How to Create a Simple Low-Pass Filter

Summary: This article shows how to create a simple low-pass filter, starting from a cutoff frequency \(f_c\) and a transition bandwidth \(b\). This article is complemented by a Filter Design tool that allows you to create your own custom versions of the example filter that is shown below, and download the resulting filter coefficients.

How to create a simple low-pass filter? A low-pass filter is meant to allow low frequencies to pass, but to stop high frequencies. Theoretically, the ideal (i.e., perfect) low-pass filter is the sinc filter. The sinc function (normalized, hence the \(\pi\)’s, as is customary in signal processing), is defined as

\[\mathrm{sinc}(x)=\frac{\sin(\pi x)}{\pi x}.\]

The sinc filter is a scaled version of this that I’ll define below. When convolved with an input signal, the sinc filter results in an output signal in which the frequencies up to the cutoff frequency are all included, and the higher frequencies are all blocked. This is because the sinc function is the inverse Fourier transform of the rectangular function. Multiplying the frequency representation of a signal with a rectangular function can be used to generate the ideal frequency response, since it completely removes the frequencies above the cutoff point. And, since multiplication in the frequency domain is equivalent with convolution in the time domain, the sinc filter has exactly the same effect.

The windowed-sinc filter that is described in this article is an example of a Finite Impulse Response (FIR) filter.

Sinc Filter

The sinc function must be scaled and sampled to create a sequence and turn it into a (digital) filter. The impulse response of the sinc filter is defined as


where \(f_c\) is the cutoff frequency. The cutoff frequency should be specified as a fraction of the sampling rate. For example, if the sampling rate is 10 kHz, then \(f_c=0.1\) will result in the frequencies above 1 kHz being removed. The central part of a sinc filter with \(f_c=0.1\) is illustrated in Figure 1.

Figure 1. Sinc filter.Figure 1. Sinc filter.

The problem with the sinc filter is that it has an infinite length, in the sense that its values do not drop to zero. This means that the delay of the filter will also be infinite, making this filter unrealizable. The straightforward solution is to simply stop computing at a certain point (effectively truncating the filter), but that produces excessive ripple. A better solution is to window the sinc filter, which results in, you guessed it, a windowed-sinc filter.


A window function is a function that is zero outside of some interval. There exist a great variety of these functions, tuned for different properties, but I’ll simply use the well-known Blackman window here, which is a good choice for general usage. It is defined as (for \(N\) points)

\[w[n]=0.42-0.5\cos\left({\frac{2\pi n}{N-1}}\right)+0.08\cos\left({\frac{4\pi n}{N-1}}\right),\]

with \(n\in[0,\,N-1]\). It is shown in Figure 2, for \(N=51\).

Figure 2. Blackman window.Figure 2. Blackman window.

Windowed-Sinc Filter

The final windowed-sinc filter is then simply the product of the two preceding expressions, as follows (with the sinc filter shifted to the range \([0,\,N-1]\)).

\[h[n]=\mathrm{sinc}\left(2f_c\left(n-\frac{N-1}{2}\right)\right)\left(0.42-0.5\cos\left({\frac{2\pi n}{N-1}}\right)+0.08\cos\left({\frac{4\pi n}{N-1}}\right)\right),\]

with \(h[n]=0\) for \(n\notin[0,\,N-1]\). I have dropped the factor \(2f_c\) from the sinc filter, since it is much easier to ignore constants at first and normalize the complete filter at the very end, by simply making sure that the sum of all coefficients is one, giving the filter unity gain, with


This results in the normalized windowed-sinc filter of Figure 3.

Figure 3. Normalized windowed-sinc filter.Figure 3. Normalized windowed-sinc filter.

Transition Bandwidth

The final task is to incorporate the desired transition bandwidth (or roll-off) of the filter. To keep things simple, you can use the following approximation of the relation between the transition bandwidth \(b\) and the filter length \(N\),


with the additional condition that it is best to make \(N\) odd. This is not really required, but an odd-length symmetrical FIR filter has a delay that is an integer number of samples, which makes it easy to compare the filtered signal with the original one. Setting \(N=51\) above was reached by setting \(b=0.08\). As for \(f_c\), the parameter \(b\) should be specified as a fraction of the sampling rate. Hence, for a sampling rate of 10 kHz, setting \(b=0.08\) results in a transition bandwidth of about 800 Hz, which means that the filter transitions from letting through frequencies to blocking them over a range of about 800 Hz. The values for \(f_c\) and \(b\) in this article were chosen to make the figures as clear as possible. The frequency response of the final filter (with \(f_c=0.1\) and \(b=0.08\)) is shown in Figure 4.

