Solution to Homework 1

Transcription

1 Solution to Homework 1 1. Exercise 2.4 in Tse and Viswanath. 1. a) With the given information we can comopute the Doppler shift of the first and second path f 1 fv c cos θ 1, f 2 fv c cos θ 2 as well as the Doppler spread D s f 1 f 2 fv c cos θ 1 cos θ 2 from which we can compute the coherence time T c 1 D s c fv cos θ 1 cos θ 2 b) There is not enough information to compute the coherence bandwidth, as it depends on the delay spread which is not given.we would need to know the difference in path length to compute the delay spread T d and use it to compute W c. 2. From part 1 we see that a larger angular range results in larger delay spread and smaller coherence time. In the richly scattered environment, cos θ i cos θ j 2. Hence, the channel would show a smaller coherence time max i,j and a larger delay spread. On the contrary, max i,j cos θ i cos θ j is smaller in the environment where the reflectors are clustered in a small angular range; consequently, the coherence time is larger and the delay spread is smaller in this environment. 1

2 2. Exercise 2.16 in Tse and Viswanath. N h l [m] a i m/w )e j2πfcτ im/w ) sinc l τ i m/w )W ) i1 Let τ 1 N N i1 τ i0) and τ i m/w ) τ i m/w ) τ. Then, h l [m] e j2πfc τ N i1 a i m/w )e j2πfc τ im/w ) sinc l τw τ i m/w )W ) Often in practice f c τ f c d/c d/λ c 1 where d is the distance between transmitter and receiver, so it is a reasonable assumption to model e j2πfc τ e jθ where θ Uniform[0, 2π] and θ is independent of everything else. Note that τ does not depend on m so a particular realization of θ is the same for all components of h. Since e jθ has uniformly distributed phase, its distribution does not change if we introduce an arbitrary phase shift φ. So e jφ e jθ e jθ. It follows that N i1 a im/w )e j2πfc τ im/w ) sinc l τw τ i m/w )W ) N e jφ h e jφ e jθ i1 a im + 1)/W )e j2πfc τ im+1)/w ) sinc l τw τ i m + 1)/W )W ). N i1 a im + n)/w )e j2πfc τ im+n)/w ) sinc l τw τ i m + n)/w )W ) N i1 a im/w )e j2πfc τ im/w ) sinc l τw τ i m/w )W ) N d e jθ i1 a im + 1)/W )e j2πfc τ im+1)/w ) sinc l τw τ i m + 1)/W )W ). N i1 a im + n)/w )e j2πfc τ im+n)/w ) sinc l τw τ i m + n)/w )W ) h where the d refers to the equality in distribution. Since this is true for all φ, under the previous assumptions h is circularly symmetric. 2

3 3. Exercise 2.17 in Tse and Viswanath. 1. hτ, t) is the response of the channel to an impulse that occurs at time t τ, i.e., δt t τ)) in the given expression we obtain: hτ, t) a δτ τ θi t)) K The projection of the velocity vector v onto the direction of the path at angle θ has a magnitude: v θ v cos θ The distance travelled by the mobile in the direction θ in time t is v θ t, which is the reduction in the distance between transmitter and receiver for the path of angle θ. Then, v cos θ t τ θ t) τ θ 0) c 2. T d 1 means that most of the paths arrive in an interval much smaller than the W sample time of the signal, indicating that all the delays fall inside one delay bin; thus, we can lump together the influence of all the paths into a single tap h 0 [m]. We assume that τ θ t) 1. Thus, W h 0 [m] [ a θi e j2πfcτ θ i m m ) ] W ) sinc τ θi W W where we use the fact that a θ t) a θ, t. Finally, we note that and since τ θ m W ) W 1 we obtain: h 0 [m] lim sinct) 1 t0 a θi e j2πfcτ θ i m W ) 3. The independence assumption requires that different paths come from difference scatters. For this to be true even for small variations in angle of arrival θ, it is necessary that the scatters be located far away from the receiver. How far depends on the size of the scatters, and the angle difference θ over which we require the paths to be independent. The identically distributed assumption requires that the lengths of the paths from transmitter to receiver be comparable for all angle θ. This occurs when r R in the following figure. 3

