void do_mdct(void ws, int c, float in, float *ou...

Question

void do_mdct(void *ws, int c, float *in, float *out) {
mdct_s *md = ws;
int N2 = md->N2;
int N4 = md->N4;
int NK = md->NK;
int ZN = md->ZN;

text
int overlap = md->overlap;

kiss_fft_scalar *dct4 = md->buf->dct4;
kiss_fft_scalar *foverlap = md->ch->overlap;
kiss_fft_cpx *cin = md->buf->cin;
kiss_fft_cpx *cout = md->buf->cout;

for (int i = 0; i < ZN; i++) {
    dct4[NK + i] = -in[N2 - overlap - 1 - i];
    dct4[N4 + i] = -in[ZN - 1 - i];
}

for (int i = 0; i < NK; i++) {
    dct4[i] = md->win[NK + i] * -in[N2 - NK - 1 - i] - in[N2 - NK + i] * md->win[NK - 1 - i];
    dct4[ZN + N4 + i] = foverlap[i] * md->win[i] - foverlap[overlap - 1 - i] * md->win[overlap - 1 - i];
}

for (int i = 0; i < overlap; i++) {
    foverlap[i] = in[N2 - overlap + i];
}

for (int i = 0; i < N4; i++) {
    cin[i].r = dct4[2 * i] * md->twid1[i].r - dct4[N2 - 1 - 2 * i] * md->twid1[i].i;
    cin[i].i = dct4[N2 - 1 - 2 * i] * md->twid1[i].r + dct4[2 * i] * md->twid1[i].i;
}

lhdc_v5_fft(md->fs, cin, cout);

for (int i = 0; i < N4; i++) {
    out[2 * i] = -(cout[i].i * md->twid2[i].i - cout[i].r * md->twid2[i].r);
    out[N2 - 1 - 2 * i] = -(cout[i].i * md->twid2[i].r + cout[i].r * md->twid2[i].i);
}

}

void do_imdct(void *ws, int c, float *in, float *out) {
mdct_s *md = ws;
int N2 = md->N2;
int N4 = md->N4;
int NK = md->NK;
int ZN = md->ZN;

text
int overlap = md->overlap;

kiss_fft_scalar *dct4 = md->buf->dct4;
kiss_fft_scalar *foverlap = md->ch->overlap;
kiss_fft_cpx *cin = md->buf->cin;
kiss_fft_cpx *cout = md->buf->cout;

for (int i = 0; i < N4; i++) {
    cin[i].i = in[2 * i] * md->twid2[i].i + in[N2 - 1 - 2 * i] * md->twid2[i].r;
    cin[i].r = in[2 * i] * md->twid2[i].r - in[N2 - 1 - 2 * i] * md->twid2[i].i;
}

lhdc_v5_fft(md->fs, cin, cout);

for (int i = 0; i < N4; i++) {
    dct4[2 * i] = cout[i].i * md->twid1[i].r - cout[i].r * md->twid1[i].i;
    dct4[N2 - 1 - 2 * i] = cout[i].r * md->twid1[i].r + cout[i].i * md->twid1[i].i;
}

for (int i = 0; i < ZN; i++) {
    out[N2 - overlap - 1 - i] = -dct4[NK + i];
    out[ZN - 1 - i] = -dct4[N4 + i];
}

}
do_imdct缺失了overlap的处理,导致out不完整,请补全

Answer 1

下面给出一种思路来「补全」do_imdct里对 overlap 的处理，使其和 do_mdct 相对应。大体思路是：

在输出 out 中，先把上一帧的 overlap 数据加回去（重叠相加，OLA）。
再从本帧 out 的末端（或者指定位置）取出新的 overlap，存入 foverlap，供下一帧使用。

在你的 do_mdct 里，最后这段是用来更新下一帧要用到的 overlap 的：

c
// do_mdct 里：将输入 in 的末尾部分保存到 foverlap，供下次使用
for (int i = 0; i < overlap; i++) {
    foverlap[i] = in[N2 - overlap + i];
}

