Ejemplo n.º 1
0
def Formant_Interpolation(u, p, fs):
    """
    插值法估计共振峰函数
    :param u:
    :param p:
    :param fs:
    :return:
    """
    ar, _ = lpc_coeff(u, p)
    U = np.power(np.abs(np.fft.rfft(ar, 2 * 255)), -2)
    df = fs / 512
    val, loc = local_maxium(U)
    ll = len(loc)
    pp = np.zeros(ll)
    F = np.zeros(ll)
    Bw = np.zeros(ll)
    for k in range(ll):
        m = loc[k]
        m1, m2 = m - 1, m + 1
        p = val[k]
        p1, p2 = U[m1], U[m2]
        aa = (p1 + p2) / 2 - p
        bb = (p2 - p1) / 2
        cc = p
        dm = -bb / 2 / aa
        pp[k] = -bb * bb / 4 / aa + cc
        m_new = m + dm
        bf = -np.sqrt(bb * bb - 4 * aa * (cc - pp[k] / 2)) / aa
        F[k] = (m_new - 1) * df
        Bw[k] = bf * df
    return F, Bw, pp, U, loc
Ejemplo n.º 2
0
def Formant_Root(u, p, fs, n_frmnt):
    """
    LPC求根法的共振峰估计函数
    :param u:
    :param p:
    :param fs:
    :param n_frmnt:
    :return:
    """
    ar, _ = lpc_coeff(u, p)
    U = np.power(np.abs(np.fft.rfft(ar, 2 * 255)), -2)
    const = fs / (2 * np.pi)
    rts = np.roots(ar)
    yf = []
    Bw = []
    for i in range(len(ar) - 1):
        re = np.real(rts[i])
        im = np.imag(rts[i])
        fromn = const * np.arctan2(im, re)
        bw = -2 * const * np.log(np.abs(rts[i]))
        if fromn > 150 and bw < 700 and fromn < fs / 2:
            yf.append(fromn)
            Bw.append(bw)
    return yf[:min(len(yf), n_frmnt)], Bw[:min(len(Bw), n_frmnt)], U
Ejemplo n.º 3
0
def pitch_Lpc(x, wnd, inc, T1, fs, p, miniL=10):
    """
    线性预测法基音周期检测函数
    :param x:
    :param wnd:
    :param inc:
    :param T1:
    :param fs:
    :param p:
    :param miniL:
    :return:
    """
    from scipy.signal import lfilter
    from chapter3_分析实验.lpc import lpc_coeff
    y = enframe(x, wnd, inc)
    fn = y.shape[0]
    if isinstance(wnd, int):
        wlen = wnd
    else:
        wlen = len(wnd)
    voiceseg, vsl, SF, Ef = pitch_vad(x, wnd, inc, T1, miniL)
    lmin = fs // 500  # 基音周期的最小值
    lmax = fs // 60  # 基音周期的最大值
    period = np.zeros(fn)
    for k in range(y.shape[0]):
        if SF[k] == 1:
            u = np.multiply(y[k, :], np.hamming(wlen))
            ar, _ = lpc_coeff(u, p)
            ar[0] = 0
            z = lfilter(-ar, [1], u)
            E = u - z
            xx = np.fft.fft(E)
            b = np.fft.ifft(2 * np.log(np.abs(xx) + 1e-20))
            lc = np.argmax(b[lmin:lmax])
            period[k] = lc + lmin
    return voiceseg, vsl, SF, Ef, period
Ejemplo n.º 4
0
from scipy.signal import filtfilt

from chapter2_基础.soundBase import *
from chapter3_分析实验.lpc import lpc_coeff

plt.rcParams['font.sans-serif'] = ['SimHei']
plt.rcParams['axes.unicode_minus'] = False

data, fs = soundBase('C6_1_y.wav').audioread()
N = len(data)
time = [i / fs for i in range(N)]  # 设置时间
p = 12
ar, g = lpc_coeff(data, p)
ar[0] = 0
est_x = filtfilt(-ar, [1], data)

plt.subplot(2, 1, 1)
plt.plot(time, data, 'k')
plt.plot(time, est_x, 'c')
plt.title('LPC解码')
plt.legend(['信号', '解码信号'])
plt.subplot(2, 1, 2)
plt.plot(est_x - data)
plt.title('误差')
plt.savefig('images/LPC解码.png')
plt.close()
data /= np.max(np.abs(data))
N = len(data)
time = [i / fs for i in range(N)]  # 设置时间
p = 12
wlen, inc = 200, 80
msoverlap = wlen - inc
y = enframe(data, wlen, inc)
fn = y.shape[0]
Acoef = np.zeros((y.shape[0], p + 1))
resid = np.zeros(y.shape)
synFrame = np.zeros(y.shape)
## 7.2.1

# 求每帧的LPC系数与预测误差
for i in range(fn):
    a, _ = lpc_coeff(y[i, :], p)
    Acoef[i, :] = a
    resid[i, :] = lfilter(a, [1], y[i, :])

# 语音合成
for i in range(fn):
    synFrame[i, :] = lfilter([1], Acoef[i, :], resid[i, :])

outspeech = Filpframe_OverlapS(synFrame, np.hamming(wlen), inc)
plt.subplot(2, 1, 1)
plt.plot(data / np.max(np.abs(data)), 'k')
plt.title('原始信号')
plt.subplot(2, 1, 2)
plt.title('还原信号-LPC与误差')
plt.plot(outspeech / np.max(np.abs(outspeech)), 'c')