digital low pass FIR filter design where the cut off frequency is stuck at around -6db

I am designing a digital FIR filter using Matlab to the specification of a 5KHz cut off frequency, however, no matter what order number(N), sampling frequency(fs) or window I use, I cannot get the cut off frequency(fc) below -6dB (to -3dB)?

Is this a quirk with MatLab or am I doing something wrong?

If I use an odd order number: 5KHz is around -15dB If I use an even order number: 5KHz is around -6dB

I have attached a screenshot of my frequency response at the moment. The code used to produce it is below:

%filter parameters 
fc = 5000; %5KHz cut off freq
fs = 20000; 
N = 18; 
h = fir1((N-1), (fc/fs)*2, 'low',bartlett(N), 'noscale'); 
fvtool(h, 'Fs', fs)

Answer 1

[Star Strider]

The cutoff frequency is defined as the half-power point, corresponding to -6 dB. If you want the filter to be -3 dB down at a certain frequency, the firls function would be worth considering.

[Lawrence Card]

I know it would depend on the intended use as such, but in general terms then, is -6dB here generally negligible and I am being picky with trying to get it right on -3dB? (i.e. would the current design generally be seen as acceptable?)

I will take a look at the firls function, thanks for the suggestion. Although, my task is academic and I think I am required to follow the method taught using the structure I have, rather than bring in other functions.

[Star Strider]

I don’t understand. the -6 dB point is the half-power point (power amplitude is 0.5 maximum), the -3 dB point power amplitude is 0.7 maximum. To get your desired passband on -3 dB this would probably mean doubling your passband frequency (the -6 dB point) for a low-pass filter, although with your particular design, that may involve some experimentation.

I don’t do a lot with FIR filters, so I don’t have a lot of experience with them. If your filter does what you want it to, by definition it would be acceptable. It’s stable and seems to have good characteristics, so I don’t see a problem with the design.

If this is an assignment and you are supposed to use fir1, then forget firls for the time being.

[Lawrence Card]

Ok, I understand what you mean, that is all fine then and answers my question as the half power point was what I meant when trying to make it -3dB. I thought that was the 'target' for all filter cut off points. I also knew that it was 0.707 which now you point it out, isn't actually 50%.

Off the back of that, why then is -3dB known as the half power point when it is 70% of the power? (ref: wikipedia half power point article ) - that's the part that is confusing me and originally led me to believe the filter I designed wasn't correct for a 5KHz cut off spec.

[Star Strider]

That is because power w.r.t. amplitude is defined (in dB) as:

dB = 20*log(amplitude);

or -6 dB. If you are already dealing in units of power:

dB = 10*log(power);

and that would be -3 dB. The conventions can be confusing. It remains the half power point regardless.

In my experience, since most often signals are filtered in terms of amplitude (and not amplitude^2, or power), the -6 dB point is considered the half-power point. That again is the result of:

P = I*E = I^2*R = E^2/R

where I is current and E is voltage, and the other relations are derived from Ohm’s law.

That’s how I learned it, anyway.

If anyone else wants to amplify on that I would welcome their comments.

Twinkle, twinkle little star,

Power equals I squared R.

[Lawrence Card]

Ok, I understand, the key thing is the difference between power and amplitude which is what I was overlooking by assuming a rule for power whilst using amplitude.

Thanks for your explanations!

[Star Strider]

My pleasure!

I’m not certain most signal processing textbooks even discuss the half-power point. The DSP text I use as a reference doesn’t specifically list it in the index. I learned about it early on in the course of studying for my Amateur Extra Class amateur radio ticket (this was back in the vacuum tube, log table and slide rule days), when it was the only way to specify filter passbands, and signals were all thought of in terms of power.

Answer 2

[Jess Stuart]

General Rule: X in dB = 10*log(Power Ratio)

DSP Filter designers typically work with voltage, so they will use:

X in dB = 20*log(Voltage Ratio)

But the end result is still in dB and represents a Power Ratio (P2/P1 = (V2/V1)^2)

even if you used the ratio of voltages squared to calculate it.

-6dB is always one-quarter power and -3dB is always half power.