而在 do_imdct 里，如果不做与之对称的处理，就会导致 out 的前半部分（或后半部分）没有正确叠加上一帧的残余(overlap)，以及没有将本帧产生的新 overlap 存起来。

补全思路

在你的 do_imdct() 结尾处，补上两段逻辑：

将上一帧的 overlap 加到本帧 out 中正确的位置。
将本帧 out 的相应部分保存到 foverlap，供下一帧使用。

根据你在 do_mdct() 中的写法(即把输入末 overlap 大小的那段存到 foverlap)，通常在 do_imdct() 中也是对称地「从 out 的末尾」或「合适位置」拿出 overlap。假设这里的 “合适位置” 就是 out[N2 - overlap + i]，那么可以写成如下（示例）：

c
void do_imdct(void *ws, int c, float *in, float *out)
{
    mdct_s *md = ws;
    int N2 = md->N2;
    int N4 = md->N4;
    int NK = md->NK;
    int ZN = md->ZN;

    int overlap = md->overlap;

    kiss_fft_scalar *dct4     = md->buf->dct4;
    kiss_fft_scalar *foverlap = md->ch->overlap;
    kiss_fft_cpx *cin  = md->buf->cin;
    kiss_fft_cpx *cout = md->buf->cout;

    // 1) 根据 in[] 恢复频域：与 twid2 结合后 FFT -> cout
    for (int i = 0; i < N4; i++) {
        cin[i].i = in[2 * i]               * md->twid2[i].i
                 + in[N2 - 1 - 2 * i]      * md->twid2[i].r;
        cin[i].r = in[2 * i]               * md->twid2[i].r
                 - in[N2 - 1 - 2 * i]      * md->twid2[i].i;
    }

    lhdc_v5_fft(md->fs, cin, cout);

    // 2) 将 FFT 结果 cout[] 与 twid1 混合, 得到时域 dct4[]
    for (int i = 0; i < N4; i++) {
        dct4[2 * i]              = cout[i].i * md->twid1[i].r - cout[i].r * md->twid1[i].i;
        dct4[N2 - 1 - 2 * i]     = cout[i].r * md->twid1[i].r + cout[i].i * md->twid1[i].i;
    }

    // 3) 将 dct4[] 的后半部分(-dct4[..])搬到 out 相应位置
    //    (这里是你原有的写法)
    for (int i = 0; i < ZN; i++) {
        out[N2 - overlap - 1 - i] = -dct4[NK + i];
        out[ZN - 1 - i]           = -dct4[N4 + i];
    }

    // -------------- 这里是缺失的 overlap 处理 ----------------

    // 4) 把上一帧遗留的 foverlap[i] 加到本帧 out 中，使得前后拼起来连续
    for (int i = 0; i < overlap; i++) {
        // 假设你要把上一帧的 overlap 加到 out 的末 overlap 段
        // 如果时域拼接位置是 N2 - overlap + i，就这样写：
        out[N2 - overlap + i] += foverlap[i];
    }

    // 5) 从本帧 out 的末 overlap 段重新取出数据，保存到 foverlap，供下一帧使用
    for (int i = 0; i < overlap; i++) {
        foverlap[i] = out[N2 - overlap + i];
    }
}

注意：

上面第 4、5 步里 out[N2 - overlap + i] 的具体下标，需与你在 do_mdct 中对 foverlap 的使用方式完全“对称”或“匹配”才行。

如果你的实际流程是希望把「输出帧」的末尾部分 (比如 out[ZN - overlap + i]) 作为下一帧的 overlap，那就把上面下标改成 ZN - overlap + i 即可，前面加 overlap 时也要对应改成 ZN - overlap + i。关键是：在 do_mdct() 与 do_imdct() 中，对 foverlap 的使用（存哪里、取哪里）要严格一致。

这样做完后，你就能把上一帧和本帧在时域上连续拼接起来，避免了「IMDCT 输出数据和上一帧不衔接」所导致的波形断裂。也能在 do_imdct 里把新的 overlap“留”给下一帧使用，从而输出「完整」的信号块。

void do_mdct(void ws, int c, float in, float *ou...

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