目录

  1  2  3  4  5  6  7  8  9  10   11  12  13  14  15  16  17  18  19  20   21  22  23  24  25  26  27

1.证明 μ [ n ] \mu[n] μ[n]的 D T F T DTFT DTFT为 X ( e j w ) = 1 1 − e − j w + ∑ k = − ∞ ∞ π δ ( w + 2 π k ) X(e^{jw})=\frac{1}{1-e^{-jw}}+\sum_{k=-\infty}^{\infty}\pi\delta(w+2\pi k) X(ejw)=1−e−jw1​+∑k=−∞∞​πδ(w+2πk)。 解：   如果直接按照定义去求的话，会发现是无穷级数不是收敛的，所以该级数一定有狄拉克函数的形式。   将 μ [ n ] \mu[n] μ[n]分为奇偶两部分，则 x e v [ n ] = 1 2 ( u [ n ] + u [ − n ] ) = 1 2 + 1 2 δ [ n ] x_{ev}[n]=\frac{1}{2}(u[n]+u[-n])=\frac{1}{2}+\frac{1}{2}\delta[n] xev​[n]=21​(u[n]+u[−n])=21​+21​δ[n] x o d [ n ] = 1 2 ( u [ n ] − u [ − n ] ) = 1 2 ( 2 μ [ n ] − ( μ [ n ] + μ [ − n ] ) ) = μ [ n ] − 1 2 − 1 2 δ [ n ] x_{od}[n]=\frac{1}{2}(u[n]-u[-n])=\frac{1}{2}(2\mu[n]-(\mu[n]+\mu[-n]))=\mu[n]-\frac{1}{2}-\frac{1}{2}\delta[n] xod​[n]=21​(u[n]−u[−n])=21​(2μ[n]−(μ[n]+μ[−n]))=μ[n]−21​−21​δ[n] 则偶数部分的DTFT为 X e v ( e j w ) = ∑ k = − ∞ ∞ π δ ( w + 2 π k ) + 1 2 X_{ev}(e^{jw})=\sum_{k=-\infty}^{\infty}\pi\delta(w+2\pi k)+\frac{1}{2} Xev​(ejw)=k=−∞∑∞​πδ(w+2πk)+21​ 因为奇数部分中含 μ [ n ] \mu[n] μ[n],稍微变一下形 x o d [ n ] − x o d [ n − 1 ] = 1 2 ( δ [ n ] + δ [ n − 1 ] ) ⇒ ( 1 − e − j w ) X o d ( e − j w ) = 1 2 ( 1 + e − j w ) ⇒ X o d ( e j w ) = 1 2 1 + e − j w 1 − e − j w = − 1 2 + 1 1 − e − j w x_{od}[n]-x_{od}[n-1]=\frac{1}{2}(\delta[n]+\delta[n-1]) \\ \Rightarrow (1-e^{-jw})X_{od}(e^{-jw})=\frac{1}{2}(1+e^{-jw}) \\ \Rightarrow X_{od}(e^{jw})=\frac{1}{2}\frac{1+e^{-jw}}{1-e^{-jw}}=-\frac{1}{2}+\frac{1}{1-e^{-jw}} xod​[n]−xod​[n−1]=21​(δ[n]+δ[n−1])⇒(1−e−jw)Xod​(e−jw)=21​(1+e−jw)⇒Xod​(ejw)=21​1−e−jw1+e−jw​=−21​+1−e−jw1​ 则 X ( e j w ) = X e v ( e j w ) + X o d ( e j w ) = 1 1 − e − j w + ∑ k = − ∞ ∞ π δ ( w + 2 π k ) \color{red}X(e^{jw})=X_{ev}(e^{jw})+X_{od}(e^{jw})=\frac{1}{1-e^{-jw}}+\sum_{k=-\infty}^{\infty}\pi\delta(w+2\pi k) X(ejw)=Xev​(ejw)+Xod​(ejw)=1−e−jw1​+k=−∞∑∞​πδ(w+2πk) 返回目录

2.证明序列 x [ n ] = 1 x[n]=1 x[n]=1的 D T F T DTFT DTFT为 X ( e j w ) = ∑ k = − ∞ ∞ 2 π δ ( w + 2 π k ) ,   − ∞ ; k ; ∞ X(e^{jw})=\sum_{k=-\infty}^{\infty}2\pi\delta(w+2\pi k),\, -\infty ; k ; \infty X(ejw)=∑k=−∞∞​2πδ(w+2πk),−∞α∣n∣,0,​∣n∣≤M其他​ X 6 ( e j w ) = ∑ n = − M M α ∣ n ∣ e − j w n = ∑ n = − M − 1 α − n e − j w n + ∑ n = 0 M α n e − j w n = ∑ m = 1 M ( α e j w ) m + ∑ n = 0 M ( α e − j w ) n = α e j w ( 1 − ( α e j w ) M ) 1 − α e j w + 1 − ( α