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/d5570b48-2971-4872-9b71-436145e832a4.jpg

X = 28(t + 2) — 25(t − 2) — rect (¹+¹). - So, dF(t) dt F2 · = F₁ (t) + F₂ (t) 2 (+)........ F₁ (1) Fit replace by F₁w (1) replace by F₂ (w) F₁(w) = -2 [¹ + ¹] Apply Fourier transform, jw F(w) = F₁(w) + F₂ (w)...... equation 2 F₁(t) = -8(t + 2) + 8(t-2)........ equation 3 now t replace by w, 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 +rect ect (²/¹) equation 1.

X = 28(t + 2) — 25(t − 2) — rect (¹+¹). - So, dF(t) dt F2 · = F₁ (t) + F₂ (t) 2 (+)........ F₁ (1) Fit replace by F₁w (1) replace by F₂ (w) F₁(w) = -2 [¹ + ¹] Apply Fourier transform, jw F(w) = F₁(w) + F₂ (w)...... equation 2 F₁(t) = -8(t + 2) + 8(t-2)........ equation 3 now t replace by w, 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 +rect ect (²/¹) equation 1.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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