# 控制工程基础第二章数学模型

(第二章）

feeding driving device and its equivalent mechanical model

The d Th dynamical skid platform of i l kid l tf f an aggregate machine tool mills

fm(t) m x (t) v (t) 参考点
2

d d f m (t ) = m v(t ) = m 2 x(t ) dt dt

x1(t) v1(t) fk(t) k x2(t) v2(t) fk(t)

f k (t ) = k [ x1 (t ) ? x2 (t ) ] = kx(t ) = k∫

[v1 (t ) ? v2 (t )] dt ?∞
t t ?∞

= k ∫ v(t )dt

v1(t) x1(t) fD(t) D v2(t) x2(t) fD(t)

f D (t ) = D [ v1 (t ) ? v2 (t )] = Dv(t ) ? dx1 (t ) dx2 (t ) ? = D? ? ? dt dt ? ? dx(t ) =D dt

fi(t) m 0 xo(t) k D fk(t) fD(t) m fi(t) fm(t) 静止（平衡）工作点作为 零点，以消除重力的影响 0 xo(t)

? d2 ? fi (t ) ? f D (t ) ? f k (t ) = m dt 2 xo (t ) ? ? kx ? f k (t ) = k o (t ) ? d ? f D (t ) = D xo (t ) ? dt ?

d2 d m 2 yo (t ) + D yo (t ) + kyo (t ) = f i (t ) dt dt

fi(t) 0 xo( ) (t) k D

f i (t ) = f D (t ) + f k (t )
d D xo (t ) + kxo (t ) = fi (t ) dt

k

θo(t) 0
J TD(t)

J —旋转体转动惯量；k —扭转刚度系数； 旋转体转动惯量 k 扭转刚度系数 D —粘性阻尼系数

? ?Tk (t ) = k [θi (t ) ? θ o (t ) ] ? d ? ?TD (t ) = D θ o (t ) dt ? ? d2 ? J 2 θ o (t ) = Tk (t ) ? TD (t ) ? dt

d2 d J 2 θ o (t ) + D θ o (t ) + kθ o (t ) = kθ i (t ) dt dt

u (t ) = R ? i (t )

1 u (t ) = ∫ i (t )dt C

di (t ) u (t ) = L dt

R-L-C无源电路网络 L

R

ui(t) ()

i(t)

C

uo(t) ()

R-L-C无源电路网络
d 1 ? ?ui (t ) = Ri (t ) + L dt i (t ) + C ∫ i (t )dt ? ? ?uo (t ) = 1 i (t )dt ∫ ? ? C

d d LC 2 uo (t ) + RC uo (t ) + uo (t ) = ui (t ) dt dt
2

d RC uo (t ) + uo (t ) = ui (t ) dt

R a ? +

i2(t)
C

uo(t)

?ua (t ) ≈ 0 ? ?i1 (t ) ≈ i2 (t )

ui (t ) duo (t ) = ?C dt R

duo (t ) 即： RC = ?ui (t ) (t dt

T (t ) = K T ia (t )
dia (t ) ei (t ) = Ra ia (t ) + La + em (t ) dt

dθ o (t ) em (t ) = K e dt

dθ o (t ) d 2θ o (t ) 牛顿第二定律 牛顿第 定律 T (t ) ? D =J 2 dt dt

&&& && & La Jθ o (t ) + ( La D + Ra J ) θ o (t ) + ( Ra D + KT K e ) θ o (t ) = KT ei (t )

& & Ra Jθ&o (t ) + (Ra D + K T K e )θ o (t ) = K T ei (t )

dn d n?1 d x (t ) + a1 n?1 xo (t ) + L + an?1 xo (t ) + an xo (t ) n o dt dt dt dm d m?1 d = b0 m xi (t ) + b1 m?1 xi (t ) + L + bm?1 xi (t ) + bm xi (t ) dt dt dt

linearization

df ( x) y = f ( x) = f ( x0 ) + ( x ? x0 ) dx d x=x
0

1 d 2 f ( x) 1 d 3 f ( x) 2 + ( x ? x0 ) + ( x ? x0 )3 + L 2! dx2 x = x 3! dx3 x = x
0 0

df ( x ) y = f ( x0 ) + ( x ? x0 ) dx d x=x
df (x ) 或：y - y0 = Δy = KΔx， 其中：K = dx x = x
0
0

