High pass filter lower amplitude of the filtered signal

Inspired by a similar question, can someone please help me understand why amplitudes of the frequencies that are in the pass zone are also decreased?

For example, I have a composite signal consists of 1 Hz and 30 Hz. Amplitude of the 30 Hz signal is 0.5. If I design a filter with cutoff frequency of 15 Hz, output signal which is frequency of 30 Hz is two times lower amplitude.

I wonder why is that. Is it because of the filter's ripple?

clear
clc
close all
fs = 100;               % sampling frequency
f1 = 1;        
f2 = 30;                % frequency of the signal
t = 5;                 % time duration
n = 0:1/fs:t;           % sample period
x1 = sin(2*pi*f1*n);    % single tone signal 1 Hz
x2 = 0.5*sin(2*pi*f2*n);   % single tone signal 30 Hz

x = x1 + x2; % composite signal contains 1 Hz and 30 Hz
subplot(4,1,1)
plot(n,x)
grid on
title('Composite signal in time domain');
xlabel("time (s)")
ylabel("Signal value")


% Perform FFT and look at the amplitude spectrum
Y = fft(x);
L = length(x);
f = fs*(0:(L/2))/L;
P2 = abs(Y/L);
P1 = P2(1:L/2+1);

Warning: Integer operands are required for colon operator when used as index.

P1(2:end-1) = 2*P1(2:end-1);
subplot(4,1,2)
plot(f,P1) 
title("Single-Sided Amplitude Spectrum of x(t)")
xlabel("f (Hz)")
ylabel("|P1(f)|")


% Now design high pass filter to pass all above 7 Hz
fc = 15; % cutoff freq 15 Hz
[b,a] = cheby1(6,5,fc/(fs/2),'high');
xfilt = filtfilt(b,a,x);

subplot(4,1,3)
plot(n,xfilt) 

% Perform FFT and look at the amplitude spectrum of filtered signal
Y = fft(xfilt);
L = length(x);
f = fs*(0:(L/2))/L;
P2 = abs(Y/L);
P1 = P2(1:L/2+1);

Warning: Integer operands are required for colon operator when used as index.

P1(2:end-1) = 2*P1(2:end-1);
subplot(4,1,4)
plot(f,P1) 
title("Single-Sided Amplitude Spectrum of x(t)")
xlabel("f (Hz)")
ylabel("|P1(f)|")

Answer 1

[Bora Eryilmaz]

Try instead a filter that is more smooth at the transition region and has less ripple at the left of the high-pass region:

fs = 100;
fc = 15;
[b,a] = butter(6,fc/(fs/2),'high');
freqz(b,a,512,fs)
a = gca;
a.YLim = [-25 0];

The Chebychev filter still has -2.81 dB attenuation at 30 Hz, so you would get an attenuation of

0.5*10^(-2.81/20)

ans = 0.3618

fs = 100;
fc = 15;
[b,a] = cheby1(6,5,fc/(fs/2),'high');
freqz(b,a,512,fs)
a = gca;
a.YLim = [-5 0];

[Askic V]

Yes, that's it. Thank you!