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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
i need the answer quickly
![|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:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbfb92b73-39b8-41bc-8ccc-f740ffb21d24%2F57e07822-1294-44a0-90ec-8ca5dbd76ab7%2Fg19s8s8_processed.jpeg&w=3840&q=75)
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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