网站建设包含内容,企业服务是做什么的,网站做担保交易,加强农业网站建设目录基本概念复数傅里叶级数冲激函数及其取样#xff08;筛选#xff09;性质连续单变量函数的傅里叶变换卷积基本概念
复数
复数CCC的定义为 CRjI(4.3)C R jI \tag{4.3}CRjI(4.3) R,IR,IR,I为实数#xff0c;RRR是实部#xff0c;III是虚部#xff0c;j−1j \sqrt{-…
目录基本概念复数傅里叶级数冲激函数及其取样筛选性质连续单变量函数的傅里叶变换卷积基本概念
复数
复数CCC的定义为
CRjI(4.3)C R jI \tag{4.3}CRjI(4.3)
R,IR,IR,I为实数RRR是实部III是虚部j−1j \sqrt{-1}j−1。复数的共轭表示为C∗C^*C∗
C∗R−jI(4.4)C^* R - jI \tag{4.4}C∗R−jI(4.4)
从几何角度来看复数可视为平面称为复平面上的一个点其横坐标是实轴纵坐标是虚轴也就是说RjIRjIRjI是得平面直角坐标系中的点(R,I)(R,I)(R,I)
极坐标表示复数
C∣C∣(cosθjsinθ)(4.5)C |C|(\text{cos}\theta \text{jsin}\theta) \tag{4.5}C∣C∣(cosθjsinθ)(4.5) ∣C∣R2I2|C|\sqrt{R^2 I^2}∣C∣R2I2是从复平面的原点延伸到点(R,I)(R,I)(R,I)的向量长度θ\thetaθ是该向量与实轴的夹角。
使用欧拉公式 ejθcosθjsinθ(4.6)e^{j\theta} \text{cos}\theta \text{jsin}\theta \tag{4.6}ejθcosθjsinθ(4.6)
可以给坐极坐标的复数表示为 C∣C∣ejθ(4.7)C |C|e^{j\theta} \tag{4.7}C∣C∣ejθ(4.7)
复函数 F(u)R(u)jI(u)F(u) R(u) jI(u)F(u)R(u)jI(u) F∗(u)R(u)−jI(u)F^*(u) R(u) - jI(u)F∗(u)R(u)−jI(u) 幅值是 ∣F(u)∣R(u)2I(u)2|F(u)| \sqrt{R(u)^2 I(u)^2}∣F(u)∣R(u)2I(u)2
a np.complex(1 2j)radian np.arctan2(a.imag, a.real)
radian1.1071487177940904degree np.rad2deg(radian)
degree63.43494882292201傅里叶级数
周期为TTT的连续变量ttt的周期函数f(t)f(t)f(t)可表示为乘以适当系数的正弦函数和余弦函数之和 f(t)∑n−∞∞cnej2πnTt(4.8)f(t) \sum_{n-\infty}^{\infty} c_n e^{j\frac{2\pi n}{T}t} \tag{4.8}f(t)n−∞∑∞cnejT2πnt(4.8)
cn1T∫−T/2T/2f(t)e−j2πnTt(4.9)c_n \frac{1}{T}\int_{-T/2}^{T/2}f(t)e^{-j\frac{2\pi n}{T}t} \tag{4.9}cnT1∫−T/2T/2f(t)e−jT2πnt(4.9)
冲激函数及其取样筛选性质
连续变量ttt在t0t0t0处的单位冲激表示为δ(t)\delta(t)δ(t)定义是 δ(t){∞,t00,t≠0(4.10)\delta(t) \begin{cases} \infty, t0 \\ 0, t \neq 0 \end{cases} \tag{4.10}δ(t){∞,0,t0t0(4.10) 它被限制为满足恒等式 ∫−∞∞δ(t)dt1(4.11)\int_{-\infty}^{\infty}\delta(t)dt 1 \tag{4.11}∫−∞∞δ(t)dt1(4.11)
自然地将ttt解释为时间时冲激就可视为幅度无限、持续时间为0、具有单位面积的尖峰信息。冲激具有关于积分的所谓取样性质
∫−∞∞f(t)δ(t)dtf(0)(4.12)\int_{-\infty}^{\infty} f(t) \delta(t)dt f(0) \tag{4.12}∫−∞∞f(t)δ(t)dtf(0)(4.12)
冲激 冲激并不是通常意义上的函数更准确的名称是分布或广义函数。
