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Signals and Systems – Time Integration Property of Fourier Transform
Fourier Transform
For a continuous-time function x(t), the Fourier transform of x(t) can be defined as,
$$\mathrm{X(\omega)=\int_{-\infty }^{\infty}x(t)\:e^{-jwt}\:dt}$$
And the inverse Fourier transform is defined as,
$$\mathrm{x(t)=\frac{1}{2\pi}\int_{-\infty }^{\infty}X(\omega)\:e^{jwt}\:d\omega}$$
Time Integration Property of Fourier Transform
Statement
The time integration property of continuous-time Fourier transform states that the integration of a function x(t) in time domain is equivalent to the division of its Fourier transform by a factor jω in frequency domain. Therefore, if,
$$\mathrm{x(t)\overset{FT}{\leftrightarrow}X(\omega )}$$
Then, according to the time integration property
$$\mathrm{\int_{-\infty }^{t}x(\tau )\:\overset{FT}{\leftrightarrow}\frac{X(\omega )}{j\omega };\:\:(if\:X(0)=0)}$$
Proof
When X(0)=0; then the time integration property of CTFT can be proved by using integration by parts.
Therefore, from the definition of inverse Fourier transform, we have,
$$\mathrm{x(t)=\frac{1}{2\pi}\int_{-\infty }^{\infty}X(\omega)\:e^{jwt}\:d\omega}$$
Replacing t by a variable τ in the above equation, we get,
$$\mathrm{x(\tau)=\frac{1}{2\pi}\int_{-\infty }^{\infty}X(\omega)\:e^{jwt}\:d\omega}$$
Taking time integration on both sides over $(-\infty)$ to (t), we get,
$$\mathrm{\int_{-\infty }^{t}x(\tau )\:d\tau=\int_{-\infty}^{t}\left [ \frac{1} {2\pi}\int_{-\infty}^{\infty }X(\omega)\: e^{j\omega\tau }\:d\tau \right]d\omega}$$
By interchanging the order of integration in RHS of the above equation, we get,
$$\mathrm{\int_{-\infty }^{t}x(\tau )\:d\tau=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega )\left [\int_{-\infty}^{t} e^{j\omega\tau}d\tau\right]d\omega}$$
$$\mathrm{\Rightarrow\int_{-\infty }^{t}x(\tau )\:d\tau=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega )\left [ \frac{e^{j\omega\tau}}{j\omega}\right]^t_{-\infty}\:d\omega}$$
$$\mathrm{\Rightarrow\int_{-\infty }^{t}x(\tau )\:d\tau=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left [\frac{X(\omega)}{j\omega} \right]e^{j\omega t} d\omega=F^{-1}\left [ \frac{X(\omega)}{j\omega} \right ]}$$
$$\mathrm{\therefore F\left [\int_{-\infty }^{t}x(\tau )\:d\tau \right ]=\frac{X(\omega)}{j\omega}}$$
Or, it can also be represented as,
$$\mathrm{\int_{-\infty }^{t}x(\tau )\:d\tau\overset{FT}{\leftrightarrow}\frac{X(\omega)}{j\omega}}$$
Note
When $X(0)
eq0;$ then, the function x(t) is not an energy function and hence the Fourier transform of $[\int_{-\infty }^{t}x(\tau )\:d\tau]$ includes an impulse function, i.e.,
$$\mathrm{F\left [\int_{-\infty }^{t}x(\tau )\:d\tau \right ]=\frac{X(\omega)}{j\omega}+\pi\:X(0)\delta(\omega)}$$
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