1 Signal Space Representation 1.1 Consider the three waveform fj (t) shown in the following figure. 2 4 −1/2 1/2 f1(t) 2 4 −1/2 1/2 f3(t) 1 3 1/2 4 f2(t) 1. Show that these waveforms are orthonormal 2. Express the waveform x(t) as a linear combination of these waveforms if x(t) =    −1, 0 ≤ t ≤ 1 1, 1 ≤ t ≤ 3 −1, 3 ≤ t ≤ 4 (1) Solution (a) To show that the waveforms fn(t), n = 1, . . . , 3 are orthogonal we have to prove that ∫ ∞ −∞ fm(t)fn(t)dt = 0, m ̸= n Clearly, c12 = ∫ ∞ −∞ f1(t)f2(t)dt = ∫ 4 0 f1(t)f2(t)dt = ∫ 2 0 f1(t)f2(t)dt + ∫ 4 2 f1(t)f2(t)dt = 1 4 ∫ 2 0 dt − 1 4 ∫ 4 2 dt = 1 4 × 2 − 1 4 × (4 − 2) = 0 Similarly, c13 = ∫ ∞ −∞ f1(t)f3(t)dt = ∫ 4 0 f1(t)f3(t)dt = 1 4 ∫ 1 0 dt − 1 4 ∫ 2 1 dt − 1 4 ∫ 3 2 dt + 1 4 ∫ 4 3 dt = 0 1 This study source was downloaded by 100000858468549 from CourseHero.com on 12-14-2022 15:05:41 GMT -06:00 https://www.coursehero.com/file/17450496/HW8-sol/ and c23 = ∫ ∞ −∞ f2(t)f3(t)dt = ∫ 4 0 f2(t)f3(t)dt = 1 4 ∫ 1 0 dt − 1 4 ∫ 2 1 dt + 1 4 ∫ 3 2 dt − 1 4 ∫ 4 3 dt = 0 Thus, the signals fn(t) are orthogonal. (b) We first determine the weighting coefficients xn = ∫ ∞ −∞ x(t)fn(t)dt, n = 1, 2, 3 x1 = ∫ 4 0 x(t)f1(t)dt = − 1 2 ∫ 1 0 dt + 1 2 ∫ 2 1 dt − 1 2 ∫ 3 2 dt + 1 2 ∫ 4 3 dt = 0 x2 = ∫ 4 0 x(t)f2(t)dt = 1 2 ∫ 4 0 x(t)dt = 0 x3 = ∫ 4 0 x(t)f3(t)dt = − 1 2 ∫ 1 0 dt − 1 2 ∫ 2 1 dt + 1 2 ∫ 3 2 dt + 1 2 ∫ 4 3 dt = 0 As it is observed, x(t) is orthogonal to the signal waveforms fn(t), n = 1, 2, 3 and thus it can not represented as a linear combination of these functions. End Solution 2 This study source was downloaded by 100000858468549 from CourseHero.com on 12-14-2022 15:05:41 GMT -06:00 https://www.coursehero.com/file/17450496/HW8-sol/ 1.2 Determine a set of orthonormal functions for the four signals shown in the following figure. 1 2 3 2 −2 2 −2 1 2 3 2 −2 1 2 3 2 −2 1 2 3 s2(t) s3(t) s4(t) s1(t) Solution As a set of orthonormal functions we consider the waveforms ψ1(t) = { 1 0 ≤ t < 1 xss=removed xss=removed xss=removed xss=removed xss=removed xss=removed xss=removed xss=removed xss=removed k=1 xss=removed xss=removed xss=removed k=1 xss=removed k=1 l=1 k=1 xss=removed k=1 xss=removed xss=removed>