任意一点t0t_0t0的取样性质为 ∫−∞∞f(t)δ(t−t0)dtf(t0)(4.13)\int_{-\infty}^{\infty} f(t) \delta(t - t_0)dt f(t_0) \tag{4.13}∫−∞∞f(t)δ(t−t0)dtf(t0)(4.13)
例如 若f(t)cos(t)f(t) \text{cos}(t)f(t)cos(t)则使用冲激δ(t−π)\delta(t-\pi)δ(t−π)得到结果f(π)cos(π)−1f(\pi) cos(\pi)-1f(π)cos(π)−1
冲激串KaTeX parse error: \tag works only in display equations
离散冲激定义为 δ(x){1,x00,x≠0(4.15)\delta(x) \begin{cases} 1, x 0 \\0, x \neq 0 \end{cases} \tag{4.15}δ(x){1,0,x0x0(4.15) ∑−∞∞δ(x)1(4.16)\sum_{-\infty}^{\infty}\delta(x) 1 \tag{4.16}−∞∑∞δ(x)1(4.16) ∑−∞∞f(x)δ(x)f(0)(4.17)\sum_{-\infty}^{\infty} f(x) \delta(x) f(0) \tag{4.17}−∞∑∞f(x)δ(x)f(0)(4.17) ∑−∞∞f(x)δ(x−x0)dtf(x0)(4.18)\sum_{-\infty}^{\infty} f(x) \delta(x - x_0)dt f(x_0) \tag{4.18}−∞∑∞f(x)δ(x−x0)dtf(x0)(4.18)
def impulse(x, x0):return np.piecewise(x, [xx0, x!x0], [1, 0])# 按定义写的不知道是否正确如不正确请指出感谢只做展示
x np.arange(10)
plt.stem(x, impulse(x, 5), )
plt.show()def impulse_serial(x):s np.ones_like(x)return s# 只做展示
x np.arange(10)
plt.stem(impulse_serial(x))
plt.show()连续单变量函数的傅里叶变换
傅里叶变换对
连续变量ttt的连续函数f(t)f(t)f(t)的傅里叶变换由J{f(t)}\mathfrak{J}\{f(t)\}J{f(t)}表示定义为 J{f(t)}∫−∞∞f(t)e−j2πμtdt(4.19)\mathfrak{J}\{f(t)\} \int_{-\infty}^{\infty} f(t) e^{-j2\pi\mu t}dt \tag{4.19}J{f(t)}∫−∞∞f(t)e−j2πμtdt(4.19) J{f(t)}F(μ)\mathfrak{J}\{f(t)\} F(\mu)J{f(t)}F(μ) 所以有 F(μ)∫−∞∞f(t)e−j2πμtdt(4.20)F(\mu) \int_{-\infty}^{\infty} f(t) e^{-j2\pi\mu t}dt \tag{4.20}F(μ)∫−∞∞f(t)e−j2πμtdt(4.20) 反变换 f(t)∫−∞∞F(μ)e−j2πμtdμ(4.21)f(t) \int_{-\infty}^{\infty} F(\mu) e^{-j2\pi\mu t}d\mu \tag{4.21}f(t)∫−∞∞F(μ)e−j2πμtdμ(4.21)
通常表示为f(t)⇔F(μ)f(t) \Leftrightarrow F(\mu)f(t)⇔F(μ)
使用欧拉公式可以写为 F(μ)∫−∞∞f(t)[cos(2πμt)−jsin(2πμt)]dt(4.22)F(\mu) \int_{-\infty}^{\infty} f(t) [\text{cos}(2\pi \mu t) - \text{jsin}(2\pi \mu t)]dt \tag{4.22}F(μ)∫−∞∞f(t)[cos(2πμt)−jsin(2πμt)]dt(4.22)
傅里叶变换是f(t)f(t)f(t)乘以正弦函数的展开式 其中正弦函数的频率是由μ\muμ值决定。因此积分后留下的唯一变量是频率因此我们说傅里叶变换域是频率域。
盒式函数的傅里叶变换
