信号期中纠错
线性时不变离散系统稳定,其单位样值响应\(h(n)\)必须满足
\[ \sum_{n=-\infty}^{\infty}|h(n)|<\infty \]线性时不变离散系统因果,其单位样值响应\(h(n)\)必须满足
\[ h(n)=h(n)u(n) \]\[ \begin{align}f(t)*\delta(t-t_0)&=f(t_0)\notag\\f(t)\cdot\delta(t-t_0)&=f(t_0)\delta(t-t_0)\notag\end{align} \]
\(f(t)\)的带宽为\(W\),则\(f(t)\sin(\omega_0t+\frac{\pi}{4})\)的带宽为
\[ 2W \]序列\(\delta(\frac{n}{2})\)可用\(\delta(n)\)表示为
\[ \delta(n) \]某线性时不变离散系统的单位样值响应为\(h(n)\),当激励为\(u(n)-u(n-2)\)时,系统的输出为
\[ h(n)+h(n-1) \]求下列序列的最小正周期
\(\cos(\frac{\pi}{3}n)+\cos(\frac{5\pi}{3}n)\)
\[ \begin{align} T_1&=\frac{2\pi}{\frac{\pi}{3}}=6\notag\\ T_2&=\frac{2\pi}{\frac{5\pi}{3}}=\frac{6}{5}\notag \end{align} \]
所以周期是两者最小公倍数,为6.\(\cos(\frac{\pi}{3}n)\cos(\frac{5\pi}{3}n)\)
需要使用积化和差
\[ \cos(\frac{\pi}{3}n)\cos(\frac{5\pi}{3}n)=\frac{1}{2}[\cos(2\pi)+\cos(-\frac{4\pi}{3})] \]
还需要对两者分别求周期
\[ \begin{align} T_1&=\frac{2\pi}{2\pi}=1\notag\\ T_2&=\frac{2\pi}{-\frac{4\pi}{3}}=-\frac{3}{2}\notag \end{align} \]
最小公倍数为\(3\),所以周期为\(3\).
已知\(f(t)=e^{-2t}[u(t)-u(t-4)]\),求其傅氏变换
\[ f(t)=e^{-2t}u(t)-e^{-8}\cdot e^{2(t-4)}u(t-4)\leftrightarrow \frac{1}{2+j\omega}-e^{-8}\frac{1}{2+j\omega}e^{-j\omega4} \]
将信号拆分,凑成了可以使用时移特性的形式.求\(u(2t-4)\)得傅氏变换
根据尺度变换性质可知
\[ f(at+b)\leftrightarrow \frac{1}{|a|}e^{j\omega(\frac{b}{a})}F(\frac{\omega}{a}) \]
将函数代入,可得
\[ u(2t-4)\leftrightarrow \frac{1}{2}e^{-j\omega(\frac{-4}{2})}F(\frac{\omega}{2}) \]
又因为
\[ F(\omega)=\pi\delta(\omega)+\frac{1}{j\omega} \]
所以
\[ \frac{1}{2}e^{-j\omega(\frac{-4}{2})}F(\frac{\omega}{2})=\frac{1}{2}e^{-j\omega(\frac{-4}{2})}[\pi\delta(\frac{\omega}{2})+\frac{1}{j\frac{\omega}{2}}]\tag{*} \]
根据冲激函数的性质
\[ \delta(at)=\frac{1}{|a|}\delta(t) \]
则\((*)\)式可以写为
\[ \frac{1}{2}e^{-j\omega(\frac{-4}{2})}2\pi\delta(\omega)+\frac{1}{2}e^{-j\omega(\frac{-4}{2})}\frac{1}{j\frac{\omega}{2}}=\pi\delta(\omega)+\frac{e^{-j\omega 2}}{j\omega} \]求信号\((1-t)\frac{d}{dt}[e^{-2t}\delta(t)]\)的傅氏变换
\[ \begin{align} (1-t)\frac{d}{dt}[e^{-2t}\delta(t)]&=(1-t)\delta^{'}(t)\notag\\ &=\delta^{'}(t)-t\delta^{'}(t)\notag \end{align} \]
由于
\[ \delta^{'}(t)\leftrightarrow j\omega \]
以及频域微分特性
\[ tf(t)\leftrightarrow j\frac{d}{d\omega}F(\omega) \]
可得
\[ \begin{align}\delta^{'}(t)-t\delta^{'}(t)&=j\omega-[j(\frac{d}{d\omega}j\omega)]\notag\\&=j\omega+1\notag\end{align} \]求信号\(e^{-(2+j5)t}u(t)\)的傅氏变换
\[ \begin{align}e^{-(2+j5)t}u(t)&=e^{-2t}u(t)e^{-j5t}\notag\\&=\frac{1}{2+j(\omega+5)}\notag\end{align} \]
此处用了单边指数的傅氏变换公式.