e − j w ) M + 1 1 − α e − j w \color{red} \begin{aligned} X_6(e^{jw});=\sum_{n=-M}^{M}\alpha^{\vert n \vert}e^{-jwn}\\ ;=\sum_{n=-M}^{-1}\alpha^{-n}e^{-jwn}+\sum_{n=0}^{M}\alpha^{n}e^{-jwn}\\ ;=\sum_{m=1}^{M}(\alpha e^{jw})^m +\sum_{n=0}^{M}(\alpha e^{-jw})^n\\ ;=\frac{\alpha e^{jw}(1-(\alpha e^{jw})^M)}{1-\alpha e^{jw}}+\frac{1-(\alpha e^{-jw})^{M+1}}{1-\alpha e^{-jw}} \end{aligned} X6​(ejw)​=n=−M∑M​α∣n∣e−jwn=n=−M∑−1​α−ne−jwn+n=0∑M​αne−jwn=m=1∑M​(αejw)m+n=0∑M​(αe−jw)n=1−αejwαejw(1−(αejw)M)​+1−αe−jw1−(αe−jw)M+1​​ 返回目录

7.求下面每个序列的 D T F T DTFT DTFT: a. x a [ n ] = μ [ n + 2 ] − μ [ n − 3 ] x_a[n]=\mu[n+2]-\mu[n-3] xa​[n]=μ[n+2]−μ[n−3] b. x b [ n ] = α n ( μ [ n − 1 ] − μ [ n − 3 ] ) , ∣ α ∣ ; 1 x_b[n]=\alpha^n(\mu[n-1]-\mu[n-3]),\vert \alpha\vert;1 xb​[n]=αn(μ[n−1]−μ[n−3]),∣α∣1,0≤n≤N0,其他​ c. y 3 [ n ] = { 1 − ∣ n ∣ N , − N ≤ n ≤ N 0 , 其 他 y_3[n]=\begin{cases}1-\frac{\vert n\vert}{N}, -N\leq n\leq N\\ 0,其他\end{cases} y3​[n]={1−N∣n∣​,−N≤n≤N0,其他​ d. y 4 [ n ] = { N + 1 − ∣ n ∣ , − N ≤ n ≤ N 0 , 其 他 y_4[n]=\begin{cases}N+1-\vert n\vert, -N\leq n\leq N\\ 0,其他\end{cases} y4​[n]={N+1−∣n∣,−N≤n≤N0,其他​ e. y 5 [ n ] = { c o s ( π n / 2 N ) , − N ≤ n ≤ N 0 , 其 他 y_5[n]=\begin{cases}cos(\pi n/2N), -N\leq n\leq N\\ 0,其他\end{cases} y5​[n]={cos(πn/2N),−N≤n≤N0,其他​ 解: a Y 1 ( e j w ) = ∑ n = − N N e − j w n = e j w N ( 1 − ( e − j w ) 2 N + 1 ) 1 − e − j w = e j w N − e − j w ( N + 1 ) 1 − e − j w = e − j w / 2 ( e j w ( N + 1 / 2 ) − e − j w ( N + 1 / 2 ) ) e − j w / 2 ( e j w / 2 − e − j w / 2 ) = s i n ( N + 1 / 2 ) w s i n ( w / 2 ) \begin{aligned} Y_1(e^{jw});=\sum_{n=-N}^{N}e^{-jwn}=\frac{e^{jwN}(1-(e^{-jw})^{2N+1})}{1-e^{-jw}}\\ ;=\frac{e^{jwN}-e^{-jw(N+1)}}{1-e^{-jw}}\\ ;=\frac{e^{-jw/2}(e^{jw(N+1/2)}-e^{-jw(N+1/2)})}{e^{-jw/2}(e^{jw/2}-e^{-jw/2})}\\ ;=\frac{sin(N+1/2)w}{sin(w/2)} \end{aligned} Y1​(ejw)​=n=−N∑N​e−jwn=1−e−jwejwN(1−(e−jw)2N+1)​=1−e−jwejwN−e−jw(N+1)​=e−jw/2(ejw/2−e−jw/2)e−jw/2(ejw(N+1/2)−e−jw(N+1/2))​=sin(w/2)sin(N+1/2)w​​ b Y 2 ( e j w ) = ∑ n = 0 N e − j w n = 1 − e − j w ( N + 1 ) 1 − e − j w \begin{aligned} Y_2(e^{jw})=\sum_{n=0}^{N}e^{-jwn}=\frac{1-e^{-jw(N+1)}}{1-e^{-jw}} \end{aligned} Y2​(ejw)=n=0∑N​e−jwn=1−e−jw1−e−jw(N+1)​​ c 考虑信号 y [ n ] = { 1 , − N / 2 ≤ n ≤ N / 2 0 , 其 他 y[n]=\begin{cases}1,-N/2 \leq n \leq N/2\\ 0,其他\end{cases} y[n]={1,−N/2≤n≤N/20,其他​ 易知， y 3 [ n ] = y [ n ] ∗ y [ n ] y_3[n]=y[n]*y[n] y3​[n]=y[n]∗y[n],所以 Y 3 ( e j w ) = Y 2 ( e j w ) Y_3(e^{jw})=Y^{2}(e^{jw}) Y3​(ejw)=Y2(ejw),而由a知 Y ( e j w ) = s i n ( N + 1 ) w / 2 s i n ( w / 2 ) Y(e^{jw})=\frac{sin(N+1)w/2}{sin(w/2)} Y(ejw)=sin(w/2)sin(N+1)w/2​ 所以 Y 3 ( e j w ) = s i n 2 ( N + 1 ) w / 2 s i n 2 ( w / 2 ) Y_3(e^{jw})=\frac{sin^2(N+1)w/2}{sin^2(w/2)} Y3​(ejw)=sin2(w/2)sin2(N+1)w/2​ d y 4 [ n ] = N y 3 [ n ] + y 1 [ n ] y_4[n]=Ny_3[n]+y_1[n] y4​[n]=Ny3​[n]+y1​[n],所以 Y 4 ( e j w ) = N Y 3 ( e j w ) + Y 1 ( e j w ) = N s i n 2 ( N + 1 ) w / 2 s i n 2 ( w / 2 ) + s i n ( N + 1 / 2 ) w s i n ( w / 2 ) \begin{aligned} Y_4(e^{jw});=NY_3(e^{jw})+Y_1(e^{jw})\\ ;=N\frac{sin^2(N+1)w/2}{sin^2(w/2)}+\frac{sin(N+1/2)w}{sin(w/2)} \end{aligned} Y4​(ejw)​=NY3​(ejw)+Y1​(ejw)=Nsin2(w/2)sin2(N+1)w/2​+sin(w/2)sin(N+1/2)w​​ e y 5 [ n ] = 1 2 ( e j π n / 2 N + e − j π n / 2 N ) , − N ≤ n ≤ N = 1 2 ( e j π n / 2 N y 1 [ n ] + e − j π n / 2 N y 1 [ n ] ) \begin{aligned} y_5[n];=\frac{1}{2}(e^{j\pi n/2N}+e^{-j\pi n/2N}),-N\leq n\leq N \\ ;=\frac{1}{2}(e^{j\pi n/2N}y_1[n]+e^{-j\pi n/2N}y_1[n])\end{aligned} y5​[n]​=21​(ejπn/2N+e−jπn/2N),−N≤n≤N=21​(ejπn/2Ny1​[n]+e−jπn/2Ny1​[n])​ 所以 Y 5 ( e j w ) = 1 2 ( Y 1 ( e j ( w − π n / 2 N ) ) + Y ( e j ( w + π n / 2 N ) ) ) = 1 2 ( s i n ( N + 1 / 2 ) ( w − π n / 2 N ) s i n ( ( w − π n / 2 N ) / 2 ) + s i n ( N + 1 / 2 ) ( w + π n / 2 N ) s i n ( ( w + π n / 2 N ) / 2 ) ) \begin{aligned} Y_5(e^{jw});=\frac{1}{2}(Y_1(e^{j(w-\pi n/2N)})+Y_(e^{j(w+\pi n/2N)}))\\ ;=\frac{1}{2}(\frac{sin(N+1/2)(w-\pi n/2N)}{sin((w-\pi n/2N)/2)}+\frac{sin(N+1/2)(w+\pi n/2N)}{sin((w+\pi n/2N)/2)}) \end{aligned} Y5​(ejw)​=21​(Y1​(ej(w−πn/2N))+Y(​ej(w+πn/2N)))=21​(sin((w−πn/2N)/2)sin(N+1/2)(w−πn/2N)​+sin((w+πn/2N)/2)sin(N+1/2)(w+πn/2N)​)​ 返回目录

9.证明 X ( e j w ) = 1 ( 1 − α e − j w ) m , ∣ α ∣ n ; 1 X(e^{jw})=\frac{1}{(1-\alpha e^{-jw})^m},\vert \alpha \vert^n;1 X(ejw)=(1−αe−jw)m1​,∣α∣n1+2δ[n],−N≤n≤N2δ[n]=0,其他​ d X d ( e j w ) = α e − j w ( 1 − α e − j w ) 2 = j d d w ( 1 1 − α e − j w ) X_d(e^{jw})=\frac{\alpha e^{-jw}}{(1-\alpha e^{-jw})^2}=j\frac{d}{dw}(\frac{1}{1-\alpha e^{-jw}}) Xd​(ejw)=(1−αe−jw)2αe−jw​=jdwd​(1−αe−jw1​) 所以 x d [ n ] = n α n μ [ n ] x_d[n]=n\alpha^n\mu[n] xd​[n]=nαnμ[n] 返回目录

