Digital Signal Processing 2 Filtering Week1 习题

课程主页：https://www.coursera.org/learn/dsp2

这部分回顾第一周的习题。

1

(Difficulty: $\star$ ) Among the choices below, select all the linear systems. (Please note that some of the choices use functions rather than discrete-time signals; the concept of linearity is identical in both cases).

Second derivative, i.e.
$y(t)=\frac{d^{2} }{d t^{2} } x(t)$
The DTFT, i.e. transform a sequence $\mathbf{x}$ into $\text{DTFT} \{\mathbf{x}\}$
Scrambling, i.e. a permutation to the input sequence, e.g.:
AM radio modulation, i.e. multiply a signal $x[n]$ by a cosine at the carrier frequency:
$y[n]=x[n] \cos \left(2 \pi \omega_{c} n\right)$
Clipping, i.e. enforce a maximum signal amplitude $M, e . g:$
$y[n]=\left\{\begin{array}{ll}x[n] & , x[n] \leq M \\ M & , \text { otherwise }\end{array}\right.$
Envelope detection (via squaring), i.e. $y[n]=|x[n]|^{2} * h[n]$ ,where $h[n]$ is the impulse response of a lowpass filter such as the moving average filter.
Time-stretch, i.e. $y(t)=x(\alpha t)$ ,e.g. if you play an old $\mathrm{LP}-45$ vinyl disc at $33 \mathrm{rpm}$, the time-stretch coefficient would be $\alpha=33 / 45$

$\begin{aligned} \mathcal{H}\left\{\alpha x_{1}(t)+\beta x_{2}(t)\right\} &=\frac {d^2}{d t^2} (\alpha x_{1}(t)+\beta x_{2}(t))\\ &= \alpha \frac {d^2}{d t^2}x_{1}(t) +\beta \frac {d^2}{d t^2}x_{2}(t) \\ &= \alpha \mathcal H\{x_1(t)\}+ \beta \mathcal H\{x_2(t)\} \end{aligned}$

所以有线性性。

$\begin{aligned} \text{DTFT} \{ \alpha \mathbf{x}_1[n] + \beta \mathbf{x}_2[n] \} &=\sum_{n=-\infty}^{\infty} \left(\alpha \mathbf{x}_1[n] + \beta \mathbf{x}_2[n]\right)e^{-j \omega n}\\ &=\alpha \sum_{n=-\infty}^{\infty} \mathbf{x}_1[n] e^{-j \omega n} +\beta \sum_{n=-\infty}^{\infty} \mathbf{x}_2[n] e^{-j \omega n}\\ &=\alpha\ \text{DTFT} \{ \mathbf{x}_1[n]\}+ \beta\ \text{DTFT} \{ \mathbf{x}_2[n]\} \end{aligned}$

所以有线性性。

不难看出有线性性。

$\begin{aligned} \mathcal{H}\left\{\alpha x_{1}[n]+\beta x_{2}[n]\right\} &=(\alpha x_{1}[n]+\beta x_{2}[n]) \cos \left(2 \pi \omega_{c} n\right)\\ &= \alpha x_1[n] \cos \left(2 \pi \omega_{c} n\right) +\beta x_2[n] \cos \left(2 \pi \omega_{c} n\right) \\ &= \alpha \mathcal H\{x_1[n]\}+ \beta \mathcal H\{x_2[n]\} \end{aligned}$

所以有线性性。

显然不具有线性性。

由于平方函数没有线性性，所以不具有线性性。

$\begin{aligned} \mathcal{H}\left\{k_1 x_{1}(t)+k_2 x_{2}(t)\right\} &=\left(k_1 x_{1}(\alpha t)+k_2 x_{2}(\alpha t)\right)\\ &=k_1 \mathcal H\{x_1(t)\}+ k_2 \mathcal H\{x_2(t)\} \end{aligned}$

所以有线性性。

2

(Difficulty: $\star$ ) Among the choices below, select all the time-invariant systems. (Please note that some of the choices use functions rather than discrete-time signals; the concept of time invariance is identical in both cases).

