As a learning exercise I am trying to simulate a simple notch filter with Matlab. For test purposes I use Matlab to simulate sinewaves of various frequencies and plot frequency vs. gain so I can see the effect of the “notch”.

I’m using the difference equation $$y(n) = x(n) - 2 \cos\theta\cdot x(n-1) + x(n-2)$$ for this simple notch filter where theta is the notch (say $\pi/3$)

I thought x(n) would be the sample input from the sinewave I generate as test input. I'm just not sure how to account for the sample rate when determining the value of y(n).

From what I expect, and with comparison to using freqz([1 -2*cos(theta) 1],1), I get a good match with a sample rate of 10 and a difference equation as shown below.

$$y(n)= f_s [x(n) -2\cos(\theta/f_s)\cdot x(n-1) + x(n-2)]$$

but, as I change the sample rate there is a huge effect on the gain.

What am I missing? My guess is I just don't understand how to account for a change to the sample rate in the difference equation but I haven't seen a good write up on this anywhere

Can anyone help?

The simple code is shown below. As I change sample rate (from 10 to 100 and 1000) the frequency response appears to be correct but the gain appears to be decreasing by 20dB for each power of 10 I increase the sample frequency. Maybe that’s the clue?

..... CODE .. (You can tell I'm quite new to Matlab also... apologies for style!)

fs=10;  %Specify Sampling Frequency

Ts=1/fs; %Sampling period.

fb=100;    % Number of frequencies to try

mb=zeros(fb,2);   %Results   (Max / Freq)

w=0.001;           %Frequency of input wave to be samples  (Don't use DC!)
fb_indx=1;         %Index into the results array

THETA=pi/3;          %For a simple test
while fb_indx < (fb+1)

    Ns=round(2*pi*fs/w);  %Nr of time samples to be plotted.

    t=[0:Ts:Ts*(Ns-1)];  %Make time array that contains Ns elements
                     %t = [0, Ts, 2Ts, 3Ts,..., (Ns-1)Ts]

     mb(fb_indx,1)=w;  %Store current test frequency

     y=zeros(1,Ns);     %Result (y(n))

     x=sin(w*t); %create sampled sinusoids at different frequencies

% Here for the notch

  for n = 3:Ns  % loop for number of samples

     y(n) = fs*(x(n)-(2*cos(THETA/fs)*x(n-1))+x(n-2));  % calculate y


% Obtain max. response
  M=max(y);                %Look for highest peak in result

  mb(fb_indx,2)=20*log10(M);  %and store in 20log10 value


      w=w+(pi/fb);  %Get next frequency  


% Plot result

figure % create new figure

subplot(2,2,1)           % first subplot

xlabel('Freq - rad/s');
ylabel('20log10 Magnitude');
axis([0 pi -100 20]);

grid on;



1 Answer 1


The sampling frequency is hidden in your variable $\theta$. The angle $\theta$ determines the frequency of the notch relative to the sampling frequency, i.e.


where $f$ is the notch frequency in Hertz, and $f_s$ is the sampling frequency (in Hertz). So if you want to change the sampling frequency while leaving the notch frequency unchanged, you need to change the angle $\theta$ according to Eq. (1).

The gain of your filter depends on the normalized frequency variable $\theta$. At DC the gain is $2(1-\cos\theta)$, and at Nyquist you get a gain of $2(1+\cos\theta)$. By normalization you can keep one of those two gain factors normalized to $1$, but never both. If you want a gain of $1$ at DC and at Nyquist you need an IIR filter or a higher order FIR filter.

  • $\begingroup$ Many thanks for the suggestion. I thought I had done just that with my sum "THETA/fs" when I try to work out y(n). In fact the notch does seem to remain constant. I'm more interested in why the Gain drops and how to compensate as a result of change to fs. Thanks again for your help! $\endgroup$
    – DaveB
    Commented Apr 6, 2015 at 18:14
  • $\begingroup$ @DaveB: If you define $\theta$ as in your first equation, then it is normalized to a certain sampling frequency $f_{s1}$. If you now change the sampling frequency to $f_{s2}$, and you want to keep the notch frequency constant, the new variable $\theta_2$ must be chosen as $\theta_2=\theta\cdot f_{s1}/f_{s2}$. I've added some information concerning the gain to my answer. $\endgroup$
    – Matt L.
    Commented Apr 6, 2015 at 18:25
  • $\begingroup$ Many thanks, Great input. Let me spend some time understanding your answer, modifiying the code and trying some examples. I really do appreciate your help! $\endgroup$
    – DaveB
    Commented Apr 7, 2015 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.