Question

7.15 Channel equalization. We suppose that u1, . . . , um is a signal (time series) that is trans- mitted (for example by radio). A receiver receives the signal y = c ∗ u, where the n-vector c is called the channel impulse response. In most applications n is small, e.g., under 10, and m is much larger. An equalizer is a k-vector h that satisfies h∗c ≈ e1, the first unit vector of length n + k − 1. The receiver equalizes the received signal y by convolving it with the equalizer to obtain z = h ∗ y.

(a) How are z (the equalized received signal) and u (the original transmitted signal) related? Hint. Recall that h∗(c∗u) = (h∗c)∗u.

(b) Numerical example. Generate a signal u of length m = 50, with each entry a random value that is either −1 or +1. Plot u and y = c ∗ u, with c = (1,0.7,−0.3). Also plot the equalized signal z = h ∗ y, with

h = (0.9, −0.5, 0.5, −0.4, 0.3, −0.3, 0.2, −0.1).

Answer 1