11.求下面每个 D T F T DTFT DTFT逆变换: a. H a ( e j w ) = s i n ( 4 w ) H_a(e^{jw})=sin(4w) Ha​(ejw)=sin(4w) b. H b ( e j w ) = c o s ( 4 w ) H_b(e^{jw})=cos(4w) Hb​(ejw)=cos(4w) c. H c ( e j w ) = s i n ( 5 w ) H_c(e^{jw})=sin(5w) Hc​(ejw)=sin(5w) d. H d ( e j w ) = c o s ( 5 w ) H_d(e^{jw})=cos(5w) Hd​(ejw)=cos(5w) 解： a s i n ( 4 w ) = 1 2 j ( e j 4 w − e − j 4 w ) sin(4w)=\frac{1}{2j}(e^{j4w}-e^{-j4w}) sin(4w)=2j1​(ej4w−e−j4w) 所以 h a [ n ] = 1 2 j ( δ [ n + 4 ] − δ [ n − 4 ] ) h_a[n]=\frac{1}{2j}(\delta[n+4]-\delta[n-4]) ha​[n]=2j1​(δ[n+4]−δ[n−4]) b c o s ( 4 w ) = 1 2 ( e j 4 w + e − j 4 w ) cos(4w)=\frac{1}{2}(e^{j4w}+e^{-j4w}) cos(4w)=21​(ej4w+e−j4w) 所以 h b [ n ] = 1 2 ( δ [ n + 4 ] + δ [ n − 4 ] ) h_b[n]=\frac{1}{2}(\delta[n+4]+\delta[n-4]) hb​[n]=21​(δ[n+4]+δ[n−4]) c   同a的解法一致 d   同b的解法一致

12.求下面每个 D T F T DTFT DTFT逆变换: a. H 1 ( e j w ) = 1 + 2 c o s w + 3 c o s 2 w H_1(e^{jw})=1+2cosw+3cos2w H1​(ejw)=1+2cosw+3cos2w b. H 2 ( e j w ) = ( 3 + 2 c o s w + 4 c o s 2 w ) c o s ( w / 2 ) e − j w / 2 H_2(e^{jw})=(3+2cosw+4cos2w)cos(w/2)e^{-jw/2} H2​(ejw)=(3+2cosw+4cos2w)cos(w/2)e−jw/2 c. H 3 ( e j w ) = j ( 3 + 4 c o s w + 2 c o s 2 w ) s i n w H_3(e^{jw})=j(3+4cosw+2cos2w)sinw H3​(ejw)=j(3+4cosw+2cos2w)sinw d H 4 ( e j w ) = j ( 4 + 2 c o s w + 3 c o s 2 w ) s i n ( w / 2 ) e j w / 2 H_4(e^{jw})=j(4+2cosw+3cos2w)sin(w/2)e^{jw/2} H4​(ejw)=j(4+2cosw+3cos2w)sin(w/2)ejw/2 解 a H 1 ( e j w ) = 3 2 e − j 2 w + e − j w + 1 + e j w + 3 2 e j 2 w H_1(e^{jw})=\frac{3}{2}e^{-j2w}+e^{-jw}+1+e^{jw}+\frac{3}{2}e^{j2w} H1​(ejw)=23​e−j2w+e−jw+1+ejw+23​ej2w 所以 h 1 [ n ] = 3 2 δ [ n + 2 ] + δ [ n + 1 ] + δ [ n ] + δ [ n − 1 ] + 3 2 δ [ n − 2 ] h_1[n]=\frac{3}{2}\delta[n+2]+\delta[n+1]+\delta[n]+\delta[n-1]+\frac{3}{2}\delta[n-2] h1​[n]=23​δ[n+2]+δ[n+1]+δ[n]+δ[n−1]+23​δ[n−2] b c o s ( w / 2 ) e − j w / 2 = 1 2 ( 1 − e − j w ) cos(w/2)e^{-jw/2}=\frac{1}{2}(1-e^{-jw}) cos(w/2)e−jw/2=21​(1−e−jw) 设 H 2 ′ ( e j w ) = 3 + 2 c o s w + 4 c o s 2 w H^{'}_2(e^{jw})=3+2cosw+4cos2w H2′​(ejw)=3+2cosw+4cos2w,则 H 2 ( e j w ) = 1 2 H 2 ′ ( e j w ) ( 1 − e − j w ) = 1 2 ( H 2 ′ ( e j w ) − H 2 ′ ( e j w ) e − j w ) H_2(e^{jw})=\frac{1}{2}H^{'}_2(e^{jw})(1-e^{-jw})=\frac{1}{2}(H^{'}_2(e^{jw})-H^{'}_2(e^{jw})e^{-jw}) H2​(ejw)=21​H2′​(ejw)(1−e−jw)=21​(H2′​(ejw)−H2′​(ejw)e−jw) 仿照a的解法知 h 2 ′ [ n ] = ( 2 δ [ n + 2 ] + δ [ n + 1 ] + 3 δ [ n ] + δ [ n − 1 ] + 2 δ [ n − 2 ] ) h^{'}_2[n]=(2\delta[n+2]+\delta[n+1]+3\delta[n]+\delta[n-1]+2\delta[n-2]) h2′​[n]=(2δ[n+2]+δ[n+1]+3δ[n]+δ[n−1]+2δ[n−2]) 所以 h 2 [ n ] = h 2 ′ [ n ] − h 2 ′ [ n − 1 ] h_2[n]=h^{'}_2[n]-h^{'}_2[n-1] h2​[n]=h2′​[n]−h2′​[n−1] c和d的解法同上。 返回目录