Second derivative, i.e.
$y(t)=\frac{d^{2} }{d t^{2} } x(t)$
The DTFT, i.e. transform a sequence $\mathbf{x}$ into $\text{DTFT} \{\mathbf{x}\}$
Scrambling, i.e. a permutation to the input sequence, e.g.:
AM radio modulation, i.e. multiply a signal $x[n]$ by a cosine at the carrier frequency:
$y[n]=x[n] \cos \left(2 \pi \omega_{c} n\right)$
Clipping, i.e. enforce a maximum signal amplitude $M, e . g:$
$y[n]=\left\{\begin{array}{ll}x[n] & , x[n] \leq M \\ M & , \text { otherwise }\end{array}\right.$
Envelope detection (via squaring), i.e. $y[n]=|x[n]|^{2} * h[n]$, where $h[n]$ is the impulse response of a lowpass filter such as the moving average filter.
Time-stretch, i.e. $y(t)=x(\alpha t)$, e.g. if you play an old $\mathrm{LP}-45$ vinyl disc at $33 \mathrm{rpm}$, the time-stretch coefficient would be $\alpha=33 / 45$

$\begin{aligned} \mathcal{H}\left\{x(t-t_0)\right\} &=\frac{d^{2} }{d t^{2} } x(t-t_0)\\ &=\frac{d^{2} }{d (t-t_0)^{2} } x(t-t_0)\\ &=y(t-t_0) \end{aligned}$

时不变。

$\begin{aligned} \text{DTFT} \{ {x}[n-n_0] \} &=\sum_{n=-\infty}^{\infty} {x}[n-n_0] e^{-j \omega n}\\ &=\sum_{n=-\infty}^{\infty} {x}[n] e^{-j \omega (n+n_0)}\\ &= e^{-j \omega n_0}\sum_{n=-\infty}^{\infty} {x}[n] e^{-j \omega n}\\ &\ne \sum_{n=-\infty}^{\infty} {x}[n-n_0] e^{-j \omega (n-n_0)}\\ &= y[n-n_0] \end{aligned}$

不是时不变。

$\begin{aligned} \mathcal{H}\left\{x[n-n_0]\right\} &=x[n-n_0] \cos \left(2 \pi \omega_{c} n\right)\\ &\neq x[n-n_0] \cos \left(2 \pi \omega_{c} (n -n_0)\right)\\ &= y[n-n_0] \end{aligned}$

不是时不变。

$\begin{aligned} \mathcal{H}\left\{x[n-n_0]\right\} &=\left\{\begin{array}{ll}x[n-n_0] & , x[n-n_0] \leq M \\ M & , \text { otherwise }\end{array}\right.\\ &= y[n-n_0] \end{aligned}$

是时不变。

$\mathcal{H}\left\{x[n]\right\}=\sum_{k=-\infty}^{\infty }|x[n-k]|^{2} h(k)$

所以

$\begin{aligned} \mathcal{H}\left\{x[n-n_0]\right\} &=\sum_{k=-\infty}^{\infty }|x[n-n_0-k]|^{2} h(k)\\ &= y[n-n_0] \end{aligned}$

是时不变。

$\begin{aligned} \mathcal{H}\left\{x(t-t_0)\right\} &=x(\alpha t -t_0)\\ &\neq x(\alpha(t-t_0))\\ &=y (t-t_0) \end{aligned}$

不是时不变。

3

(Difficulty: $\star$ ) The impulse response of a room can be recorded by producing a sharp noise (impulsive sound source) in a silent room, thereby capturing the scattering of the sound produced by the walls.
The impulse response $h[n]$ of Lausanne Cathedral was measured by Dokmanic et al. by recording the sound of balloons being popped ( hear it! ).

The balloon is popped at time $n=0$ and after a number of samples $N$, the reverberations die out, i.e. $h[n]=$ 0 for $n<0$ or $n>N$
The acoustic of this large space can then be artificially recreated by convolving any audio recording with the impulse response, e.g. this cello recording becomes this.
What are the properties of $h[n] ?$ (tick all the correct answers)

Anticausal
FIR
BIBO stable

只有在$0\le n \le N$时脉冲响应才有非零值，所以满足FIR,BIBO stable。

4

(Difficulty: $\star$ ) Let

$\begin{array}{l} h[n]=\delta[n]-\delta[n-1] \\ x[n]=\left\{\begin{array}{cc} 1 & , n \geq 0 \\ 0 & , \text { else } \end{array}\right. \\ y[n]=x[n] * h[n] \end{array}$

Compute $y[-1], y[0], y[1], y[2]$ and write the result as space-separated values. E.g.: If you find $y[-1]=-2$ $y[0]=-1, y[1]=0, y[2]=1,$ you should enter

解：

$\begin{aligned} y[n] &=x[n] * h[n]\\ &= x[n] * (\delta[n]-\delta[n-1] )\\ &=x[n] *\delta[n] - x[n] *\delta[n-1]\\ &= x[n] - x[n-1] \end{aligned}$

