Parseval's Theorem & Parseval’s Identity of Fourier Transform

---

---

Fourier Transform

For a continuous-time function x(t), the Fourier transform of x(t) can be defined as,

$$\mathrm{X(\omega) \:=\: \int_{-\infty }^{\infty }\:x(t)\:e^{-j\omega t}dt}$$

And the inverse Fourier transform is defined as,

$$\mathrm{x(t)\:=\:\frac{1}{2\pi}\int_{-\infty }^{\infty }\:X(\omega)\:e^{j\omega t}d\omega} $$

Parseval's Theorem of Fourier Transform

Statement – Parseval's theorem states that the energy of signal x(t) [if x(t) is aperiodic] or power of signal x(t) [if x(t) is periodic] in the time domain is equal to the energy or power in the frequency domain.

Therefore, if,

$$\mathrm{x_{1}(t)\:\overset{FT}{\leftrightarrow} \:X_{1}(\omega)\:\:and\:\: x_{2}(t)\overset{FT}{\leftrightarrow} X_{2} (\omega)}$$

Then, Parseval's theorem of Fourier transform states that

$$\mathrm{\int_{-\infty }^{\infty }\:x_{1}(t)\:x_{2}^{*}(t)dt \:=\: \frac{1}{2\pi}\int_{-\infty }^{\infty }\:X_{1}(\omega) X_{2}^{*}(\omega)\:d\omega} $$

Where, $\mathrm{x_{1}(t)}$ and $\mathrm{x_{2}(t)}$ are complex functions.

Proof

Parseval's relation is given by,

$$\mathrm{\int_{-\infty}^{\infty}\:x_{1}(t)x_{2}^{*}(t)dt = \frac{1}{2\pi}\int_{-\infty }^{\infty }X_{1}(\omega) X_{2}^{*}(\omega)\:d\omega}$$

From the definition of inverse Fourier transform, we have,

$$\mathrm{LHS \:=\: \int_{-\infty }^{\infty }\:x_{1}(t)x_{2}^{*}(t)\:dt\:=\:\int_{-\infty}^{\infty}\left[\frac{1}{2\pi } \int_{-\infty }^{\infty }\:X_{1}(\omega)\:e^{j\omega t}\:d\omega \right ]x_{2}^{*}(t)\:dt}$$

By interchanging the order of integration in RHS of the above expression, we get,

$$\mathrm{\int_{-\infty}^{\infty}\:x_{1}(t)x_{2}^{*}(t)\:dt \:=\: \frac{1}{2\pi}\int_{-\infty}^{\infty}\:X_{1}(\omega) \left [\int_{-\infty}^{\infty}\:x_{2}^{*}(t)\:e^{j\omega t}\:dt \right ]\:d\omega}$$

$$\mathrm{\Rightarrow\: \int_{-\infty}^{\infty}\:x_{1}(t)x_{2}^{*}(t)\:dt\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty} X_{1}(\omega)\left [\int_{-\infty}^{\infty}\:x_{2}(t)\:e^{-j\omega t}\:dt \right ]^{*}\:d\omega}$$

$$\mathrm{\therefore\: \int_{-\infty}^{\infty}\:x_{1}(t)x_{2}^{*}(t)\:dt\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty}\:X_{1} (\omega) X_{2}^{*}(\omega)\:d\omega \:=\: RHS}$$

Parseval's Identity of Fourier Transform

The Parseval's identity of Fourier transform states that the energy content of the signal x(t) is given by,

$$\mathrm{E\:=\:\int_{-\infty}^{\infty}\:\left |x(t) \right |^{2}\:dt\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty}\:\left | X(\omega) \right |^{2}\:d\omega}$$

• The Parseval's identity is also called energy theorem or Rayleigh's energy theorem.
• The quantity $\mathrm{\left[ \left | X\left ( \omega \right ) \right |^{2}\right]}$ is called the energy density spectrum of the signal x(t).

Proof

If $\mathrm{x_{1}(t) \:=\: x_{2}(t) \:=\: x(t)}$; then the energy of the signal is given by,

$$\mathrm{E \:=\:\int_{-\infty}^{\infty}x(t)x^{*}(t)\:dt\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty}\:X(\omega)X^{*}(\omega) \:d\omega}$$

$$\mathrm{\because\: x(t)x^{*}(t) \:=\: \left |x(t) \right |^{2}\: \: and\: \: X(\omega)X^{*}(\omega)\:=\:\left |X(\omega) \right |^{2}}$$

Therefore,

$$\mathrm{E \:=\: \int_{-\infty}^{\infty}\:\left | x(t) \right |^{2}\:dt \:=\: \frac{1}{2\pi}\int_{-\infty}^{\infty}\: \left |X(\omega) \right |^{2}d\omega\: \: \:(Hence \:Proved)}$$