13.证明 D T F T DTFT DTFT的如下定理: (a)线性定理 (b)时间反转定理 (c)时移定理 (d)频移定理 解：请参考这篇文章 返回目录

14.计算并作图表示级联LTI离散时间系统的 D T F T DTFT DTFT。其冲激响应分别为双边序列 h 1 [ n ] = δ [ n ] − s i n w 1 n π n h_1[n]=\delta[n]-\cfrac{sinw_1n}{\pi n} h1​[n]=δ[n]−πnsinw1​n​以及 h 2 [ n ] = s i n w 2 n π n , 0 ; w 1 ; w 2 ; π h_2[n]=\cfrac{sinw_2n}{\pi n},0;w_1;w_2;\pi h2​[n]=πnsinw2​n​,0n2,0,​−N≤n≤N其他​ c. x 3 [ n ] = s i n w c n π n x_3[n]=\cfrac{sinw_cn}{\pi n} x3​[n]=πnsinwc​n​ d. x 4 [ n ] = { 0 , n 为 偶 数 2 π n , n 为 奇 数 x_4[n]=\begin{cases}0,;n为偶数\\ \cfrac{2}{\pi n},;n为奇数\end{cases} x4​[n]=⎩⎨⎧​0,πn2​,​n为偶数n为奇数​ e. x 5 [ n ] = { 0 , n = 0 c o s π n n , ∣ n ∣ ; 0 x_5[n]=\begin{cases}0,;n=0\\ \cfrac{cos\pi n}{n},;\vert n\vert;0\end{cases} x5​[n]=⎩⎨⎧​0,ncosπn​,​n=0∣n∣>0​ 解： 偶序列对应实值 D T F T DTFT DTFT,b,c是偶序列，有实值 D T F T DTFT DTFT 奇序列对应虚值 D T F T DTFT DTFT,a,d,e是虚序列，对应虚值 D T F T DTFT DTFT 返回目录

20.不计算 D T F T DTFT DTFT,确定下面的 D T F T DTFT DTFT中，哪个有偶序列的逆，哪个有奇序列的逆: a. Y 1 ( e j w ) = { ∣ w ∣ , 0 ≤ ∣ w ∣ ≤ w c 0 , 其 他 Y_1(e^{jw})=\begin{cases}\vert w \vert,;0\leq \vert w \vert \leq w_c\\ 0,;其他\end{cases} Y1​(ejw)={∣w∣,0,​0≤∣w∣≤wc​其他​ b. Y 2 ( e j w ) = j w , 0 ≤ ∣ w ∣ ≤ π Y_2(e^{jw})=jw,0\leq \vert w \vert \leq \pi Y2​(ejw)=jw,0≤∣w∣≤π c. Y 3 ( e j w ) = { j , − π ; w ; 0 − j , 0 ; w ; π Y_3(e^{jw})=\begin{cases}j,;-\pi ; w ; 0\\ -j,;0 ; w ; \pi\end{cases} Y3​(ejw)={j,−j,​−π2,3,−1,0,−4,3,1,2,4},−2≤n≤6 其 D T F T DTFT DTFT为 X ( e j w ) X(e^{jw}) X(ejw)。不通过自身变换来计算下面的 X ( e j w ) X(e^{jw}) X(ejw)函数: a. X ( e j 0 ) X(e^{j0}) X(ej0) b. X ( e j π ) X(e^{j\pi}) X(ejπ) c. ∫ − π π X ( e j w ) d w \int_{-\pi}^{\pi}X(e^{jw})dw ∫−ππ​X(ejw)dw d. ∫ − π π ∣ X ( e j w ) ∣ 2 d w \int_{-\pi}^{\pi}\vert X(e^{jw})\vert^2dw ∫−ππ​∣X(ejw)∣2dw e. ∫ − π π ∣ d X ( e j w ) d w ∣ 2 d w \int_{-\pi}^{\pi}\vert\cfrac{dX(e^{jw})}{dw}\vert^2dw ∫−ππ​∣dwdX(ejw)​∣2dw 解： a X ( e j 0 ) = ∑ n = − ∞ ∞ x [ n ] X(e^{j0})=\sum_{n=-\infty}^{\infty}x[n] X(ej0)=n=−∞∑∞​x[n] b X ( e j π ) = ∑ n = − ∞ ∞ ( − 1 ) n x [ n ] X(e^{j\pi})=\sum_{n=-\infty}^{\infty}(-1)^nx[n] X(ejπ)=n=−∞∑∞​(−1)nx[n] c ∫ − π π X ( e j w ) d w = 2 π x [ 0 ] \int_{-\pi}^{\pi}X(e^{jw})dw=2\pi