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x(t) So, (t) = F₁ (t) + F₂ ³ (+1) + P₂ (t).….…..….…... dt = = 28(t + 2) — 28(t− 2) - rect(+¹) + rect( +rect (¹¹) F₁t replace by F₁w F₂ (1) replace by F₂ (w) F2 equation 1. IN Apply Fourier transform, jw F(w) = F₁ (w) + F2₂(w)...... equation 2 F₁ (t) = -8(t+2)+8(t-2)........ equation 3 now t replace by w, F₁ (w) = -2 [+] × 2 F₁ (w) = -4 [c²-e*] F1(w) = 4jwSin²w. and F2 (w) = 2 x 4Sin (2) F₂ (w) = 8Sin (2) F2 (w) = -8Sin2w jw

x(t) So, (t) = F₁ (t) + F₂ ³ (+1) + P₂ (t).….…..….…... dt = = 28(t + 2) — 28(t− 2) - rect(+¹) + rect( +rect (¹¹) F₁t replace by F₁w F₂ (1) replace by F₂ (w) F2 equation 1. IN Apply Fourier transform, jw F(w) = F₁ (w) + F2₂(w)...... equation 2 F₁ (t) = -8(t+2)+8(t-2)........ equation 3 now t replace by w, F₁ (w) = -2 [+] × 2 F₁ (w) = -4 [c²-e*] F1(w) = 4jwSin²w. and F2 (w) = 2 x 4Sin (2) F₂ (w) = 8Sin (2) F2 (w) = -8Sin2w jw

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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|x (t) - - - X = 26(t + 2) — 28(t − So, dF(t) dt F2 · = F₁ (t) + F₂ (t) 2 (+)... ..... equation 1. F₁ (w) Fit replace by F₁w F₁ (1) (1) replace by F₂ (w) F2 Apply Fourier transform, jw F(w) = F₁ (w) + F2₂(w)...... equation 2 F₁(t) = −6(t + 2) + 6(t - 2)........ equation 3 now t replace by w, F₁(w) = -2 [+] F₁(w) F1(w) = 4jwSin²w. and F2(w) = 2 x 4Sin (2) = 2) — rect(+¹) +rect (¹¹) x 2 |F2(w) = 8Sin (2) F2(w) = -8Sin2w jw How did he get the value:
Transcribed Image Text:|x (t) - - - X = 26(t + 2) — 28(t − So, dF(t) dt F2 · = F₁ (t) + F₂ (t) 2 (+)... ..... equation 1. F₁ (w) Fit replace by F₁w F₁ (1) (1) replace by F₂ (w) F2 Apply Fourier transform, jw F(w) = F₁ (w) + F2₂(w)...... equation 2 F₁(t) = −6(t + 2) + 6(t - 2)........ equation 3 now t replace by w, F₁(w) = -2 [+] F₁(w) F1(w) = 4jwSin²w. and F2(w) = 2 x 4Sin (2) = 2) — rect(+¹) +rect (¹¹) x 2 |F2(w) = 8Sin (2) F2(w) = -8Sin2w jw How did he get the value:
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