8-1试求下列函数的z变换

= lim
[
]
= ( −1) n −1 n −1 1 + (−1) n −1 4 ∞ ⎧n −1 c * (t ) = ∑ ⎨ 1 + (−1) n −1 n =0 ⎩ 4 c( nT ) =
n −1 4
[
]
[
]⎫ ⎬δ (t − nT )
⎭
2 2
(3)
[z C ( z ) − z c(0) − z c(1) − zc(2)] + 6[z C ( z ) − z c(0) − zc(1)]
8-2 试分别用部分分式法、幂级数法和反演积分法求下列函数的z反变换。
10 z ( z − 1)( z − 2 ) − 3 + z −1 (2) E( z) = 1 − 2 z −1 + z − 2 10 z 解 （1） E ( z ) = ( z − 1)( z − 2) (1) E ( z) =
3 3 2
对差分方程两端取 z 变换得
+ 11[zC ( z ) − zc(0)] + 6C ( z ) = 0
代入初条件整理得
( z 3 + 6 z 2 + 11z + 6) ⋅ C ( z ) = z 3 + 7 z 2 + 17 z C ( z) = z 3 + 7 z 2 + 17 z z 3 + 6 z 2 + 11z + 6 C ( z ) 11 1 1 5 1 = ⋅ −7⋅ + ⋅ z 2 z +1 z+2 2 2+3 11 5 5 ⎡11 ⎤ c(n) = (−1) n − 7(−2) n + (−3) n = (−1) n ⎢ − 7 ⋅ 2 n + ⋅ 3 n ⎥ 2 2 2 ⎣2 ⎦
E( z) =
10 z 10 z = 2 = 10 z −1 + 30 z −2 + 70 z −3 + Λ ( z − 1)( z − 2) z − 3z + 2
e* (t ) = 10δ (t − T ) + 30δ (t − 2T ) + 70δ (t − 3T ) + Λ
③ 反演积分法
10 z n = −10 z →1 z − 2 10 z n Re s[E ( z ) ⋅ z n −1 ]z →2 = lim = 10 × 2 n z →2 z − 1 e(nT ) = −10 × 1 + 10 × 2 n = 10(2 n − 1) e * (t ) = ∑10(2 n − 1)δ (t − nT )
8-6 试由以下差分方程确定脉冲传递函数。
c(n + 2) − (1 + e −0.5T ) ⋅ c(n + 1) + e −0.5T c(n) = (1 − e −0.5T ) ⋅ r (n + 1)
解 对上式实行 z 变换，并设所有初始条件为 0 得
z 2 C ( z ) − (1 + e −0.5T ) zC ( z ) + e −0.5T C ( z ) = (1 − e −0.5T ) z ⋅ R( z )
( 4)
解
c( k + 2) + 5c ( k + 1) + 6c (k ) = cos( kπ / 2)
c (0) = c(1) = 0
（1） 令 t = −T ，代入原方程可得： c (T ) = 0 。对差分方程两端取 z 变换，整理得
C ( z) =
1 1 z R( z ) = ⋅ ( z − 2)( z − 4) z − 1 z − 6z + 8 C ( z) 1 1 1 1 1 1 = ⋅ − ⋅ + ⋅ z 3 z −1 2 z − 2 6 z − 4 z 1 z 1 1 z C ( z) = ⋅ − ⋅ + ⋅ 3 z −1 2 z − 2 6 z − 4 1 1 1 c(nT ) = × 1n − × 2 n + × 4 n 3 2 6
ess = lim( z − 1) E ( z )
z →1
（2）
= lim
0.792 z 2 0.792 = =1 2 z →1 z − 0.416 z + 0.208 1 − 0.416 + 0.208
8-4 已知差分方程为
c( k ) − 4c( k + 1) + c( k + 2) = 0
初始条件：c(0)=0,c(1)=1。试用迭代法求输出序列c(k),k=0，1，2，3，4。 解 依题有
