A message from our CEO about the future of Stack Overflow and Stack Exchange. Read now.

# Tag Info

25

I'd like to know how to remove environmental noise from a speech recording. Well it's stored digitally now, right? so are you planning on putting your microphone next to the speaker after an analog filter to re-record it? Enough messing around, I'll be serious. In order to make a filter attenuate more in a smaller range of frequencies, aka making the ...

16

The deciding factor between stand-alone analyzer or soundcard would be the frequencies you're interested in. I will record the sound it makes with a microphone (it sounds differently if it is calibrated correctly than if it is not), If the difference is audible, this means the frequencies you want to measure are lower than 20kHz, which means audio gear ...

14

There are several parts to this answer. I base this answer on the characteristics of the FFT algorithm. I am not familiar with the specific LTSpice implementation, but the behavior you report is exactly what I would expect. The most common FFT implementations operate on an integer power of 2 data points. So, most implementations would pad your 1,000,000 ...

13

I assume for "high speed" you mean a small delay from data collection to the resultant FFT. With a low sample rate, your computational ability isn't the limiting factor, given modern computers. The delay problem lies in having enough data for analysis. If you want your 1Hz bin to be different from DC/0Hz, you have to accumulate enough signal data to capture ...

12

They say a picture is worth a thousand words. And a handful of animated pictures... All of these from Wikipedia. You can see these in context in the articles for Fourier transform, square wave, and Fourier series. As it turns out, any periodic waveform can be represented by a Fourier series, although to represent it exactly may require an infinitely ...

12

This really looks like a sampling artifact on your end, not something the crystal is doing. Expand the scope time scale (lower time/division) until you only have a cycle or two per division at most. If it's a sampling aliasing problem, then the artifact should disappear. Alternatively, look at the signal with a Ye Olde Analog scope. If it turns out it's ...

10

One usually needs to acquire multiple samples per waveform period to get good results from an FFT. The Nyquist limit of 2 samples per period is a lower bound but usually 10 samples per period or more is what is practically used. So to analyze a 64Hz signal you probably want to acquire samples at a rate of 640Hz or more. Also (up to a point) you will get ...

9

The main reason that frequency-domain processing isn't done directly is the latency involved. In order to do, say, an FFT on a signal, you have to first record the entire time-domain signal, beginning to end, before you can convert it to frequency domain. Then you can do your processing, convert it back to time domain and play the result. Even if the two ...

9

Two observations: 12.28 and 12.72 are exactly symmetrical about 12.50 MHz. The displayed wave form seems to have "beats" in it Beats are either real (you would see beats if you had a mixture of two frequencies present) or they are a sampling artifact. It is not necessary for the sampling frequency to be too low (in the sense of the Nyquist criterion) - it ...

8

The basic physics of the Cooley-Tukey radix-2 Fast Fourier Transform are well-known. It does log2 N "layers" of butterfly operations. Each layer does N/2 butterflies. Each butterfly does 2 multiplications and 2 additions. For a binary million (2^20) point sample, that's 20 million multiplications and 20 million additions. You also need a million-point (...

7

There is a direct, and actually quite simple, relationship between all the figures. Let's start with the sample size. You get half the bins (or "buckets") as you have samples in your set. For instance, if you have 1024 samples, then you get 512 bins. As simple as that. Now for the sample rate. The maximum frequency is half the sample rate (see Nyquist-...

7

But why can't you use a classical electronic filter to remove the noise frequencies? Who says you can't? It is how this was done in the days before digital signal processing. The problem is that filtering noise is always a compromise between keeping your wanted signal (speech, music) untouched while lowering the noise. For cassette tapes and other analog ...

7

I would personally recommend using a PC for several reasons. From a monetary stand point, if there is sufficient information currently to identify issues using a trained ear, there is no need for an expensive hardware analyser with extended bandwidth. Most microphones are only designed to measure 20 Hz - 20 kHz anyway. You can probably buy a modest ...

6

The easy solution is to get an RF wattmeter. Those will measure transmit power directly. Alternately, you can transmit into a $50\Omega$ dummy load, measure the RMS voltage, and calculate power as $P = V^2/50\Omega$. This will give you total power. To calculate the spectral density, divide this by the bandwidth of your signal, which you should know ...

6

10 Log10 of the magnitude squared yields dB, but not dBm. dB is a relative measure. Since dB are logarithmic, you can add some constant to a pile of dB numbers and the pile still has the same meaning because separate dB numbers are only relative to each other. However, dBm defines 0 as representing 1 mW. Now there is a absolute reference, so individual ...

