by andete. Original documents available at:


Last time we looked at how 2 YM2413 operators combined form one channel with frequency modulation. In this post we'll complete that picture and look at the 'feedback' stuff.

The YM2413 application manual contains the following diagram:

   |                         |
   |  +-------------------+  |  +-------------------+
   |  |           +-----+ |  |  |           +-----+ |
   v----->(+)---->| SIN |---------->(+)---->| SIN |----> F(t)
      |    ^      +-----+ |     |    ^      +-----+ |
      |    |         ^    |     |    |         ^    |
      |    |         | I  |     |    |         | A  |
      | +-----+   +-----+ |     | +-----+   +-----+ |
      | | PG  |   | EG  | |     | | PG  |   | EG  | |
      | +-----+   +-----+ |     | +-----+   +-----+ |
      +-------------------+     +-------------------+

This is much like the diagram in the previous post, with the addition that the output of the modulator operator is looped back to the input of that same operator, so this forms a feedback loop. The feedback signal is still multiplied by a feedback-factor (FB).

Notice how the inside of the operator-block is identical for the modulator and carrier. On the real chip there is also only one instance of this operator. It is time-multiplexed (i.e. reused in sequence over time) for all 18 operators. So it's logical that the operator block has the same features when used as modulator or as carrier.

So from a high level we only need to figure out how the FB-bits (register R#3, bits 2-0) control the feedback factor. Though in practice it's more complicated:


We'll investigate the feedback mechanism very much like how we investigated the modulation stuff last time. So let's start by making some measurements. I captured waveforms using the following settings. These are identical to last time's settings, except that now also FB varies:

carrier0010000 0 15000015
reg#0x20 = 0x00key-off
reg#0x10 = 0x40fnum-low=0x40
reg#0x30 = 0x00max volume / custom instrument
reg#0x20 = 0x10key-on / block=0


Note that for FB=0 the measurements are identical to those of last time. From this we can already conclude that FB=0 really means no feedback because last time we did not take feedback into account and our model could already 100% predict the measured data.

I did not capture all combinations of TL/FB (that would take way too long). Instead I only measured:

You can see the results in the following tables:

One full period



Model the observed behavior.

To reverse engineering the feedback behavior, I again took a lazy/smart approach. Instead of trying to figure out everything myself, I first checked what information was already available:

    FB (R#3 bits 2-0)|  0  |  1  |  2  |  3  |  4  |  5  |  6  |  7
    Modulation index |  0  |pi/16|pi/8 |pi/4 |pi/2 | pi  | 2pi | 4pi

This confirms that for FB=0 there is no feedback at all. It also shows the amount of feedback increases exponentially with increasing FB register values. This suggests FB controls a shift-amount instead of a (linear) multiplication factor.

is about feedback on OPL3. It also talks about the sum of the last two samples. Likely in OPLL (=YM2413) it works the same.

With this information in mind and after some iterations I got the following model. (After quite a few iterations actually, because I explicitly also wanted to try (and dismiss) various other alternatives, I won't go into more detail on these alternatives):

    uint16_t logsinTable[256];
    uint16_t expTable[256];

    void initTables() {
        for (int i = 0; i < 256; ++i) {
            logsinTable[i] = round(-log2(sin((double(i) + 0.5) * M_PI / 256.0 / 2.0)) * 256.0);
            expTable[i] = round((exp2(double(i) / 256.0) - 1.0) * 1024.0);

    uint16_t lookupSin(uint16_t val) {
        bool sign   = val & 512;
        bool mirror = val & 256;
        val &= 255;
        uint16_t result = logsinTable[mirror ? val ^ 0xFF : val];
        if (sign) result |= 0x8000;
        return result;

    int16_t lookupExp(uint16_t val) {
        bool sign = val & 0x8000;
        int t = (expTable[(val & 0xFF) ^ 0xFF] << 1) | 0x0800;
        int result = t >> ((val & 0x7F00) >> 8);
        if (sign) result = ~result;
        return result;

    int main() {

        int16_t p0 = 0;
        int16_t p1 = 0;
        int TL = 37; // 0..63
        int FB = 7; // 0..7

        for (int i = 0; i < 16 * 1024; ++i) {
            auto f = FB ? (p0 + p1) >> (8 - FB) : 0;
            auto s = lookupSin((i - 1) / 16 + f);
            auto m = lookupExp(s + 32 * TL) >> 1;
            p1 = p0;
            p0 = m;

            auto s2 = lookupSin(i / 16 +  2 * m);
            auto c = lookupExp(s2) >> 4;
            cout << 255 - c << endl;

This model can exactly predict the measured data. Though with these limitations:

In blue is the measured and in green is the predicted graph for TL=37, FB=7. There's almost no blue in this image, meaning the 2 graphs exactly overlap, including the 'blobs' with rapid oscillations. The only small blue spot is encircled in red, but it's clearly measurement noise (it disappears or moves when re-measuring).

Again the measured data is shown in blue and the predicted values in green. There are 3 main parts in this graph, in all 3 parts both curves are chaotic. In the 1st part the blue and green curves are very different. In the 2nd and 3rd part the curves exactly overlap (and are both chaotic). We can better see this by subtracting one curve from the other:

In the first part the difference wildly varies between -512 and 512, meaning the predictions are completely off. In the later parts the difference is zero (actually sometimes +/-1 because of measurement noise), meaning the exact same chaos was produced.

In this mode (TL=0, FB=7) the 'p0' and 'p1' values have a very large effect. So if they are only slightly off, the predicted values will be completely different. But for some reason, after a while, the predicted and actual 'px' values do get in sync with each other, and from that point onwards the model exactly predicts the measurements.

The fact that (after this initial sync problem) this chaos is predicted exactly gives me great confidence that the underlying model is correct. We only need to more accurately set the initial conditions.

Die-shot, feedback

It's always nice to look at the same problem from a few different angles. So let's look at it from a more hardware oriented point of view. What hardware features would be required to implement the feedback mechanism as described above?

Maybe the most obvious feature is the storage for the previous two outputs. So how much storage does that require? One output value (-2047..+2047) requires 12 bits. We store 2 such values for all 9 modulators. So that's 12 x 9 x 2 bits.

Now, guess what I found in the YM2413[1] die-shot:

Vertically near the top, horizontally in the center, there are two blocks close to each other, each containing 12 rows of 9 bits arranged in a shift register. That's exactly what we need here. So this again increases my confidence in the model.

Die-shot, summary

In the previous posts I already identified other regions in this die-shot. Let's repeat/summarize that information here. Take a look at this annotated image. (It's a snapshot generated from the link above, I manually added the colored annotations. Though I still suggest to actually visit that link because it allows to freely zoom/pan the image in much more detail):

With this all the major areas containing storage are identified. The remaining minor bits that I can identify are (though still very much speculation at this point):

There must also be a 'global counter' somewhere that controls the envelopes and the LFO AM and PM stuff, I did not yet find it (though it's not a shift register, so I don't recognize it). I also didn't calculate yet how many bits this counter requires.

The other area is mostly filled with logic functions(*). I can sort-of recognize adders, but the other logic components remain a mystery to me. (*) Actually most area is filled with wires instead of logic. But it's easy enough to distinguish between the two.

Next steps

Not sure yet. I think all major components have been looked at. Maybe I should go over all previous posts and tie up some loose ends? Or maybe I should go over the YM2413 registers and check whether all bits have been covered?

Some of the bigger remaining TODOs are:

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