F(μ)∫−∞∞f(t)e−j2πμtdt∫−W/2W/2f(t)Ae−j2πμtdt−Aj2πμ[e−j2πμt]−W/2W/2−Aj2πμ[e−jπμW−ejπμW]Aj2πμ[ejπμW−e−jπμW]AWsin(πμW)(πμW)\begin{aligned} F(\mu) \int_{-\infty}^{\infty} f(t) e^{-j2\pi\mu t}dt \int_{-W/2}^{W/2} f(t) A e^{-j2\pi\mu t}dt\\ \frac{-A}{j2\pi\mu}[e^{-j2\pi\mu t}]_{-W/2}^{W/2} \frac{-A}{j2\pi\mu}[e^{-j\pi\mu W} - e^{j\pi\mu W}]\\ \frac{A}{j2\pi\mu}[e^{j\pi\mu W} - e^{-j\pi\mu W}] \\ AW \frac{sin(\pi \mu W)}{(\pi \mu W)} \\ \end{aligned}F(μ)∫−∞∞f(t)e−j2πμtdt∫−W/2W/2f(t)Ae−j2πμtdtj2πμ−A[e−j2πμt]−W/2W/2j2πμ−A[e−jπμW−ejπμW]j2πμA[ejπμW−e−jπμW]AW(πμW)sin(πμW)
最后的结果是sinc\text{sinc}sinc函数 sinc(m)sin(πm)(πm)(4.23)\text{sinc}(m) \frac{\text{sin}(\pi m)}{(\pi m)} \tag{4.23}sinc(m)(πm)sin(πm)(4.23)
通常傅里叶变换中包含复数项这是为了显示变换的幅值一个实量的约定。这个幅值称为傅里叶频谱或频谱 ∣F(μ)∣AW∣sin(πμW)(πμW)∣|F(\mu)| AW \bigg|\frac{sin(\pi \mu W)}{(\pi \mu W)}\bigg|∣F(μ)∣AW∣∣∣∣(πμW)sin(πμW)∣∣∣∣
def box_function(x, w):w_2 w / 2y np.where(x, x -w_2, 0)y np.where(x w_2, y, 0)return yimport mpl_toolkits.axisartist as axisartist
def setup_axes(fig, rect):ax axisartist.SubplotZero(fig, rect)fig.add_axes(ax)for direction in [xzero, yzero]:# adds arrows at the ends of each axisax.axis[direction].set_axisline_style(-|)# adds X and Y-axis from the originax.axis[direction].set_visible(True)for direction in [left, right, bottom, top]:# hides bordersax.axis[direction].set_visible(False)return ax# 盒式函数的傅里叶变换
x np.arange(-5, 5, 0.1)
y box_function(x, 6)fig plt.figure(figsize(15, 5))
ax_1 setup_axes(fig, 131)
ax_1.plot(x, y), ax_1.set_title(f(t), loccenter, y1.05), ax_1.set_ylim([0, 2]), ax_1.set_yticks([])f_u np.sinc(x)
ax_2 setup_axes(fig, 132)
ax_2.plot(x, f_u), ax_2.set_title(F(u), loccenter, y1.05), ax_2.set_yticks([]), #ax_2.set_ylim([-1, 2]),f_u_absolute abs(f_u)
ax_3 setup_axes(fig, 133)
ax_3.plot(x, f_u_absolute), ax_3.set_title(|F(u)|, loccenter, y1.05), ax_3.set_yticks([]), #ax_3.set_ylim([-1, 2]),plt.tight_layout()
plt.show()冲激和冲激串的傅里叶变换
J{δ(t)}F(μ)∫−∞∞δ(t)e−j2πμtdt∫−∞∞e−j2πμtδ(t)dte−j2πμ0e01\begin{aligned} \mathfrak{J}\{\delta(t)\} F(\mu) \int_{-\infty}^{\infty} \delta(t) e^{-j2\pi\mu t}dt \\ \int_{-\infty}^{\infty} e^{-j2\pi\mu t} \delta(t)dt \\ e^{-j2\pi\mu_0} \\ e^0 1 \end{aligned}J{δ(t)}F(μ)∫−∞∞δ(t)e−j2πμtdt∫−∞∞e−j2πμtδ(t)dte−j2πμ0e01