?f y = f ( x10 , x20 ) + ?x1

x1 = x10 x2 = x20

?f ( x1 ? x10 ) + ?x2

x1 = x10 x2 = x20

( x2 ? x20 ) + L

?f 其中： K1 = ?x1 ?f , K2 = x1 = x10 ?x2
x2 = x20

x1 = x10 x2 = x20

0 y0 y f(x) y=f(x) Δy’ Δ ’

A

α Δx

Δy

x0 非线性关系线性化

x

Ti (t ) ? mgl sin θ o (t ) = ml θ o (t )
2

..

ml θ o (t ) + mglθ o (t ) = Ti (t )
2

..

QL0 = f ( pL0 ,x0 ) ,

? ?f ( p L ,x ) ? QL = f ( p L0 , 0 ) + ? ,x ? x = x0 ? Δx ? ?x ? p L = p L0 ? ?f ( p L ,x ) ? +? ? x = x ? Δp L + L ? ?p L ? p L =0p L0
Δ QL = K q Δx ? K c Δ p L
? ?f ( pL ,x ) ? Kq = ? ? x = x0 , ? ?x ? pL = pL0 ? ?f ( pL ,x )? Kc = ?? ? x= x ? ?pL ? pL =0pL0

d (Δy ) 液压腔工作腔流动连续性方程为： Q = A Δ dt 液压腔力平衡方程为：

d (Δy ) d (Δy ) ΔpL A = M +D 2 dt dt
2

K c M d (Δy ) ? K c D ? d ( Δy ) ? +? + A? = K q (Δx) 2 A dt ? A ? dt
2

Kc M ? Kc D ? & ? &&(t ) + ? y + A ? y (t ) = K q x(t ) A ? A ?

y (t ) = f ( x (t ))

df ( x ) 1 d 2 f (x ) y = f ( x) = f ( x ) + (x ? x ) + ( x ? x )2 + L dx 2 dx 2

Δy = K Δx

dx

x= x

out

out

0 in 真实特性 饱和非线性 out

0 死区非线性

in

out

0 间隙非线性

in

0

in

Definition of Laplace transform

lim e
t→ ∞ ?σt

f (t ) → 0

F ( s ) = L[ f (t )] ≡ ∫ f (t )e dt
∞ 0 ? st

∫0

e ? st dt 称为拉普拉氏积分；

F(s)称为函数f(t)的拉普拉氏变换或象函数， 它是一个复变函数；f(t)称为F(s)的原函数； L为拉氏变换的符号。 L为拉氏变换的符号
Laplace inverse transform

1 σ + j∞ st f (t ) = L [F ( s )] = ∫σ ? j∞ F ( s )e ds , t > 0 2πj
?1

1为拉氏反变换的符号。 L－1为拉氏反变换的符号

Laplace transforms of simple functions p p Unit step function 1(t) f(t) 1

?0 1(t ) = ? ?1
∞ 0

t<0 t≥0
? st

L [ ( t ) ] = ∫ 1( t ) e 1

dt

0

t

∞ 1 ? st 1 = ? e = s s 0

(Re( s ) > 0 )

Exponential function E ti l f ti

f ( t ) = e ? at （a为常数）

f(t) 1

L[e

? at

]= ∫ e =∫ e
∞ 0

∞ 0

? at

? e dt dt d
0 指数函数 t

? st

?( s + a )t

1 = , (R ( s + a) > 0) (Re( s+a

Sine f Si function and Cosine function ti dC i f ti f(t) ∞ 1 L[ ω t ] = ∫0 cos ω t ? e ?st dt [cos

L[sin ω t ] = ∫0 sin ω t ? e ? st dt

f(t)=sinωt

1 jω t ? jω t e ?e sin ω t = 2j 1 jω t cos ω t = e + e ? jω t 2

0

t

(

)

-1 f(t)=cosωt 正弦及余弦函数

(

)

1 ∞ jωt ? st ∞ ? jωt ? st L[sin ω t ] = ∫0 e e dt ? ∫0 e e dt 2j 1 ? 1 1 ? ? = ? s ? jω ? s + jω ? ? 2j? ?