所以

$\begin{aligned} y[-1]&= x[-1]-x[-2]= 0\\ y[0]&= x[0]-x[-1]= 1\\ y[1]&= x[1]-x[0]= 0\\ y[2]&= x[2]-x[1]= 0\\ \end{aligned}$

所以结果为

0 1 0 0

5

(Difficulty: $\star \star)$ Consider the filter $h[n]=\delta[n]-\delta[n-1]$,the signal

$x[n]=\left\{\begin{array}{ll} n & , n \geq 0 \\ 0 & , \text { else } \end{array}\right.$

and the output $y[n]=x[n] * h[n]$.
Compute $y[-1], y[0], y[1], y[2]$ and write the result as space-separated values. E.g.: If you find $y[-1]=-2$ $y[0]=-1, y[1]=0, y[2]=1,$ you should enter

由上一题可得

$\begin{aligned} y[n] &= x[n] - x[n-1] \end{aligned}$

所以

$\begin{aligned} y[-1]&= x[-1]-x[-2]= 0\\ y[0]&= x[0]-x[-1]= 0\\ y[1]&= x[1]-x[0]= 1\\ y[2]&= x[2]-x[1]= 1\\ \end{aligned}$

结果为

0 0 1 1

6

(Difficulty: $\star \star \star)$ Which of the following filters are BIBO-stable?
Assume $N \in \mathbb{N}$ and $0<\omega_{c}<\pi$

The ideal low pass filter with a cutoff frequency $\omega_{c}: H\left(e^{j \omega}\right)=\left\{\begin{array}{ll}1 & |\omega| \leq \omega_{c} \\ 0 & \text { otherwise }\end{array}\right.$
The moving average: $h[n]=\frac{\delta[n]+\delta[n-1]}{2}$
The following smoothing filter: $h[n]=\sum_{k=0}^{\infty} \frac{1}{k+1} \delta[n-k]$
The filter $h[n]=\sum_{k=0}^{N-1} \delta[n-k] \sin \left(2 \pi \frac{k}{N}\right)$

由课件结论可得

$h[n]=\frac{\sin \omega_{C} n}{\pi n}$

所以不是BIBO

$\begin{aligned} \sum_{n}|h[n]| &= \sum_{n} \frac{\delta[n]+\delta[n-1]}{2}\\ &= \frac 1 2\sum_{n}\delta[n] + \frac 1 2\sum_{n}\delta[n-1]\\ &= 1 \end{aligned}$

所以BIBO

$\begin{aligned} \sum_{n}|h[n]| &= \sum_{n} \sum_{k=0}^{\infty} \frac{1}{k+1} \delta[n-k]\\ &=\sum_{k=0}^{\infty}\sum_{n}\frac{1}{k+1} \delta[n-k]\\ &= \sum_{k=0}^{\infty} \frac 1 {k+1}\\ &\to \infty \end{aligned}$

所以不是BIBO

$\begin{aligned} \sum_{n}|h[n]| &= \sum_{n} \sum_{k=0}^{N-1} \left|\delta[n-k] \sin \left(2 \pi \frac{k}{N}\right)\right|\\ &=\sum_{k=0}^{N-1}\sum_{n} \left|\delta[n-k] \sin \left(2 \pi \frac{k}{N}\right)\right|\\ &= \sum_{k=0}^{N-1} \left| \sin \left(2 \pi \frac{k}{N}\right)\right|\\ &\le N \end{aligned}$

$h[n]$绝对可和，所以BIBO

7

(Difficulty: $\star \star)$ Consider an LTI system $\mathcal{H}$. When the input to $\mathcal{H}$ is the following signal

then the output is

Assume now the input to $\mathcal{H}$ is the following signal

Which one of the following signals is the system’s output?

输入

$x[n]=\delta [n]- \delta[n - 1]$

输出

$y[n] = \delta[n - 1]+\delta[n + 1]-\delta [n]$

新的输入

$\begin{aligned} x_1[n] &= -2\delta[n + 1] + \delta[n]-2\delta[n - 1] + 3\delta[n - 2]\\ &= (-2\delta[n + 1] + 2\delta [n])-(\delta[n]-\delta[n-1]) -(3\delta[n - 1] - 3\delta[n - 2])\\ &= -2x[n+1]-x[n]-3x[n-1] \end{aligned}$

所以

$\begin{aligned} y_1[n] &= -2y[n+1]-y[n]-3y[n-1]\\ &= -2(\delta[n]+\delta[n + 2]-\delta [n+1])-(\delta[n - 1]+\delta[n + 1]-\delta [n]) -3(\delta[n - 2]+\delta[n]-\delta [n-1])\\ &= -2\delta[n + 2] +\delta[n+1]-4\delta[n]+2\delta[n-1]-3\delta[n-2] \end{aligned}$

所以选第三项

8

(Difficulty: $\star$ ) Consider the system shown below, consisting of a cosine modulator at frequency $\omega_{0}$ followed by an ideal bandpass filter $h[n]$ whose frequency response is also shown in the figure; assume that the input to the system is the signal $x[n],$ whose spectrum is shown below.

