Download Final 2004 answers

EEL 6545 Final Exam ANSWERS Dec. 8, 1:00 – 3:00. 2 hours. Open books and notes.
1. A zero-mean stationary random process d(n) has an autocorrelation function rd(k) =
5/(k2+1) . A noisy measurement of each d(n) is available, x(n) = d(n) + v(n), where v is
zero-mean white noise with power 0.5. But you need to estimate d(100), and the only
measurement data available to you is x(98) and x(100).
(a) If x(98) = 1 and x(100) = 2, state a range of numbers
? < d(100) < ?
that you would consider reasonable for the estimate. Briefly explain your reasoning.
(b) Construct the Wiener filter that estimates d(100) from the measurements x(100) and
x(98); set up the matrix equations for the weights, with the numerical values of the matrix
data, but do not solve them.
| 5.5 1 | | w(0) | = | 5 |
| 1 5.5 | | w(2) |
|1|
2. Again d(n) is a zero-mean stationary random process, and a noisy measurement of each
d(n) is available, x(n) = d(n) + v(n), where v is zero-mean white noise with power 0.5.
But now the autocorrelation function of the measurements rx(k) is 5/(k2+1) . Construct
the Wiener filter that estimates d(100) from the measurements x(100) and x(98); set up
the matrix equations for the weights, with the numerical values of the matrix data, but do
not solve them.
| 5 1 | | w(0) | = | 4.5 |
| 1 5 | | w(2) | | 1 |
3. Suppose that in one day a radioactive substance decays to half its weight (its half-life is
one day). Yesterday at noon a sample was measured to weigh 1005 grams. You may
assume that 100 grams was an unbiased estimate of its true weight and that 5 grams was
the square root of this estimate’s expected squared error.
Today at noon, its weight was measured to be 493 grams. Again, you may assume that
this specifies an unbiased estimate and its root-mean-square error.
A note has just been discovered stating that vandals had altered the sample by 1 gram just
before today’s weighing; we don’t know whether the 1g was added or subtracted, and
either possibility seems equally likely.
What is the Kalman estimate of the sample’s true weight at noon today? What is the
expected value of the squared error in this estimate?
d^old = 100, E(e2old) = 25, dnew = .5 dold + v1, 12 = .512+.512=1,
xnew = 49 = dnew + v2, 22 = 9 . K = {.2525 + 1} / {.2525 + 1+9} = .4462
d^new = .446249 + .5538[.5100] = 49.5538
E(enew2) = .44629 = 4.015
4. Working backward, use the data of problem 3 to calculate the Kalman estimate of the
original weight of the sample, at noon yesterday.
d^old = 49, E(e2old) = 9, dnew = 2 [dold - v1], (-2v1)2 = 412=4,
xnew = 100 = dnew + v2, 22 = 25 . K = {49 + 4} / {49 + 4+25} = .6154
d^new = .6154100 + .3846[249] = 99.23
5. Can the Kalman gain ever be negative? no
6. Can the Kalman gain ever be greater than one?no