0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ If the integral diverges to infinity, then there's your answer; the signal in question has infinite energy. $\endgroup$
    – Jason R
    Jul 1, 2016 at 12:56
  • $\begingroup$ This integral is not divergent; the integrand decays as $1/t^2$. $\endgroup$
    – Matt L.
    Jul 1, 2016 at 21:20

1 Answer 1

2
$\begingroup$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.