# Downsampling a high pass filter

If I have an ideal high pass filter as the one in the picture:

And I downsample it by 2, e.g. $h_{new}[n] = h[2n]$, is my understanding correct that the resulting filter would look like this in the frequency domain:

Your understanding in not correct now. And it can be seen from the equation which states the relation between DTFTs of $h[n]$ and $h_{new}[n] = h[2n]$ as follows : $$H_{new}(e^{j\omega}) = \frac{1}{2} \left( H(e^{j\frac{\omega}{2}}) + H(e^{j\frac{\omega-2\pi}{2}}) \right)$$ The general form of this relation for $h_{new} = h[Mn]$ is

$$H_{new}(e^{j\omega}) = \frac{1}{M} \sum_{k=0}^{M-1} H(e^{j\frac{\omega+2\pi k}{M}})$$

It's up to you to sketch the resulting expanded and shifted spectrums that add up to create the final spectrum, to see that it's a lowpass filter with cutoff frequency of $\omega_c = \pi/2$

The following MATLAB/Octave code also helps to see it practically:

h = fir1(256, 0.75, 'high');    % Create a HIHPASS filter with cutoff wc = 3*pi/4
h2 = h(1:2:end);                % Create h_new[n] = h[2n] , downsampled filter

figure, plot(linspace(-1,1,1024), abs(fftshift(fft(h,1024))));
hold on
plot(linspace(-1,1,1024), abs(fftshift(fft(h2,1024))),'r');
title('DTFT magnitudes for h[n] and h2[n] = h[2n]');
legend('h[n]','h[2n]')


The resulting plot is: • Thank you, I was actually coming to this realization myself, just could not find out how it was supposed to be 1/2. You're a lifesaver! :) Dec 3 '17 at 0:17