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What is the Fourier series for the signal (7) - 6 NOTE. If you use trigonometric identities, you don't have to solve any integrals. x(1) =3 cos (1001) + 3√√3sin (1001) -√3cos (2001) + sin(2001) x(t) = 3√√3 cos (100t) + 3sin (100r) + cos (2001) -√3 sin (2001) Ox(t) = 3 cos (100t) - 3√3sin (100)+cos (2001) -√√3sin (2001) x(t) = 3√3 cos (100r) - 3sin (100t) + √3cos (2001) - sin (2001) x(t) = 6cos 100t + - 5)² 2sin 200t -

What is the Fourier series for the signal (7) - 6 NOTE. If you use trigonometric identities, you don't have to solve any integrals. x(1) =3 cos (1001) + 3√√3sin (1001) -√3cos (2001) + sin(2001) x(t) = 3√√3 cos (100t) + 3sin (100r) + cos (2001) -√3 sin (2001) Ox(t) = 3 cos (100t) - 3√3sin (100)+cos (2001) -√√3sin (2001) x(t) = 3√3 cos (100r) - 3sin (100t) + √3cos (2001) - sin (2001) x(t) = 6cos 100t + - 5)² 2sin 200t -

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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What is the Fourier series for the signal 6cos (100+) - 2sin ( 200 - ² I ? 3 6 NOTE. If you use trigonometric identities, you don't have to solve any integrals. Ox(t) = 3 cos (100t) + 3√√3sin (100t) -√√3 cos (2001) + sin(2001) x(t) = 3√√3 cos (100t) + 3sin (1001) + cos(2001) -√3 sin (2001) O x(t) =3 cos (100t) – 3√3sin(100t) + cos(2001) -√3sin (2001) Ox(t) = 3√√3 cos (100t) - 3sin(100t) + √3 cos (2001) - sin (2001) x(t) =
Transcribed Image Text:What is the Fourier series for the signal 6cos (100+) - 2sin ( 200 - ² I ? 3 6 NOTE. If you use trigonometric identities, you don't have to solve any integrals. Ox(t) = 3 cos (100t) + 3√√3sin (100t) -√√3 cos (2001) + sin(2001) x(t) = 3√√3 cos (100t) + 3sin (1001) + cos(2001) -√3 sin (2001) O x(t) =3 cos (100t) – 3√3sin(100t) + cos(2001) -√3sin (2001) Ox(t) = 3√√3 cos (100t) - 3sin(100t) + √3 cos (2001) - sin (2001) x(t) =
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