Determine the value of $\omega_{0} \in[0,2 \pi]$ that maximizes the energy of the output $y[n]$ when the input is $x[n]$
Remember that $\pi$ must be entered in the answer box as pi.

利用如下事实即可

$\begin{aligned} \text { DTFT }\left\{x[n] \cos \left(\omega_{c} n\right)\right\} &=\text { DTFT}\left\{\frac{1}{2} e^{j \omega_{c} n} x[n]+\frac{1}{2} e^{-j \omega_{c} n} x[n]\right\} \\ &=\frac{1}{2}\left[X\left(e^{j\left(\omega-\omega_{c}\right)}\right)+X\left(e^{j\left(\omega+\omega_{c}\right)}\right)\right] \end{aligned}$

所以结果为

0.5* pi

9

(Difficulty: $\star$ ) Consider a lowpass filter with the following frequency response.

What is the output $y[n]$ when the input to this filter is $x[n]=\cos \left(\frac{\pi}{5} n\right)+\sin \left(\frac{\pi}{4} n\right)+0.5 \cos \left(\frac{3 \pi}{4} n\right) ?$

利用

以及

$\begin{aligned} \cos(\omega n)&= \frac{e^{j\omega n}+e^{-j\omega n} }{2}\\ \sin(\omega n)&= \frac{e^{j\omega n}-e^{-j\omega n} }{2j} \end{aligned}$

可得

$\begin{aligned} y[n] &= \frac{H(\frac \pi 5)e^{j\frac \pi 5 n}+H(-\frac \pi 5)e^{-j\frac \pi 5 n} }{2} +\frac{H(\frac \pi 4)e^{j\frac \pi 4 n}-H(-\frac \pi 4)e^{-j\frac \pi 4 n} }{2j} \\ &= H(\frac \pi 5) \cos (\frac \pi 5 n) +H(\frac \pi 4)\sin \left(\frac{\pi}{4} n\right)\\ &=\frac 3 5 \cos (\frac \pi 5 n) +\frac 1 2 \sin \left(\frac{\pi}{4} n\right) \end{aligned}$

结果为

3 / 5* cos(pi / 5 * n)+1 /2*sin(pi / 4 * n)

10

(Difficulty: $\star$ ) Consider a filter with real-valued impulse response $h[n] .$ The filter is cascaded with another filter whose impulse response is $h^{\prime}[n]=h[-n],$ i.e. whose impulse response is the time-reversed version of $h[n]$

The cascade system can be seen as a single filter with impulse response $g[n]$
What is the phase of $G(e^{j \omega}) ?$

第一个系统

$\begin{aligned} x[n] &\stackrel{\mathrm{DTFT} }\longleftrightarrow H\left(e^{j \omega}\right)H^{\star}\left(e^{j \omega}\right) X\left(e^{j \omega}\right) =|H\left(e^{j \omega}\right)|^2 X\left(e^{j \omega}\right) \end{aligned}$

所以相位为$0$。

11

(Difficulty: $\star$ ) Let $x[n]=\cos \left(\frac{\pi}{2} n\right)$ and $h[n]=\frac{1}{5} \operatorname{sinc}\left(\frac{n}{5}\right) .$ Compute the convolution $y[n]=x[n] * h[n],$ and write the value of $y[5]$
Hint: First find the convolution result in the frequency domain.

