Question

. Suppose that the signal x(t) = 2e −t/3 + e −t + 3e −t/2 is measured in the presence of noise distributed uniformly in the interval [−0.25, 0.25]. Are the decay rates well separated? Are measurements dominated by the component with the slowest decay rate on any time interval?

Answer 1

Step-by-step explanation:

An exponentially decaying signal is of the form x(t) = Ce^(-αt) in terms of an initial value C and a decay rate α > 0. The signal equals a fraction 1/e of its initial value after the characteristic time scale t = 1/α.

Given

x(t) = 2e^(-t/3) + e^(-t) + 3e^(-t/2)

The decay rates are: 1/3, 1, and 1/2.

The slowest decay rate α is the minimum of {1/3, 1, 1/2} = 1/3.

The corresponding time scale

is only 2/3 times larger than the next faster decay rate 1/2, so the decay rates are NOT well separated. The slowest component is larger than 0.5 as long as t < (ln3)/0.5 ≈ 2.2.

The other two components add

up to more than the value of the slowest component. We conclude that the component with slowest decay rate dominates measurements on any interval even in the presence of noise.