How to convert the passband ripple from dB to linear in the Filter Designer?

Hi everyone,

I'm using the Filter Designer in Matlab 2018b. I designed a low pass FIR filter using the equiripple design method. In the filter specifications, I set the passband ripple as Apass = 1dB and the stopband attenuation as Astop = 40dB. Then I generate the matlab code by selecting Filter Design Function.

Here is a section of the generated Matlab code:

% All frequency values are in MHz.
Fs = 1600;  % Sampling Frequency

Fpass = 90;              % Passband Frequency
Fstop = 110;             % Stopband Frequency
Dpass = 0.057501127785;  % Passband Ripple
Dstop = 0.01;            % Stopband Attenuation
dens  = 20;              % Density Factor

I knew that the values of Apass and Astop have been converted from dB to linear, so I get Dpass = 0.057501127785 and Dstop = 0.01. Here I figured out the linear value of stopband attenuation can be caculated using Dstop = 10^(-Astop/20), which is the inverse of Astop = -20*log10(Dstop).

I also assume Apass = -20*log10(1-Dpass) and the inverse is Dpass = 1 - 10^(-Apass/20). However, I'm unable to get Dpass = 0.057501127785 by substituting Apass = 1dB.

Can anyone tell me how Dpass is calculated? I.e. how to convert the passband ripple from dB to linear?

Thank you very much.

Regards,

Tong

[Jaikrishnan Mahalekshmi Sahasranamam]

Dear Xu,

This was confusing for me too. Thory wise what you are telling is right.

But then I figured out that, the matlab calculates it in this way

Apass =1dB

In linear scale it is 1+D and 1-D above and below 1

so it comes out to be. Apass is the difference between 1+D and 1-D

20log(1+D)-20log(1-D) =1dB

hence D=0.057501127785

[Tong Xu]

Dear Jaikrishnan,

Thank you for answering my question. I've also considered about this possibility but I was unable to solve it.

Thank you!

Regards,

Tong

[Paul Martin]

I too was trying to find how Dpass was calculated. Rearraging the above equation and putting it into matlab gives

x = 1; D = (power(10,x/20)-1) / (power(10,x/20)+1)

D =

0.0575

[paravai s]

20log base 10 [AP]=-0.1505

pls tell this ans and how to implement to write step by step