2
（2） 对差分方程两端取 z 变换，整理得
C( z) =
1 z z ⋅ = 2 2 ( z + 1) ( z − 1) 2 z + 2 z + 1 ( z − 1)
2
Re s C ( z ) ⋅ z n −1
[
]
z →1
=
⎤ 1 d ⎡ z ⋅ z n −1 ⎥ lim ⎢ 2 1! z →1 dz ⎣ ( z + 1) ⎦ d ⎡ zn ⎤ nz n −1 ( z + 1) − 2 − 2( z + 1) −3 ⋅ z n ⎥ = lim ⎢ z →1 dz ( z + 1) 2 z →1 ⎣ ⎦
e(nT ) = Re s E ( z ) ⋅ z n −1
s →1
∞
1 d [ ] =1 lim [(−3 z ! dz = lim[− 3(n + 1) z + nz ] = −2n − 3
z →1 s →1
2
+ z ) ⋅ z n −1
]
n
n −1
e * (t ) = ∑ (−2n − 3)δ (t − nT )
② 幂级数法：用长除法可得
−3 z 2 + z = −3 − 5 z −1 − 7 z −2 − 9 z −3 − Λ 2 z − 2z + 1 * e (t ) = −3δ (t ) − 5δ (t − T ) − 7δ (t − 2T ) − 9δ (t − 3T ) − Λ E( z) =
③ 反演积分法
8-1
试求下列函数的z变换
(1)
(2)
e(t ) = a
t T
(3)
( 4)
解
e( t ) = t 2 e −3t s +1 E ( s) = 2 s
E (s) =
s+3 s ( s + 1)( s + 2)
（1） E ( z ) = （2） Z
2
∑a
n =0
2
∞
n
z −n =
z 1 = −1 z−a 1 − az
① 部分分式法
− 10 E ( z) − 10 10 = = + z ( z − 1)( z − 2) z − 1 z − 2 − 10 z 10 z E( z) = + ( z − 1) ( z − 2) e(nT ) = −10 × 1 + 10 × 2 n = 10(2 n − 1)
② 幂级数法：用长除法可得
n =0 ∞
Re s[E ( z ) ⋅ z n −1 ]z →1 = lim
z (−3z + 1) z (−3z + 1) − 3 + z −1 = = 2 （2） E ( z ) = −2 1 − 2z + z z − 2z + 1 ( z − 1) 2
① 部分分式法
E( z) 1 − 3z −2 3 = = − 2 2 z z −1 ( z − 1) ( z − 1) − 2z 3z E ( z) = − 2 z −1 ( z − 1) −2 e(t ) = t − 3 × 1(t ) T ∞ ∞ ⎡− 2 ⎤ e * (t ) = ∑ ⎢ nT − 3⎥δ (t − nT ) = ∑ (−2n − 3)δ (t − nT ) ⎦ n =0 ⎣ T n =0
[t ] = T
[ ]
z ( z + 1) ( z − 1) 3
由移位定理：
T 2 ze 3T ( ze 3T + 1) T 2 ze −3T ( z + e −3T ) = ( ze 3T − 1) 3 ( z − e −3T ) 3 s +1 1 1 （3） E ( s ) = 2 = + 2 s s s z Tz E( z) = + z − 1 ( z − 1) 2 c c c （4） E ( s ) = 0 + 1 + 2 s s +1 s + 2 s+3 3 c0 = lim = s →0 ( s + 1)( s + 2) 2 s+3 2 c1 = lim = = −2 s → −1 s ( s + 2) −1 s+3 1 c 2 = lim = s → −2 s ( s + 1) 2 32 12 2 = − + s s +1 s + 2 3z 2z z E( z) = − + −T 2( z − 1) z − e 2( z − e − 2 T ) Z t 2 e − 3t =
