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Fourier Transform Basics-1

Authored by Suganthi G

Mathematics

University

Used 108+ times

Fourier Transform Basics-1
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12 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The Fourier Transform of f(x) is  F[f(x)]=F[f(x)]=  

 12π∫−∞∞f(x) dx = F[s]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)\ dx\ =\ F\left[s\right]  

 12π∫−∞∞f(x)eisx dx = F[s]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)e^{isx}\ dx\ =\ F\left[s\right]  

 12π∫−∞∞f(x) ds = F[s]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)\ ds\ =\ F\left[s\right]  

 12π∫−∞∞f(x)  cos⁡sx dx = F[s]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)\ \ \cos sx\ dx\ =\ F\left[s\right]  

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The  inverse Fourier Transform of   F[f(x)] F[f(x)]\  is  f(x) =f\left(x\right)\ =   

 12π∫−∞∞ F[s] dx = f(x)\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\ F\left[s\right]\ dx\ =\ f\left(x\right)  

 12π∫−∞∞F[s]eisx ds =f(x)\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F\left[s\right]e^{isx}\ ds\ =f\left(x\right)  

 12π∫−∞∞F[s] e−isxds =f(x)\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F\left[s\right]\ e^{-isx}ds\ =f\left(x\right)  

 12π∫−∞∞F[s]  cos⁡sx ds = f(x)\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F\left[s\right]\ \ \cos sx\ ds\ =\ f\left(x\right)  

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The complex form  eisx e^{isx\ }  and  e−isxe^{-isx}   is 

 eisx  = cos⁡sx + sin⁡sx  and e−isx  = cos⁡sx − sin⁡sx  e^{isx\ }\ =\ \cos sx\ +\ \sin sx\ \ and\ e^{-isx\ }\ =\ \cos sx\ -\ \sin sx\ \   

 eisx  = cos⁡sx +isin⁡sx  and e−isx  = cos⁡sx −isin⁡sx  e^{isx\ }\ =\ \cos sx\ +i\sin sx\ \ and\ e^{-isx\ }\ =\ \cos sx\ -i\sin sx\ \   

 eisx  = cos⁡sx  and e−isx  =  sin⁡sx  e^{isx\ }\ =\ \cos sx\ \ and\ e^{-isx\ }\ =\ \ \sin sx\ \   

None of the above

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The Real part of  e^{isx\ }  is

sinsx

i sinsx

cossx

cosx

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The Imaginary part of  e^{isx\ }  is

sinsx

i sinsx

cossx

cosx

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The Fourier sine Transform of f(x) is  Fs[f(x)]=F_s[f(x)]=  

 12π∫−∞∞f(x) dx = Fs[s]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)\ dx\ =\ F_s\left[s\right]  

 12π∫0∞f(x) sin⁡sx dx = Fs[s]\frac{1}{\sqrt{2\pi}}\int_0^{\infty}f\left(x\right)\ \sin sx\ dx\ =\ F_s\left[s\right]  

 2π∫0∞f(x) sin⁡sx dx = Fs[s]\sqrt{\frac{2}{\pi}}\int_0^{\infty}f\left(x\right)\ \sin sx\ dx\ =\ F_s\left[s\right]  

 12π∫−∞∞f(x)  cos⁡sx dx = Fs[s]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)\ \ \cos sx\ dx\ =\ F_s\left[s\right]  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The inverse Fourier sine Transform of   F−1{Fs[f(x)]} = f(x)F^{-1}\left\{F_s[f(x)]\right\}\ =\ f\left(x\right)  

 12π∫−∞∞f(x) ds= Fs[s]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\left(x\right)\ ds=\ F_s\left[s\right]  

 12π∫0∞ Fs[s]sin⁡sx ds=f(x) \frac{1}{\sqrt{2\pi}}\int_0^{\infty}\ F_s\left[s\right]\sin sx\ ds=f\left(x\right)\   

 2π∫0∞Fs[s]sin⁡sx ds = f(x) \sqrt{\frac{2}{\pi}}\int_0^{\infty}F_s\left[s\right]\sin sx\ ds\ =\ f\left(x\right)\   

 12π∫−∞∞ Fs[s]  cos⁡sx dx =f(x)\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\ F_s\left[s\right]\ \ \cos sx\ dx\ =f\left(x\right)  

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