# Bandwith of a energy signal

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

I have this energy spectral density : Ψ(f)=rect(f) , and I need to find the bandwith, 3 dB band and the band with 90 % power.

About the 90 % case, do I need to find the total energy and then ∫W0Ψ(f)=0.9×TotalEnergy?

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 ∫W0Sx(f)=0.5×TotalPower ?

• all these terms are well-googleable, and a drawing of the spectral density would instantly answer your questions – what really is your question? – Marcus Müller Dec 28 '19 at 22:26
• You do realize that this is 3 entirely separate questions? Just take them one by one. You know the bandwidth; it was given in your knowns. 90% is 90% of that. 3 DB is where it crosses 3 DB; which in your case is at the band limit. Of course, this presumes that you have a coherent signal; if the signal is real noise then the calculations differ. Basically noise adds as the square root of bandwidth. – rrogers Dec 31 '19 at 22:37

For simple filters, one definition of bandwidth is the -3 dB frequency (for a lowpass filter) $$f_c$$ where the filter gain at $$f_c$$ is $$H(f_c) = H(0)/\sqrt(2)$$; i.e. $$f_c$$ is the half-power or cut-off frequency. This definition could be applied for a bandpass filters too but not for highpass filters. Note again that this definition of bandwidth requires that the filter frequency response be a continuous waveform, otherwise the the frequency will be undefined.
For signals, a 3dB bandwidth concept will not work and instead something like a 90 % energy (or power) definition is invoked. In this case the frequency spectrum of the signal is used to find the minimum frequency $$f_b$$ at which 90% of the total enegy of the (baseband, real) signal is contained.