Digital Signal Processing (A) 2022

1. Fill-in-the-blanks

1

The system function of an LTI system is $H(z)=(1+1.2z^{-1})/(1+0.1z^{-1}-0.06z^{-2})$$H(z)=(1+1.2z^{-1})/(1+0.1z^{-1}-0.06z^{-2})$ and thus its difference equation is $\underset{―}{y[n]+0.1y[n-1]-0.06y[n-2]=x[n]+1.2x[n-1]}$$\underline{y[n]+0.1y[n-1]-0.06y[n-2]=x[n]+1.2x[n-1]}$. It is an IIR system. Please plot the parallel realization of the system in the following area.

2

Given two sequences $x_1(n)=\{1,2,-1\}$$x_1(n)=\{1,2,-1\}$ and $x_2(n)=\{-1,3,2,1\}$$x_2(n)=\{-1,3,2,1\}$, the linear convolution between them is $\underset{―}{\{\underset{―}{-1},1,9,2,0,-1\}}$$\underline{\{\underline{-1},1,9,2,0,-1\}}$ and the circular convolution $L=6$$L=6$ is $\{\underset{―}{-1},1,9,2,0,-1\}$$\{\underline{-1},1,9,2,0,-1\}$.

3

Which of the following statements are correct?

• $y(n)=3x(n)+2$$y(n)=3x(n)+2$ is a linear system. (增量线性系统 )

• $y(n)=nx(2n)$$y(n)=nx(2n)$ is a time-invariant system. (不是时不变系统)

• $y(n)=3^{n}u(n)$$y(n)=3^n u(n)$ is a stable and causal system. (不稳定)

• $y(n)=\delta (n+4)$$y(n)=\delta(n+4)$ is a stable system. (是稳定系统)

4

The system function of a causal LTI system is $H(z)=(3-4z^{-1})/(1-3.5z^{-1}+1.5z^{-2})$$H(z)=(3-4z^{-1})/(1-3.5z^{-1}+1.5z^{-2})$ and thus the ROC of $H(z)$$H(z)$ is $\underset{―}{|z|>3}$$\underline{|z|>3}$.

零点为 $z=3,z=0.5$$z=3,z=0.5$. 因为因果所以 ROC 必须包含 $\mathrm{\infty }$$\infin$，也就是 $|z|>3$$|z|>3$,

5

Supposing that the system function of an analog low-pass filter is $H_a(s)$$H_a(s)$, we convert it to a digital filter using the relation ${\displaystyle H(z)=H_a(s){|}_{s=\frac{z+1}{z-1}}}$$\displaystyle H(z)=H_a(s)|_{s=\frac{z+1}{z-1}}$ and thus the digital filter is high-pass.

6

The sequence $x(n)$$x(n)$ with period $N$$N$ and its discrete Fourier transform is $X(k)$$X(k)$. Hence, the discrete Fourier transform of $x(n+l)$$x(n+l)$ can be expressed in terms of $X(k)$$X(k)$ as $\underset{―}{X(k)W_N^{-lk}}$$\underline{X(k)W_N^{-lk}}$.

利用：

可得：

7

Which of the following statements are correct?

• The convolution of two minimum-phase sequence is always a minimum-phase sequence.（时域卷积对应频域点乘，零点和极点都位于单位圆内，正确）

• For a non-causal and stable system, when all zeros are outside the unit circle, the system is a minimum-phase system.（正确，注意是 non-causal）

• If the impulse response function of a lowpass filter is $h(n)$$h(n)$, $(-1{)}^{n}h(n)$$(-1)^n h(n)$ is the impulse response function of a high-pass filter.（正确，相当于 $z\to -z$$z\to -z$）

• Using a window to truncate a sequence induces frequency leakage.（错误，也有可能没有频谱泄露）

• It is easy for window method to determine the cutoff frequency exactly in design of an FIR filter.（错误）

• When we design a digital filter, all poles should be placed outside the unit circle.（希望稳定，错误）

8

We use an all-pass filter to stabilize a system with two unstable poles $z=\frac{1}{r}{e}^{\pm j\theta },r<1$$z=\frac{1}{r}e^{\pm j\theta},r<1$, and thus the system function of the all-pass filter should be written as $\underset{―}{H_{ap}(z)=\frac{z^{-1}-r{e}^{-j\theta }}{1-r{e}^{j\theta }{z}^{-1}}\cdot \frac{z^{-1}-r{e}^{j\theta }}{1-r^{-j\theta }{z}^{-1}}}$$\underline{H_{ap}(z)=\frac{z^{-1}-re^{-j\theta}}{1-re^{j\theta}z^{-1}}\cdot \frac{z^{-1}-re^{j\theta}}{1-r^{-j\theta}z^{-1}}}$.

