View source for ASN6 a,b- Prove given Fourier Transform property

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'''Problem Statement'''

6(a) Show <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \mbox{ if } S(f_0) = 0 </math>.

6(b) If <math> S(f_0) \neq 0 </math> can you find <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] </math> in terms of <math> \displaystyle S(f_0) </math>?

'''Answer'''

a)Remember that dummy variable <math> \lambda \!</math>  was used in substitution such that <math> \lambda= t-t_0 \! </math> 

Then <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!</math> 

and <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f)- S(f_0) \right] \,d\lambda \! </math>

The problem statement says to make <math>S(f_0)=0 \!</math> that makes the above equation simplify to 

<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f) \right] \,dt \! </math>

Taking the inverse Fourier Transform and changing the order of intgration

<math> \int_{- \infty}^{t} s(\lambda ) \,d\lambda = \int_{- \infty}^{t} e^{j2 \pi f t} \,dt \int_{- \infty}^{\infty}   S(f)\,df = \frac{ e^{j2 \pi f t}} {j2 \pi f }\int_{- \infty}^{\infty} S(f) \,df \! </math> 

Then

<math>\int_{- \infty}^{t} s(\lambda ) \,d\lambda = \int_{\infty}^{\infty} S(f)\frac{ e^{j2 \pi f t}} {j2 \pi f }\,df = \mathcal{F }^{-1}\left[  \frac{S(f)}{j2 \pi f} \right] \! </math>

Therefore it is demonstrated that <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \!</math>


b) If <math>S(f_0)\neq 0</math>  

Then

<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}^{-1}\left[ S (f)- S(f_0) \right] \,d\lambda = \int_{- \infty}^{t}\int_{- \infty}^{\infty} e^{j2 \pi f t} [S (f)- S(f_0)] \,d\lambda\,d\lambda  \! </math>

<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{\infty} e^{j2 \pi f t}S (f)\,d\lambda\,d\lambda - \int_{- \infty}^{t}\int_{- \infty}^{\infty} e^{j2 \pi f t} S(f_0) \,d\lambda\,d\lambda \! </math>

<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{\infty}\frac{ e^{j2 \pi f t}} {j2 \pi f } S (f)\,d\lambda - \int_{- \infty}^{\infty}\frac{ e^{j2 \pi f t}} {j2 \pi f } S (f_0)\,d\lambda \! </math>

Using <math> dt=d\lambda \!</math> and taking the Fourier transform of the equation

answer is <math>\mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] =  \frac{ e^{j2 \pi f t}} {j2 \pi f }S(f) - \frac{ e^{j2 \pi f t}} {j2 \pi f }S(f_0) \! </math>