Question
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The derivative property of Fourier transforms states that if X(jw) is the Fourier transform of x(t), then j X (jw) is the Fourier transform of d/dt x(t). This is readily proved by writing down the inverse Fourier transform formula and taking the derivative with respect to t of both sides. Let's try proving this with another approach. Remember from your Freshman calculus class that a derivative could be defined as where the + superscript indicates that approaches 0 from "above." (We will assume this limit exists.) Take the Fourier transform of [x(t) - x(t - )]/ using the time-shift property of Fourier transforms, and then replace the exponential in the resulting expression with the first two terms of its Taylor series expansion, i.e., exp(a) 1 + a. This is legitimate since we are letting go to zero. The derivative shown in part (a) was, technically speaking, a "left derivative." Repeat the procedure in part (a), except this time apply it to the "right derivative," given by The Fourier transforms of cos(w0t) and sin(w0t) are readily derived by rewriting both of them with their "inverse Euler's formulas" and applying the Fourier transform pair Suppose we had derived the Fourier transform for cos(w0t) but didn't know the "inverse Euler's formula" for sin(w0t). Derive the Fourier transform of sin(w0) by applying the time-derivative property to the Fourier transform pair and remembering what the derivative of cos(w0t) is from your calculus classes.
Explanation / Answer
a) If X(w) is fourier transform of x(t),
then,X(w)*exp(-jwa) is fourier transform of x(t-a).
Let Y(w) be the fourier transform of d(x(t))/dt .
Therefore Y(w) = lim a->0 *( X(w) - X(w)* exp(-jwa) ) / a.
Y(w) = X(w)*( 1- exp(-jwa) ) / a . -(1)
using taylor series expansion, exp(x) = 1 + x + x^2/2 ....... ;
thus, lim x-> 0 ,( 1-exp(x) ) = -x .
Using the same in equation (1),Y(w) = X(w) * -1*(-jwa) / a = X(w) * jw .
b) Here Y(w) =( X(w)* exp(-jw(-a)) - X(w) )/a (fourier transform of x(t+a ) is X(w)*exp(-jw*(-a))
= X(w) * (exp(jwa)-1)/a = X(w)*jw .
c) since sin(a*t) is differential of -cos(a*t) / a ,therefore
fourier( sin(a*t) ) = -(1/a)*jw*pi*(delta(w+a) + delta(w-a) )
= - (1/a)*pi*j*(w*delta(w+a) + w*delta(w-a))
w*delta(w-a) = a*delta(w-a) and w*delta(w+a) = -a*delta(w+a) (this is due to the fact that delta(w-a) is only non zero at a)
thus fourier( sin(a*t) ) = j*pi*(delta(w+a) - delta(w-a) )