Figure 4. Frequency response on a linear (left) and logarithmic (right) scale.Figure 4. Frequency response on a linear (left) and logarithmic (right) scale.

Python Code

In Python, all these formulas can be implemented concisely.

import numpy as np
fc = 0.1  # Cutoff frequency as a fraction of the sampling rate (in (0, 0.5)).
b = 0.08  # Transition band, as a fraction of the sampling rate (in (0, 0.5)).
N = int(np.ceil((4 / b)))
if not N % 2: N += 1  # Make sure that N is odd.
n = np.arange(N)
# Compute sinc filter.
h = np.sinc(2 * fc * (n - (N - 1) / 2.))
# Compute Blackman window.
w = 0.42 - 0.5 * np.cos(2 * np.pi * n / (N - 1)) + \
    0.08 * np.cos(4 * np.pi * n / (N - 1))
# Multiply sinc filter with window.
h = h * w
# Normalize to get unity gain.
h = h / np.sum(h)

Applying the filter \(h\) to a signal \(s\) by convolving both sequences can then be as simple as writing the single line:

s = np.convolve(s, h)

In the Python script above, I compute everything in full to show you exactly what happens, but, in practice, shortcuts are available. For example, the Blackman window can be computed with w = np.blackman(N).

In the follow-up article How to Create a Simple High-Pass Filter, I convert this low-pass filter into a high-pass one using spectral inversion. Both kinds of filters are then combined in How to Create Simple Band-Pass and Band-Reject Filters

Filter Design Tool

This article is complemented with a Filter Design tool. Experiment with different values for \(f_c\) and \(b\), visualize the resulting filters, and download the filter coefficients. Try it now!

Filter designer.Filter designer.
Submitted by Tom Roelandts on 15 April 2014


Thanks! That's a very useful article.
Could you explain, what can I do with delay between source and filtered signals?

The nice thing about these kinds of filters is that it is easy to compensate for their delay. Symmetrical FIR filters, of which the presented windowed-sinc filter is an example, delay all frequency components in the same way. This means that the delay can be characterized by a single number. The delay of a filter of length M equals (M-1)/2. Hence, the shown filter with 51 coefficients has a delay of exactly 25 samples. So, if you want to overlay the original signal with the filtered one to compare them, you just need to shift one of them by 25 samples!

Really appreciate how concise this article is. Thanks!

Thank you! It's a very nice article. Could you maybe explain why it is necessary to normalize each element in h by the sum(h)?

It’s not really necessary, but if you make sure that the sum of the coefficients is one, then the gain of the filter at DC is also one. Additionally, it allows you to make the gain of the filter whatever you want simply by multiplying the coefficients of the normalized filter by the required gain factor.

Thanks very much for your article... I had considered this topic whole week and now my mind is clear!

Thanks, this is really helpful !

Great article!
How do you make your plots?

Thanks! The plots for most of the articles, including this one, were made with Python (using matplotlib).

Very interesting and clear article, thanks Tom! Question: You say that the number of coefficients is approx 4/b. If i use your fir designer tool and design a lowpass (windowed sinc) with Samplerate=44100, Cutoff=1000, Bandwidth=1000, i would expect approx 4/(1000/44100) = 176.4 thus 177 coefficients, but it gives 203 coefficients. I have been searching for the logic in that but i don't understand it. How do you define the specific number of coefficients? Thanks.

That’s a good observation. The reason for this is that the number of coefficients in the tool depends on the window function. This is because the window has a large influence on the transition bandwidth, so that, e.g., the rectangular window can get by with much less coefficients than the Blackman window. Maybe I'll write a separate article on this with more details, because it’s quite interesting stuff. In practice, the tool uses 4.6/b for Blackman, 3.1/b for Hamming, and 0.91/b for rectangular. From your example, I can tell that you’ve used a Blackman window, since 4.6/(1000/44100) = 202.86.

Thanks for your tutorial, well detailed!
Is the signal that you filter an array containing the amplitude of the signal?

Thanks! And yes, the variable s in the example is an array of numbers. I would say that this is the signal, and not just its amplitude.

Thanks for this great article. I have a question regarding Figure 4. How did you create the frequency response diagram?

This is another great idea for a follow-up article! I'll do one where I explain this in detail. In short, you first pad the filter with zeros to increase the resolution of the frequency plot, then take an fft, compute the power, and plot the result, either on a linear scale or in dB.

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