考虑下式

$\operatorname{rect}\left(\frac{\omega}{2 \omega_{c} }\right) \quad \stackrel{\mathrm{DTFT} }{\longrightarrow} \quad \frac{\omega_{c} }{\pi} \operatorname{sinc}\left(\frac{\omega_{c} }{\pi} n\right)$

显然这里有

$\omega_c =\frac {\pi }5$

所以

$H\left(e^{j \omega}\right) =\operatorname{rect}\left( \frac{5\omega}{2 \pi}\right)$

另一方面

$\text{DTFT}\left\{\cos \omega_{0} n\right\}=\left[{\delta}\left(\omega-\omega_{0}\right)+{\delta}\left(\omega+\omega_{0}\right)\right] / 2$

所以做DTFT可得

$\begin{aligned} Y\left(e^{j \omega}\right) &=X\left(e^{j \omega}\right) H\left(e^{j \omega}\right)\\ &=\frac 1 2 \left[{\delta}\left(\omega-\frac \pi 2\right)+{\delta}\left(\omega+\frac \pi 2\right)\right] \operatorname{rect}\left( \frac{5\omega}{2 \pi}\right) \end{aligned}$

取逆变换可得

$\begin{aligned} h[n] &= \frac 1 {2\pi} \int_{-\pi}^\pi \frac 1 2 \left[{\delta}\left(\omega-\frac \pi 2\right)+{\delta}\left(\omega+\frac \pi 2\right)\right] \operatorname{rect}\left( \frac{5\omega}{2 \pi}\right) d\omega\\ &=0 \end{aligned}$

12

(Difficulty: $\star \star)$ Consider the system below, where $H\left(e^{j \omega}\right)$ is an ideal lowpass filter with cutoff frequency $\omega_{c}=$ $\pi / 4$

Consider two input signals to the system:

$x_{1}[n]$ is bandlimited to $[-\pi / 4, \pi / 4]$
$x_{2}[n]$ is band-limited to $[-\pi,-3 \pi / 4] \cup[3 \pi / 4, \pi]$
Which of the following statements is correct?

Which of the following statements is correct?

$x_{2}[n]$ is not modified by the system while $x_{1}[n]$ is eliminated.
Both $x_{1}[n]$ and $x_{2}[n]$ are not modified by the system.
$x_{1}[n]$ is not modified by the system while $x_{2}[n]$ is eliminated.
Both $x_{1}[n]$ and $x_{2}[n]$ are eliminated by the system.

注意

$(-1)^n = e^{j\pi n}$

所以

$\begin{aligned} \text { DTFT }\left\{x[n] (-1)^n\right\} &=\text { DTFT }\left\{x[n] e^{j\pi n}\right\} \\ &=X\left(e^{j\left(\omega-\pi\right)}\right) \end{aligned}$

因此$x_1$的频域属于

$[3\pi / 4, 5\pi / 4]$

$x_2$的频域属于

$[0, \pi / 4] \cup[7 \pi / 4, 2\pi]$

所以选择

$x_{2}[n]$ is not modified by the system while $x_{1}[n]$ is eliminated.

13

(Difficulty: \star) $x[n]$ and $y[n]$ are two square-summable signals in $\ell_{2}(\mathbb{Z}) ; X\left(e^{j \omega}\right)$ and $Y\left(e^{j \omega}\right)$ are their corresponding DTFTs.
We want to compute the value

$\sum_{n=-\infty}^{\infty} x[n] y^{*}[n]$

in terms of $X\left(e^{j \omega}\right)$ and $Y\left(e^{j \omega}\right) .$ Select the correct expression among the choices below.

$X\left(e^{j \omega}\right) * Y\left(e^{-j \omega}\right)$
$\frac{1}{2 \pi} \int_{-\pi}^{\pi} X\left(e^{j \omega}\right) Y^{*}\left(e^{j \omega}\right) d \omega$
$\frac{1}{2 \pi} X\left(e^{j \omega}\right) Y\left(e^{-j \omega}\right)$
$\frac{1}{2 \pi} \int_{-\pi}^{\pi} X\left(e^{j \omega}\right) Y\left(e^{-j \omega}\right) d \omega$
$\frac{1}{2 \pi} X\left(e^{j \omega}\right) Y^{*}\left(e^{j \omega}\right)$
$X\left(e^{j \omega}\right) Y\left(e^{-j \omega}\right)$

首先

$\begin{aligned} X\left(e^{j \omega}\right) &=\sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n}\\ Y^{*}\left(e^{j \omega}\right) &=\sum_{m=-\infty}^{\infty} y^{*}[m] e^{j\omega m}\\ \end{aligned}$

所以

$\begin{aligned} X\left(e^{j \omega}\right)Y^{*}\left(e^{j \omega}\right) &=\left(\sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} \right) \left(\sum_{m=-\infty}^{\infty} y^{*}[m] e^{j\omega m}\right)\\ &= \sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}x[n]y^{*}[m] e^{-j\omega (n- m)}\\ &= \sum_{n=-\infty}^{\infty} x[n] y^{*}[n] + \sum_{n=-\infty}^{\infty}\sum_{m\neq n} x[n] y^{*}[m]e^{-j\omega (n- m)} \end{aligned}$