x[0] ∫−ππ​X(ejw)dw=2πx[0] d ∫ − π π ∣ X ( e j w ) ∣ 2 d w = 2 π ∑ n = − ∞ ∞ ∣ x [ n ] ∣ 2 \int_{-\pi}^{\pi}\vert X(e^{jw})\vert^2dw=2 \pi\sum_{n=-\infty}^{\infty}\vert x[n]\vert^2 ∫−ππ​∣X(ejw)∣2dw=2πn=−∞∑∞​∣x[n]∣2 e ∫ − π π ∣ d X ( e j w ) d w ∣ 2 d w = 2 π ∑ n = − ∞ ∞ ∣ n x [ n ] ∣ 2 \int_{-\pi}^{\pi}\vert\cfrac{dX(e^{jw})}{dw}\vert^2dw=2\pi \sum_{n=-\infty}^{\infty}\vert nx[n]\vert^2 ∫−ππ​∣dwdX(ejw)​∣2dw=2πn=−∞∑∞​∣nx[n]∣2 返回目录

25. a.序列 x [ n ] x[n] x[n]的时间延时常常通过其重心来度量，重心定义为 C g = ∑ n = − ∞ ∞ n x [ n ] ∑ n = − ∞ ∞ x [ n ] C_g=\cfrac{\sum_{n=-\infty}^{\infty}nx[n]}{\sum_{n=-\infty}^{\infty}x[n]} Cg​=∑n=−∞∞​x[n]∑n=−∞∞​nx[n]​ 试以 x [ n ] x[n] x[n]的 D T F T DTFT DTFT的 X ( e j w ) X(e^{jw}) X(ejw)来表示 C g C_g Cg​ b.计算序列 x [ n ] = α n μ [ n ] x[n]=\alpha^n\mu[n] x[n]=αnμ[n]的重心。 解： a ∑ n = − ∞ ∞ x [ n ] = X ( e j 0 ) = X ( e j w ) ∣ w = 0 \sum_{n=-\infty}^{\infty}x[n]=X(e^{j0})=X(e^{jw})\vert_{w=0} n=−∞∑∞​x[n]=X(ej0)=X(ejw)∣w=0​ 同理 ∑ n = − ∞ ∞ n x [ n ] = j d X ( e j w ) d w ∣ w = 0 \sum_{n=-\infty}^{\infty}nx[n]=j\frac{dX(e^{jw})}{dw}\vert_{w=0} n=−∞∑∞​nx[n]=jdwdX(ejw)​∣w=0​ 所以 C g = j d X ( e j w ) d w ∣ w = 0 X ( e j w ) ∣ w = 0 C_g=\frac{j\cfrac{dX(e^{jw})}{dw}\vert_{w=0}}{X(e^{jw})\vert_{w=0}} Cg​=X(ejw)∣w=0​jdwdX(ejw)​∣w=0​​ b X ( e j w ) = 1 1 − α e − j w ⇒ X ( e j 0 ) = 1 1 − α X(e^{jw})=\frac{1}{1-\alpha e^{-jw}} \Rightarrow X(e^{j0})=\frac{1}{1-\alpha} X(ejw)=1−αe−jw1​⇒X(ej0)=1−α1​ j d X ( e j w ) d w = α e − j w ( 1 − α e − j w ) 2 ⇒ j d X ( e j w ) d w ∣ w = 0 = α ( 1 − α ) 2 j\cfrac{dX(e^{jw})}{dw}=\frac{\alpha e^{-jw}}{(1-\alpha e^{-jw})^2} \Rightarrow j\cfrac{dX(e^{jw})}{dw}\vert_{w=0}=\frac{\alpha}{(1-\alpha)^2} jdwdX(ejw)​=(1−αe−jw)2αe−jw​⇒jdwdX(ejw)​∣w=0​=(1−α)2α​ 所以 C g = α 1 − α C_g=\frac{\alpha}{1-\alpha} Cg​=1−αα​ 返回目录

26.令 y [ n ] y[n] y[n]表示两个序列 h [ n ] h[n] h[n]和 x [ n ] x[n] x[n]的线性卷积，即 y [ n ] = x [ n ] ∗ h [ n ] y[n]=x[n]*h[n] y[n]=x[n]∗h[n]。证明: a ∑ n = − ∞ ∞ y [ n ] = ( ∑ n = − ∞ ∞ x [ n ] ) ( ∑ n = − ∞ ∞ h [ n ] ) \sum_{n=-\infty}^{\infty}y[n]=(\sum_{n=-\infty}^{\infty}x[n])(\sum_{n=-\infty}^{\infty}h[n]) n=−∞∑∞​y[n]=(n=−∞∑∞​x[n])(n=−∞∑∞​h[n]) b ∑ n = − ∞ ∞ ( − 1 ) n y [ n ] = ( ∑ n = − ∞ ∞ ( − 1 ) n x [ n ] ) ( ∑ n = − ∞ ∞ ( − 1 ) n h [ n ] ) \sum_{n=-\infty}^{\infty}(-1)^ny[n]=(\sum_{n=-\infty}^{\infty}(-1)^nx[n])(\sum_{n=-\infty}^{\infty}(-1)^nh[n]) n=−∞∑∞​(−1)ny[n]=(n=−∞∑∞​(−1)nx[n])(n=−∞∑∞​(−1)nh[n]) 