(

)

ω = 2 2 s +ω

(Re( s) > 0)

s 同理： L [cos ω t ] = 2 s +ω2

unit-impulse function δ(t)

?0 ? δ (t ) = ? 1 lim ?li→0 ε ?ε
∞ 0 ε →0

( t < 0且 t > ε ) (0 < t < ε )

f(t) 1 ε

L[δ (t )] = ∫ li lim ? e ? st d dt 1

ε

1 = li lim (1 ? e ?ε s ) ε →0 ε s

0 ε 单位脉冲函数 单

t

1 (1 ? e ?ε s )′ lim lim 由洛必达法则： li→0 (1 ? e ?ε s ) = li→0 由洛必达法则 ε ε εs (ε s )′

ε ? e ?ε s =1 所以： L[δ ( t )] = lim ε →0 ε

unit-ramp function

?0 t < 0 f (t ) = ? ?t t ≥ 0
L[ f (t )] = ∫ te ? st dt
∞ 0 ? st ∞

f(t) 1

e =t ?s 1 = 2 s

0

e ? st ∞ ? ∫0 dt ?s

0

1 单位速度函数

t

(Re(s) > 0)

Parabolic function

?0 ? f (t ) = ? 1 2 ?2 t ?
∞ 0

t<0 t≥0

f(t)

1 2 ? st L[ f (t )] = ∫ t e dt 2 1 (Re(s) > 0) = 3 s

0 t 单位加速度函数

Power function

t n ? 1(t )

n! L ?t ?1( t ) ? = ∫ t e dt = n +1 ∫ u e du = n +1 ? ? s 0 s 0
n n ? st

1

n ?u

L+ [ f (t )] = ∫ f (t )e dt
∞ 0+ ? st

L? [ f (t )] = ∫ f (t )e ? st dt
∞ 0?

= L+ [ f (t )] + ∫ f (t )e ? st dt
0+ 0?

Characters of Laplace transform Ch t fL l t f
Superposition principle

Differentiation theorem （实微分定理）

? df (t ) ? L? ? = sF ( s ) ? f (0), ? dt ?

f ( 0) = f ( t ) t = 0

? ? d 2 f (t ) ? = s 2 F ( s ) ? sf (0) ? f ′(0) ?L ? ? dt 2 ? ? ? ? ?LL ? ? d n f (t ) ? ?L = s n F ( s ) ? s n ?1 f (0) ? s n ? 2 f ′(0) ? L ? f ( n ?1) (0) ? ? ? dt n ? ? ?

? ? df (t ) ? ? L ? dt ? = sF ( s ) ? ? ? ? ? d 2 f (t ) ? 2 ? L ? dt 2 ? = s F ( s ) ? ? ? ?LL ? ? ? d n f (t ) ? n ?L? d n ? = s F ( s) ? ? dt ?

? df (t ) ? L+ ? = sF ( s ) ? f (0 + ) ? dt ? ? ? df (t ) ? ? L? ? ? = sF ( s ) ? f (0 ) ? dt ?

d F ( s ) = ? L[tf (t )] f ds d2 F ( s ) = L t 2 f (t ) ds 2 LL

[

] ]
(n = 1, 2, 3, L)

dn F ( s ) = (?1) n L t n f (t ) ds d n

[

Integral theorem g
F ( s ) f ( ? 1) ( 0 ) L ∫ f (t )dt = + , s s

[

]

f ( ?1) (0) = ∫ f (t )dt

t=0

L

[∫

1 f ( t ) dt = F ( s ) s

]

F ( s ) f ( ?1) (0 + ) L+ [∫ f (t )dt ] = + s s F ( s ) f ( ?1) (0 ? ) L? [∫ f (t )dt ] = + s s

? ? 1 1 ( ?1) 1 ( ? n +1) L ? ∫ L∫ f ( t ) dt ? = n F ( s ) + n ?1 f ( 0) + L + f ( 0) s s ? s ?{ ? n ?

