Knowing the integral property: $$\int_{-\infty}^tx(\tau)\,d\tau\,\overset{F}\longleftrightarrow\,\frac{X(\omega)}{j\omega}\,+\,\pi X(0)\delta(\omega)$$ and the inversion theorem: $$f(t)\,\overset{F}\longleftrightarrow\,g(\omega)\,\Rightarrow\,g(t)\,\overset{F}\longleftrightarrow\,2\pi f(-\omega)$$
How can we prove that: $$\frac{-x(t)}{jt}\,+\,\pi x(0) \delta(t)\,\overset{F}\longleftrightarrow\,\int_{-\infty}^\omega X(\eta)\,d\eta$$
Here the $\,\overset{F}\longleftrightarrow\,$ denotes the Fourier transform defined as: $$X(\omega)\,=\,\int_{-\infty}^\infty\,x(t)\,e^{-j\omega t}\,dt$$So:$$\Rightarrow\,x(t)\,=\,\frac{1}{2\pi}\,\int_{-\infty}^\infty\,X(\omega)\,e^{j\omega t}\,d\omega$$ My question stems from the signals and systems course in the electrical engineering field, so here the Fourier transform should give us all of the available frequencies in the signal $x(t)$ ($x(t)$ models the signal's behavior in the time dimension): $$X(\omega)=2\pi.(\frac{available\,frequency}{d\omega})=2\pi.(frequency\,density)$$ Also the $\delta(x)$ is the Dirac delta function.
Here are two other properties that might be helpful in solving this question: $$x(-t)\,\overset{F}\longleftrightarrow\,X(-\omega)$$ $$x^*(t)\,\overset{F}\longleftrightarrow\,X^*(-\omega)$$ By $x^*(t)$ I mean the complex conjugate of $x(t)$.
Let me know if anything needs further explanations.
- $\begingroup$ I have no idea what any of this means. In particular, what does $\overset{F}\longleftrightarrow$ mean? $\endgroup$David C. Ullrich– David C. Ullrich2020-09-26 11:35:54 +00:00Commented Sep 26, 2020 at 11:35
- $\begingroup$ @DavidC.Ullrich I apologize for the ambiguity. I have edited the question. $\endgroup$Friendly Fella– Friendly Fella2020-09-27 00:38:55 +00:00Commented Sep 27, 2020 at 0:38
1 Answer
I assume that $f(t)\,\overset{F}\longleftrightarrow\,g(\omega)$ means that the Fourier transform of $f(t)$ is $g(\omega)$: $$ g(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt $$ By the Fourier inversion theorem one has $$ f(t)\,\overset{F}\longleftrightarrow\,g(\omega)\,\Rightarrow\,g(t)\,\overset{F}\longleftrightarrow\,2\pi\,f(-\omega) $$ so since $$ \int_{-\infty}^tx(\tau)\,d\tau \,\overset{F}\longleftrightarrow\, \frac{X(\omega)}{j\omega}+\pi X(0)\delta(\omega) $$ where $x(t)\,\overset{F}\longleftrightarrow\,X(\omega)$ you get $$ \frac{X(t)}{jt}+\pi X(0)\delta(t) \,\overset{F}\longleftrightarrow\, 2\pi\int_{-\infty}^{-\omega} x(\tau)\,d\tau $$ Now we want to replace $X(t)$ with $x(t)$, but we need a Fourier transform in the left hand side. Luckily, since $x(t)\,\overset{F}\longleftrightarrow\,X(\omega)$, by the Fourier inversion theorem we have $X(t)\,\overset{F}\longleftrightarrow\,2\pi\,x(-\omega)$ so by replacing $X(t)$ with $2\pi\,x(-t)$ in the left hand side we can replace $x(\tau)$ with $X(\tau)$ in the right hand side: $$ \frac{2\pi\,x(-t)}{jt}+\pi\,2\pi x(-0)\delta(t) \,\overset{F}\longleftrightarrow\, 2\pi\int_{-\infty}^{-\omega} X(\tau)\,d\tau $$ or after division with $2\pi$: $$ \frac{x(-t)}{jt}+\pi\,x(-0)\delta(t) \,\overset{F}\longleftrightarrow\, \int_{-\infty}^{-\omega} X(\tau)\,d\tau $$ Finally, changing $t$ to $-t$ in the left hand side is equivalent to changing $\omega$ to $-\omega$ in the right hand side, so we get $$ \frac{x(t)}{j(-t)}+\pi\,x(0)\delta(-t) \,\overset{F}\longleftrightarrow\, \int_{-\infty}^{\omega} X(\tau)\,d\tau $$ i.e. $$ -\frac{x(t)}{jt}+\pi\,x(0)\delta(t) \,\overset{F}\longleftrightarrow\, \int_{-\infty}^{\omega} X(\tau)\,d\tau $$
- $\begingroup$ Thanks, it seems to be correct but shouldn't it be $x(-\tau)$ instead of $x(\tau)$ when you first apply the inversion theorem to the integral property? Maybe that's why you get $-\omega$ in the upper bound of the integral in the end instead of $\omega$. $\endgroup$Friendly Fella– Friendly Fella2020-09-27 01:03:04 +00:00Commented Sep 27, 2020 at 1:03
- $\begingroup$ Also when you apply the transform and replace $X(t)$ with $2\pi\,x(-t)$ shouldn't it be replaced with $2\pi\,x(-\omega)$ and shouldn't the integral bound in the right hand side change from $\omega$ to $t$? I mean shouldn't it go back to the time dimension? This was my problem. $\endgroup$Friendly Fella– Friendly Fella2020-09-27 01:04:59 +00:00Commented Sep 27, 2020 at 1:04
- $\begingroup$ You've found the place where I made an error. But it shouldn't be $x(-\tau)$. Instead it should be $-\omega$ as a limit of integration. $\endgroup$md2perpe– md2perpe2020-09-27 08:04:59 +00:00Commented Sep 27, 2020 at 8:04
- $\begingroup$ I have now fixed that sign error. Regarding your second question, I want $t$ on the left hand side and $\omega$ on the right hand side, so I don't think that you are correct there. $\endgroup$md2perpe– md2perpe2020-09-27 08:09:10 +00:00Commented Sep 27, 2020 at 8:09
- $\begingroup$ Excuse me sir, it still doesn't make sense. When you negate what's inside the definite integral and want it to instead affect the integral limits, then by what we know the bounds should be substituted so $\int_{b}^{a} (-x)\,dx$ becomes $\int_{a}^{b} x\,dx$ or $\int_{-b}^{-a} x\,dx$ instead of $\int_{b}^{-a} x\,dx$. $\endgroup$Friendly Fella– Friendly Fella2020-10-03 06:54:47 +00:00Commented Oct 3, 2020 at 6:54