取樣和取樣函數的傅里葉變換

取樣

$$\bar{f}(t)=f(t)s_{\Delta T}(t)=\sum _{n=?\infty }^{\infty }f(t)\delta (t?n\Delta T)\text{\hspace{1em}(4.27)}$$
$$f_k={\int }_{?\infty }^{\infty }f(t)\delta (t?k\Delta T)dt=f(k\Delta T)\text{\hspace{1em}(4.28)}$$

# 取樣
x = np.arange(-10, 10, 0.01)
y_1 = 1.5 - np.sin(x)fig = plt.figure(figsize=(9, 6))
ax_1 = setup_axes(fig, 211)
ax_1.plot(x, y_1), ax_1.set_title('f(t)', loc='center', y=1.05), ax_1.set_ylim([0, 4]), ax_1.set_yticks([]), ax_1.set_xticks([])x_2 = x[::50]
y_2 = y_1[::50]
ax_2 = setup_axes(fig, 212)
ax_2.scatter(x_2, y_2), ax_2.set_title('f_k', loc='center', y=1.05), ax_2.set_ylim([0, 4]), ax_2.set_yticks([]), ax_2.set_xticks([])plt.tight_layout()
plt.show()

取樣后的函數的傅里葉變換

取樣后的函數$\bar{f}(t)$的傅里葉變換$\bar{F}(\mu )$是
Fˉ(μ)=J{fˉ(t)}=J{f(t)sΔT(t)}=(F?S)(μ)(4.29)\bar F(\mu) = \mathfrak{J} \{ \bar f(t) \} = \mathfrak{J} \{ f(t) s_{\Delta T}(t)\} = (F \star S)(\mu) \tag{4.29}Fˉ(μ)=J{fˉ?(t)}=J{f(t)sΔT?(t)}=(F?S)(μ)(4.29)

$$S(\mu )=\frac{1}{\Delta T}\sum _{n=?\infty }^{\infty }\delta (\mu ?\frac{n}{\Delta T})\text{\hspace{1em}(4.30)}$$

所以有
KaTeX parse error: No such environment: align at position 7: \begin{?a?l?i?g?n?}? \bar F(\mu) & …

# 過取樣、臨界取樣和欠取樣
x = np.arange(0, 6, 0.01)
y = 2 - x
y = np.where(x <= 2, y, 0)
x_1 = np.concatenate((-x[::-1], x), axis=0)
y_1 = np.pad(y, (y.shape[0], 0), mode='reflect')fig = plt.figure(figsize=(10, 8))
# 帶限函數的傅里葉變換
ax_1 = setup_axes(fig, 411)
ax_1.plot(x_1, y_1), ax_1.set_title('F(u)', loc='center', y=1.05), ax_1.set_ylim([0, 3]), ax_1.set_yticks([]), ax_1.set_xticks([])# 過取樣
y_2 = y_1[600-250:600+250]
y_2 = np.tile(y_2, 5)
x_2 = np.linspace(-6, 6, y_2.shape[0])ax_2 = setup_axes(fig, 412)
ax_2.plot(x_2, y_2), ax_2.set_title('F(u)', loc='center', y=1.05), ax_2.set_ylim([0, 3]), ax_2.set_yticks([]), ax_2.set_xticks([])# 臨界取樣
y_3 = y_1[600-200:600+200]
y_3 = np.tile(y_3, 5)
x_3 = np.linspace(-6, 6, y_3.shape[0])ax_3 = setup_axes(fig, 413)
ax_3.plot(x_3, y_3), ax_3.set_title('F(u)', loc='center', y=1.05), ax_3.set_ylim([0, 3]), ax_3.set_yticks([]), ax_3.set_xticks([])# 欠取樣
y_4 = y_1[600-200:600+200]
y_4 = np.where(y_4 > 1, y_4, 1)
y_4 = np.where(y_4 > -1, y_4, 1)
y_4 = np.tile(y_4, 5)
x_4 = np.linspace(-6, 6, y_4.shape[0])ax_4 = setup_axes(fig, 414)
ax_4.plot(x_4, y_4), ax_4.set_title('F(u)', loc='center', y=1.05), ax_4.set_ylim([0, 3]), ax_4.set_yticks([]), ax_4.set_xticks([])plt.tight_layout()
plt.show()

