Digital Signal Processing 2 Filtering Week3 习题

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

这部分回顾第三周的习题。

1

(Difficulty: $\star$ ) Assume $x[n]$ is a WSS random process with power spectral density $P_{x}\left(e^{j \omega}\right) .$ Which of the following properties are true?

• $P_{x}\left(e^{j \omega}\right)$ is constant
• $P_{x}\left(e^{j \omega}\right)$ is symmetric around $\omega=0$
• $P_{x}\left(e^{j \omega}\right) \geq 0$
• $P_{x}\left(e^{j \omega}\right)$ is real-valued.

回顾定义

$$P_{X}\left(e^{j \omega}\right)=\sum_{k=-\infty}^{\infty} r_{X}[k] e^{-j \omega k}$$

所以显然有

$$P_{X}\left(e^{j \omega}\right)=P_{X}\left(e^{-j \omega}\right)$$

另一方面

$$r_X[k]=E[X[n+k]X[n]]=E[X[n]X[n+k]]=r_X[-k]$$

所以

$$\begin{aligned} P_{X}\left(e^{j \omega}\right) &=\sum_{k=-\infty}^{\infty} r_{X}[k] e^{-j \omega k}\\ &=r_{X}[0] + \sum_{k=1}^{\infty}r_{X}[k] (e^{-j \omega k} + e^{j \omega k})\\ &= r_{X}[0] + \sum_{k=1}^{\infty}2r_{X}[k] \cos (\omega k)\\ &\in \mathbb R \end{aligned}$$

最后，利用定义

$$P\left(e^{j \omega}\right)=\lim _{M \rightarrow \infty}\left\{\frac{1}{2 M+1}\left|X_{M}\left(e^{j \omega}\right)\right|^{2}\right\} \ge 0$$

所以选2,3,4

2

(Difficulty: $\star$ ) Consider a white noise process $w[n]$ with variance $\sigma_{w}^{2}=4 .$ The process goes through a filter with transfer function

$$H(z)=\frac{1}{1-\sqrt{3} z^{-1}}$$

to produce a WSS process $x[n]$。
Compute the value of the power spectral density of $x[n]$ in $\omega=\pi / 2$

由定义可得

$$\begin{aligned} P_{X}\left(e^{j \omega}\right) &=\left|H\left(e^{j \omega}\right)\right|^{2} \sigma_w ^2\\ &= \left | \frac{1}{1-\sqrt{3} (e^{j \omega})^{-1}}\right |^2 4 \end{aligned}$$

令$\omega = \pi /2$可得

$$P_{X}\left(e^{j \omega}\right) \big|_{\omega =\pi /2}=1$$

3

(Difficulty: $\star$) Consider the stochastic process defined as

$$[n]=x[n]+\beta x[n-1]$$

where $\beta \in \mathbb{R}$ and $x[n]$ is a zero-mean wide-sense stationary process with autocorrelation

$$r_{x}[k]=\sigma^{2} \alpha^{|k|}, \quad|\alpha| \leq 1$$

$y[n]$ can also be expressed as filtered version of $x[n]$ where the filter’s impulse response $h[n]$ is:

• $h[n]=\delta[n]-\beta \delta[n-1]$
• $h[n]=\delta[n]+\beta \delta[n+1]$
• $h[n]=\delta[n]+\beta \delta[n-1]$
• $h[n]=\delta[n]-\beta \delta[n+1]$
• $h[n]=\delta[n+1]+\beta \delta[n]$

利用定义

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

所以

$$\begin{aligned} h[0] & =1\\ h[1] &=\beta \\ h[n] &=0, n\neq 0,1 \end{aligned}$$

所以

$$h[n]=\delta[n]+\beta \delta[n-1]$$

4

(Difficulty: $\star \star \star)$ Using the same setup as in the previous question, select the correct expression for the power spectral density $P_{y}\left(e^{j \omega}\right)$

