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(2) Given that "dw and ewz +e twz = 2 cos(wz), we obtain by applying the inverse Fourier Transform: w cos(wr) too a. e-ll = -dw * Jo 1+w t0 w cos(wr) dw 2 b. e-=| 1+ w² * Jo +* cos(wr). 1+ w² c. e-#l = -dw * Jo d. None of the above (3) Based on part (2), we can deduce that: dw a. 1+ w? 2 mp 1+ w²" b. mp dw dw C. %3D 1+ w2 d. None of the above

(2) Given that "dw and ewz +e twz = 2 cos(wz), we obtain by applying the inverse Fourier Transform: w cos(wr) too a. e-ll = -dw * Jo 1+w t0 w cos(wr) dw 2 b. e-=| 1+ w² * Jo +* cos(wr). 1+ w² c. e-#l = -dw * Jo d. None of the above (3) Based on part (2), we can deduce that: dw a. 1+ w? 2 mp 1+ w²" b. mp dw dw C. %3D 1+ w2 d. None of the above

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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inverse fourier transform part 2 3
Let f be the function defined on R by f(x) = e-lxl, and consider the following boundary problem (G): TE R, t> 0 (1) I u(x, 0) = S(x), r€R (2) (1) The Fourier Transform of f is given by: 2 1 11+ w² ' a. F(f(x)) = V w b. F(f(1)) = V + w² * 1 c. F(f(x)) = d. None of the above %3! T1+ w +oo (2) Given that iuz dw = -tuz dw and ew - 2 cos (wr), we obtain by applying +e the inverse Fourier Transform: w cos(wr) dw a. erl = 1+w Jo rtoo 2 w cos(wr) 1+ w² t∞ cos(wr) 1+ w² b. e-| dw T Jo 2 -dw * Jo d. None of the above (3) Based on part (2), we can deduce that: rtoo dw a. 1+ w² dw Jo mp 1+ w²' rtoo dw b. dw = 1+ w2 C. d. None of the above (4) Let v(w,t) be the Fourier Transform of u(r, t) acting to the variable r. By Applying the Fourier Transform to the first equation of (G), we obtain: a. vr(w,t) – iwv(w, t) = 0 b. vz(w, t) – w²v(uw,t) = 0 c. vr(w,t) – w*v(w, t) = 0 d. None of the above (5) The general solution of (G) is: rtoo ptwa ewt -dw 1+ w 1 a. u(x, t) =- / %3D b. u(r,t) = -dw 1+ w? rto0 piwa put -dw 1+ w3 -00 c. u(x, t) d. None of the above -00
Transcribed Image Text:Let f be the function defined on R by f(x) = e-lxl, and consider the following boundary problem (G): TE R, t> 0 (1) I u(x, 0) = S(x), r€R (2) (1) The Fourier Transform of f is given by: 2 1 11+ w² ' a. F(f(x)) = V w b. F(f(1)) = V + w² * 1 c. F(f(x)) = d. None of the above %3! T1+ w +oo (2) Given that iuz dw = -tuz dw and ew - 2 cos (wr), we obtain by applying +e the inverse Fourier Transform: w cos(wr) dw a. erl = 1+w Jo rtoo 2 w cos(wr) 1+ w² t∞ cos(wr) 1+ w² b. e-| dw T Jo 2 -dw * Jo d. None of the above (3) Based on part (2), we can deduce that: rtoo dw a. 1+ w² dw Jo mp 1+ w²' rtoo dw b. dw = 1+ w2 C. d. None of the above (4) Let v(w,t) be the Fourier Transform of u(r, t) acting to the variable r. By Applying the Fourier Transform to the first equation of (G), we obtain: a. vr(w,t) – iwv(w, t) = 0 b. vz(w, t) – w²v(uw,t) = 0 c. vr(w,t) – w*v(w, t) = 0 d. None of the above (5) The general solution of (G) is: rtoo ptwa ewt -dw 1+ w 1 a. u(x, t) =- / %3D b. u(r,t) = -dw 1+ w? rto0 piwa put -dw 1+ w3 -00 c. u(x, t) d. None of the above -00
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