Back to my Home Page
1 / T ⟶ d f {\displaystyle 1/T\longrightarrow df}
n / T ⟶ f {\displaystyle n/T\longrightarrow f}
∑ n = − ∞ ∞ 1 T ⟶ ∫ − ∞ ∞ ( ) d f {\displaystyle \sum _{n=-\infty }^{\infty }{\frac {1}{T}}\longrightarrow \int _{-\infty }^{\infty }()df\!}
Can we make an unperiodic signal and make it periodic by taking the limit?
x ( t ) = lim T → ∞ 1 T ( ∫ − T 2 T 2 x ( t ′ ) e − j 2 π n t ′ T d t ′ ) e j 2 π n t T {\displaystyle x(t)=\lim _{T\to \infty }{\frac {1}{T}}(\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t')e^{\frac {-j2\pi nt'}{T}}dt')e^{\frac {j2\pi nt}{T}}\!}
note that X ( F ) = F [ x ( t ) ] ∫ − T 2 T 2 x ( t ′ ) e − j 2 π n t ′ T d t ′ {\displaystyle X(F)={\mathcal {F}}[x(t)]\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t')e^{\frac {-j2\pi nt'}{T}}dt'\!}
becomes as the limit is taken n/t becomes f x ( t ) = ∫ − ∞ ∞ [ ∫ − ∞ ∞ x ( t ′ ) e − j 2 π f t ′ d t ′ ] e j 2 π f t d f {\displaystyle x(t)=\int _{-\infty }^{\infty }[\int _{-\infty }^{\infty }x(t')e^{-j2\pi ft'}dt']e^{j2\pi ft}df\!}
x ( t ) = ∫ − ∞ ∞ x ( t ′ ) [ ∫ − ∞ ∞ e j 2 π f ( t ′ − t ) d f ] d t ′ {\displaystyle x(t)=\int _{-\infty }^{\infty }x(t')[\int _{-\infty }^{\infty }e^{j2\pi f(t'-t)}df]dt'\!}
note that the defination of the delta function is ∫ − ∞ ∞ e j 2 π f ( t ′ − t ) d f {\displaystyle \int _{-\infty }^{\infty }e^{j2\pi f(t'-t)}df\!}
x ( t ) = ∫ − ∞ ∞ x ( t ′ ) δ ( t ′ − t ) d t ′ {\displaystyle x(t)=\int _{-\infty }^{\infty }x(t')\delta _{(}t'-t)dt'\!}
T H E G A M E {\displaystyle THEGAME}
L T I ( L i n e a r T i m e I n v a r i a n t S y s t e m ) {\displaystyle LTI(LinearTimeInvariantSystem)}
I n p u t L T I O u t p u t R e a s o n {\displaystyle InputLTIOutputReason}
x ( t ) ⟶ ∫ − ∞ ∞ x ( t ′ ) δ ( t ′ − t ) d t ′ {\displaystyle x(t)\longrightarrow \int _{-\infty }^{\infty }x(t')\delta _{(}t'-t)dt'\!} Superposition X ( f ) ⟶ ∫ − ∞ ∞ X ( f ′ ) δ ( f ′ − f ) d f ′ {\displaystyle X(f)\longrightarrow \int _{-\infty }^{\infty }X(f')\delta _{(}f'-f)df'\!} "