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54 f(t)= -e eq if t <0 if t>0 (a < 0) HEL

54 f(t)= -e eq if t <0 if t>0 (a < 0) HEL

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter6: Linear Transformations
Section6.CR: Review Exercises
Problem 40CR: For T:P5P3 and nullity(T)=4, find rank(T).
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Find the fourier transform pair for the given function (Solve 54)

 

One example is given for reference . Must follow the same rule :

Example 6.10: Fourier transform pair Find the Fourier transform pair for Solution By Eq. (6.30), we have Ch.6 Fourier Analysis [0 if t<0 0 - - if :>0 f(t)=< 1 f(a)=e-are-l dt = 1 f(0) = -√2! * √√2.T √√2r The inverse Fourier transform is determined via Eq. (6.31), as f(t)= 1 f(t)= √√2.r 1 1 = a +i 1 (a-iw)(cosat +i sin cot) a² +0² _jax dw= do- 1 X (a > 0) 0 This is the complex Fourier integral representation of f(t). The real form of the Fourier integral is easily derived from here as follows. Multiply and divide the integrand by the conjugate of a +ia, and use Euler's formula, to obtain 1 2л 18 1 a+ia which is the real form of the Fourier integral for f(t). I 1 a+io 1a cosat +ia sin at-iacos ax + @sin axt 2r 21-0 a² +0² doo el do Since sin ot and acos x are odd functions of , and cos at and csin or are even functions of this reduces to f(t)= a cosax+cosin ox a² + a² 199 do
Transcribed Image Text:Example 6.10: Fourier transform pair Find the Fourier transform pair for Solution By Eq. (6.30), we have Ch.6 Fourier Analysis [0 if t<0 0 - - if :>0 f(t)=< 1 f(a)=e-are-l dt = 1 f(0) = -√2! * √√2.T √√2r The inverse Fourier transform is determined via Eq. (6.31), as f(t)= 1 f(t)= √√2.r 1 1 = a +i 1 (a-iw)(cosat +i sin cot) a² +0² _jax dw= do- 1 X (a > 0) 0 This is the complex Fourier integral representation of f(t). The real form of the Fourier integral is easily derived from here as follows. Multiply and divide the integrand by the conjugate of a +ia, and use Euler's formula, to obtain 1 2л 18 1 a+ia which is the real form of the Fourier integral for f(t). I 1 a+io 1a cosat +ia sin at-iacos ax + @sin axt 2r 21-0 a² +0² doo el do Since sin ot and acos x are odd functions of , and cos at and csin or are even functions of this reduces to f(t)= a cosax+cosin ox a² + a² 199 do
of u 54 f(t)=- -at if t <0 if t>0 eq (a < 0)
Transcribed Image Text:of u 54 f(t)=- -at if t <0 if t>0 eq (a < 0)
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