# Digital Signal Processing (A) 2022
## 1. Fill-in-the-blanks
### 1
The system function of an LTI system is $H(z)=(1+1.2z^{-1})/(1+0.1z^{-1}-0.06z^{-2})$ and thus its difference equation is $\underline{y[n]+0.1y[n-1]-0.06y[n-2]=x[n]+1.2x[n-1]}$. It is an IIR system. Please plot the parallel realization of the system in the following area.
$$
H(z)=\frac{-1.8}{1+0.3z^{-1}}+\frac{2.8}{1-0.2z^{-1}}
$$
### 2
Given two sequences $x_1(n)=\{1,2,-1\}$ and $x_2(n)=\{-1,3,2,1\}$, the linear convolution between them is $\underline{\{\underline{-1},1,9,2,0,-1\}}$ and the circular convolution $L=6$ is $\{\underline{-1},1,9,2,0,-1\}$.
### 3
Which of the following statements are correct?
- [ ] $y(n)=3x(n)+2$ is a linear system. (增量线性系统 )
- [ ] $y(n)=nx(2n)$ is a time-invariant system. (不是时不变系统)
- [ ] $y(n)=3^n u(n)$ is a stable and causal system. (不稳定)
- [x] $y(n)=\delta(n+4)$ is a stable system. (是稳定系统)
### 4
The system function of a causal LTI system is $H(z)=(3-4z^{-1})/(1-3.5z^{-1}+1.5z^{-2})$ and thus the ROC of $H(z)$ is $\underline{|z|>3}$.
零点为 $z=3,z=0.5$. 因为因果所以 ROC 必须包含 $\infin$，也就是 $|z|>3$,
### 5
Supposing that the system function of an analog low-pass filter is $H_a(s)$, we convert it to a digital filter using the relation $\displaystyle H(z)=H_a(s)|_{s=\frac{z+1}{z-1}}$ and thus the digital filter is high-pass.
### 6
The sequence $x(n)$ with period $N$ and its discrete Fourier transform is $X(k)$. Hence, the discrete Fourier transform of $x(n+l)$ can be expressed in terms of $X(k)$ as $\underline{X(k)W_N^{-lk}}$.
利用：
$$
X(k)=\sum_{n=0}^{N-1} x(n) W_{N}^{nk}
$$
可得：
$$
\sum_{n=0}^{N-1} x(n+l) W_{N}^{nk}=W_{N}^{-lk} X(k)
$$
### 7
Which of the following statements are correct?
- [x] 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）
- [x] 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）
- [x] If the impulse response function of a lowpass filter is $h(n)$, $(-1)^n h(n)$ is the impulse response function of a high-pass filter.（正确，相当于 $z\to -z$）
- [ ] 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 $z=\frac{1}{r}e^{\pm j\theta},r<1$, and thus the system function of the all-pass filter should be written as $\underline{H_{ap}(z)=\frac{z^{-1}-re^{-j\theta}}{1-re^{j\theta}z^{-1}}\cdot \frac{z^{-1}-re^{j\theta}}{1-r^{-j\theta}z^{-1}}}$.
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：
$$
H_{ap}(z)=K\frac{z^{-1}-a^*}{1-az^{-1}}\cdot \frac{z^{-1}-a}{1-a^*z^{-1}}
$$
零点为 $(a^*)^{-1}$ 和 $a^{-1}$. 对应 $z=\frac{1}{r}e^{\pm j\theta}$，取 $a=re^{j\theta}$.
因此可以设计全通滤波器：
$$
H_{ap}(z)=\frac{z^{-1}-re^{-j\theta}}{1-re^{j\theta}z^{-1}}\cdot \frac{z^{-1}-re^{j\theta}}{1-r^{-j\theta}z^{-1}}\\=
\frac{z^{-2}-2r\cos \theta z^{-1}+r^2}{1-2r\cos \theta z^{-1}+r^2 z^{-2}}
$$
### 9
A discrete-time sinusoid is periodic only if its frequency $f$ is a rational number.
### 10
If we sample $x_1(t)=\cos (2\pi(10)t)$ and $x_2(t)=\cos (2\pi(50)t)$ at a rate 40Hz, the sampled sequences are written as $x_1(n)=\underline{\cos(2\pi(0.25)n)}$ and $x_2(n)=\underline{\cos(2\pi(1.25)n)}$. We say they are 相同？. The sampling frequency should be increased to 100Hz at least such that we can reconstruct the signal $x_2(t)$ from the samples.
## 2. Calculation and Design
### 11
Determine the coefficients of a linear-phase FIR filter
$$
y(n)=b_0 x(n)+b_1x(n-1) +b_2(n-2)
$$
such that
1. It rejects completely a frequency component at $\omega_0=2\pi/3$;
2. Its frequency response is normalized so that $H(0)=1$;（$H(0)$ 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.
$$
Y(z)=b_0 X(z)+b_1 z^{-1} X(z)+b_2 z^{-2} X(z)
$$
$$
H(z)=\frac{Y(z)}{X(z)}=b_0+b_1z^{-1}+b_2 z^{-2}
$$
满足条件 1. 也就是
$$
H(e^{j\omega_0})=0\\
\Rightarrow b_0+b_1 e^{-j\omega_0}+b_2 e^{-j2\omega_0}=0
$$
满足 $b_0=b_1=b_2$.
