I tend to believe that if we multiply an energy signal with a power signal the product is an energy signal.

My reasoning is that the energy signal will have finite bandwidth and consequenty the product will have finite bandwith and will be an energy signal.

Can anybody confirm that I am right? Or am I missing something?

Thanks in advance.

  • $\begingroup$ What about a signal whose spectrum is infinite but with finite area? A Gaussian, for instance. $\endgroup$
    – MBaz
    Sep 22, 2017 at 16:13
  • 1
    $\begingroup$ I see your point. But again, a gaussian multiplied by a power signal is an energy signal, isn't it? $\endgroup$
    – gdaras
    Sep 22, 2017 at 16:38
  • $\begingroup$ What about the product of a signal with Gaussian spectrum and finite area, times a power signal whose spectrum is 1/Gaussian? (or any other spectrum which grows as $f$ grows). $\endgroup$
    – MBaz
    Sep 22, 2017 at 16:47
  • $\begingroup$ (BTW, i had never thought of this question and I don't know the answer, I'm just throwing questions out there in case it helps you) $\endgroup$
    – MBaz
    Sep 22, 2017 at 16:48

1 Answer 1


If you are not looking for a rigorous proof, then the following would suffice for a particular class of power signals which has the property that their magnitude is bounded; i.e, $ |x_p(t)| \leq K$ for all $t$.

Numerous examples exist for such power signals like sinusoidal waves, complex exponentials, all kinds of periodic continuous-time signals which are defined via convergent Fourier series sums... Then for such a power signal $x_p(t)$ and a given energy signal $x_e(t)$ you can show that their product $y(t) = x_p(t) x_e(t)$ will be an energy signal as follows; given

$$ E_x = \int_{-\infty}^{\infty} |x_e(t)|^2 dt $$

and $ |x_p(t)| \leq K $ for all $t$ then the energy of y(t) is

$$ E_y = \int_{-\infty}^{\infty} |y(t)|^2 dt = \int_{-\infty}^{\infty} |x_p(t) x_e(t)|^2 dt \leq \int_{-\infty}^{\infty} K^2 |x_e(t)|^2 dt = K^2 E_x$$

Hence $y(t)$ will also be an energy signal.


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