# Bandwidth of lowpass filter?

This article here on the Window method for FIR filter design speaks of

the total normalized bandwidth of the lowpass filter in $\textrm{Hz}$ (counting both negative and positive frequencies)

What is it?

• Thumbs up for Matt and Peter. If it is possible then please mark answers to your questions as accepted. That will help other users to choose the right answer. But also it will prevent your questions from occasionally appearing on the main page as unanswered. (P.S. You get extra reputation points for that!). Thanks! – jojek Jun 29 '16 at 16:51

If you consider an ideal low-pass filter with cut-off frequency of $f_c$, all frequencies greater than $f_c$ will be removed. Then it's bandwidth is equal to $f_c\;\mathrm{Hz}$ (from $0$ up to $f_c$).

The total bandwidth $B_T$ is simply twice that: $B_T=2f_c$, since we are also considering negative frequencies, from $-f_c$ up to $f_c$.

Here is a picture of that: • Also, what does the normalization refer to? – mavavilj Jun 29 '16 at 17:00
• +1 (also) for the figure, which helped me formulate my own answer (in which I tried to clear up what is meant by normalization). – Matt L. Jun 29 '16 at 18:08

Note that the bandwidth of a low pass filter always refers to the part at positive frequencies. So, referring to jojek's answer, $f_c$ is the bandwidth.

Generally, there is no such thing as a "total bandwidth" of a low pass filter, it just has a bandwidth, and that's $f_c$. The author of that article you refer to just wanted to make a distinction between the actual bandwidth and the constant $B$ in Equation (5.9), which simply equals $2f_c$ (and calling it the "total bandwidth" does not really help to make it any clearer).

Finally, concerning the term "normalized bandwidth (or frequency)", whenever we talk about cut-off frequencies of discrete-time filters, we usually mean frequencies that are normalized by the sampling frequency. E.g., any digital low pass filter has a cut-off frequency that is fixed with respect to the sampling frequency, but it's not fixed as an absolute frequency! If you leave the filter coefficients unchanged and you change the sampling frequency, also the cut-off frequency will change.

In Equation (5.9) of that article, $B$ is indeed twice the normalized bandwidth. If $f_c$ is the cut-off frequency in Hertz, and $f_s$ is the sampling frequency in Hertz, then $B$ is given by

$$B=\frac{2f_c}{f_s}\tag{1}$$

So the sentence

[...] where $B=2f_c$ is the total normalized bandwidth of the lowpass filter in Hz (counting both negative and positive frequencies), and $f_c$ denotes the cut-off frequency in Hz.

is non-sense because a normalized frequency cannot have the unit Hertz, but it is dimensionless. $B=2f_c$ is only valid if $f_c$ is already normalized, and then both $B$ and $f_c$ are dimensionless; or, if $f_c$ is the cut-off frequency in Hertz, then the equation $B=2f_c$ is not valid, but you have to use $(1)$ instead.

basically when you work in digital domain you have to work with the so-called digital frequency, or normalized frequency. Inside the book "Applied Digital Signal Processing", written by Dimitris G. Manolakis and Vinay K. Ingle, digital frequency is defined as :

In the discrete-time case, if we assume that the units of the dimensionless integer index n are “samples” (sampling intervals would be a more appropriate term), then the units of “normalized” frequency, f, are cycles per sample and the units of normalized angular frequency, ω, are radians per sample. In the literature, this normalized frequency is also known as digital frequency. The frequency is normalized in the sense that it does not depend on the sampling interval.

Ok seems pretty, but let me introduce you to the mathematical details. We are in discrete domain then: $$t = n \cdot T$$ where $t$ is the continuous time and $T$ is the period of a sample.

Then the continuous frequency is given by $$F_{signal} = \frac{1}{t_{signal}} = \frac{1}{n \cdot T}$$ where $t_{signal}$ is the continuous period of our signal. Now we know that $1/T$ is the sampling frequency $f_{sampling}$, then: $$\frac{F_{signal}}{f_{sampling}} = f_{digital}= \frac{1}{n}$$ and $$\omega_{digital} = 2\pi \cdot f_{digital}$$

This result means that in discrete world the period is defined by how much samples, see the value of $n$, are needed to describe it. Now try to think when normalized frequency is considered high or low, because is not trivial, to see that we have to maximize the $f_{digital}$. To do this task we set $f_{sampling} = 2 \cdot F_{signal}$, this relation is given by Nyquist sampling theory, in which is stated that in order to avoid aliasing is necessary a sampling frequency at least equal to two times the signal frequency. If we substitute the minimum sampling frequency in the above equation we obtain: $$\frac{F_{signal}}{2 \cdot F_{signal}} = \frac{1}{n}$$

So that means that we obtain the highest frequency when the period of signal is described by 2 samples.

This is all the story behide normalized frequency, I hope that this can help you in some way.