# Cut-off frequency in Fourier-Spectrum of a noise message

I have a problem involving cut-off frequency.

I have a voice message in an audio file .mat which have an eletromagnetic noise. I am trying to recover the original message by applying digital Butterworth filter with MatLab. To this, besides of design the Butterworth filter, also I have to find the cut-off frequency to eliminate all, or partially, the electromagnetic noise.

The picture below shows the Fourier Spectrum of the voice message with the electromagnetic noise. How could I find the cut-off frequency to eliminate the noise? Thanks in advance!

Sampling frequency = 44.8 kHz.

• um, that spectrum doesn't look good at all. I don't think a simple butterworth will do here (it's questionable a butterworth design is appropriate at all – you probably want some multi band-pass filter, but it's hard to say without knowing the properties of your noise). Looking at this rectangular nightmare, I'd presume you have something very peculiarly shaped in the time-domain signal, not something you'd cancel in frequency domain. – Marcus Müller Nov 26 '19 at 18:54
• @MarcusMüller that noise? seems from a modulated sinc() function in time. Why is it there ? homework/project may be :-). I assumed that the noisy looking part towards the DC from the pulse was the message. Otherwise the message is a sinc() :-) – Fat32 Nov 26 '19 at 19:25
• yesss, and the Gibb's on that signal looks incredibly textbook. And that means the thing is, even if it's not just a single impulse/burst (might be an elegantly arranged comb...), easy to spot in time domain – Marcus Müller Nov 26 '19 at 19:27

I think you consider that pulse in the spectrum as the noise. Then you should remove it by a lowpass filter of cutoff frequency $$f_c = 5 ~\text{kHz.}$$

Slightly less than this should be used for guarding against the transition width of the lowpass filter. So you better select something like $$4.8$$ kHz for the lowpass filter cutoff frequency.

Note that your sampling frequency $$F_s$$, probably, is $$44.1$$ kHz and not $$44.8$$ kHz. Then your digital frequency for cutoff into the Matlab functions will be: $$f_{cd} = \frac{ 2 f_c }{F_s}$$

with the given values of $$F_s$$ and $$F_c$$ your digital cutoff will be:

$$f_{cd} = \frac{ 2 \times 5 \times 10^3 }{44.1 10^3} = 0.2268$$

or better

$$f_{cd} = \frac{ 2 \times 4.8 \times 10^3 }{44.1 10^3} = 0.2177$$

• Shouldn't it be $\frac{2\pi f_c}{F_s}$? – David Nov 26 '19 at 18:37
• @David No. $f_c$ and $F_s$ are in Hz. and $f_{cd}$ is between $0$ and $1$. – Fat32 Nov 26 '19 at 18:39
• Oh of course. Glossed over the 'into the Matlab functions' part. – David Nov 26 '19 at 19:05
• Thanks @Fat32 your help was fundamental. Thank you so much! – user4642 Nov 26 '19 at 20:12