i need the answer quickly |x (t) - - -X = 26(t + 2) — 28(t −So,dF(t)dtF2· = F₁ (t) + F₂ (t)2 (+)... ..... equation 1.F₁ (w)Fit replace by F₁wF₁ (1)(1) replace by F₂ (w)F2Apply Fourier transform,jw F(w) = F₁ (w) + F2₂(w)...... equation 2F₁(t) = −6(t + 2) + 6(t - 2)........ equation 3now t replace by w,F₁(w) = -2 [+]F₁(w)F1(w) = 4jwSin²w.andF2(w) = 2 x 4Sin (2)=2) — rect(+¹) +rect (¹¹)x 2|F2(w) = 8Sin (2)F2(w) = -8Sin2w jwHow did he get thevalue:

Answered: Image /qna-images/answer/478cdd5a-c687-4d4b-8fe9-2c77f44e4b7d.jpg

|x (t) - X So, dF(t) dt = 28(t + 2) — 28(t − 2) — rect(+¹) +rect (¹¹) 2 (+)........ + = F₁ (t) + F₂ (t) F₁(1): Fit replace by F₁w F2 2(1) replace by F₂ (w) Apply Fourier transform, |jw F(w) = F₁ (w) + F₂ (w)...... equation 2 F₁(t) = −6(t + 2) + 6(t− 2)........ equation 3 now t replace by w, F₁(w) = -2 [¹ + ¹] x 2 F₁ (w) = -4 [c²-e ] F1(w) = 4jwSin²w. and F2(w) = 2 x 4Sin (2) |F2(w) = 8Sin (2) F2(w) = -8Sin2w jw equation 1. How did he get t value

|x (t) - X So, dF(t) dt = 28(t + 2) — 28(t − 2) — rect(+¹) +rect (¹¹) 2 (+)........ + = F₁ (t) + F₂ (t) F₁(1): Fit replace by F₁w F2 2(1) replace by F₂ (w) Apply Fourier transform, |jw F(w) = F₁ (w) + F₂ (w)...... equation 2 F₁(t) = −6(t + 2) + 6(t− 2)........ equation 3 now t replace by w, F₁(w) = -2 [¹ + ¹] x 2 F₁ (w) = -4 [c²-e ] F1(w) = 4jwSin²w. and F2(w) = 2 x 4Sin (2) |F2(w) = 8Sin (2) F2(w) = -8Sin2w jw equation 1. How did he get t value

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.6: Additional Trigonometric Graphs

Problem 77E

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