Derivações dos Fourier Transforms

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Comments on Derivations of the Fourier Transforms of the Unit Step and Signum Signals 
 
Using the notion of the generalized derivative, we have for the Heaviside unit step signal 
 
0,
1
)}({ ≠= ω
ωj
tuF h 
Formula (3.54) gives the Fourier transform for the signum signal 
0,
2
)}{sgn( ≠= ω
ωj
tF 
Note that for 0=ω , for the signum signal we obtain 




=
≠
=⇒== ∫
∞
∞−
=
0,0
0,
2
)}{sgn(0)sgn()}{sgn( 0
ω
ω
ωω jtFdtttF 
However, for the unit step signal, we have 
 




=
≠
=⇒=×=×== ∫ ∫
∞ ∞
∞−
=
0),(
0,
1
)}({)()(2
2
1
1
2
1
1)}({
0
0
ωωπδ
ω
ωωπδωπδω jtuFdtdttuF h 
 
Having obtained the Fourier transform for the signum signal, we can use relationship (2.4) to find the 
Fourier transform for the unit step signal as given in (3.55). 
 
 
 
 
Bracewell’s derivations of the Fourier Transform for the Heaviside Unit Step Signal 
 
He postulates the following result 
)(
1
)}({1)}({ ωδ
ω
ω k
j
tuFtuFj +=⇒= 
To find the unknown constant, he uses the following arguments 
 
∫∫
∞
∞−
∞
∞−
− +=== ωωδ
ωπ
ωω
π
ω ωω dek
j
dejFjFFtu tjtj
h ))(
1
(
2
1
)(
2
1
)}({)( 1 
 
For 0=t 
 
π
ππ
ω
ωπ
=⇒=+== ∫
∞
∞−
k
kk
d
j
uh 22
1
2
1
2
1
)0(