全通滤波器极点位于单位圆内，零点位于单位圆外，零极点可以对消，设计全通滤波器：

零点为 $(a^{\ast }{)}^{-1}$$(a^*)^{-1}$ 和 $a^{-1}$$a^{-1}$. 对应 $z=\frac{1}{r}{e}^{\pm j\theta }$$z=\frac{1}{r}e^{\pm j\theta}$，取 $a=r{e}^{j\theta }$$a=re^{j\theta}$.

因此可以设计全通滤波器：

9

A discrete-time sinusoid is periodic only if its frequency $f$$f$ is a rational number.

10

If we sample $x_1(t)=\cos (2\pi (10)t)$$x_1(t)=\cos (2\pi(10)t)$ and $x_2(t)=\cos (2\pi (50)t)$$x_2(t)=\cos (2\pi(50)t)$ at a rate 40Hz, the sampled sequences are written as $x_1(n)=\underset{―}{\cos (2\pi (0.25)n)}$$x_1(n)=\underline{\cos(2\pi(0.25)n)}$ and $x_2(n)=\underset{―}{\cos (2\pi (1.25)n)}$$x_2(n)=\underline{\cos(2\pi(1.25)n)}$. We say they are 相同？. The sampling frequency should be increased to 100Hz at least such that we can reconstruct the signal $x_2(t)$$x_2(t)$ from the samples.

2. Calculation and Design

11

Determine the coefficients of a linear-phase FIR filter

such that

1. It rejects completely a frequency component at ${\omega }_0=2\pi /3$$\omega_0=2\pi/3$;

2. Its frequency response is normalized so that $H(0)=1$$H(0)=1$;（$H(0)$$H(0)$ 是幅度函数）

Also, compute and sketch the magnitude and phase response（幅频响应和相频响应） of the filter.

满足条件 1. 也就是

满足 $b_0=b_1=b_2$$b_0=b_1=b_2$.

满足条件 2. 也就是

因此 $b_0=b_1=b_2=1/3$$b_0=b_1=b_2=1/3$，$H(z)=1/3(1+z^{-1}+z^{-2})$$H(z)=1/3(1+z^{-1}+z^{-2})$.

画出幅频响应和相频响应，注意相频响应在零点处需要 $\pm \pi$$\pm \pi$.

12

Given the following signal graph for an FFT algorithm

1. The graph represents a decimation-in-time algorithm.

2. Complete the graph with arrowed lines and coefficients.

   

3. Using the above graph, compute the 4-point DFT of the sequence $x(n)=\delta (n)+\delta (n-1)$$x(n)=\delta(n)+\delta(n-1)$.

   $$X(k)=\sum _{n=0}^{3}x(n)W_4^{nk}$$

   • $X(0)=x(0)+x(1)=2$$X(0)=x(0)+x(1)=2$.

   • $X(1)=x(0)+x(1)W_4^1=x(0)+jx(1)=1+j$$X(1)=x(0)+x(1)W_{4}^1=x(0)+jx(1)=1+j$.

   • $X(2)=x(0)+x(1)W_4^2=x(0)-x(1)=0$$X(2)=x(0)+x(1)W_{4}^2=x(0)-x(1)=0$.

   • $X(3)=x(0)+x(1)W_4^3=1-j$$X(3)=x(0)+x(1)W_{4}^3=1-j$.

13

Use Butterworth filter with impulse invariance to design a low-pass filter with

• Passband edge frequency: $0.2\pi$$0.2\pi$

  Passband attenuation: $1\mathrm{dB}$$1\mathrm{~dB}$

• Stopband edge frequency: $0.3\pi$$0.3\pi$

  Stopband attenuation: $10\mathrm{dB}$$10\mathrm{~dB}$

The sampling interval is $T=1\mathrm{s}$$T=1\mathrm{~s}$.

