How to apply Convolution Reverb¶
Mono version¶
Input/Output data: See how to pass data to KFR
univector<float> audio;
univector<float> impulse_response;
convolve_filter<float> reverb(impulse_response);
reverb.apply(audio);
Note
convolve_filter uses Filter API and preserves its internal state between calls to apply.
Audio can be processed in chunks.
Use reset function to reset its internal state.
convolve_filter has zero latency.
Internally, FFT is used for performing convolution
True stereo version¶
Formula
where \(*\) is convolution operator.
Input/Output data:
univector<float, 2> stereo_audio;
univector<float, 4> impulse_response;
// impulse_response[0] is left to left
// impulse_response[1] is right to left
// impulse_response[2] is left to right
// impulse_response[3] is right to right
// Prepare filters
convolve_filter<float> reverb_LL(impulse_response[0]);
convolve_filter<float> reverb_LR(impulse_response[1]);
convolve_filter<float> reverb_RL(impulse_response[2]);
convolve_filter<float> reverb_RR(impulse_response[3]);
// Allocate temp data
univector<float> tmp1(stereo_audio[0].size());
univector<float> tmp2(stereo_audio[0].size());
univector<float> tmp3(stereo_audio[0].size());
univector<float> tmp4(stereo_audio[0].size());
// Apply convolution
reverb_LL.apply(tmp1, audio[0]);
reverb_RL.apply(tmp2, audio[1]);
reverb_LR.apply(tmp3, audio[0]);
reverb_RR.apply(tmp4, audio[1]);
// final downmix
audio[0] = tmp1 + tmp2;
audio[1] = tmp3 + tmp4;
Implementation Details¶
The convolution filter efficiently computes the convolution of two signals. The efficiency is achieved by employing the FFT and the circular convolution theorem. The algorithm is a variant of the overlap-add method. It works on a fixed block size \(B\) for arbitrarily long input signals. Thus, the convolution of a streaming input signal with a long FIR filter \(h[n]\) (where the length of \(h[n]\) may exceed the block size \(B\)) is computed with a fixed complexity \(O(B \log B)\).
More formally, the convolution filter computes \(y[n] = (x * h)[n]\) by partitioning the input \(x\) and filter \(h\) into blocks and applies the overlap-add method. Let \(x[n]\) be an input signal of arbitrary length. Often, \(x[n]\) is a streaming input with unknown length. Let \(h[n]\) be an FIR filter with \(M\) taps. The convolution filter works on a fixed block size \(B=2^b\).
First, the input and filter are windowed and shifted to the origin to give the \(k\)-th block input \(x_k[n] = x[n + kB] , n=\{0,1,\ldots,B-1\},\forall k\in\mathbb{Z}\) and \(j\)-th block filter \(h_j[n] = h[n + jB] , n=\{0,1,\ldots,B-1\},j=\{0,1,\ldots,\lfloor M/B \rfloor\}\). The convolution \(y_{k,j}[n] = (x_k * h_j)[n]\) is efficiently computed with length \(2B\) FFTs as [ y_{k,j}[n] = \mathrm{IFFT}(\mathrm{FFT}(x_k[n])\cdot\mathrm{FFT}(h_j[n])) . ]
The overlap-add method sums the "overlap" from the previous block with the current block. To complete the \(k\)-th block, the contribution of all blocks of the filter are summed together to give [ y_{k}[n] = \sum_j y_{k-j,j}[n] . ] The final convolution is then the sum of the shifted blocks [ y[n] = \sum_k y_{k}[n - kB] . ] Note that \(y_k[n]\) is of length \(2B\) so its second half overlaps and adds into the first half of the \(y_{k+1}[n]\) block.
Maximum efficiency criterion¶
To avoid excess computation or maximize throughput, the convolution filter should be given input samples in multiples of the block size \(B\). Otherwise, the FFT of a block is computed twice as many times as would be necessary and hence throughput is reduced.