【FZU-MU】EE406 Assignment1

数学：人类精神虐待(ﾟДﾟ)ﾉ

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分类专栏：福州带专，梅院职高文章标签：笔记

于 2023-11-04 09:06:11 首次发布

本文链接：https://blog.youkuaiyun.com/Gauss_code/article/details/134213450

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这篇博客详细解答了EE406课程作业中的五个问题，涉及Z变换的展开与简化，系统函数的表示，以及系统的稳定性条件。在Q1中，解析了Z变换的系数求解；Q2部分讨论了带有特定零点的系统函数表达式；Q3至Q5继续深入探讨了系统的其他方面。通过这个作业，读者可以深入理解Z变换在数字信号处理中的应用。

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Q1

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$\begin{aligned} \frac{F(z)}{z} &= \frac {b}{(z-p_1)^2(z-p_2)} \\ &= \frac{A_1}{z-p_1} + \frac{A_2}{z-p_2} + \frac{A_3}{(z-p_1)^2} \\ &= \frac {A_1(z-p_1)(z-p_2) + A_2(z-p_1)^2 + A_3(z-p_2)}{(z-p_1)^2(z-p_2)} \\ &= \frac {(A_1+ A_2)z^2 + [-A_1(p_1+p_2)-2A_2p_1+A_3]z + (A_1p_1p_2+A_2p_1^2-p_2A_3)}{(z-p_1)^2(z-p_2)} \\ \end{aligned} \\$
令 $z = p_1$ ，则 $A_3(p_1-p_2) = b$ ，显然 $A3=bp1−p2A_3 = \frac b{p_1-p_2}$
又 $\left \{ \begin{aligned} A_1+ A_2 &= 0 \\ -(p_1+p_2)A_1-2p_1A_2+A_3 &= 0\\ p_1p_2A_1+p_1^2A_2-p_2A_3 &= b \end{aligned} \right. \Rightarrow \left \{ \begin{aligned} (p_1+p_2)A_1 + 2p_1A_2 &= A_3\\ p_2A_1+p_1A_2 &= \frac{b+p_2A_3}{p_1} \end{aligned} \right.$
化简得
$\left \{ \begin{aligned} A_1 + A_2 &= 0\\ p_2A_1+p_1A_2 &= \frac{b+p_2A_3}{p_1} \end{aligned} \right.$
解得
$\left \{ \begin{aligned} A_1 &= -\frac{b}{(p_1-p_2)^2} \\ A_2 &= \frac{b}{(p_1-p_2)^2}\\ A_3 &= \frac b{p_1-p_2} \end{aligned} \right.$
故
$y(k) = A_1 p_1^k + A_2 p_2^k+ A_3kp_1^{k-1}$

Q2

在这里插入图片描述
(1)
$\left [ z^2 F(z) - z^2y_0 - zy_1 \right ] + a_1\left [ z F(z) - zy(0) \right ] + a_2F(z) = \frac{z}{z-b_0} \\ (z^2 + a_1z + a_2)F(z) = \frac{z}{z-b_0} + z^2y(0) + z\left[y(1) +y(0) \right]$
代入数据 $a_1,a_2$ ，则 $z^2 + a_1z + a_2$ 有零点 $p_1,p_2$ ，所以
$\frac{y_0z^3 + (y_0+y_1-b_0y_0)z^2 + [1-b_0(y_0+y_1)]z}{(z-b_0)(z -p_1)(z -p_2)} \\ = \frac{y_0z^3 + (y_0+y_1-b_0y_0)z^2 + [1-b_0(y_0+y_1)]z}{z^3 -(p_1+p_2+b_0)z^2 + (p_1p_2 + p_1 b_0 + p_2 b_0 ) z -p_1p_2 b_0}$
(2)
$\begin{aligned} \frac{F(z)}{z} &= \frac{az^2+bz+c}{(z-p_1)(z -p_2)(z -p_3)} \\ &= \frac{A_1}{z-p_1} + \frac{A_2}{z-p_2} + \frac{A_3}{z-p_3} \\ &= \frac {A_1(z-p_2)(z-p_3) + A_2(z-p_1)(z-p_3) + A_3(z-p_1)(z-p_2)}{(z-p_1)(z-p_2)(z-p_3)} \\ \end{aligned}$
令 $z = p_1$ ，则 $A_1(p_1-p_2)(p_1-p_3) = ap_1^2+bp_1+c$ ，显然 $A1=ap12+bp1+c(p1−p2)(p1−p3)A_1 = \frac {ap_1^2+bp_1+c}{(p_1-p_2)(p_1-p_3)}$
同理可得， $A2=ap22+bp2+c(p2−p1)(p2−p3),A3=ap32+bp3+c(p3−p1)(p3−p2)A_2 = \frac {ap_2^2+bp_2+c}{(p_2-p_1)(p_2-p_3)},A_3 = \frac {ap_3^2+bp_3+c}{(p_3-p_1)(p_3-p_2)}$
所以
$y(k) = A_1 p_1^k + A_2 p_2^k + A_3 p_3^k$
(3) 通过通项式和递推式求解
(4)

如果所有极点都在单位圆内( $p_1|,|p_2|,|b_0|<1$ )，则稳定；
不稳定，如果任何极点是在单位圆外( $p_1|,|p_2|,|b_0|>1$ )或单位圆上有一个重复的极点( $存在|p_i|=1，且p_1=p_2或p_i=b_0$ )；
如果一个或多个不重复的极点在单位圆上，并且所有其他极点都在单位圆内，则边际稳定。

Q3

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Q4

%%The area for parameter
b2 = 20;
b3 = 14;
a1 = 9.7;
a2 = 30;
a3 = 28;
n = 49;
T = 0.09;

%% Question 1 find the system discrete-time transfer function H(z).
Gs = tf([b2 b3],[1 a1 a2 a3]);
Gz = c2d(Gs,T,'zoh');
Dz =1;
Hz = Dz*Gz %the system discrete-time transfer function H(z)

%% Question 2 simulate and plot the unit step response for n samples.
t = 0:T:T*n;
e = ones(size(t));
y = lsim(Hz,e,t);
plot(t, e, '--g', t, y, '--r')
legend('e(k): input','y(k): output')

%% Question 3 Find the zeros, poles and DC gain of your open-loop system.
%pzmap(Hz) % plot the pole and zero on the z-plane which can used in the report
zp = zero(Hz); % zero point
pp = pole(Hz); % pole point
gain = dcgain(Hz); % DC gain

Q5

%% The area for parameter
b2 = 20;
b3 = 14;
a1 = 9.7;
a2 = 30;
a3 = 28;
n = 49;
T = 0.09;

%% Question 1 find the system discrete-time transfer function H(z).
Gs = tf([b2 b3],[1 a1 a2 a3]);
Gz = c2d(Gs,T,'zoh');
Dz = tf([2 -1],[1 -1],T);
Hz = feedback(Dz*Gz,1); %the system discrete-time transfer function H(z)

%% Question 2 simulate and plot the unit step response for n samples.
t = 0:T:T*n;
e = ones(size(t));
y = lsim(Hz,e,t);
plot(t, e, '--g', t, y, '--r')
legend('e(k): input','y(k): output')

%% Question 3 Find the zeros, poles and DC gain of your open-loop system.
pzmap(Hz) % plot the pole and zero on the z-plane which can used in the report
zp = zero(Hz);
pp = pole(Hz);
gain = dcgain(Hz);