Question

PROBLEM 4.1 The parts of this problem are independent of each other (a) The derivative property of Fourier transforms states that if X(jw) is the Fourier transform of r(t), then jwX(ju) is the Fourier transform of (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 e approaches 0 from "above." (We will assume this limit exists.) Take the Fourier transform of (t) e)/e 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 e go to zero. (b) 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 (c) The the Fourier transforms of cos(o and sin(o) 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(wot) but didn't know the "inverse Euler's formula for sin(wot). Derive the Fourier transform of sin(uo) by applying the time- derivative property to the Fourier transform pair and remembering what the derivative of cos(wot) is from your calculus classes.

Answer 1

(a)let's providing this d/dx x(t) = limit_e rightarrow d x(t) - x(t - e)/ e x(jw) = integral^infinity_-infinity x(t) e^-jt dt d/ dt x(t) rightarrow limit_e rightarrow 0^+ [integral^infinity_-infinity x(t) e^-jwt/t dt - integral^infinity_-infinity x(t) e^-jwt/e dt] = limit_e rightarrow 0^+ [integral^infinity_-infinity x(t) e^-jwt/ e dt - e^-jw t integral^infinity_-infinity e^-jwt/t dt] = limit_ e rightarrow 0^+ [integral^infinity_-infinity x(t) e^-jwt dt (1- e^-jwt/e)] = limit_e rightarrow 0^+ x(jw) [1-(ijwt)/ e] = limit_e rightarrow 0^+ x(jw) [1-1 + jwt/ e] limit_e rightarrow 0^+ x(jw) [jw] (b)for right hand derivation type d dt x(t) = limit_e rightarrow 0 x(t + e) - /e using direct def of fourier transform x(t) rightarrow x(jw) x(t + e) rightarrow e^jwe times (jw) limit_e rightarrow 0^+ [e^jwt times (jw) - x(jw)/ e] \r\n limit_e rightarrow 0^+ x^-(jw) [e^jwt - 1/ e] therefore = x(jw) [1 + jwe-1/ e] = jw times (jw) (c) cos (w_0 t) rightarrow pi (delta (w - w_0)+ delta (w + w_0)) d/ dt [cos (w_0 t)] = -w_0 sin (w_0 t)