2f T F 1j 2fj 2ft T2= sgn(t) sgn(t T ) = 2Th(t) = F 1 [H (f )] = F 1where sgn(x) is the signum signal (1 if x &gt; 0, -1 if x &lt; 0, and 0 if x = 0) and (x) i
NYU Poly - EE - 501
CHAPTER 6Problem 6.1 :Nn=1 rn fn (t)Using the relationship r (t) =and s(t; ) =[r (t) s(t; )]2 dt =1N0=1N0=1N0Nn=1=1N0Nn=1 (rn=1N0122Nn=1 (rnNn=1Nn=1sn ( )fn (t) we have :2 sn ( )fn (t) dtNm=1 (rnNm=1 (rn sn ( )(rm s
NYU Poly - EE - 501
CHAPTER 7Problem 7.1 :I (xj ; Y ) =SinceQ 1i=0 P (yi |xj ) logP (yi|xj )P (y i )q 1j =0we have :P (xj ) =P ( Xj ) = 1P (xj )=0q 1I (X ; Y ) =P (xj )=0 CP (xj )j =0 P (xj )I (xj ; Y ) == C P (xj )=0 P (xj ) = C = maxP (xj ) I (X ; Y )Thu
NYU Poly - EE - 501
CHAPTER 8Problem 8.1 :(a) Interchanging the rst and third rows, we obtain the systematic form :1001110G= 0 1 0 0 1 1 1 0011101(b)H = P T |I4 = 1110011111011000010000100001(c) Since we have a (7,3) code, there are 23 = 8
NYU Poly - EE - 501
CHAPTER 9Problem 9.1 :We want y (t) = Kx(t t0 ). Then :X (f ) =Y (f ) =j 2f tdt x(t)ey (t)ej 2f tdt =K exp(j 2f t0 )X (f )Therefore :A(f ) = K,for all f(f ) = 2f t0 n, n = 0, 1, 2, .A(f )ej(f ) = Kej 2f t0 Note that n, n odd, results in a s
NYU Poly - EE - 501
CHAPTER 10Problem 10.1 :Suppose that am = +1 is the transmitted signal. Then the probability of error will be :Pe|1 = P (ym &lt; 0|am = +1)= P (1 + nm + im &lt; 0)1P (1/2 + nm &lt; 0) +=41311=+QQ42n42n11P (3/2 + nm &lt; 0) + P (1 + nm &lt; 0)421
NYU Poly - EE - 501
CHAPTER 11Problem 11.1 :(a)4 3 112z + z 1+ z X (z ) = F (z )F (z 1 ) = 1 +5525F (z ) =Hence :=112250and :Copt12250122513/5 = 4/5 012251c 11 a2 aa23/51= c0 = 1 = a1a 4/5 220c1a 1 aawhere a = 0.48 and = 1 2a2 = 0.
NYU Poly - EE - 501
CHAPTER 12Problem 12.1 :(a)NU=n=1NE [U ] =n=1XnE [Xn ] = Nm2u = E U 2 E 2 [U ] = E22m+ N (N 1) 2 N 2 m2 = N 2= N +mHence :nXn Xm N 2 m2N 2 m2N m2(SNR)u ==2N 22 2(b)NV=n=1NE [V ] =n=12Xn2E Xn = N 2 + m2For the varianc
NYU Poly - EE - 501
CHAPTER 13Problem 13.1 :Tc1 6E ccos 2t,3TcTc2g (t) =G (f ) ==But cos 2 Tc t Tc2=12G(0) =12g (t)ej 2f t dt16Ec3Tc1 + cos 2 t Tc0 t TcTc0Tc2cos 2 Tc t Tc2ej 2f t dt. Then1 6E cTc =3Tc4Ec Tc4Ec Tc |G(0)|2 =33and1
NYU Poly - EE - 501
CHAPTER 14Problem 14.1 :Based on the info about the scattering function we know that the multipath spread is Tm = 1 ms,and the Doppler spread is Bd = 0.2 Hz.(a) (i) Tm = 103 sec(ii) Bd = 0.2 Hz1(iii) (t)c Bd = 5 sec(iv) (f )c T1 = 1000 Hzm(v) Tm
NYU Poly - EE - 501
CHAPTER 15Problem 15.1 :L1gk (t) = ejkn=0ak (n)p(t nTc )The unit energy constraint is :T0gk (t)gk (t)dt = 1We also dene as cross-correlation :ij ( ) =T0gi (t)gj (t )dt(a) For synchronous transmission, the received lowpass-equivalent signal
NYU Poly - EE - 501
1 .2I.T he s ignalsa re s ketchedn F igure S l.2l.iz ( +-t/z)-iLt-r)Figure 5 1.21I'L)I-Ut'(c)Figure 5 1.231 .25.( a) P eriodic, p eriod : 2r/(4): r /2.( b ) P e r i o d i c ,p e r i o d : 2 r / ( r )=2.(c) c (t) : [ t + c os(4r _ 2 r/B)]/2
NYU Poly - EE - 501
2 .2t.( a) T he d esired c onvolution i sslnl :o[nl* h [nl@c [t]tr,[n ' t]!/c=-mig&quot;|.lo,lilkfornZo/c=0tP+l-o*1.,.-ty,l u l n lf o r c l + p .0_a(d) T he d esired c onvolution i s@sl n l :! r[k]h ' [n- r ]lc= _ o o::- 3 ] r lalhln-
NYU Poly - EE - 501
13 . 2 2 .( &quot; ) ( i ) T : l , o o = 0 , a r : 1 \$ f ,k+0.(ii) H ere,c ( r :)cfw_T :6,( t+2,r,l2-t,-2&lt;t &lt; -r_t&lt;r&lt;lr&lt;t&lt;2oo: Ll2, and^. - | o ,r ke ven7fi2s in(t')sin(t'),(iii) ?:k o dd3, as : 1, andoo :s in(k2n/J)zAkr/3sin(,tzrl3)], k *
NYU Poly - EE - 501
3.28. (a)N=7,&quot;r=tffit kto,\-./.a; = p * 4 cos(nft/3) 2 cos(2n'kl\)U6(c)N=6'pg3.31. ( &quot;) S [r] i s a s s howni n F igure 5 3.31. C lea,rly, [r] h as a f undamental eriod o f 1 0.?[nJ4v6?qr-onir13lvu'161+-lFigure 5 3.31(b) T he F ourier s eri
NYU Poly - EE - 501
4 '23'F or t he g iven s ignalr s(t), w e u se t he F ouriert ransforma nalysise q. ( a.g) t o e valuatet hecorrespondingFourier transform- .&quot;,t l*'&quot;'.xoUd: tL+JU(i)W e k now t hato1(t):r6(t)+ro(-t).Using the linearity and time reversal properti
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