of University Illinois 1.(a)
Problem Set #5: Solutions Page 1 of 4
ECE 361 Spring 2001
The signals are illustrated below for the two cases 0 T/2 and T/2 < T. They are obviously orthogonal if = 0 or = T. They are not antipodal for any . 0 T/2, T/2 T
s0(t)
s0(t)
s1(t)
T
s1(t)
T
h(t) h(t)
(b) Almost by inspection (see figure above), E 0 = A2 , E 1 = A2 T, and
A2 , s0 , s 1 = 2 A (T), A2 (T), if 0 T/2, if 0 T/2, Hence, E0 + E 1 2s0 , s 1 = 2 if T/2 T. A (3T), if T/2 T, and the optimum minimax error probability achievable is
(c)
(d)
A2 (T ) A2 (3 T) if 0 T/2, and Q 2N0 2N0 if T/2 T. E0 + E 1 2s0 , s 1 decreases linearly from A2 T at = 0 to A2 T/2 at = T/2, and then increases linearly to 2A 2 T at = T. Hence, the least value of the minimax error probability occurs when = T. The signals are orthogonal (with E0 + E 1 = A2 T) in this case. The signals are also orthogonal if = 0 but E0 + E 1 is only A 2 T/2 in this case. Thus, = 0 gives a local minimum but not the global minimum for the minimax error probability. Similarly, the maximum error probability occurs if = T/2. Note that these results cannot be derived by the usual method of finding where the derivative is zero (why not?), but are obvious if you sketch (SNR)2 as a function of . I take h(t) = [s0 (T t) s1 (T t)]. The impulse response functions are as shown above. For this choice of matched lter response, the minimax threshold can be expressed as (E0 E 1 )/2. Thus, we have
Q
(e)
A2 (T )/2, if 0 T/2, m = (E0 E 1 )/2 = 2 if T/2 T. +A (T )/2, The filter outputs are as sketched on the next page. If 0 T / 2 , then T 0 = 3T/2 at which time, ^0 (T0 ) = 0, ^1 (T0 ) = AT/2 and the minimax threshold s s is AT/4. Notice that this will also hold if = 0 (in which case ^0 (t) = 0 for all t.) Since the noise s ^ (T )^ (T ) N N T s s AT/4 variance is 2 = 0 h2 (t)dt = 0 , the minimax error probability is Q 0 0 1 0 = Q 2 2 2 N0 T/2
= Q
A2 T 8N0
independent of the value of . In contrast, the optimum receiver has P = Q e Q
A2 T 2N0
A2 T at = T/2. 4N0 Note that at = 0, the optimum receiver has a 6 dB advantage over the (simpler?) nonoptimum receiver (in terms of the value of A required to achieve a given error probability). The improvement decreases to a 3dB advantage at = T/2. The optimum receiver can also use a sampling time of T0 = T whereas the at = 0 increasing to
University of Illinois
Problem Set #5: Solutions Page 2 of 4
ECE 361 Spring 2001
nonoptimum receiver uses a sampling time of 3T/2 (i.e. has a greater delay). 0 T/2, T/2 T
s0(t)
s0(t)
s1(t)
T
s1(t)
T
+
2
2 +
If T / 2 T , any T 0 , T T 0 3T/2, is an optimum sampling time! Choosing T0 = T, ^0 (T0 ) = A, ^1 (T0 ) = 0, and the minimax threshold is A/2. s s The minimax error probability is Q This has value A = Q 2 N0 T/2 A2 2 2N0 T
. at = T. In comparison, the
Q
A2 T 8N0
at = T/2 decreasing to Q
A2 T 2N0
A2 T A2 T at = T/2 decreasing to Q at = T. Thus, the 4N0 N0 optimum receiver has a 3dB advantage over the nonoptimum receiver for all , T/2 T. optimum receiver achieves Q P 0 P and P 1 P. Under these constraints, which of the above signal sets will give the smallest error probability? and what is this smallest error probability? E0 + E 1 2 s0 ,s 1 The signals are equally likely, and the error probability is Q 2N0 . We can find the values of P0 = E0 /T, P 1 = E1 /T, P = (P 0 + P 1 )/2, and E0 + E 1 - 2s0 ,s 1 for each of the signal sets. The results are as follows: P0 P1 P s 0 , s 1 E 0 + E 1 - 2 s 0 , s 1 2 2 2 2T I: ( 3/2) (2 3) 2 T = (2 3)PT II: 2 2 2 0 22 T = 2PT III: 2 2 2 2 T 4 2 T = 4PT IV: 2 2 /2 32 /4 2 T/ 2 2 T(3 + 2 2)/2 = 2PT(3+2 2)/3 Clearly, 2 = 2 = 2 = 32 /4 = P. Since 2 3 < 2 < 2(3+2 2)/3 < 4, Set III has the smallest error 2P T probability which is Q . N0 Now we have 2 = 2 = 2 = 2 = P. As in part (a), set III has E0 + E 1 2s0 ,s 1 = 4PT, while set IV
2.