取樣定理

• 帶限函數

  • 對于以原點為中心的有限區間(帶寬)$[?{\mu }_{max},{\mu }_{max}]$外的頻率值，傅里葉變換為零的函數

• 奈奎斯特定理

  • 如果以超過函數最高頻率2倍的取樣率來得到樣本，那么連續帶限函數就能夠完全由其樣本集復原

$$\frac{1}{\Delta T}>2{\mu }_{max}\text{\hspace{1em}(4.32)}$$

• 低通濾波器
  $$H\mu =\{\begin{array}{ll}\Delta T,& ?{\mu }_{max}\le \mu \le {\mu }_{max}\\ 0,& \text{others}\end{array}\text{\hspace{1em}(4.33)}$$

• 濾波器乘以傅里葉變換后的函數
  $$F(\mu )=H(\mu )\tilde{F}(\mu )\text{\hspace{1em}(4.34)}$$

• 傅里葉反變換復原$f(t)$:
  $$f(t)={\int }_{?\infty }^{\infty }F(\mu )e^{j2\pi \mu t}d\mu \text{\hspace{1em}(4.35)}$$

# 采樣、濾波
x = np.arange(0, 6, 0.01)
y = 2 - x
y = np.where(x <= 2, y, 0)
x_1 = np.concatenate((-x[::-1], x), axis=0)
y_1 = np.pad(y, (y.shape[0], 0), mode='reflect')fig = plt.figure(figsize=(10, 8))# 取樣
y_2 = y_1[600-250:600+250]
y_2 = np.tile(y_2, 5)
x_2 = np.linspace(-6, 6, y_2.shape[0])# 帶限函數的傅里葉變換
ax_1 = setup_axes(fig, 311)
ax_1.plot(x_2, y_2), ax_1.set_title(r'\tilde F(u)', loc='center', y=1.05), ax_1.set_ylim([0, 3]), ax_1.set_yticks([]), ax_1.set_xticks([])x_1 = np.linspace(-6, 6, y_2.shape[0])
y_3 = np.where(x_1, x_1 >= -1, 0)
y_3 = np.where(x_1 <= 1, y_3, 0)ax_3 = setup_axes(fig, 312)
ax_3.plot(x_1, y_3), ax_3.set_title('H(u)', loc='center', y=1.05), ax_3.set_ylim([0, 3]), ax_3.set_yticks([]), ax_3.set_xticks([])# 濾波
y_4 = y_2 * y_3ax_4 = setup_axes(fig, 313)
ax_4.plot(x_1, y_4), ax_4.set_title(r'F(u) = H(u)\tilde F(u)', loc='center', y=1.05), ax_4.set_ylim([0, 3]), ax_4.set_yticks([]), ax_4.set_xticks([])plt.tight_layout()
plt.show()

混疊

在信號處理領域，混疊是指取樣后不同信號變得彼此無法區分的取樣現象，或者一個信號“偽裝”成另一個信號的現象。

這種稱為混疊對， 這種混疊對在取樣后是無法區分的。出現這種函數混疊的原因是我們所用的取樣率太粗，也就是說欠取樣

帶限函數的取樣率小于奈奎斯特率的話，不管使用何種濾波器器，都不可能分離出來一個周期。也就不可能完美的復原函數。

# 混疊
x = np.arange(0, 6*np.pi, 0.1)y_1 = 2 - np.sin(4 * x)
y_2 = 2 - np.sin(x)sample_1_f = 21
sample_2_f = 21
sample_1 = y_1[::sample_1_f]
sample_2 = y_2[::sample_2_f]
fig = plt.figure(figsize=(14, 8))
plt.subplot(2, 2, 1), plt.plot(x, y_1), plt.xticks([]), plt.yticks([])
plt.subplot(2, 2, 3), plt.plot(x, y_2), plt.xticks([]), plt.yticks([])plt.subplot(2, 2, 2), plt.stem(x[::sample_1_f], sample_1, linefmt='--', markerfmt='o', basefmt='C0-')
plt.xticks([]), plt.yticks([])
plt.subplot(2, 2, 4), plt.stem(x[::sample_2_f], sample_2, linefmt='--', markerfmt='o', basefmt='C0-')
plt.xticks([]), plt.yticks([])
plt.tight_layout()
plt.show()

由取樣后的數據重建（復原）函數

f(t)=J{F(μ)}=J{H(μ)F~(μ)}=h(t)?f~(t)(4.37)f(t) = \mathfrak{J} \{ F(\mu) \} = \mathfrak{J} \{ H(\mu) \tilde F(\mu) \} = h(t) \star\tilde f(t)\tag{4.37}f(t)=J{F(μ)}=J{H(μ)F~(μ)}=h(t)?f~?(t)(4.37)

得到空間域表達式：
$$f(t)=\sum _{?\infty }^{\infty }f(n\Delta T)sinc[(t?n\Delta T)/\Delta T]\text{\hspace{1em}(4.38)}$$