【FZU-MU】EE406 Assignment1

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

在这里插入图片描述
F(z)z=b(z−p1)2(z−p2)=A1z−p1+A2z−p2+A3(z−p1)2=A1(z−p1)(z−p2)+A2(z−p1)2+A3(z−p2)(z−p1)2(z−p2)=(A1+A2)z2+[−A1(p1+p2)−2A2p1+A3]z+(A1p1p2+A2p12−p2A3)(z−p1)2(z−p2) \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} \\ zF(z)=(zp1)2(zp2)b=zp1A1+zp2A2+(zp1)2A3=(zp1)2(zp2)A1(zp1)(zp2)+A2(zp1)2+A3(zp2)=(zp1)2(zp2)(A1+A2)z2+[A1(p1+p2)2A2p1+A3]z+(A1p1p2+A2p12p2A3)
z=p1z = p_1z=p1,则A3(p1−p2)=bA_3(p_1-p_2) = bA3(p1p2)=b,显然A3=bp1−p2A_3 = \frac b{p_1-p_2}A3=p1p2b
{A1+A2=0−(p1+p2)A1−2p1A2+A3=0p1p2A1+p12A2−p2A3=b⇒{(p1+p2)A1+2p1A2=A3p2A1+p1A2=b+p2A3p1 \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. A1+A2(p1+p2)A12p1A2+A3p1p2A1+p12A2p2A3=0=0=b(p1+p2)A1+2p1A2p2A1+p1A2=A3=p1b+p2A3
化简得
{A1+A2=0p2A1+p1A2=b+p2A3p1 \left \{ \begin{aligned} A_1 + A_2 &= 0\\ p_2A_1+p_1A_2 &= \frac{b+p_2A_3}{p_1} \end{aligned} \right. A1+A2p2A1+p1A2=0=p1b+p2A3
解得
{A1=−b(p1−p2)2A2=b(p1−p2)2A3=bp1−p2 \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. A1A2A3=(p1p2)2b=(p1p2)2b=p1p2b

y(k)=A1p1k+A2p2k+A3kp1k−1 y(k) = A_1 p_1^k + A_2 p_2^k+ A_3kp_1^{k-1} y(k)=A1p1k+A2p2k+A3kp1k1

Q2

在这里插入图片描述
(1)
[z2F(z)−z2y0−zy1]+a1[zF(z)−zy(0)]+a2F(z)=zz−b0(z2+a1z+a2)F(z)=zz−b0+z2y(0)+z[y(1)+y(0)] \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] [z2F(z)z2y0zy1]+a1[zF(z)zy(0)]+a2F(z)=zb0z(z2+a1z+a2)F(z)=zb0z+z2y(0)+z[y(1)+y(0)]
代入数据a1,a2a_1,a_2a1,a2,则z2+a1z+a2z^2 + a_1z + a_2z2+a1z+a2有零点p1,p2p_1,p_2p1,p2,所以
F(z)=y0z3+(y0+y1−b0y0)z2+[1−b0(y0+y1)]z(z−b0)(z−p1)(z−p2)=y0z3+(y0+y1−b0y0)z2+[1−b0(y0+y1)]zz3−(p1+p2+b0)z2+(p1p2+p1b0+p2b0)z−p1p2b0 F(z) = \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} F(z)=(zb0)(zp1)(zp2)y0z3+(y0+y1b0y0)z2+[1b0(y0+y1)]z=z3(p1+p2+b0)z2+(p1p2+p1b0+p2b0)zp1p2b0y0z3+(y0+y1b0y0)z2+[1b0(y0+y1)]z
(2)
F(z)z=az2+bz+c(z−p1)(z−p2)(z−p3)=A1z−p1+A2z−p2+A3z−p3=A1(z−p2)(z−p3)+A2(z−p1)(z−p3)+A3(z−p1)(z−p2)(z−p1)(z−p2)(z−p3) \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} zF(z)=(zp1)(zp2)(zp3)az2+bz+c=zp1A1+zp2A2+zp3A3=(zp1)(zp2)(zp3)A1(zp2)(zp3)+A2(zp1)(zp3)+A3(zp1)(zp2)
z=p1z = p_1z=p1,则A1(p1−p2)(p1−p3)=ap12+bp1+cA_1(p_1-p_2)(p_1-p_3) = ap_1^2+bp_1+cA1(p1p2)(p1p3)=ap12+bp1+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)}A1=(p1p2)(p1p3)ap12+bp1+c
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)}A2=(p2p1)(p2p3)ap22+bp2+c,A3=(p3p1)(p3p2)ap32+bp3+c
所以
y(k)=A1p1k+A2p2k+A3p3k y(k) = A_1 p_1^k + A_2 p_2^k + A_3 p_3^k y(k)=A1p1k+A2p2k+A3p3k
(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(∣p1∣,∣p2∣,∣b0∣<1|p_1|,|p_2|,|b_0|<1p1,p2,b0<1),则稳定;
  • 不稳定,如果任何极点是在单位圆外(∣p1∣,∣p2∣,∣b0∣>1|p_1|,|p_2|,|b_0|>1p1,p2,b0>1)或单位圆上有一个重复的极点(存在∣pi∣=1,且p1=p2或pi=b0存在|p_i|=1,且p_1=p_2或p_i=b_0存在pi=1,且p1=p2pi=b0);
  • 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

在这里插入图片描述
在这里插入图片描述

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);
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