60 cycles of 10 kHz is at least 10000/(60/2) = 333 Hz separation. You cannot filter down as far as 100 Hz with that.
For 100 Hz resolving power you would need at least 2/100 seconds of data, but you only have 0.3 of that 2/100 seconds .
You would need 200 samples of 10 kHz to be able to resolve 100 Hz.
Fs = 10000;
Hz = 100;
N = 60;
t = (0:N-1)/Fs;
x = sin(2*pi*Hz*t);
filter1 = designfilt('lowpassiir', 'FilterOrder', 6, 'HalfPowerFrequency', 100, 'SampleRate', 10000);
filtered = filtfilt(filter1, x);
plot(t, x, 'b', t, filtered, 'r');
legend({'original', 'filtered'})
Not no effect, but nothing too useful.
And since you want lowpass, then you need a lot more samples to be able to resolve those frequencies. The duration at a given sampling frequency controls not only the upper frequency that can be resolved, but also the lowest frequency that can be resolved.
