i need the answer quickly 50,dex(+) = -25 (+ + 2) + 25 (1-0) + 15 (t-1)at 2d2 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) = jw2x1w)1=W2-402w2jw2W2ما لم +20 +-j24+2+e2+ €2002م-pj2w]ازم2 + (-1- + (-pj ₂0 + p-j2w)][e-j²¹0-e-j240)]x1w) [2-2j sin (220)]How did he get thevalue inside the circle

Answered: Image /qna-images/answer/ae04d9f3-6a88-4883-a737-936ba331ed84.jpg

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

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter5: Inner Product Spaces

Section5.5: Applications Of Inner Product Spaces

Problem 91E

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