I came across this confusion in Chapter 16,"Book - The Scientist and Engineers guide to DSP" and Topic is "Windowed Sinc-Filters". I am doing some calculations for designing the sinc filter and came across these doubts:

                               M = 4/BW

where "M" is the filter kernel length and "BW" is the transition bandwidth. The author says that "The BW must be expressed as a fraction of sampling rate with a value between 0 to 0.5 and same goes with the cut off frequency". For the values of M = 20, 40, 200, the BW is 0.2, 0.1, 0.02 and like we can see, it satisfies the condition of 0 to 0.5.

After I searched online, I understood that the value of cut off frequency and BW to exist between 0 to 0.5 has to do something with the power (dB) and presumed that 0 is 0db and 0.5 is -3 dB, since 10log(0.5) = -3dB.

My question is, what is the relationship between the cutoff of frequency and BW being a fraction of the sampling rate and between the values 0 to 0.5? What does it mean?. Is it something to do with the frequency components present beyond the cut off freq or BW being attenuated?

For example: If my sampling rate is 100sps, so the cut off frequency and BW must be a fraction of 100 sps with a value between 0 to 0.5?

  • $\begingroup$ Check that book for 'Nyquist frequency' and 'sampling theorem', this should make it clear. Note that the cut-off frequency (and transition bandwidth) of a fixed discrete-time filter change if you change the sampling rate, so all frequencies are relative to the sampling rate. $\endgroup$
    – Matt L.
    Commented Feb 20, 2015 at 14:02
  • $\begingroup$ Oh I see, it has something to do with Sampling theorem. Ok I will check it out. Thanks! $\endgroup$
    – PsychedGuy
    Commented Feb 20, 2015 at 14:15
  • $\begingroup$ Or, from a different (plain DSP) perspective, the cut-off frequency (and transition bandwidth) is not related to the sampling rate at all. The filter operates on a sequence of samples, nothing in the filter description relates to the rate of these samples. It is only during the sampling and reconstruction steps that the sampling rate is a factor (and, of, course figuring out which fraction of the sample rate is required for the particular application). Hence, it makes sense to specify filters in relation to the sampling rate, 1, whatever the actual number may be. $\endgroup$
    – Oscar
    Commented Feb 20, 2015 at 15:09
  • $\begingroup$ Guys,the cutoff frequency and BW changes relative to the sampling rate, So, when I am designing the filter, should I keep the sampling rate as constant throughout? $\endgroup$
    – PsychedGuy
    Commented Feb 21, 2015 at 11:59

1 Answer 1


It is just a normalization so that the apparent sampling frequency is 1 Hz. Say for example you have a 101 tap lowpass filter with with a cutoff of 10 Hz at a sampling frequency of 50 Hz.

If you don't change the filter coefficients but apply the exact same filter at a sampling rate of 500 Hz then the cutoff will be 100 Hz (not 10 Hz) because the frequencies scale according to the sampling rate.

An important point to note - consider the original filter has a transition width of 10 Hz i.e. from 10 Hz to 20H at the 50 Hz sampling rate. If the filter coefficients remain unchanged when applied in a system with a sampling rate is 500 Hz, the resulting transition width is now from 100Hz - 200 Hz, so the transition width has also increased. This is the reason that if you are trying to design a filter with a narrow transition width at high sampling rates it requires more filter taps i.e. the impulse response of the filter gets longer in terms of the number of samples.

In most of the filter design software they accept normalized frequency values for the specification - but be careful some use slightly different normalizations e.g. I believe Matlab expects the frequencies to be between 0 and 1 rather than 0 and 0.5, their filter design interface has changed over the years.

Also, if you look at the formulas in the literature for estimation of the FIR filter length for routines like the Parks- McClellan algorithm, you'll see that the frequencies are normalized with respects to the sampling frequency.

  • $\begingroup$ I just read about the sampling theorem and Nyquist rate/frequency and correlated with your example. So basically, we consider the normalized sampling rate to be 1 and from Nyquist theorem the input signal must contain frequencies below one-half of the sampling rate i.e. 0.5. So, if we have an input signal with a frequency of 200Hz or 200cycles/sec, then the proper sampling rate for this signal is 400 sps or 400 Hz, thus satisfying 200/400 = 0.5, and for 200/600 = 0.33.This is what I understood, Pl correct me if wrong. Thanks! $\endgroup$
    – PsychedGuy
    Commented Feb 21, 2015 at 6:36

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