Looking for a succinct example showing convolution processing (in the time domain) is notoriously computationally intensive; e.g., the typical reverberation time of a room is approximately 0.3 seconds which corresponds to 2400 samples, i.e., taps (filter coefficients), for an 8 kHz sampled sound. Because the sound is sampled at 8 kHz, the "delay steps" are each of length 1/8000 (0.000125 seconds). Assuming that the sampled sound and the impulse response have a length of 0.3 seconds, the number of samples is equal to 2400 (0.3 seconds/0.000125 seconds). A more complete example (from an anonymous source at the University of Miami) showing that time-domain convolution processing is notoriously computationally intensive follows: "For typical audio signals, three second impulse response sampled at 44.1 kHz requires around 2 * (3 * 44,100)^2 or 35 billion computations to convolve with the same length input." I can understand most of where the 35 billion number comes from. It comes from this: Delay steps for 44.1 kHz signal is equal to 2.675737 E-5 seconds. 3.0 seconds / 2.675737 E-5 seconds = 132300 increments. 2*(132300^2) = 2 * (17,503,290,000) = 35,006,580,000. What I don't understand are the purposes of the “2” and the “^2” expressions. I suppose that the “^2” expression represents each convolution operation (from convolving the impulse response with the sound). Trying to figure out the purpose of the “2” i.e., the “2” multiplied with “(132300)”. A Professor Emeritus at Penn State recently suggested that the “2” might be needed to evaluate whether or not each convolution was positive or negative; but he wasn’t sure. --------------------------------------- Posted through http://www.DSPRelated.com

# Example Digital Convolution in the Time Domain is Computationally Expensive

Started by ●January 7, 2016

Reply by ●January 7, 20162016-01-07

On 07.01.16 18.06, johnnmonroe wrote:> "For typical audio signals, three > second impulse response sampled at 44.1 kHz requires around 2 * (3 * > 44,100)^2 or 35 billion computations to convolve with the same length > input."> What I don't understand are the purposes of the “2” and the “^2” > expressions. I suppose that the “^2” expression represents each > convolution operation (from convolving the impulse response with the > sound).You need to multiply 3 * 44100 samples of the FIR filter with the same number of input samples to get exactly one sample of output. For 3 seconds of output you need 3 * 44100 times as much operations. More useful in fact is the number of computations per second, i.e. 2 * 3 * 44100� (per channel) ~ 1.2 Gflops. But no one who is not off one's head would implement a 130k FIR filter in the time domain. With FFT convolution even a raspberry pi 1 is able to do the job. If you already have several seconds latency the additional latency of the FFT convolution does not count. Marcel

Reply by ●January 7, 20162016-01-07

>On 07.01.16 18.06, johnnmonroe wrote: >> "For typical audio signals, three >> second impulse response sampled at 44.1 kHz requires around 2 * (3 * >> 44,100)^2 or 35 billion computations to convolve with the same length >> input." > >> What I don't understand are the purposes of the â€œ2â€ and theâ€œ^2â€>> expressions. I suppose that the â€œ^2â€ expression representseach>> convolution operation (from convolving the impulse response with the >> sound). > >You need to multiply 3 * 44100 samples of the FIR filter with the same >number of input samples to get exactly one sample of output. For 3 >seconds of output you need 3 * 44100 times as much operations. > >More useful in fact is the number of computations per second, i.e. >2 * 3 * 44100² (per channel) ~ 1.2 Gflops. > >But no one who is not off one's head would implement a 130k FIR filter >in the time domain. With FFT convolution even a raspberry pi 1 is able >to do the job. If you already have several seconds latency the >additional latency of the FFT convolution does not count. > > >MarcelMarcel, I really appreciate your reply. Could you clarify 44100� (per channel) please? I think the andsign and the numbersign might obscure the 2013266098 and not sure where the 2013266098 is coming from. Thanks again. John --------------------------------------- Posted through http://www.DSPRelated.com

Reply by ●January 7, 20162016-01-07

On 07.01.16 19.18, johnnmonroe wrote:> I really appreciate your reply. Could you clarify 44100� (per > channel) please?Seems that one of us has some problems with the encoding. From my point of view my post seems to be clean with respect to special characters (ISO/IEC 8859-15).> I think the andsign and the numbersign might obscure the > 2013266098 and not sure where the 2013266098 is coming from.No idea. It was a superscript 2. Probably your reader is not capable of ISO 8895-15 (Latin-9) encoding. Look here: https://groups.google.com/d/msg/comp.dsp/0nZ9L7h99mw/NsnEHHiNDgAJ Marcel

Reply by ●January 7, 20162016-01-07

Marcel, I can see the uncorrupted message using the link you provided. Thanks again. John --------------------------------------- Posted through http://www.DSPRelated.com