Lowpass filter not making any difference
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I'm new to filtering, trying to use a low-pass filter to filter a sine wave with another high frequency sine wave on top of it. Using highpass(valArray, .03) I can effectively isolate the higher frrequency. But lowpass(valArray, .03) just appears to return the original signal (or something close to it).
It's possible my setup is incomplete. I'm simply using the lowpass() and highpass() commands. But seems odd that the highpass filter works fine while the lowpass filter fails.
lowpass(valArray, .03);
highpass(valArray, .03)
4 个评论
Cris LaPierre
2023-4-5
Please include the code you use to create your signals.
Nick Li
2023-4-5
So your data is something like this.
millis = linspace(1,33850,17000);
valArray = 3600 * sin(mod(millis,3600)/3600 * 6.28) + 200 * sin(mod(millis,100)/100 * 6.28);
figure
lowpass(valArray, .03);
figure
highpass(valArray, .03);
Nick Li
2023-4-6
采纳的回答
The reason your lowpass filter is not working is because your filter settings are not aggressive enough to both pass your low frequency signal and block your high frequency signal (they are very close together for a filter to separate them). It took a little investigation to be able to explain why.
Normalized frequency can make it a bit harder to undestand the cutoff frequencies, so I started by figuring out your approximate sample frequency (~500 Hz).
You can then convert normalized frequency back to Hz by multiplying it by fs/2. You are using a wpass of 0.03.
fs = 500;
wpass = 0.03;
fpass = wpass*fs/2
As you can see, your corner frequency (fpass) is around 7.5 Hz. Next we need to determine the frequencies of your two sine waves. Inspecting the signals confirmed that they are approximately 1/3.6 Hz and 1/0.1 Hz.
LF = 1/3.6 % low frequency wave, Hz
HF = 1/.1 % high frequency wave, Hz
The first thing to notice is that the frequency does not just shut off after your corner frequency is reached. Instead, there is a transition period until you reach the specified StopbandAttenuation (60 dB by default). The steepness of this transition is set by the Steepness parameter (85% by default).
Since you are using all the default settings for lowpass, 
, and W is 15% of 
:
, and W is 15% of :fstop = fpass + 0.15*(fs/2 - fpass)
As you can see, your high frequency signal is well within this window. To exclude it, you can lower fpass, but that alone won't do it, as you can only reduce by around 7 Hz. You need to adjust Steepness so that 
is below 10 Hz. Playing around with this equation, you can see that, if you keep fpass the same, your Steepness will need to be 99% or higher, and even then, it's not perfect.
is below 10 Hz. Playing around with this equation, you can see that, if you keep fpass the same, your Steepness will need to be 99% or higher, and even then, it's not perfect.fstop = fpass + 0.01*(fs/2 - fpass)
millis = linspace(1,33850,17000);
valArray = 3600 * sin(mod(millis,3600)/3600 * 6.28) + 200 * sin(mod(millis,100)/100 * 6.28);
figure
lowpass(valArray, wpass,"Steepness",0.99);
Fortunately, though, this approach does mean you do not have to modify your highpass function code, as wpass stayed the same.
You can go through the same exercise with high pass filters by using the corresponding info found here.
7 个评论
Nick Li
2023-4-7
Ran
2024-8-2
Thank you!
Jing
2024-10-2
I have a question, how do you get the sampling frequency for 500 Hz?
From the plot, there's 2000 samples in one cycle, so the sampling rate should be 2000 Hz, correct?
If wrong, please let me know how you find out the 500 Hz sampling rate.
Cris LaPierre
2024-10-2
Samples per second (sample frequency) is not the same thing as samples per cycle. Ultimately, I believe I figured out it was 500 Hz by reversing what fpass would have to be to give the results observed.
Jing
2024-10-3
I still not clear about how you get the 500 Hz sampling rate, fpass is 7.5 Hz from the calculation, how to get the sampling rate from the fpass?
Trial and error. I knew the approximate number of samples (17000), and from the arduino code could assume millis was milliseconds. Since the user shared the expression for the waveform, I just had to find a value for milliseconds that duplicated the waveform.
So, for example, if the plot contained 10 seconds of data, this is what it would look like:
start = 1; % milliseconds
stop = 10000; % millisecons
millis = linspace(start,stop,17000);
valArray = 3600 * sin(mod(millis,3600)/3600 * 6.28) + 200 * sin(mod(millis,100)/100 * 6.28);
plot(valArray)
xlim([0,17000])
This does not contain the same number of cycles, so I need to increase the stop time. This contains about 2.75 cycles. The original signal contains about 9.35 cycles, so I need to increase by a factor of 9.35/2.75
stop = stop*9.35/2.75
millis = linspace(start,stop,17000);
valArray = 3600 * sin(mod(millis,3600)/3600 * 6.28) + 200 * sin(mod(millis,100)/100 * 6.28);
plot(valArray)
xlim([0,17000])
That's a little too much, but hopeefully you see the process. Once I have the correct stop time, I can figure out the sample frequency by inverting the time step.
dt = (millis(2)-millis(1))/1000 % convert to seconds
fs = 1/dt
Hope that helps.
Jing
2024-10-8
Thanks so much for this detailed explanation! It really helps me to understand this process. Thanks for your time and your patience!
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