ECE UCSD 154C Communications Systems III Prof. Jack Keil Wolf
Solutions to ECE 154C Problem Set 3
1. Solution: Since the window size is 32, we need 5 bits to encode the position of the matches. Refer to following pages for the procedure. 2. Solution: We will compute the case for 2 bits. The other cases can be done in the same way. However it can also be done by using Matlab as follows. Let the initial values for ai s and bi s as the following. a = [-1.66, -0.589, 0.589, 1.665] b = [-, -1.177, 0.1.177, ] Since fX (x) is symmetric, we only need to compute one side of the region. Using the following formula for the Lloyd Algorithm, we acquire the iterated values of the ai s and bi s. aj =
bj xfX (x)dx bj-1 bj f xdx bj-1 X
=
- 1 e-1/2x |bj j-1 2 Q(bj ) - Q(bj-1 )
b
bj = The converged values are
aj+1 + aj 2
a = [-1.5104, -0.4528, 0.4528, 1.5104] b = [-, -0.9816, 0, 0.9816, ] Now we compute average MSE.
4 2 bj bj-1
=
j=1
(x - aj )2 fX (x)dx
= 0.0328 + 0.0259 + 0.0259 + 0.0328 = 0.117
1
Let's check if this value satisfies the following inequality, where R = 2 and 2 = 1. 1 2 R log2 ( 2 ) 2 Then 22R
2
2
2
2 = 1/16 22R Hence the computed MSE satisfies the rate distortion function. We can compute the case for 3, 4, 5 bits in the same way. The following values were found with Matlab. For R=3, a = [-2.1519, -1.3439, -0.7560, -0.2451, 0.24...
06-HW-3-SOLUTIONS