6

I was under the impression that the FFT was inherently calculated at every frequency 0->(Sampling Frequency)/2 distributed in bins of width Fs/(2*N). This is roughly correct. A discrete fourier transform (DFT) produces output for frequencies between $-F_s/2$ and $F_s/2$. If the input data is real-valued (rather than complex) the negative-frequency ...

6

For a band limited power spectral density $S(f)$ to obtain the total power all you need to do is integrate over the spectrum. In measured data the power is measured in discrete frequency steps, so despite the fact that it may show that the measurement is in dBm it is actually in dBm/Hz. The first step is to convert your power measurement into a linear ...

6

Welcome to the real world! The "mathematically perfect" transform you show at the top, with the "discrete" harmonics is generated assuming that the rise and fall times of the waveform are zero, and that you're doing a continuous transform — no discrete sampling in the time domain. It assumes that the integrations are over an infinite span — or ...

6

The multitude of spurious frequencies in your FF transform are reflections of undersampled higher harmonics of your signal. These are aliases, as you rightfully noted. See this EDN article for better explanations. You are not using any cut-off filters on your signal, so all harmonics above the Nyquist frequency are "folded" back into the main frequency ...

6

There isn’t enough information in 10 mS of sampled audio to reliably tell the difference in frequency between a low E and a low F or D# notes. 10 mS is barely more than 1 full pitch cycle at those low frequencies, and there’s tons of non-pitched noise from the attack (pluck or strum) interfering with any measurement of the rate of phase change of the higher ...

5

To my understanding, the magnitude squared is equivalent to the power, The magnitude squared is proportional to the power. Think of it this way, if you measured the voltage across a resistor and squared it, you have the numeric value of the power normalized to 1 ohm, i.e., the power that would be associated with that voltage across a 1 ohm resistor. ...

5

My confusion is arising from coming to know that the window function has its origins in the time domain This is correct. Normally when we talk about a window function we're talking about something that's applied in the time domain. and not the frequency domain and the window function is convolved in time with our main signal to filter it! This is ...

5

The first entry [1] in the FFT is your fundamental, but the zeroth entry [0] is DC! A standard FFT will have the same number of output samples as input samples, and they will all be complex. The results aren't necessarily symmetrical about DC. However in the special case of a real only input, they are complex conjugate about DC, which looks for all the ...

5

Well, the first step to understanding why we need FFT is to understand how digital filtering works. So basically, you have a structure, like a shift register, with a number of memory elements, an input and an output. A sample value goes into the input, gets shifted through the register and moves to the output. At each stage in the register, it's multiplied ...

5

Since your guitar string can only produce a handful of pitches, a full FFT is wasteful. Take the lower E string. The tones are approximately 5Hz apart. Assume a sampling rate of 11025 Hz. To get your FFT bins spaced closely enough together, you will need an FFT length of at least 2048 resulting in 1024 bins. You calculate 1024 bins, then discard 1000 ...

5

I'd make this a comment, but I don't have enough points to do that yet. You should plot your FFT data starting at 0 Hz and go up to, say, 500 Hz. That will give you 10 or so harmonics. You are probably zoomed too far out in the frequency domain(x-axis) to get much detail and realize what's going on here. And yes, the fundamental should be at 25 Hz.

5

Your signal is a square wave with its base at 0V and its peak at 2.7V or so. So it has an average voltage of 1.35V. In the frequency domain, the overall average of a signal is its content at DC or 0Hz -- so that's why there's a peak at 0Hz. The FFT of a square wave that is centered on 0V has energy at every odd harmonic, starting at 1. So there's energy ...

4

I have a pure sinusoidal test function. I already have a sufficiently high enough SNR to not need to worry too much about noise If you know the frequency (as you say you do), then any fractional part of the sine wave can be analysed to give you the peak and RMS values of that pure sine wave. It's all in the math. No fractional part of a sine wave ...

4

Since you are working with a fixed sample rate, your FFT length (which will require your window to be at the same width) will increase your frequency resolution. The benefit of having a finer frequency resolution is twofold: the apparent one is that you get a finer freqeuecy resolution, so that you might be able to distinguish two signals that are very close ...

4

If you do a FFT in MATLAB, you get one complex valued output (bin) for every input sample. If you then take the magnitude of this complex vector, assuming your original input was real valued, you will see that half the magnitudes will be mirrored. However, the number of bins has nothing to do with the sample rate of the original waveform, only the number of ...

Only top voted, non community-wiki answers of a minimum length are eligible