I'm studying for my midterm , and I'm struggling with finding bandwidths.

I have this energy spectral density : $\Psi(f)= { 1 \over \sqrt { 1 + ({f \over B }) ^2} }$ , and I need to find the $3$ $dB$ band and the band with $90%$ % power.

About the $90%$ % case, do I need to find the total energy and then $\int_0^W \Psi(f) = 0.9 \times Total Energy$?

One think that doesn't make sense for me is that I'm dealing with an energy signal but I need to find something related to power.

If I had a power signal and wanted to find the 3 dB bandwidth, would I need to find the average power and then solve $\int_0^W S_x(f) = 0.5 \times TotalPower$ ?

Thanks in advance, and sorry for the bad english.

  • $\begingroup$ Are you sure that the energy spectral density is given as in your question with the square root in the denominator? Could it be that it is just the magnitude of the Fourier transform, i.e. $|X(f)|$ (and not $|X(f)|^2$ )? Then the whole problem would make much more sense. $\endgroup$ – Matt L. Oct 31 '14 at 16:50

I'm using the following assumption: the magnitude of the spectrum is given by the formula


Then the energy spectral density is


The 3 dB bandwidth is simply the frequency for which (1) becomes $0.5$, which is $W_{3dB}=B$. The bandwidth according to the 90% energy/power definition is


Equation (2) is the reason why I assume the definition as given in (1), because otherwise the integral on the right-hand side of (2) does not converge, so the 90% energy/power definition would not make sense. With the definition (1) the integrals in (2) are easily solved using the $\arctan$ function, from which $W=const\cdot B$ follows, where $const=6.31$ (if I'm not mistaken), but I'm sure you can do the calculation yourself.

  • $\begingroup$ You assumed correctly. Thanks a lot. But the answer is that, because the total energy is 1, right? $\endgroup$ – Joao Reis Oct 31 '14 at 23:49
  • $\begingroup$ Another question about the dBs. With energy / power spectral density everything works "smoothly", I mean, 3 dB means really 0.5, but why is that with amplitude, 3 dB means 1/sqrt (2) ? $\endgroup$ – Joao Reis Oct 31 '14 at 23:57
  • $\begingroup$ @JoaoReis: The total energy is just the integral with infinite limits on the right-hand side of (2) (without the factor $0.9$). Its value doesn't matter, and here it is actually not equal to $1$. And the factor $1/\sqrt{2}$ appears with amplitudes because power/energy = amplitude$^2$. $\endgroup$ – Matt L. Nov 1 '14 at 9:31
  • $\begingroup$ Ok, so why is the method for solving 3 dB (which is 50%) and 90% different? I mean isn't 3 dB a different way to express 50%? $\endgroup$ – Joao Reis Nov 1 '14 at 11:11
  • $\begingroup$ @JoaoReis: It's a matter of convention. It explicitly says '90% power', so here you know for sure that they're talking about integrals. When talking about a 3dB cut-off frequency for filters, then one means the frequency at which a signal is attenuated by 3dB. $\endgroup$ – Matt L. Nov 1 '14 at 11:15

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