50, dex(+) = -25 (+ + 2) + 25 (1-0) + 15 (t-1) at 2 = -√(x + 2) + 251f) + 5lt-2) By using differentiation in time domain property, the fourier tramform is (jw) ² x 1₂0) = -1.e d2x(1) dt 2 (j)² w²x1w) = rejw2 x1w) 1 = W2 - 402 w2 jw2 W2 ما لم +20 + -j24 +2+e 2+ e-J200 2 م -pj2w] ازم 2 + (- : + (-p) ₂20 + p-j²w)] J [e-j²¹0-e-j240)] x1w) [2-2j sin (220)] How did he get the value inside the circle
50, dex(+) = -25 (+ + 2) + 25 (1-0) + 15 (t-1) at 2 = -√(x + 2) + 251f) + 5lt-2) By using differentiation in time domain property, the fourier tramform is (jw) ² x 1₂0) = -1.e d2x(1) dt 2 (j)² w²x1w) = rejw2 x1w) 1 = W2 - 402 w2 jw2 W2 ما لم +20 + -j24 +2+e 2+ e-J200 2 م -pj2w] ازم 2 + (- : + (-p) ₂20 + p-j²w)] J [e-j²¹0-e-j240)] x1w) [2-2j sin (220)] How did he get the value inside the circle
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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i need the answer quickly
![50,
dex(+) = -25 (+ + 2) + 25 (1-0) + 15 (t-1)
at 2
d2 x (1)
dt 2
= -√(x + 2) + 251f) + 5lt-2)
By using differentiation in time domain property,
the fourier tramform is
(jw) ² x 120) = -1.e
(j)² w²x1w) = jw2
x1w)
1
=
W2
-
402
w2
jw2
W2
ما لم +20 +
-j24
+2+e
2+ €200
2
م
-pj2w]
ازم
2 + (-1
- + (-pj ₂0 + p-j2w)]
[e-j²¹0-e-j240)]
x1w) [2-2j sin (220)]
How did he get the
value inside the circle](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe57beecf-da32-494c-b3c8-16fe86914832%2Fae04d9f3-6a88-4883-a737-936ba331ed84%2Flxvwf3f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:50,
dex(+) = -25 (+ + 2) + 25 (1-0) + 15 (t-1)
at 2
d2 x (1)
dt 2
= -√(x + 2) + 251f) + 5lt-2)
By using differentiation in time domain property,
the fourier tramform is
(jw) ² x 120) = -1.e
(j)² w²x1w) = jw2
x1w)
1
=
W2
-
402
w2
jw2
W2
ما لم +20 +
-j24
+2+e
2+ €200
2
م
-pj2w]
ازم
2 + (-1
- + (-pj ₂0 + p-j2w)]
[e-j²¹0-e-j240)]
x1w) [2-2j sin (220)]
How did he get the
value inside the circle
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