F ( z ) → H ( z ) x = 0 → Y f ( z ) F(z) \to \boxed{\underset{x=0}{H(z)}} \to Y_f(z) F(z)→x=0H(z)→Yf(z)
H ( z ) = Y z s ( z ) F ( z ) H(z) = \frac{Y_{zs}(z)}{F(z)} H(z)=F(z)Yzs(z)
h ( k ) ↔ H ( z ) = ∑ k = 0 ∞ h ( k ) z − k h(k) \leftrightarrow H(z) = \sum^{\infty}_{k=0} h(k) z^{-k} h(k)↔H(z)=k=0∑∞h(k)z−k
H ( z ) = Z [ h ( k ) ] H(z) = \mathcal{Z}[h(k)] H(z)=Z[h(k)]
计算方法:
应用:
分解:
f ( k ) = 1 2 π j ∮ c F ( z ) z z k d z , − ∞ < k < ∞ f({\color{red}k}) = \frac{1}{2\pi j} \oint_c \frac{F(z)}{z} z^{ {\color{red}k}} dz, \; -\infty < k < \infty f(k)=2πj1∮czF(z)zkdz,−∞<k<∞
z 0 k → h ( k ) → z 0 k ⋅ H ( z 0 ) z_0^k \to \boxed{h(k)} \to z_0^k \cdot H(z_0) z0k→h(k)→z0k⋅H(z0)
f ( k ) → h ( k ) → y f ( k ) f(k) \to \boxed{h(k)} \to y_f(k) f(k)→h(k)→yf(k)
1 2 π j F ( z ) z ⋅ z k → 1 2 π j F ( z ) z ⋅ z k H ( z ) \frac{1}{2\pi j} \frac{F(z)}{z}\cdot z^k \to \frac{1}{2\pi j} \frac{F(z)}{z} \cdot z^k H(z) 2πj1zF(z)⋅zk→2πj1zF(z)⋅zkH(z)
∮ c 1 2 π j F ( z ) z ⋅ z k d z → ∮ c 1 2 π j F ( z ) z ⋅ z k H ( z ) d z \oint_c\frac{1}{2\pi j} \frac{F(z)}{z}\cdot z^k dz \to \oint_c\frac{1}{2\pi j} \frac{F(z)}{z} \cdot z^k H(z) dz ∮c2πj1zF(z)⋅zkdz→∮c2πj1zF(z)⋅zkH(z)dz
∮ c 1 2 π j F ( z ) z ⋅ z k d z → ∮ c 1 2 π j F ( z ) ⋅ H ( z ) z ⋅ z k d z \oint_c\frac{1}{2\pi j} \frac{F(z)}{z}\cdot z^k dz \to \oint_c\frac{1}{2\pi j} \frac{ {\color{blue}F(z)\cdot H(z)} }{z} \cdot z^k dz ∮c2πj1zF(z)⋅zkdz→∮c2πj1zF(z)⋅H(z)⋅zkdz
Y f ( z ) = F ( z ) ⋅ H ( z ) Y_f(z) = F(z) \cdot H(z) Yf(z)=F(z)⋅H(z)
H ( z ) = B ( z ) A ( z ) = b m z m + b m − 1 z m − 1 + ⋯ + b 1 z + b 0 z n + a n − 1 z n − 1 + ⋯ + a 1 z + a 0 = b m ( z − ζ 1 ) ( z − ζ 2 ) ⋯ ( z − ζ m ) ( z − P 1 ) ( z − P 2 ) ⋯ ( z − P n ) = b m ∏ j = 1 m ( z − ζ j ) ∏ i = 1 n ( z − P i ) , m ≤ n \begin{aligned}H(z) & = \displaystyle \frac{B(z)}{A(z)}\\ & = \displaystyle \frac{b_m z^m + b_{m-1}z^{m-1} + \cdots + b_1 z + b_0}{z^n + a_{n-1}z^{n-1} + \cdots +a_1z+a_0}\\ & = \frac{b_m(z-\zeta_1)(z-\zeta_2)\cdots(z-\zeta_m)}{(z-P_1)(z-P_2)\cdots(z-P_n)} \\ &= \frac{b_m \prod^{m}_{j=1}(z-\zeta_j)}{\prod^{n}_{i=1}(z-P_i)}, \; m\leq n \end{aligned} H(z)=A(z)B(z)=zn+an−1zn−1+⋯+a1z+a0bmzm+bm−1zm−1+⋯+b1z+b0=(z−P1)(z−P2)⋯(z−Pn)bm(z−ζ1)(z−ζ2)⋯(z−ζm)=∏i=1n(z−Pi)bm∏j=1m(z−ζj),m≤n
ζ i , i = 1 , 2 , ⋯ , m \zeta_i, \; i =1,2,\cdots, m ζi,i=1,2,⋯,m
P i , i = 1 , 2 , ⋯ , m P_i, \; i =1,2,\cdots, m Pi,i=1,2,⋯,m
极点在单位圆内
在实轴上:
不在实轴上:
结论: 对应 h ( k ) h(k) h(k)按指数规律衰减;
极点在单位圆上
在实轴上:
不在实轴上:
结论: 一阶极点对应 h ( k ) h(k) h(k)为稳态分量;二阶及二阶以上极点对应 h ( k ) h(k) h(k)增长。
极点在单位圆外
在实轴上:
不在实轴上:
结论: 对应 h ( k ) h(k) h(k)按指数规律增长。
∑ k = − ∞ ∞ ∣ h ( k ) ∣ < ∞ \sum^{\infty}_{k=-\infty} \lvert h(k)\rvert < \infty k=−∞∑∞∣h(k)∣<∞
离散系统稳定性的Z域充要条件:
离散因果系统稳定性判定--朱里准则(Jury stability criterion)
H ( z ) = B ( z ) A ( z ) = b m z m + b m − 1 z m − 1 + ⋯ + b 1 z + b 0 a n z n + a n − 1 z n − 1 + ⋯ + a 1 z + a 0 \begin{aligned}H(z) = \displaystyle \frac{B(z)}{A(z)} = \displaystyle \frac{b_m z^m + b_{m-1}z^{m-1} + \cdots + b_1 z + b_0}{a_nz^n + a_{n-1}z^{n-1} + \cdots +a_1z+a_0}\end{aligned} H(z)=A(z)B(z)=anzn+an−1zn−1+⋯+a1z+a0bmzm+bm−1zm−1+⋯+b1z+b0
要判断 A ( z ) = 0 A(z)=0 A(z)=0 的所有根的绝对值是否都小于 1 1 1 。
朱里准则指出: A ( z ) = 0 A(z)=0 A(z)=0 的所有根都在单位圆内的充要条件是:
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