|author||Yu-Hsuan Hsu <firstname.lastname@example.org>||Wed Oct 07 14:45:53 2020|
|committer||Commit Bot <email@example.com>||Thu Oct 15 17:17:52 2020|
audiofuntest: Terminate child process after parent exits If the process exit abnormally, it will not call recorder.Terminate and player.Terminate to terminate its child process. Add prctl(PR_SET_PDEATHSIG, SIGHUP) to terminate children after parent exits. BUG=b:168281719 TEST=The clild process stops successfully after parent exits Change-Id: Ia89631ff631c5fe6f2c05c559bcc243cc43bf962 Reviewed-on: https://chromium-review.googlesource.com/c/chromiumos/platform/audiotest/+/2457206 Reviewed-by: Cheng-Yi Chiang <firstname.lastname@example.org> Commit-Queue: Yu-Hsuan Hsu <email@example.com> Tested-by: Yu-Hsuan Hsu <firstname.lastname@example.org>
This package intends to be a fast and powerful automated audio test tool. It performs polyphonic tone synthesis while simultaneously capturing the incoming sound in some form of external audio loopback. The external Mic/Headphone jack can easily be tested but the built-in speakers and microphone can also be tested by placing the ChromeBook/Laptop in a box and arranging for the built-in speaker sound to bounce off a surface and be directed to the built-in microphone. USB/audio dongles can also be tested and the audio output present in a HDMI port can likewise be tested with an appropriate dongle while looping back to an incoming microphone port.
The incoming sound is convolved through a Fourier Transform and the resulting energy peaks are filtered, sorted and compared to the known expected tones. Any missing or unexpected tones are reported. This test is fast- usually running in 1/2 second per port combination tested. During that time up to 7 sinsusoidal tones are polyphonically played, captured, Fourier analyzed, a report given and an optional spectrogram created. More than 7 tones can be used but peak discrimination of the convolved signal becomes more difficult. This could be handled by using a larger transform space and taking a little more time in processing.
It is a command line tool which lends itself to being script driven. Scripts would be different for different laptop/chromebook devices as the configuration of audio inputs and outputs can be very different across different platforms. This tool can be run on any Linux machine, not just a ChromeBook, providied the ALSA audio subsystem is available, and it usually will be.
Additional features in the tool:
Additional features planned for inclusion very shortly:
ALSA Conformance Test - A tool to verify audio drivers.
ALSA API Test - Test basic ALSA API function.
ALSA Helper - Get basic information for PCM devices.
AudioFunTest - A tool to test loopback, comparing output streaming and input streaming with a special designed algorithm.
CRAS API Test - Test basic CRAS API function.
Loop Test - Test loopback function for PCM devices.
Loopback Latency - Test loopback latency for PCM devices.
Test Tones - A tool to play tone from PCM devices.
Properties of the DFT DFT stands for ‘Discrete Fourier Transform’. A pure Fourier Transform is continuous; DFTs were invented so Fourier Transforms could be used in the real world. The reverse of the DFT is calculated simply by applying the Fourier transform again, and reversing the resulting buffer (to satisfy normalization, also divide resulting samples by N).
The components of the DFT are complex numbers. They have a modulus and a phase: The modulus of V[k], sqrt(V[k]V[k]), describes the intensity of the particular frequency corresponding to k (because of normalization, divide modulus by sqrt(N))
The phase of a given component describes the phase shift of this component--at what angle it starts its oscillations.
What is the relation between k, the index of the Fourier component, and its frequency? The function describing the k-th basis vector is e-2πikt/N. Sample it at t= 0,1,2,..,N-1. Its real part is cos(2πkt/N) (cosine is symmetric, so sign of argument doesn't matter). Cosine starts repeating after 2π, which corresponds to kt/N = 1 or t = N/k. So how many such repetitions fit in a segment of length N? Exactly k. In other words, the k-th basis vector has k periods per buffer. What time interval corresponds to one buffer? That depends on sampling speed. If sps is number of samples/sec, then sps/N is number of buffers/second. Therefore k*sps/N is number of periods/second, or frequency.