Richard Lyons
• Member for 6 years, 9 months
• Last seen this week
• Northern California, United States

The word “resolution” is used in two different ways in DSP. And this causes confusion. Some people use the word resolution to refer to the FFT’s bin spacing (measured in Hz). I don’t like that use of ...

Looking at the 875 Hz spectral component, it looks like its DFT magnitude is 30. For real-valued signals, the DFT magnitude ‘M’ is M = AN/2 where A is the peak amplitude of the 875 Hz sinusoid within ...

You may be confusing "transient response" with "group delay". I'm not sure what is your definition of transient response but it's important to know that for a linear-phase, N-tap, ...

@Jazim Schail. Do you really want to window your signal? Is your goal to use a window sequence to reduce FFT spectral leakage? Because window sequences are so heavily discussed in the tutorial ...

@Dan Boschen What I call "frequency sampling filters" (FSFs) is a fascinating subject. (FSF filters are very briefly mentioned in five or six DSP textbooks that I have on my bookshelf.) As ...

The question, as written, is ambiguous. Instead of "sinusoidal signal" the question's author should have written either "continuous sinusoidal signal" or "discrete sinusoidal ...

You're plotting the real parts of your FFT results. Rather than doing that, you should plot the absolute value of your FFT results. For example, instead of 'stem(y_normal_fft)' use 'stem(abs(...

For tapped-delay line FIR filters, if you want to know "What is the constraint on real- and complex-valued FIR filters that guarantee linear phase behavior in the frequency domain?", see the ...

@Mour_Ka Your first figure (which looks astoundingly similar to figures in my publications) is a three-dimensional spectral plot showing the phase relationships of the spectral components in real-...

If you implement the Goertzel algorithm P times to detect P different spectral samples, Goertzel is more efficient (fewer multiplies) than the N-point FFT when P < log2(N).

I believe your thinking is correct. For bandpass filters, for each z-plane pole in the positive-frequency range there's a conjugate pole in the z-plane's negative-frequency range. So for bandpass ...

I thank everyone for their comments and Answers. And the links given herein are very interesting. I confess that I did not word my question as well as I should have. I wasn’t asking why people believe ...

The filter you desire is often called a “DC removal” or a “DC cancelation” filter. Unfortunately there’s no way to design a useful IIR DC removal filter that has no “settling time”. Your designed ...

The best quantitative (using signal measurements) way to compare an information-carrying signal contaminated with noise and an information-carrying signal contaminated with less noise is to compare ...

I can't interpret numpy code, but I don't think you should define an 'N' variable. For a 2-second signal you should have a total of 16000 samples. Here's how to do this in MATLAB (notice how I defined ...

For an N-point FFT the frequency-domain real-valued zero Hz (DC) bin value will be the sum of your N sinc input samples. The sum of your N = 2001 sinc input samples is 9.9799. However, you didn’t ...

@Nimit. I recommend that you study and learn to design and implement what are called 2nd-order IIR lowpass filters. Once you thoroughly understand such filters you should design a 2nd-order IIR ...

@Ali23. If by "How can I get a blue lines?" you mean how to generate samples of the your red curve, try the following: m = [1, -1, -1, 1 1, -1]; m = [m m m m]; N = length(m); M = 10; mup=...

@Ali23. your 'binary_data = rand(1,M) > 0.5;' command generates "logical" data. Try using 'binary_data = int8(rand(1,M));' and see if that helps you.

If you have a filter impulse response represented by H(z), replacing z with z^2 is equivalent to inserting a zero-valued sample in between each original impulse response sample. Such an operation ...

@bjornhartmann. The material at the following web page explains, and demonstrates using MATLAB, how beat notes are generated when we sum two sine wave sequences: dsprelated.com/showarticle/189.php

The standard deviation of your voltage samples will only be equal to the RMS of your samples if the average of your samples is zero. And that is not your situation here.

The period of $\omega_0 = 999\pi/1000$ is 2.0202. The period $\omega_0 = \pi/2$ is 4. So the period of $\omega_0 = 999\pi/1000$ is roughly half the period $\omega_0 = \pi/2$.

@geo. To show an example of why there's so much confusion regarding "DFT periodicity", below is a paragraph from a famous college DSP textbook: Think, now, of what the book is saying. It's ...

I believe I answered your original 'DFT versus DTFT' question. I see from your Comment following Matt L's Answer that you are contemplating a DFT topic that is commonly misunderstood. An input ...

All your algebra is for infinite-length sequences, and you cannot compute a DFT of an infinite-length sequence. So let's talk about finite-length sequences. Please realize that the DTFT of a finite-...

Giles' answer applies to tapped delay-line digital filters. A more general answer is: The DC gain of a digital system is the sum of the system's impulse response.

Perhaps standard deviation is what you want. Standard deviation $$\sigma= \sqrt{\frac{1}{n}\sum_{k=1}^n (x_k-x_{ave})^2}$$ (where $x_{ave}$ is the average of your sequence) quantifies the fluctuations ...