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Problem Statement
6(a) Show F [ ∫ − ∞ t s ( λ ) d λ ] = S ( f ) j 2 π f if S ( 0 ) = 0 {\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {S(f)}{j2\pi f}}{\mbox{ if }}S(0)=0} . Hint: S ( 0 ) = S ( f ) | f = 0 = ∫ − ∞ ∞ s ( t ) e − j 2 π ( f → 0 ) t d t = ∫ − ∞ ∞ s ( t ) d t {\displaystyle S(0)=S(f)\vert _{_{f=0}}=\int _{-\infty }^{\infty }s(t)e^{-j2\pi (f\rightarrow 0)t}\,dt=\int _{-\infty }^{\infty }s(t)\,dt}
6(b) If S ( 0 ) ≠ 0 {\displaystyle S(0)\neq 0} can you find F [ ∫ − ∞ t s ( λ ) d λ ] {\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]} in terms of S ( 0 ) {\displaystyle \displaystyle S(0)} ?
Answer
a)
Remember dummy variable λ = t − t 0 {\displaystyle \lambda =t-t_{0}\!} Then s ( λ ) = s ( t − t 0 ) = F [ S ( f ) − S ( f 0 ) ] {\displaystyle s(\lambda )=s(t-t_{0})={\mathcal {F}}\left[S(f)-S(f_{0})\right]\!} and ∫ − ∞ t s ( λ ) d λ = ∫ − ∞ t F [ S ( f ) − S ( f 0 ) ] d λ {\displaystyle \int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}\left[S(f)-S(f_{0})\right]\,d\lambda \!}
f 0 = 0 {\displaystyle f_{0}=0\!} where S ( 0 ) = S ( f ) | f = 0 = ∫ − ∞ ∞ s ( t ) e − j 2 π f t d t = ∫ − ∞ ∞ s ( t ) d t {\displaystyle S(0)=S(f)|_{f=0}=\int _{-\infty }^{\infty }s(t)e^{-j2\pi ft}dt=\int _{-\infty }^{\infty }s(t)dt\!}
if S ( 0 ) = 0 ∫ − ∞ ∞ s ( t ) d t = 0 {\displaystyle {\mbox{ if }}S(0)=0\,\,\,\int _{-\infty }^{\infty }s(t)dt=0\!}
∫ − ∞ t s ( λ ) d λ = ∫ − ∞ t F [ S ( f ) ] d t {\displaystyle \int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}\left[S(f)\right]\,dt\!}
F − 1 [ S ( f ) − S ( f 0 ) ] = ∫ − ∞ t e j 2 π f t d t ∫ − ∞ ∞ S ( f ) d f = e j 2 π f t j 2 π f ∫ − ∞ ∞ S ( f ) d f = {\displaystyle {\mathcal {F}}^{-1}\left[S(f)-S(f_{0})\right]=\int _{-\infty }^{t}e^{j2\pi ft}\,dt\int _{-\infty }^{\infty }S(f)\,df={\frac {e^{j2\pi ft}}{j2\pi f}}\int _{-\infty }^{\infty }S(f)\,df=\!}
∫ − ∞ t s ( λ ) d λ = ∫ ∞ ∞ S ( f ) e j 2 π f t j 2 π f d f = F − 1 [ S ( f ) j 2 π f ] {\displaystyle \int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{\infty }^{\infty }S(f){\frac {e^{j2\pi ft}}{j2\pi f}}\,df={\mathcal {F}}^{-1}\left[{\frac {S(f)}{j2\pi f}}\right]\!}
Therefore F [ ∫ − ∞ t s ( λ ) d λ ] = S ( f ) j 2 π f {\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {S(f)}{j2\pi f}}\!}