解： Y ( e j w ) = X ( e j w ) H ( e j w ) ⇒ ∑ n = − ∞ ∞ y [ n ] e − j w n = ( ∑ n = − ∞ ∞ x [ n ] e − j w n ) ( ∑ n = − ∞ ∞ h [ n ] e − j w n ) Y(e^{jw})=X(e^{jw})H(e^{jw})\Rightarrow \sum_{n=-\infty}^{\infty}y[n]e^{-jwn}=(\sum_{n=-\infty}^{\infty}x[n]e^{-jwn})(\sum_{n=-\infty}^{\infty}h[n]e^{-jwn}) Y(ejw)=X(ejw)H(ejw)⇒n=−∞∑∞​y[n]e−jwn=(n=−∞∑∞​x[n]e−jwn)(n=−∞∑∞​h[n]e−jwn) a 令 w = 0 w=0 w=0得到 ∑ n = − ∞ ∞ y [ n ] = ( ∑ n = − ∞ ∞ x [ n ] ) ( ∑ n = − ∞ ∞ h [ n ] ) \sum_{n=-\infty}^{\infty}y[n]=(\sum_{n=-\infty}^{\infty}x[n])(\sum_{n=-\infty}^{\infty}h[n]) n=−∞∑∞​y[n]=(n=−∞∑∞​x[n])(n=−∞∑∞​h[n]) b 令 w = π w=\pi w=π ∑ n = − ∞ ∞ ( − 1 ) n y [ n ] = ( ∑ n = − ∞ ∞ ( − 1 ) n x [ n ] ) ( ∑ n = − ∞ ∞ ( − 1 ) n h [ n ] ) \sum_{n=-\infty}^{\infty}(-1)^ny[n]=(\sum_{n=-\infty}^{\infty}(-1)^nx[n])(\sum_{n=-\infty}^{\infty}(-1)^nh[n]) n=−∞∑∞​(−1)ny[n]=(n=−∞∑∞​(−1)nx[n])(n=−∞∑∞​(−1)nh[n]) 返回目录

27.设 x [ n ] x[n] x[n]表示 D T F T DTFT DTFT为 X ( e j w ) X(e^{jw}) X(ejw)的绝对可和的因果实序列。若 X r e ( e j w ) X_{re}(e^{jw}) Xre​(ejw)和 X i m ( e j w ) X_{im}(e^{jw}) Xim​(ejw)表示 X ( e j w ) X(e^{jw}) X(ejw)的实部和虚部，试证明它们的关系为 X i m ( e j w ) = 1 2 π ∫ − π π X r e ( e j v ) c o t ( w − v 2 ) d v X_{im}(e^{jw})=\frac{1}{2\pi}\int_{-\pi}^{\pi}X_{re}(e^{jv})cot(\frac{w-v}{2})dv Xim​(ejw)=2π1​∫−ππ​Xre​(ejv)cot(2w−v​)dv X r e ( e j w ) = 1 2 π ∫ − π π X i m ( e j v ) c o t ( w − v 2 ) d v + x [ 0 ] X_{re}(e^{jw})=\frac{1}{2\pi}\int_{-\pi}^{\pi}X_{im}(e^{jv})cot(\frac{w-v}{2})dv+x[0] Xre​(ejw)=2π1​∫−ππ​Xim​(ejv)cot(2w−v​)dv+x[0] 上面的等式称为离散希尔伯特变换关系。 解： 若 x [ n ] x[n] x[n]为因果实序列，则下列表达式成立 (1) x [ n ] = 2 x e v [ n ] μ [ n ] − x e v [ 0 ] δ [ n ] x[n]=2x_{ev}[n]\mu[n]-x_{ev}[0]\delta[n] \tag{1} x[n]=2xev​[n]μ[n]−xev​[0]δ[n](1) (2) x [ n ] = 2 x o d [ n ] μ [ n ] + x [ 0 ] δ [ n ] x[n]=2x_{od}[n]\mu[n]+x[0]\delta[n] \tag{2} x[n]=2xod​[n]μ[n]+x[0]δ[n](2) 首先对(1)两边进行傅里叶变换，得到 X ( e j w ) = 1 π ∫ − π π X r e ( e j v ) U ( e j ( w − v ) ) d v − x e v [ 0 ] X(e^{jw})=\frac{1}{\pi}\int_{-\pi}^{\pi}X_{re}(e^{jv})U(e^{j(w-v)})dv-x_{ev}[0] X(ejw)=π1​∫−ππ​Xre​(ejv)U(ej(w−v))dv−xev​[0] 由于 U ( e j w ) = 1 1 − e − j w + π ∑ k = − ∞ ∞ δ ( w + 2 π k ) = 1 2 ( 1 − j c o t ( w 2 ) ) + π ∑ k = − ∞ ∞ δ ( w + 2 π k ) U(e^{jw})=\frac{1}{1-e^{-jw}}+\pi\sum_{k=-\infty}^{\infty}\delta(w+2\pi