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I know how to determine the energy of $\text{sinc}^2(kt)$, but how does this change when I need to find the energy of $(kt)\cdot \text{sinc}^2(kt)$.

If the $(kt)$ in front is squared (in energy integral) does that make the energy of this infinity ? the energy integral in time domain would be $$E_{\infty}=\int_{-\infty}^{\infty}(k^2t^2) \cdot \text{sinc}^4(kt)dt$$

But then Wolfram Alpha returned that the integral does not converge?

If multiplication in the time domain is convolution in freq domain

So, how do I handle the $(k^2 t^2)$ portion out front?

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  • $\begingroup$ If the integral diverges to infinity, then there's your answer; the signal in question has infinite energy. $\endgroup$ – Jason R Jul 1 '16 at 12:56
  • $\begingroup$ This integral is not divergent; the integrand decays as $1/t^2$. $\endgroup$ – Matt L. Jul 1 '16 at 21:20
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The energy of the signal $x(t)=kt\,\text{sinc}^2(kt)$ is finite. Note that with $\text{sinc}(kt)=\sin(kt)/(kt)$, $x(t)$ can be written as

$$x(t)=\sin(kt)\frac{\sin(kt)}{kt}\tag{1}$$

The energy of $x(t)$ is

$$E_x=\int_{-\infty}^{\infty}x^2(t)dt=\int_{-\infty}^{\infty}\left(\sin(kt)\frac{\sin(kt)}{kt}\right)^2dt$$

Using Parseval's theorem, $E_x$ can also be written in terms of $X(j\omega)$, the Fourier transform of $x(t)$:

$$E_x=\frac{1}{2\pi}\int_{-\infty}^{\infty}|X(j\omega)|^2d\omega\tag{2}$$

With the Fourier transform relation

$$f(t)g(t)\Longleftrightarrow\frac{1}{2\pi}F(j\omega)\star G(j\omega)$$

where $\star$ denotes convolution, we get from $(1)$

$$\begin{align}X(j\omega)&=\frac{1}{2\pi}\frac{\pi}{j}\left[\delta(\omega-k)-\delta(\omega+k)\right]\star\frac{\pi}{k}\text{rect}(\omega,k)\\&=\frac{\pi}{2kj}\left[\text{rect}(\omega-k,k)-\text{rect}(\omega+k,k)\right]\tag{3}\end{align}$$

where $\text{rect}(\omega,k)$ is a rectangular function with support $[-k,k]$. From $(3)$ we get

$$|X(j\omega)|^2=\frac{\pi^2}{4k^2}\text{rect}(\omega,2k)\tag{4}$$

Plugging $(4)$ into $(2)$ finally yields

$$E_x=\frac{1}{2\pi}\frac{\pi^2}{4k^2}\int_{-2k}^{2k}d\omega=\frac{\pi}{2k}\tag{5}$$

where I have implicitly assumed that $k>0$ holds. Since $x^2(t)$ is independent of the sign of $k$, $E_x$ must also be independent of the sign of $k$. So if negative values of $k$ are allowed, the result is

$$E_x=\frac{\pi}{2|k|},\quad k\neq 0\tag{6}$$

Even WolframAlpha agrees with me:

enter image description here

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