根据 $\omega =\mathrm{\Omega }T$$\omega=\Omega T$ 可得 ${\mathrm{\Omega }}_p=0.2\pi ,{\mathrm{\Omega }}_{st}=0.3\pi$$\Omega_p=0.2\pi,\Omega_{st}=0.3\pi$，因为低通滤波器频带位于 $[-\pi /T,\pi /T]$$[-\pi/T,\pi/T]$ 之间，所以没有混叠。

计算巴特沃斯滤波器阶数和中心频率：

• $N\ge \lg \left(\frac{{10}^{0.1A_s}-1}{{10}^{0.1R_p}-1}\right)/2\lg ({\mathrm{\Omega }}_{st}/{\mathrm{\Omega }}_p)$$N\ge \lg \left.\left(\frac{10^{0.1A_s}-1}{10^{0.1R_p}-1}\right)\right/2\lg (\Omega_{st}/\Omega_p)$. 得到 $4.38$$4.38$ 取 $N=5$$N=5$.

• ${\mathrm{\Omega }}_c={\mathrm{\Omega }}_p/\sqrt[2N]{{10}^{0.1R_p}-1}=0.719$$\Omega_c=\Omega_p/\sqrt[2N]{10^{0.1R_p}-1}=0.719$.

确定归一化巴特沃斯滤波器 $H_{an}(s)$$H_{an}(s)$，去归一化得到 $H_a(s/{\mathrm{\Omega }}_c)$$H_{a}(s/\Omega_c)$.

使用冲激响应不变法，设计数字滤波器。

14

1. Derive the decimation-in-time FFT algorithm with $N=4$$N=4$ and plot its signal flow graph.

   $$\begin{array}{rl}X[k]& =\sum _{n=0}^{N-1}x[n]W_N^{nk}\\ & =\sum _{m=0}^{N/2-1}(x[2m]W_N^{2mk}+x[2m+1]W_N^{2mk+k})\\ & =X_1[k]+W_N^k{X}_2[k]\end{array}$$
   $$X[k+N/2]=X_1[k]-W_N^k{X}_2[k]$$

   

2. Using the results in (1), plot the signal flow graph for the IFFT with $N=4$$N=4$.

   直接调换 $X,x$$X,x$ 顺序，并且把 $W_N^k$$W_{N}^{k}$ 改成 $W_N^{-k}$$W_{N}^{-k}$. 并且乘以 $1/N=1/4$$1/N=1/4$.

3. Given the DFT for a 8-point sequence $x(n)$$x(n)$ is $X(k)=\{11,3,-5,3,3,3,-5,3\}$$X(k)=\{11,3,-5,3,3,3,-5,3\}$, use the result in (2) to calculate the sequence $x(2n)$$x(2n)$.

   假设 $X_1[k]$$X_1[k]$ 是 $x(2n)$$x(2n)$ 的 DFT，$X_2[k]$$X_2[k]$ 是 $x(2n+1)$$x(2n+1)$ 的 DFT.

   • 前一半有：

     $$X[k]=X_1[k]+W_N^k{X}_2[k]$$

   • 后一半有：

     $$X[k+N/2]=X_1[k]-W_N^k{X}_2[k]$$

   因此，$X_1[k]=(X[k]+X[k+N/2])/2$$X_1[k]=(X[k]+X[k+N/2])/2$

   $$X_1[k]=\{7,3,-5,3\}$$

   计算其 IDFT 为：

   $$\{2,3,-1,3\}$$

15

Given the lowpass-to-bandpass transform in the analog domain is

where ${\bar{\mathrm{\Omega }}}_p$$\overline{\Omega}_p$ is the passband edge frequency for the lowpass filter and ${\mathrm{\Omega }}_{p1}$$\Omega_{p1}$ and ${\mathrm{\Omega }}_{p1}$$\Omega_{p1}$ are the two passband edge frequencies for the bandpass filter, please derive the lowpass-to-bandpass transformation in the digital domain using bilinear transformation.

代入双线性变换（取 $T=2$$T=2$）：

一般取归一化低通滤波器，$\overline{{\mathrm{\Omega }}_p}=1$$\overline{\Omega_p}=1$，令 ${\mathrm{\Omega }}_{p0}=\sqrt{{\mathrm{\Omega }}_{p1}{\mathrm{\Omega }}_{p2}}$$\Omega_{p0}=\sqrt{\Omega_{p1}\Omega_{p2}}$，$B_p={\mathrm{\Omega }}_{p2}-{\mathrm{\Omega }}_{p1}$$B_p=\Omega_{p2}-\Omega_{p1}$.

根据，

带进去化简即可. 再代入 $z=e^{j\omega }$$z=e^{j\omega}$ 可以得到频率间的关系。