How to apply Convolution Reverb

Mono version

Input/Output data: See how to pass data to KFR

univector<float> audio;
univector<float> impulse_response;

Code

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

$\begin{aligned} out_{L} &= in_{L} * ir_{LL} + in_{R} * ir_{RL}\\ out_{R} &= in_{L} * ir_{LR} + in_{R} * ir_{RR} \end{aligned}$

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

Code

// 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.