4 4. 5. R 0 [n + m, m] E [h 0 [n + m]h 0[m]] [ ] E a θi e j2πfcτ θ i n+m W ) a θk e j2πfcτ θ k m W ) k0 k0 k0 E [ [ ] ] a θi a j2πf θ i e c τ θi 0) v cos θ i n+m) τ cw θk 0)+ v cos θ k m) cw a 2 [ ] τ δ[i θi 0) v cos θ i n+m) τ cw θk 0)+ v cos θ k m) cw k]e j2πfc K a2 vn e j2πfc cw cos θ i) 2π 2π K K a2 2π 2π 0 vn e j2πfc cw cos θ) dθ R 0 [n] which is indeed stationary. Further, given the Doppler spread D s 2f c v/c and the zeroth-order Bessel function of the first kind J 0 x) 1 π π 0 ejx cos θ dθ, we yield R 0 [n] a2 π Sf)e j2πfn df π 0 a2 π Ds/2W 2fcv jnπ e 1 c W cos θ dθ a 2 J 0 nπd s /W ) D s/2w π 0 1 2π 4a2 W/D s 1 2fW/Ds ) 2 )e j2πfn df e jnπds cosθ)/w dθ R 0 [n] where we use the substitution cosθ) 2fW/D s and sin θdθ 2W D s df. 6. To calculate the power inside [f, f + df], we consider the mapping T : f θ given by the eqaulity cosθ) 2fW/D s. Note that this is a one-to-two mapping since for each f satisfying D s /2W < f < D s /2W there are two corresponding θ for the equation to hold true. Thus, P ower [f, f + df]) 2 P ower [θ, θ + dθ]) 2a 2 dθ 2π 1 2a2 2π 2W D s sin θ df 1 2π 4a 2 W df Sf)df D s 1 2fW/Ds ) 2 This is not surprising because the equality follows from the definition of power spectral density. 4

5 4. Exercise A.2 in Tse and Viswanath. z i ) M i1 is a linear transformation of a Gaussian process, so it has a jointly Gaussian distribution, which is completely specified by the first and second moments. [ ] E [z i ] E wt)s i t)dt E [wt)] s i t)dt 0 [ E [z i z j ] E ] wt)wτ)s i t)s j τ)dtdτ N 0 2 δt τ)s it)s j τ)dtdτ N 0 2 s iτ)s j τ)dτ N 0 2 δ i,j Therefore E [ zz T ] N 0 /2)I M and z N0, N 0 /2)I M )) 5

6 5. Exercise A.3 in Tse and Viswanath. For simplicity, let us assume that x is zero mean. 1. The covariance matrix, K, of x is given by K E [ xx ] E [ R[x] + ji[x]) R[x] T ji[x] T )] E [ R[x]R[x] T ] + E [ I[x]I[x] T ] je [ R[x]I[x] T ] + je [ I[x]R[x] T ] where the dagger means the conjugate transpose of a matrix. Similarly, the pseudo-covariance matrix of x is given by J E [ xx T ] E [ R[x] + ji[x]) R[x] T + ji[x] T )] E [ R[x]R[x] T ] E [ I[x]I[x] T ] + je [ R[x]I[x] T ] + je [ I[x]R[x] T ] The covariance of matrix of [R[x], I[x]] T is given by [ [ ] E R[x]R[x] T E [ R[x]I[x] ] ] T E [ I[x]R[x] ] T E [ I[x]I[x] ] T 1 [ ] R[K + J] I[J K] 2 I[K + J] R[K J] 2. For a circularly symmetric x, J 0. To show this, note that for an arbitrary φ, we have J e jφ x J x since e jφ x has the same distribution with x for a circularly symmetric x. But we also have J e jφ x E [ e jφ x ) e jφ x ) T ] e jφ J x The only possibility is that J x 0. Thus, the covariance of matrix of [R[x], I[x]] T is given by [ [ ] E R[x]R[x] T E [ R[x]I[x] ] ] T E [ I[x]R[x] ] T E [ I[x]I[x] ] T 1 [ ] R[K] I[ K] 2 I[K] R[K] 6

