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通俗理解时域采样与频域延拓
2022-06-12 08:41:00 【ZEERO~】
讲连续信号 x a ( t ) x_{a}(t) xa(t)与冲激串信号 p s ( t ) p_{s}(t) ps(t)相乘,即可得到离散时间信号 x ( n ) x(n) x(n),因此,有如下公式:
x ( n ) = x a ( t ) ∣ t = n T s = x a ( t ) p s ( t ) = x a ( t ) ∑ n = − ∞ ∞ δ ( t − n T s ) x(n)=x_{a}(t) |_{t=nT_{s}}=x_{a}(t)p_{s}(t)=x_{a}(t)\sum_{n=-\infty}^{\infty}\delta(t-nT_{s}) x(n)=xa(t)∣t=nTs=xa(t)ps(t)=xa(t)n=−∞∑∞δ(t−nTs)
采样周期为 T s T_{s} Ts,我们用 X a ( j Ω ) X_{a}(j\Omega) Xa(jΩ)来表示模拟信号 x a ( t ) x_{a}(t) xa(t)的频谱,用 X ( e j w ) X(e^{jw}) X(ejw)来表示离散信号 x ( n ) x(n) x(n)的频谱,根据Fourier变换公式,
X a ( j Ω ) = ∫ − ∞ ∞ x a ( t ) e − j Ω t d t X ( e j w ) = ∑ n = − ∞ ∞ x ( n ) e − j w n X_{a}(j\Omega)=\int_{-\infty}^{\infty}x_{a}(t)e^{-j\Omega t}dt\\X(e^{jw})=\sum_{n=-\infty}^{\infty}x(n)e^{-jwn} Xa(jΩ)=∫−∞∞xa(t)e−jΩtdtX(ejw)=n=−∞∑∞x(n)e−jwn
示意图如下所示:
上图中,我们假设 x a ( t ) x_{a}(t) xa(t)为带限信号,就是频带有限长的信号,我们希望能找出 X a ( j Ω ) X_{a}(j\Omega) Xa(jΩ)与 X ( e j w ) X(e^{jw}) X(ejw)之间的区别和联系。
我们知道,离散信号可以视为一种特殊的连续信号,将其Fourier变换记为 X s ( j Ω ) X_{s}(j\Omega) Xs(jΩ)。由模拟频率与数字频率的关系, X s ( j Ω ) X_{s}(j\Omega) Xs(jΩ)与 X ( e j w ) X(e^{jw}) X(ejw)之间的关系为
X ( e j w ) = X s ( j Ω ) ∣ Ω = w / T s X(e^{jw})=X_{s}(j\Omega) |_{\Omega =w/T_{s}} X(ejw)=Xs(jΩ)∣Ω=w/Ts
有关模拟频率和数字频率的关系,看这篇文章模拟频率与数字频率之间的关系。
我们接着往下讲,我们都知道,时域相乘等于频域的卷积,由数学公式表示
X s ( j Ω ) = X a ( j Ω ) ∗ P s ( j Ω ) X_{s}(j\Omega)=X_{a}(j\Omega)*P_{s}(j\Omega) Xs(jΩ)=Xa(jΩ)∗Ps(jΩ)
,其中 P s ( j Ω ) P_{s}(j\Omega) Ps(jΩ)为冲激信号串的Fourier变换。
参考资料
《学以致用:深入浅出数字信号处理》 江志宏
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