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[英]Mixer installed on system are different on Safari and Firefox using java AudioSystem APIs
[英]Why glissando frequency goes up too high using java Audiosystem
我嘗試創建一個從開始音符到結束音符(下面的 Java 代碼)的滑音(平滑的音高上升)。 我像這樣從起始音符頻率線性上升到停止音符頻率
for (i = 0; i < b1.length; i++) {
instantFrequency = startFrequency + (i * deltaFreq / nrOfSamples);
b1[i] = (byte) (127 * Math.sin(2 * Math.PI * instantFrequency * i / sampleRate));
}
在生成的音頻片段中,滑音結尾的音調明顯高於停音符。 是我的數學有問題還是有聽力學原因導致這個上升正弦波似乎過沖? 非常感謝任何想法!
public static void main(String[] args) throws IOException {
int sampleRate = 44100;
int sampleSizeInBits = 8;
int nrOfChannels = 1;
byte[] sine220 = createTimedSine(220, sampleRate, 0.5);
byte[] gliss220to440 = createTimedGlissando(220, 440, sampleRate, 4);
byte[] sine440 = createTimedSine(440, sampleRate, 2);
byte[] fullWave = concatenate(sine220, gliss220to440, sine440);
AudioInputStream stream = new AudioInputStream(new ByteArrayInputStream(fullWave),
new AudioFormat(sampleRate, sampleSizeInBits, nrOfChannels, true, false), fullWave.length);
File fileOut = new File(path, filename);
Type wavType = AudioFileFormat.Type.WAVE;
try {
AudioSystem.write(stream, wavType, fileOut);
} catch (IOException e) {
System.out.println("Error writing output file '" + filename + "': " + e.getMessage());
}
}
public static byte[] createTimedSine(float frequency, int samplingRate, double duration) {
int nrOfSamples = (int) Math.round(duration * samplingRate);
return (createSampledSine(nrOfSamples, frequency, samplingRate));
}
public static byte[] createSampledSine(int nrOfSamples, float frequency, int sampleRate) {
byte[] b1 = new byte[nrOfSamples];
int i;
for (i = 0; i < b1.length; i++) {
b1[i] = (byte) (127 * Math.sin(2 * Math.PI * frequency * i / sampleRate));
}
System.out.println("Freq of sine: " + frequency);
return b1;
}
public static byte[] createTimedGlissando(float startFrequency, float stopFrequency, int samplingRate,
double duration) {
int nrOfSamples = (int) Math.round(duration * samplingRate);
return (createGlissando(nrOfSamples, startFrequency, stopFrequency, samplingRate));
}
public static byte[] createGlissando(int nrOfSamples, float startFrequency, float stopFrequency, int sampleRate) {
byte[] b1 = new byte[nrOfSamples];
float deltaFreq = (stopFrequency - startFrequency);
float instantFrequency = 0;
int i;
for (i = 0; i < b1.length; i++) {
instantFrequency = startFrequency + (i * deltaFreq / nrOfSamples);
b1[i] = (byte) (127 * Math.sin(2 * Math.PI * instantFrequency * i / sampleRate));
}
System.out.println("Start freq glissando :" + startFrequency);
System.out.println("Stop freq glissando :" + instantFrequency);
return b1;
}
static byte[] concatenate(byte[] a, byte[] b, byte[] c) throws IOException {
ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
outputStream.write(a);
outputStream.write(b);
outputStream.write(c);
byte d[] = outputStream.toByteArray();
return d;
}
控制台 output:
Freq of sine: 220.0
Start freq glissando :220.0
Stop freq glissando :439.9975
Freq of sine: 440.0
出現問題是因為每個幀的相鄰間距太寬。 instantFrequency
的計算很好,但通過將它乘以i
得出的值是可疑的。 當你 go 從i到i+1時,前進的距離如下:
distance = ((n+1) * instantFrequency[n+1]) - (n * instantFrequency[n])
這大於所需的 delta 值,該值應等於新的instantFrequency
值,例如:
distance = ((n+1) * instantFrequency[n]) - (n * instantFrequency[n])
下面的代碼幫我弄清楚了這個讓我疑惑了幾個小時的問題。 只有在睡過頭之后,我才能得到上面簡潔的解釋(在編輯中添加)。
這是一個更簡單的案例來說明這個問題。 由於問題發生在 sin function 計算之前,我排除了它們以及所有在 trig 計算之后的操作。
public class CuriousSeries {
public static void main(String[] args) {
double aa = 1; // analogous to your 220
double bb = 2; // analogous to your 440
double delta = bb - aa;
int steps = 10;
double[] travelVals = new double[steps + 1];
// trip aa
for (int i = 0; i <= 10; i++) {
travelVals[i] = aa * i;
System.out.println("aa trip. travelVals[" + i + "] = " + travelVals[i]);
}
// trip ab
for (int i = 0; i <= 10; i++) {
double instantFreq = aa + (i / 10.0) * delta;
travelVals[i] = instantFreq * i;
System.out.println("ab trip. travelVals[" + i + "] = " + travelVals[i]);
}
// trip bb
for (int i = 0; i <= 10; i++) {
travelVals[i] = bb * i;
System.out.println("bb trip. travelVals[" + i + "] = " + travelVals[i]);
}
// trip cc
travelVals[0] = 0;
for (int i = 1; i <= 10; i++) {
double travelIncrement = aa + (i / 10.0) * delta;
travelVals[i] = travelVals[i-1] + travelIncrement;
System.out.println("cc trip. travelVals[" + i + "] = " + travelVals[i]);
}
}
}
讓我們將aa
視為類似於 220 Hz,將bb
視為類似於 440 Hz。 在每個部分中,我們從 0 和 go 開始到 position 10。我們 go 向前的金額計算與您的計算類似。 對於“固定利率”,我們只需將步驟的值乘以i
(行程aa和bb )。 在 trip ab中,我使用了與您類似的計算。 它的問題是最后的步驟太大了。 如果您檢查 output 行,您可以看到這一點:
ab trip. travelSum[9] = 17.099999999999998
ab trip. travelSum[10] = 20.0
“步”移動的距離接近 3,而不是所需的 2!
在最后一個示例 trip cc中, instantFrequency
的計算與travelIncrement
的計算相同。 但在這種情況下,增量只是簡單地添加到之前的 position。
事實上,出於音頻合成的目的(當計算創建 wave forms 時),使用加法來最小化 cpu 成本是有意義的。 按照這些思路,我通常會做一些更像下面的事情,盡可能多地從內部循環中刪除計算:
double cursor = 0;
double prevCursor = 0;
double pitchIncrement = 2 * Math.PI * frequency / sampleRate;
for (int i = 0; i < n; i++) {
cursor = prevCursor + pitchIncrement;
audioVal[i] = Math.sin(cursor);
prevCursor = cursor;
}
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