• $P_{y}\left(e^{j \omega}\right)=\sigma(1-\alpha) \frac{1+\beta+2 \cos (\omega)}{1+\alpha-2 \cos (\omega)}$
• $P_{y}\left(e^{j \omega}\right)=\sigma^{2}\left(1-\alpha^{2}\right) \frac{1+\beta^{2}+2 \beta \cos (\omega)}{1+\alpha^{2}-2 \alpha \cos (\omega)}$
• $P_{y}\left(e^{j \omega}\right)=\alpha(1-\sigma) \frac{1-\beta^{2}+2 \beta \cos (\omega)}{1+\alpha^{2}-2 \alpha \cos (\omega)}$
• $P_{y}\left(e^{j \omega}\right)=\frac{1+\beta^{2}+2 \beta \cos (\omega)}{1+\alpha^{2}-2 \alpha \cos (\omega)}$
• $P_{y}\left(e^{j \omega}\right)=\alpha^{2} \frac{\beta^{2}+2 \beta \cos (\omega)}{\alpha^{2}-2 \alpha \cos (\omega)}$
• $P_{y}\left(e^{j \omega}\right)=\sigma^{2} \frac{1+\alpha^{2}+2 \alpha \cos (\omega)}{1+\beta^{2}-2 \beta \cos (\omega)}$

首先：

$$\begin{aligned} H\left(e^{j w}\right)&= 1+\beta e^{-j\omega}\\ |H\left(e^{j w}\right)|^2&= (1+\beta \cos (\omega))^2 + \beta^2\sin^2(\omega)\\ &= 1+\beta^2 +2\beta \cos (\omega) \end{aligned}$$

其次：

$$\begin{aligned} P_{X}\left(e^{j \omega}\right) &=\sum_{k=-\infty}^{\infty} r_{X}[k] e^{-j \omega k}\\ &= \sum_{k=-\infty}^{\infty} \sigma^{2} \alpha^{|k|}e^{-j \omega k} \\ &= \sigma^2 +\sigma^2\sum_{k=1}^{\infty}\alpha^k e^{-j \omega k} + \sum_{k=1}^{\infty}\alpha^k e^{j \omega k}\\ &=\sigma^2 +\sigma^2 \left(\frac{\alpha e^{-j\omega}}{ 1- \alpha e^{-j\omega}}+ \frac{\alpha e^{j\omega}}{ 1-\alpha e^{j\omega}}\right)\\ &= \sigma^2 + \alpha\sigma^2 \frac{e^{-j\omega}(1-\alpha e^{j\omega})+e^{j\omega}(1-\alpha e^{-j\omega})}{(1-\alpha e^{-j\omega})(1-\alpha e^{j\omega})}\\ &= \sigma^2 + \alpha\sigma^2 \frac{2\cos (\omega) -2\alpha}{(1-\alpha \cos (\omega))^2 + \alpha^2 \sin^2(\omega)}\\ &= \sigma^2 +\sigma^2 \frac{2\alpha \cos(\omega) -2\alpha^2}{1+\alpha^2 -2\alpha \cos(\omega)}\\ &= \sigma^2\frac{1-\alpha^2}{1+\alpha^2 -2\alpha \cos(\omega)} \end{aligned}$$

因为

$$P_{Y}\left(e^{j \omega}\right)=\left|H\left(e^{j \omega}\right)\right|^{2} P_{X}\left(e^{j \omega}\right)$$

所以

$$P_{Y}\left(e^{j \omega}\right)= \sigma^2 (1-\alpha^2)\frac{1+\beta^2 +2\beta \cos (\omega)}{1+\alpha^2 -2\alpha \cos(\omega)}$$

5

(Difficulty: $\star \star)$ Using the same setup as in the previous questions, assume that the output $y[n]$ turns out to be a white noise sequence. Which of the following statements are necessarily true?

• The samples of $y[n]$ must be correlated.
• The samples of $y[n]$ must be uncorrelated.
• The power spectrum $P_{y}\left(e^{j \omega}\right)$ must be constant.
• $\beta=-\alpha$
• $\beta=\alpha$
• $\beta=2 \alpha$

因为$y[n]$是WSS，所以$P_{y}\left(e^{j \omega}\right)$为常数，因此选2,3,4