k)=\frac{1}{2}(1-jcot(\frac{w}{2}))+\pi\sum_{k=-\infty}^{\infty}\delta(w+2\pi k) U(ejw)=1−e−jw1​+πk=−∞∑∞​δ(w+2πk)=21​(1−jcot(2w​))+πk=−∞∑∞​δ(w+2πk) 其中 1 1 − e − j w = 1 e − j w / 2 ( e j w / 2 − e − j w / 2 ) = 1 2 − j c o s ( w / 2 ) + s i n ( w / 2 ) s i n ( w / 2 ) = 1 2 ( 1 − j c o t ( w 2 ) ) \frac{1}{1-e^{-jw}}=\frac{1}{e^{-jw/2}(e^{jw/2}-e^{-jw/2})}=\frac{1}{2}\frac{-jcos(w/2)+sin(w/2)}{sin(w/2)}=\frac{1}{2}(1-jcot(\frac{w}{2})) 1−e−jw1​=e−jw/2(ejw/2−e−jw/2)1​=21​sin(w/2)−jcos(w/2)+sin(w/2)​=21​(1−jcot(2w​)) 则 X ( e j w ) = 1 2 π ∫ − π π X r e ( e j v ) d v − j 1 2 π ∫ − π π X r e ( e j v ) c o t ( w − v 2 ) d v + X r e ( e j w ) − x r e [ 0 ] X(e^{jw})=\frac{1}{2\pi}\int_{-\pi}^{\pi}X_{re}(e^{jv})dv-j\frac{1}{2\pi}\int_{-\pi}^{\pi}X_{re}(e^{jv})cot(\frac{w-v}{2})dv+X_{re}(e^{jw})-x_{re}[0] X(ejw)=2π1​∫−ππ​Xre​(ejv)dv−j2π1​∫−ππ​Xre​(ejv)cot(2w−v​)dv+Xre​(ejw)−xre​[0] 根据 X ( e j w ) = X r e ( e j w ) + j X i m ( e j w ) ⇒ X i m ( e j w ) = − 1 2 π ∫ − π π X r e ( e j v ) c o t ( w − v 2 ) d v X(e^{jw})=X_{re}(e^{jw})+jX_{im}(e^{jw}) \Rightarrow X_{im}(e^{jw})=-\frac{1}{2\pi}\int_{-\pi}^{\pi}X_{re}(e^{jv})cot(\frac{w-v}{2})dv X(ejw)=Xre​(ejw)+jXim​(ejw)⇒Xim​(ejw)=−2π1​∫−ππ​Xre​(ejv)cot(2w−v​)dv 对(2)两边进行傅里叶变换，得到 X ( e j w ) = j π ∫ − π π X i m ( e j v ) U ( e j ( w − v ) ) d v + x [ 0 ] X(e^{jw})=\frac{j}{\pi}\int_{-\pi}^{\pi}X_{im}(e^{jv})U(e^{j(w-v)})dv+x[0] X(ejw)=πj​∫−ππ​Xim​(ejv)U(ej(w−v))dv+x[0] 代入 U ( e j w ) = 1 2 ( 1 − j c o t ( w 2 ) ) + π ∑ k = − ∞ ∞ δ ( w + 2 π k ) U(e^{jw})=\frac{1}{2}(1-jcot(\frac{w}{2}))+\pi\sum_{k=-\infty}^{\infty}\delta(w+2\pi k) U(ejw)=21​(1−jcot(2w​))+πk=−∞∑∞​δ(w+2πk) X ( e j w ) = j 2 π ∫ − π π X i m ( e j v ) + 1 2 π ∫ − π π X i m ( e j w ) c o t ( w − v 2 ) d v + j X i m ( e j w ) + x [ 0 ] X(e^{jw})=\frac{j}{2\pi}\int_{-\pi}^{\pi}X_{im}(e^{jv})+\frac{1}{2\pi}\int_{-\pi}^{\pi}X_{im}(e^{jw})cot(\frac{w-v}{2})dv+jX_{im}(e^{jw})+x[0] X(ejw)=2πj​∫−ππ​Xim​(ejv)+2π1​∫−ππ​Xim​(ejw)cot(2w−v​)dv+jXim​(ejw)+x[0] 根据 X ( e j w ) = X r e ( e j w ) + j X i m ( e j w ) ⇒ X r e ( e j w ) = 1 2 π ∫ − π π X i m ( e j v ) c o t ( w − v 2 ) d v + x [ 0 ] X(e^{jw})=X_{re}(e^{jw})+jX_{im}(e^{jw}) \Rightarrow X_{re}(e^{jw})=\frac{1}{2\pi}\int_{-\pi}^{\pi}X_{im}(e^{jv})cot(\frac{w-v}{2})dv+x[0] X(ejw)=Xre​(ejw)+jXim​(ejw)⇒Xre​(ejw)=2π1​∫−ππ​Xim​(ejv)cot(2w−v​)dv+x[0] 得证。 返回目录