7 6. Exercise 3.4 in Tse and Viswanath. To keep the same probability of error, the separation between consecutive points should be the same for both PAM and QAM. Let this separation be 2a. Then, the average energy for a PAM with 2 k points is given by: E av 2 k PAM ) 1 2 k 1 2i 1) 2 a 2 2 k 1 a2 3 i1 2 2k 1 ) Since a 2 k QAM can be thought as two independent 2 k/2 PAMs, we get that the average energy for a QAM with 2 k points is: E av 2 k PAM ) 2E av 2 2k PAM ) 2a2 3 Thus, the loss in energy is given by: ) 2 k log 2 which grows linearly in k. 2 k 1 ) 7

8 7. Exercise 3.11 in Tse and Viswanath P {x A x B } 1 + SNR where d is the Hamming distance between the binary codewords x A and x B. The diversity gain of the code is the minimum Hamming distance d min between the codewords. 2. It s 2. Same diversity gain as the repetition code but higher rate 3/2 rather than 1/2. 3. The probability a symbol gets decoded incorrectly is of the order of SNR 1. If d min /2 errors are made, then there is a significant probability i.e., the probability does not decay with SNR) that an overall error is made, as an incorrect codeword may be closer to ĉ than the transmitted codeword. We are ok if fewer than that is made. Hence, the diversity gain is d min /2. For the example, the diversity is only The typical error event for each symbol is when the channel is in a deep fade. If we declare an erasure whenever the channel is in a deep fade, then the typical error is that there are only erasures and no hard decision errors. We can decode whenever the number of erasures is less than d min, since there is at most one codeword that is consistent with the erasure pattern. Hence the diversity gain is back to d min, same as soft decision decoding. How do we know the channel is in deep fade? Heuristically, when h l 2 < 1/SNR. More rigorously, we can fix a ɛ > 0 and use the threshold h l 2 < 1/SNR 1 ɛ to decide on an erasure. This will give us a diversity gain of d min 1 ɛ). But we can choose ɛ arbitrarily close to zero so we can get close to the desired diversity gain. ) d 8

9 8. Exercise 3.19 in Tse and Viswanath. 1. Let H be the fading matrix for the MIMO channel. Then the channel model can be written as: Y HX + W Now, this channel model can be rewritten as a MISO channel with block-length n r N. Let X and h be X 0 0 X X h [ H1, 1) H1, L) H2, 1) Hn r, L) ] Then the received signal can be rewritten as y h X + w with y and w appropriately defined in terms of Y and W. Then the probability of pairwise error can be written as: E Q SNR h X A X B ) X A X B ) h 2 2. Since we have reduced the MIMO problem with i.i.d. Rayleigh fading to a MISO probelm with i.i.d. Rayleigh fading, probability of pairwise error can be upper bounded as: 4 Lnr P {X A X B } SNR Lnr det X A X B ) X A X ) B ) [ 4 L SNR L det X A X B )X A X B ) ) where the last step follows from the diagonal structure of X A and X B. 3. Thus, the code design criterion of maximizing the minimum determinant remains unchanged. ] nr 9

10 9. Exercise 3.27 in Tse and Viswanath. 1. The channel matrix is [ ] h0 0 H h 1 h 0 2. The ZF equalizer takes the form of [ ] [ x[0] h0 0 x[1] h 1 h 0 ] 1 [ ] y[0] 1 [ h0 0 y[1] 2 h 0 h 1 h 0 ] [ ] y[0] y[1] It follows that the receiver detects x[0] only based on the information of y[0]. But then it makes an error whenever h 0 is in deep fade. Hence, its diversity is 1. This is less than the diversity gain of 2 in the MLSD approach. 10