满足条件 2. 也就是
$$
H(0)=H(e^{j0})=b_0+b_2+b_3=1
$$
因此 $b_0=b_1=b_2=1/3$，$H(z)=1/3(1+z^{-1}+z^{-2})$.
画出幅频响应和相频响应，注意相频响应在零点处需要 $\pm \pi$.
### 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 $x(n)=\delta(n)+\delta(n-1)$.
$$
X(k)=\sum_{n=0}^{3} x(n)W_{4}^{nk}
$$
- $X(0)=x(0)+x(1)=2$.
- $X(1)=x(0)+x(1)W_{4}^1=x(0)+jx(1)=1+j$.
- $X(2)=x(0)+x(1)W_{4}^2=x(0)-x(1)=0$.
- $X(3)=x(0)+x(1)W_{4}^3=1-j$.
### 13
Use Butterworth filter with impulse invariance to design a low-pass filter with
- Passband edge frequency: $0.2\pi$
Passband attenuation: $1\mathrm{~dB}$
- Stopband edge frequency: $0.3\pi$
Stopband attenuation: $10\mathrm{~dB}$
The sampling interval is $T=1\mathrm{~s}$.
根据 $\omega=\Omega T$ 可得 $\Omega_p=0.2\pi,\Omega_{st}=0.3\pi$，因为低通滤波器频带位于 $[-\pi/T,\pi/T]$ 之间，所以没有混叠。
计算巴特沃斯滤波器阶数和中心频率：
- $N\ge \lg \left.\left(\frac{10^{0.1A_s}-1}{10^{0.1R_p}-1}\right)\right/2\lg (\Omega_{st}/\Omega_p)$. 得到 $4.38$ 取 $N=5$.
- $\Omega_c=\Omega_p/\sqrt[2N]{10^{0.1R_p}-1}=0.719$.
确定归一化巴特沃斯滤波器 $H_{an}(s)$，去归一化得到 $H_{a}(s/\Omega_c)$.
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 $N=4$ and plot its signal flow graph.
$$
\begin{aligned}
X[k]&=\sum_{n=0}^{N-1} x[n] W_{N}^{nk}\\
&=\sum_{m=0}^{N/2-1} (x[2m] W_{N}^{2mk}+x[2m+1]W_{N}^{2mk+k})\\
&=X_1[k]+W_{N}^{k} X_2[k]
\end{aligned}
$$
$$
X[k+N/2]=X_1[k]-W_{N}^{k} X_2[k]
$$
2. Using the results in (1), plot the signal flow graph for the IFFT with $N=4$.
直接调换 $X,x$ 顺序，并且把 $W_{N}^{k}$ 改成 $W_{N}^{-k}$. 并且乘以 $1/N=1/4$.
3. Given the DFT for a 8-point sequence $x(n)$ is $X(k)=\{11,3,-5,3,3,3,-5,3\}$, use the result in (2) to calculate the sequence $x(2n)$.
假设 $X_1[k]$ 是 $x(2n)$ 的 DFT，$X_2[k]$ 是 $x(2n+1)$ 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，$X_1[k]=(X[k]+X[k+N/2])/2$
$$
X_1[k]=\{7,3,-5,3\}
$$
计算其 IDFT 为：
$$
\{2,3,-1,3\}
$$
### 15
Given the lowpass-to-bandpass transform in the analog domain is
$$
\overline{s}=\overline{\Omega}_p \frac{s^2+\Omega_{p2}\Omega_{p1}}{(\Omega_{p2}-\Omega_{p1})s}
$$
where $\overline{\Omega}_p$ is the passband edge frequency for the lowpass filter and $\Omega_{p1}$ and $\Omega_{p1}$ 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 $T=2$）：
$$
s=\frac{1-z^{-1}}{1+z^{-1}},\Omega=\tan\left(\frac{\omega}{2}\right)
$$
一般取归一化低通滤波器，$\overline{\Omega_p}=1$，令 $\Omega_{p0}=\sqrt{\Omega_{p1}\Omega_{p2}}$，$B_p=\Omega_{p2}-\Omega_{p1}$.
$$
\begin{aligned}
\overline{s}&=\frac{s^2+\Omega_{p0}^2}{B_ps}\\
&=\frac{(1-z^{-1})^2+\Omega_{p0}^2(1+z^{-1})^2}{B_p(1-z^{-2})}\\
&=\frac{1+\Omega_{p0}^2}{B_p}\frac{1-z^{-2}+2\frac{\Omega_{p0}^2-1}{\Omega_{p0}^2+1}z^{-1}}{1-z^{-2}}
\end{aligned}
$$
根据，
$$
\frac{\Omega_{p0}^2-1}{\Omega_{p0}^2+1}=\frac{\tan \left(\frac{\omega_1}{2}\right)\tan \left(\frac{\omega_2}{2}\right)-1}{\tan \left(\frac{\omega_1}{2}\right)\tan \left(\frac{\omega_2}{2}\right)+1}=\frac{\cos \left(\frac{\omega_1+\omega_2}{2}\right)}{\cos \left(\frac{\omega_1-\omega_2}{2}\right)}\doteq \cos \omega_0
$$
$$
\frac{1+\Omega_{p0}^2}{B_p}=\cot \left(\frac{\omega_2-\omega_1}{2}\right)
$$
带进去化简即可. 再代入 $z=e^{j\omega}$ 可以得到频率间的关系。