Fourier Transform Step Function [2021]
1jωthe fraction with numerator 1 and denominator j omega end-fraction
: The Laplace transform of ( u(t) ) is ( 1/s ) (for ( \textRe(s)>0 )). Substituting ( s = i\omega ) gives ( 1/(i\omega) ), but the Fourier version also requires the ( \pi \delta(\omega) ) term because the Fourier integral uses the imaginary axis, which passes through the pole at ( s=0 ). The delta captures the contribution of that pole. fourier transform step function
u(t)=12+12sgn(t)u open paren t close paren equals one-half plus one-half sgn open paren t close paren 12one-half is a DC constant (even). is the signum function (odd), defined as -1negative 1 4. Transform Individual Components Using Fourier transform pairs and properties: : The transform of a constant Signum function : The transform of 1jωthe fraction with numerator 1 and denominator j
(it’s 0 half the time and 1 half the time), it possesses a DC component. Without this term, the transform would be incomplete and mathematically inconsistent. Why Does This Matter? u(t)=12+12sgn(t)u open paren t close paren equals one-half
[ \int_0^\infty e^-\alpha t e^-i\omega t dt = \int_0^\infty e^-(\alpha + i\omega) t dt = \frac1\alpha + i\omega ]
). The addition of the delta function in the Fourier domain accounts for the fact that we are evaluating the signal right on the boundary of stability ( Summary Table Frequency Domain DC Component Present (represented by the Delta function) Decay Rate (Magnitude decreases as frequency increases)