Question

4. This is an exercise about the geometry of signals, and a possible exam-type question. All signals here are over the interval 0 < t S 1. We want to obtain mutually perpendicular polynomial signals. We start with fo(t) 1 for all t. (a) Find the value of the constant a so that the signal fi (t)t+a is perpendicular to the signal fo perpendicular to both fo and fi is perpendicular to each of fo, fı, and f. (b) Find the values of constants bi and b so that the signf(t2 + bit + b2 is e values of constants b1 an 2 1S (c) Find the values of constants c, 2, and c so that the signal fa(t) +ct+tc C3

Answer 1

We will say two functions
f and g
defined in {0,1] are perpendicular if the inner product
f, g >= | f(x)g(x)dx = 0
(This is the usual inner product defined for functions in [0,1])

For answer (a)

fo.jpe li faltfi(f) It-o

$\implies a_1 = -\frac{1}{2}$

For answer (b)

$<f_0,f_2> = \int_0^1f_0(t)f_2(t)dt= 0\implies\int_0^1(t^2+b_1t+b_2)dt =0$ and

$<f_1,f_2> = \int_0^1f_1(t)f_2(t)dt= 0\implies\int_0^1(t-\frac{1}{2})(t^2+b_1t+b_2)dt =0$

bi 0 andbi

2

For answer (c)

$<f_0,f_3> = \int_0^1f_0(t)f_3(t)dt= 0\implies\int_0^1(t^3+c_1t^2+c_2t+c_3)dt =0$

$<f_1,f_3> = \int_0^1f_1(t)f_3(t)dt= 0\implies\int_0^1(t-\frac{1}{2})(t^3+c_1t^2+c_2t+c_3)dt =0$

$<f_2,f_3> = \int_0^1f_2(t)f_3(t)dt= 0\implies\int_0^1(t^2-t+\frac{1}{6})(t^3+c_1t^2+c_2t+c_3)dt =0$

9 C1 + C2 =-10

Solving the above equations we get

5