两边关于$-\pi ,\pi$积分可得

$\frac{1}{2 \pi} \int_{-\pi}^{\pi} X\left(e^{j \omega}\right) Y^{*}\left(e^{j \omega}\right) d \omega = \sum_{n=-\infty}^{\infty} x[n] y^{*}[n]$

14

(Difficulty: $\star \star \star) h[n]$ is the impulse response of an ideal lowpass filter with cutoff frequency $\omega_{c}<\frac{\pi}{2} .$ Select the correct description for the system represented in the following figure?

Hint: Use the trigonometric identity $\cos (x)^{2}=\frac{1}{2}(1+\cos (2 x))$

A highpass filter with gain $\frac{1}{2}$ and cutoff frequency $\pi-\omega_{c}$
A lowpass filter with gain 1 and cutoff frequency $\omega_{c}$
A highpass filter with gain $\frac{1}{4}$ and cutoff frequency $\omega_{c}$
A lowpass filter with gain 1 and cutoff frequency $\omega_{c} / 2$
A highpass filter with gain 1 and pass band $\left[\omega_{c}, \pi-\omega_{c}\right]$
A lowpass filter with gain $\frac{1}{2}$ and cutoff frequency $2 \omega_{c}$

注意到

$\left(\cos \left(\frac{\pi}{2} n\right)\right)^{2}=\frac{1}{2}(1+\cos (\pi n))$

以及

所以

$\begin{aligned} \text{DTFT}\left\{\left(\cos \left(\frac{\pi}{2} n\right)\right)^{2} h[n]\right\} &= \frac 1 2\text{DTFT}\left\{(1+\cos (\pi n)) h[n]\right\}\\ &=\frac 1 2 H(e^{j\omega}) + \frac{1}{4}\left[H\left(e^{j\left(\omega-\pi\right)}\right)+H\left(e^{j\left(\omega+\pi\right)}\right)\right] \end{aligned}$

如果在通过一个脉冲响应为$h[n]$的系统，那么叠加后的滤波器的DTFT

$\frac 1 2 H(e^{j\omega})$

所以根据上图，总滤波器的DTFT为

$\frac{1}{4}\left[H\left(e^{j\left(\omega-\pi\right)}\right)+H\left(e^{j\left(\omega+\pi\right)}\right)\right]$

不难看出该滤波器为高通滤波器，所以选择

A highpass filter with gain $\frac{1}{2}$ and cutoff frequency $\pi-\omega_{c}$

备注：这里感觉系数应该是$\frac 1 4$

15

(Difficulty: $\star \star \star)$ Consider the following system, where $H\left(e^{j \omega}\right)$ is a half-band filter, i.e. an ideal lowpass with cutoff frequency $\omega_{c}=\pi / 2$

Assume the input to the system is $x[n]=\delta[n]$. Compute
$\sum_{n=-\infty}^{\infty} y[n]$
Hint: Perform the derivations in the frequency domain.

由图可得

$\begin{aligned} y[n] &= (1 + (-1)^n) (x[n] * h[n])\\ &= (1 + e^{j\pi n}) (x[n] * h[n]) \end{aligned}$

另一方面，我们有

$H\left(e^{j \omega}\right)=\operatorname{rect}\left(\frac{\omega}{\pi }\right)$

所以取DTFT可得

$\begin{aligned} Y(e^{j\omega}) &= X(e^{j\omega}) H(e^{j\omega}) + X(e^{j(\omega-\pi)}) H(e^{j(\omega-\pi)}) \\ \end{aligned}$

注意到

$Y(e^{j\omega}) =\sum_{n=-\infty}^{\infty} y[n]e^{-j\omega n}$

所以在前一个式子中取$\omega=0$可得

$\begin{aligned} Y(e^{j\omega}) |_{\omega=0} &= \sum_{n=-\infty}^{\infty} y[n]\\ &= X(e^{j\omega}) H(e^{j\omega}) + X(e^{j(\omega-\pi)}) H(e^{j(\omega-\pi)}) |_{\omega=0}\\ &=\sum_{n=-\infty}^{\infty}x[n] \end{aligned}$

所以如果$x[n]=\delta[n]$，那么

$\sum_{n=-\infty}^{\infty} y[n] = \sum_{n=-\infty}^{\infty} \delta[n] =1$