Set Set Set Set (a)
(b)
University of Illinois
Problem Set #5: Solutions Page 3 of 4
ECE 361 Spring 2001
now has E 0 + E 1 2s0 ,s 1 = PT(3+2 2)/3 < 2PT < 4PT. Hence, Set III has the smallest error 2PT probability which is Q . N0 (c) We now must equate the maximum amplitudes 3, 2, 2 , and respectively to A. The values of E0 + E 1 2s0 ,s 1 are (2 3)A2 T/3, A 2 T, 2A2 T, and [(3+2 2)/2]A2 T respectively. Hence, set IV has A2 T(3+2 2) . 4N0 The threshold is zero if E0 = E1 . Sets I-III satisfy the condition. the least error probability which is Q (d) 3.(a) Since t 3 is an odd function while 0 and 2 are even functions, it follows that c3,0 = t3 ,0 = 0 = c3 , 2 =
1
t3 ,2 . c3 , 1 =
(b) (c)
-1 1 2 Since ||t3 ||2 = t 6 dt = 7 , the MMSE is ||t3 ||2 (c 3,i )2 = 2/7 6/25 = 8/175. -1 3 = (3/5)t if t = 0 or t = 3/5. Geez, that was easy! t
t3
3 tdt = 2
-1 3 t5 6 6 = 5 . Hence, the MMSE representation of t3 is 5 2 5 1
|
3 3 t = t. 2 5
3 = t 3 (3/5)t, and ||3 ||2 = 8/175. Hence, 3 (t) = [t3 - (3/5)t]/ 8/175 = 7/2(1/2)[5t3 - The 3t]. reason for the peculiar way of writing the result is that the Legendre polynomial P 3 (t) is commonly defined as (1/2)[5t3 - 3t]. More generally, Pn (t) = polynomials. The orthonormal polynomial is n (t) = [(2n+1)/2]1/2 P n (t).
1 dn 2 n nn! dtn(t -1) . 2
The Pn s are orthogonal (but not orthonormal)
(d)(e) We already found c3,0 = c3,2 = 0, and c3,1 = 6/5. It should be obvious from all of the above (and from the answer to Problem 2(c)!) that c3,3 = 8/175 (f)
||^0 ||2 = ||t3 ||2 = t 6 dt = 2 , so that we set 0 (t) = 7/2t3 . Since ^1 (t) = t2 , and t2 , t 3 = 0, we get s s 7
-1
1
immediately that 1 = ^1 (t) = t2 with ||1 ||2 = 2/5. Hence 1 (t) = s
1
5/2t2 . Next, ^2 (t) = t is orthogonal to s
t2 while c 2,0 =
-1
t ( 7/2)t3 dt = 7/2(2/5) = 14/5. Hence, 2 = t (7/5)t3 , ||2 ||2 = ||t||2 14/25
3/2(1/2)[5t7t3 ]. Finally, ^3 (t) = 1 s
1
(why?) = (2/3) 14/25 = 8/75. Thus, 2 (t) = [t (7/5)t 3 ]/ 8/75 = is orthogonal to t3 and t, while c 3,1 =
-1 2 = ||1||2 - 10/9 = 8/9, and hence (t) = (1 5t 2 /3 2)/ 8/9 = ||3 || 3
1 ( 5/2)t2 dt = 10/3 giving 3 = 1 - 5t2 /3 2.
1/2(1/2)[3 5t 2 ]. Next, some 50 (t)]/3,
algebra shows that 0 (t) = 7/2t3 2 (t) =
= [23 (t) +
211 (t)]/5,
1 (t) =
5/2t2 = [22 (t) +
3/2(1/2)[5t 7t3 ] = [ 213 (t) + 21 (t)]/5,
(g)
3 (t) = 1/2(1/2)[3 5t 2 ] = [ 52 (t) + 2 0 (t)]/3. Thus, [0 (t), 1 (t), 2 (t), 3 (t)] = [0 (t), 1 (t), 2 (t), 3 (t)]A 0 2/3 5/3 0 2/5 0 0 where A = 21/5 . Note that ATA = I, the identity matrix (why?). 0 0 2/3 5/3 2/5 0 21/5 0 Since the 's and 's span the same four dimensional space (of all polynomials of degree 3 or less), the approximation of any s(t) is the same function but expressed in different bases. For example, t4 2/(5 2)0 (t) + 8/(7 10)2 (t) = 6t2 /7 3/35 and t4 ( 10/7)1 (t) - (4/(35 2))3 (t) = 6t2 /7 3/35 also. Obviously, the MMSE is the same in either
University of Illinois
Problem Set #5: Solutions Page 4 of 4
ECE 361 Spring 2001
case. To describe the relationship between the two representations,write b = [b0 , b 1 , b 2 , b 3 ], c = [c0 , c 1 , c 2 , c 3 ], = [0 (t), 1 (t), 2 (t), 3 (t)] and = [0 (t), 1 (t), 2 (t), 3 (t)] as vectors. Then the two approximations are bT and cT. Since = A, we get that cT = AcT= bT, where bT = AcT. Note that A TAcT = ATbT, i.e. cT = ATbT or c = bA. Exercise: verify that [0, 2/(5 2), 0, 8/(7 10)]A = [0, 4.(a) 10/7, 0, 4/(35 2)]
(b) (c)
(d) (e) (f)
(g) (h) (i)
The nearest point must lie on the straight line through (0,0) and (+1,+1) and hence must be of the form (x,x) where (x1)2 + (x1) 2 = 1 2 . It follows that x = 1 1/ 2. The nearest points on the other circles are (x,x), (x,x), and (x,x). The small circle passes through the above four points and thus has radius given by r 2 = x 2 + x 2 = 3 2 2 = ( 2 1) 2 . Hence, r = 2 1 < 1. The nearest point must lie on the straight line through (0,0,0) and (+1,+1,+1) and hence must be of the form (x,x,x) where (x1)2 + (x1) 2 + (x1) 2 = 1 2 . It follows that x = 1 1/ 3. The other nearest points on the other circles are of the form (x,x,x). The ...