Equivalent Noise Bandwidth 1 comment

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Equivalent noise bandwidth(ENBW) is defined as the bandwidth of a brickwall filter which produce same integrated noise power as that of an actual filter. This is also refered to as noise bandwidth or effective noise bandwidth.

Given a filter with transfer function H(j\omega), the equivalent noise bandwidth (\omega_{enbw}) is defined as

(1)   \begin{equation*} \omega_{enbw} = \int\limits_0^{\infty}{\left|H(j\omega) \over H_{max}}\right|^2 d\omega \end{equation*}

where H_{max} is the maximum value of H(j\omega) or magnitude of brickwall filter in passband.

In other words, it is the frequency of brickwall filter at which both filters(real filter and brickwall filter) have same integrated noise power.

Figure 1. Filter transfer function and Equivalent Noise Bandwidth of a brickwall filter

Figure 1. Filter transfer functionH(j\omega) and Equivalent Noise Bandwidth of a brickwall filter

As an example let us consider two filters, first order lowpass and second order lowpass, and find the relationship between thier 3dB bandwidth(\omega_{3dB}) and equivalent noise bandwidth\omega_{enbw}.

ENBW of First Order Lowpass filter

Transfer function of I-order lowpass filter is

(2)   \begin{equation*} H(j\omega) = {H_{max} \over {1+j{\omega\over\omega_{c} }}} \end{equation*}

where \omega_{c} is the corner frequency of the filter. It is same as -3dB frequency for first order filter.

The equivalent noise bandwidth of I-order lowpass filter, substituting Eq-(2) in Eq-(1), is given by

(3)   \begin{eqnarray*} \omega_{enbw} &=& \int\limits_0^\infty \left|{H(j\omega)\over H_{max}}\right|^2 d\omega = \int\limits_0^\infty {1\over {1+{\left(\omega\over\omega_{c}\right)^2}}}d\omega \\ &=& {\omega_{c}} \tan^{-1} \left.{\omega\over\omega_{c}}\right|_0^\infty = {\pi \over 2}{\omega_{c}}\\ \end{eqnarray*}

The 3-dB bandwidth (\omega_{3dB}) is found by evaluating the value of (\omega) where, 20 \log|H(j\omega)| - 20 \log|H_{max}| = -3dB. Hence

    \begin{equation*} 20 \log \left| {1 \over 1 + j{\omega\over\omega_c}}\right| = -3dB \end{equation*}

(4)   \begin{equation*} \boxed{\omega_c = \omega_{3dB}} \end{equation*}

Therefore, for first order lowpass filter the corner frequency(\omega_c) is same as 3dB frequency (\omega_{3dB}).

From Eq-(4) and Eq-(3), the I-order passive low-pass filter ENBW is related to 3dB frequency as

(5)   \begin{equation*} \boxed{ \omega_{enbw} = {\pi \over 2} \omega_{3dB} = 1.57 ~\omega_{3dB}} \end{equation*}

ENBW of Second Order(cascaded) Lowpass Filter

Transfer function of II-order lowpass filter with two poles at \omega=\omega_c is

(6)   \begin{equation*} H(j\omega) = {H_{max} \over \left({1+j{\omega\over\omega_{c}} }\right)^2} \end{equation*}

The equivalent noise bandwidth of II-order lowpass filter, substituting Eq-(2) in Eq-(1), is given by

(7)   \begin{eqnarray*} \omega_{enbw} &=& \int\limits_0^\infty \left|{H(j\omega)\over H_{max}}\right|^2 d\omega = \int\limits_0^\infty {1\over \left({1+{\left(\omega\over\omega_{c}\right)^2}}\right)^2}d\omega = {\pi \over 4}{\omega_{c}}\\ \end{eqnarray*}

Wolfram calculator for evaluating the integral

When the second order filter is realized by cascading of two first order filters, \omega_{3dB}\neq \omega_c. At -3dB bandwidth |H(j\omega_{-3dB})| = 1/\sqrt{2}. Solving for -3dB frequency \omega_c=1.554 ~\omega_{3dB}. Now substituting in Eq-(7), we get

(8)   \begin{equation*} \boxed{ \omega_{enbw} = 1.22 ~\omega_{3dB}} \end{equation*}

ENBW of Second Order Butterworth Lowpass Filter

Transfer function of II-order lowpass filter is

(9)   \begin{equation*} H(j\omega) = {H_{max} \over {1+j\left({\omega\over\omega_{c}}\right)^2 }} \end{equation*}

where \omega_{c} is the corner frequency of the pole

The equivalent noise bandwidth of II-order lowpass filter, substituting Eq-(2) in Eq-(1), is given by

(10)   \begin{eqnarray*} \omega_{enbw} &=& \int\limits_0^\infty \left|{H(j\omega)\over H_{max}}\right|^2 d\omega \\ &=& \int\limits_0^\infty {1\over {1+{\left(\omega\over\omega_{c}\right)^4}}}d\omega = {\pi \over 2\sqrt{2}}{\omega_{c}\over 2\pi}\\ \end{eqnarray*}

Wolfram calculator for evaluating the integral

For butterworth filter the pole frequency is same as -3dB frequency(for any order)

(11)   \begin{equation*} \boxed{ \omega_{enbw} = {\pi \over 2\sqrt{2}} \omega_{3dB} = 1.11 \omega_{3dB}} \end{equation*}


Noise Bandwidth Calculator

To find the ENBW of any other filter use the WolframAlpha and follow the steps below
WolframAlpha_Calculator


  1. Find the \omega_{enbw} / \omega_c : Feed the filter transfer function in “function to integrate” box


  2. Find the realtion between \omega_{3dB} / \omega_c


  3. Substitue (2) in (1) and get \omega_{enbw} / \omega_{3dB}


Filter Order vs ENBW

Filter Order \omega_{enbw} / \omega_{3dB}
1 1.57
2 1.22
3 1.15
4 1.13
5 1.11
Table 1. Filter order vs Multiplication factor (\omega_{enbw} / \omega_{3dB} )

Table-1 shows the filter order vs (\omega_{enbw} / \omega_{3dB}). Here n^{th}-order filter mean “n” 1-order filters are cascaded to form n^{th}-order filter. The transfer function of n-order filter is

(12)   \begin{equation*} H(j\omega) = {H_{max} \over \left({1+j{\omega\over\omega_{c} }}\right)^n} \end{equation*}

As the order of the filter increases, the filter roll-off increases and approach that of brick-wall filter. This is why the multiplication factor approach “1” as the filter order increases.

 

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