c(k + 2) = 4c(k + 1) − c(k ) c(0) = 0 , c(1) = 1 c(2) = 4 ×1 − 0 = 4 c(3) = 4 × 4 − 1 = 15 c(4) = 4 ×15 − 4 = 56
8-5 试用z变换法求解下列差分方程：
(1)
(2)
c( k + 2) − 6c(k + 1) + 8c(k ) = r ( k ) r ( k ) = 1( k ) , c( k ) = 0
= lim
[
]
= n 2 − 2 − 2 ⋅ 2 −3 = Re s C ( z ) ⋅ z n −1
n −1 4
[
]
z →1
=
wenku.baidu.com
⎤ 1 d ⎡ z lim ⎢ ⋅ z n −1 ⎥ 2 z 1 → − 1! dz ⎣ ( z − 1) ⎦ d ⎡ zn ⎤ nz n −1 ( z − 1) − 2 − 2( z − 1) −3 ⋅ z n ⎢ ⎥ = zlim z → −1 dz ( z − 1) 2 → −1 ⎣ ⎦
n =0
8-3 试确定下列函数的终值
(1)
E ( z) =
Tz −1 (1 − z −1 ) 2
(2)
解
E( z) =
0.792 z 2 ( z − 1)( z 2 − 0.416 z + 0.208)
−1 z →1
（1） e ss = lim(1 − z )
Tz −1 =∞ (1 − z −1 ) 2
z ( z − cos ) 2 z 2 − 2 z cos
（4） 由原方程可得
π
( z 2 + 5 z + 6) ⋅ C ( z ) =
π
=
2
+1
z2 z2 +1
C ( z) =
z2 z2 = ( z 2 + 5 z + 6)( z 2 + 1) ( z + 2)( z + 3)( z 2 + 1) C ( z) z −2 1 3 1 1 z −1 = = ⋅ + ⋅ + ⋅ 2 2 z ( z + 2)( z + 3)( z + 1) 5 z + 2 10 z + 3 10 z + 1
c(nT ) =
π π⎤ −2 3 1 ⎡ ⋅ (−2) n + ⋅ (−3) n + ⎢cos n − sin n ⎥ 5 10 10 ⎣ 2 2⎦
π π⎤ 1 ⎡1 ⎤ 1 ⎡ = (−1) n +1 ⎢ ⋅ 2 n +1 − ⋅ 3 n +1 ⎥ + ⎢cos n − sin n ⎥ 10 2 2⎦ ⎣5 ⎦ 10 ⎣
5 ⎤ 10 z (e − 2T − e −5T ) ⎡ 2 G( z) = Z ⎢ ⋅ = ⋅ − 2T −5T ⎣ s + 2 s + 5⎥ ⎦ 3 ( z − e )( z − e )
根据定义有
C ( z) z (1 − e −0.5 / T ) G( z) = = R ( z ) z 2 − (1 + e −0.5T ) z + e −0.5T
8-7 设开环离散系统分别如图8-1（a）,（b）,（c）所示，试求开环脉冲传递函数 G ( z ) 。
解
（a）Z⎢
2z ⎡ 2 ⎤ = − 2T ⎥ ⎣s + 2⎦ z − e 5z ⎡ 5 ⎤ Z⎢ = − 5T ⎥ ⎣ s + 5⎦ z − e 10 z 2 ⎡ 2 ⎤ ⎡ 5 ⎤ G( z) = Z ⎢ ⋅ Z = − 2T −5T ⎣s + 2⎥ ⎦ ⎢ ⎣ s + 5⎥ ⎦ ( z − e )( z − e ) 5 ⎤ 10 1 ⎤ 10 z (e −2T − e −5T ) ⎡ 2 ⎡10 1 Z = ⋅ ⋅ = ⋅ − ⋅ （b ） Z⎢ − 2T −5T ⎢ ⎣ s + 2 s + 5⎥ ⎦ ⎣ 3 s + 2 3 s + 5⎥ ⎦ 3 ( z − e )( z − e )
c ( k + 2) + 2c (k + 1) + c ( k ) = r ( k ) c(0) = c(T ) = 0 r (n) = n , ( n = 0 , 1, 2 , Λ )
( k ≤ 0)
( 3)
c( k + 3) + 6c( k + 2 ) + 11c( k + 1) + 6c( k ) = 0 c( 0) = c(1) = 1 , c( 2 ) = 0