J{δ(t−t0)}F(μ)∫−∞∞δ(t−t0)e−j2πμtdt∫−∞∞e−j2πμtδ(t−t0)dte−j2πμt0\begin{aligned} \mathfrak{J}\{\delta(t - t_0)\} F(\mu) \int_{-\infty}^{\infty} \delta(t - t_0) e^{-j2\pi\mu t}dt \\ \int_{-\infty}^{\infty} e^{-j2\pi\mu t} \delta(t - t_0)dt \\ e^{-j2\pi\mu t_0} \\ \end{aligned}J{δ(t−t0)}F(μ)∫−∞∞δ(t−t0)e−j2πμtdt∫−∞∞e−j2πμtδ(t−t0)dte−j2πμt0
冲激串的傅里叶变换
S(μ)J{SΔT(t)}J{1Δ∑n−∞∞ej2πnΔTt}1ΔTJ{∑n−∞∞ej2πnΔTt}1ΔT∑n−∞∞δ(μ−nΔT)\begin{aligned} S(\mu) \mathfrak{J}\{S_{\Delta T}(t)\} \mathfrak{J}\{\frac{1}{\Delta } \sum_{n-\infty}^{\infty} e^{j\frac{2\pi n}{\Delta T}t}\} \\ \frac{1}{\Delta T} \mathfrak{J}\{\sum_{n-\infty}^{\infty} e^{j\frac{2\pi n}{\Delta T}t}\} \\ \frac{1}{\Delta T} \sum_{n-\infty}^{\infty} \delta(\mu - \frac{n}{\Delta T}) \\ \end{aligned}S(μ)J{SΔT(t)}J{Δ1n−∞∑∞ejΔT2πnt}ΔT1J{n−∞∑∞ejΔT2πnt}ΔT1n−∞∑∞δ(μ−ΔTn)
卷积
(f⋆h)(t)∫−∞∞f(τ)h(t−τ)dτ(4.24)(f\star h)(t) \int_{-\infty}^{\infty} f(\tau) h(t - \tau) d\tau \tag{4.24}(f⋆h)(t)∫−∞∞f(τ)h(t−τ)dτ(4.24)
J{(f⋆h)(t)}∫−∞∞[∫−∞∞f(τ)h(t−τ)dτ]e−j2πμtdt∫−∞∞f(τ)[∫−∞∞h(t−τ)e−j2πμtdt]dτH(μ)∫−∞∞f(τ)e−j2πμτdτH(μ)F(μ)(H∙F)(μ)\begin{aligned} \mathfrak{J}\{(f \star h)(t) \} \int_{-\infty}^{\infty} \bigg[\int_{-\infty}^{\infty} f(\tau) h(t - \tau) d\tau \bigg] e^{-j2\pi \mu t} dt\\ \int_{-\infty}^{\infty} f(\tau) \bigg[\int_{-\infty}^{\infty} h(t - \tau) e^{-j2\pi \mu t} dt\bigg] d\tau \\ H(\mu) \int_{-\infty}^{\infty} f(\tau) e^{-j2\pi \mu \tau} d\tau \\ H(\mu)F(\mu) (H \bullet F)(\mu) \end{aligned}J{(f⋆h)(t)}∫−∞∞[∫−∞∞f(τ)h(t−τ)dτ]e−j2πμtdt∫−∞∞f(τ)[∫−∞∞h(t−τ)e−j2πμtdt]dτH(μ)∫−∞∞f(τ)e−j2πμτdτH(μ)F(μ)(H∙F)(μ) 傅里叶正变换 (f⋆h)(t)⇔(H∙F)(μ)(4.25)(f\star h)(t) \Leftrightarrow (H \bullet F)(\mu) \tag{4.25}(f⋆h)(t)⇔(H∙F)(μ)(4.25) 傅里叶反变换 (f∙h)(t)⇔(H⋆F)(μ)(4.26)(f\bullet h)(t) \Leftrightarrow (H \star F)(\mu) \tag{4.26}(f∙h)(t)⇔(H⋆F)(μ)(4.26)