? ? 1 L ? ∫ L∫ f (t )dt ? = n F ( s ) ?{ ? s ? n ?

Time-delay theorem f(t) f(t) f(t-τ)

0

τ

t

L[x(t ? a ) ? 1(t ? a )] = e

? as

X (s )

Attenuation theorem

L[e ? at f (t )] = F ( s + a )
ω 例：L[sin ωt ] = 2 s +ω2
Le

s L[cos ωt ] = 2 s +ω2

( s + a) 2 + ω 2 ( s + a) ? at t L e cos ωt = ( s + a) 2 + ω 2

[ [

? at

sin ωt =

]

ω

]

Initial value theorem
t →0

lim f (t ) = f (0 + ) = lim sF ( s ) +
s →∞

Final value theorem

lim f (t ) 存在。则 t →∞
t →∞

lim li f (t ) = f ( ∞ ) = li sF ( s ) lim F
s →0

Imagine function of convolution integral

L[ f (t ) ? g (t )] = F ( s )G ( s )

f (t ) * g (t ) ≡ ∫0 f (t ? τ ) g (τ )dτ = ∫0 f (τ ) g (t ? τ )dτ
t t

?t? f? ? Image function of ?a?

? ? t ?? L ? f ? ?? = aF (as) ? ? a ??

a = constant > 0

Le

1 s +1

[ ]
?t / a

a = aF (as) = as + 1

● inverse transform
The method of partial fraction expansion

B( s) b0 s m + b1s m ?1 + L + bm ?1s + bm = F ( s) = A( s) a0 s n + a1s n ?1 + L + an ?1s + an ( n ≥ m)

B( s ) c0 s m + c1s m?1 + L + cm?1s + c0 F ( s) = = A( s ) ( s + p1 )( s + p2 ) L( s + pn )

There are only single real poles in F(s)
n B( s ) A1 A2 An Ai F ( s) = = + +L+ =∑ A( s ) s + p1 s + p2 s + pn i =1 s + pi

Ai = [F ( s ) ? ( s + p i ) ]s = ? p i

Ai ? n L [ F ( s )] = L ? ∑ = ∑ Ai e ? pi t s + pi ? i =1 ?i =1 ?
?1 n ?1 ?

s2 ? s + 2 例：求 F ( s ) = 2 的原函数。 s ( s ? s ? 6)
s2 ? s + 2 s2 ? s + 2 A1 A2 A3 = = + + 解：F ( s ) = 2 s( s ? s ? 6) s( s ? 3)( s + 2) s s ? 3 s + 2
A1 = [sF ( s )]s = 0 ? s2 ? s + 2 ? 1 =? =? ? ( s ? 3)( s + 2) ? s = 0 3 ?

? s2 ? s + 2 ? 8 A2 = [( s ? 3) F ( s )]s = 3 = ? = ? s ( s + 2) ? s = 3 15 ?

A3 = [( s + 2) F ( s )]s = ?2

? s2 ? s + 2 ? 4 = =? ? s ( s ? 3) ? s = ?2 5 ?

1 1 8 1 4 1 + ? 即： 即 F ( s) = ? ? + ? 3 s 15 s ? 3 5 s + 2

1 8 3t 4 ? 2 t f (t ) = L [ F ( s)] = ? + e + e 3 15 5
?1

(t ≥ 0)

There are complex-conjugate poles in F(s) p j g p ()

F ( s) =

B( s ) A1s + A2 A3 An = + +L+ A( s ) ( s + p1 )( s + p2 ) s + p3 s + pn

[F ( s)( s + p1 )( s + p2 )]s = ? p1或s = ? p2 = [A1s + A2 ]s = ? p1或s = ? p2

n B( s ) A1 A2 An Ai F ( s) = = + +L+ =∑ A( s ) s + p1 s + p2 s + pn i =1 s + pi

Ai = [F ( s ) ? ( s + p i ) ]s = ? p i

s + 1 的原函数。 2 s ( s + s + 1)

A0 A1s + A2 s +1 F = + 2 解： ( s ) = ? 1 3 ?? 1 3 ? s s + s +1 ?? s + ? j ? s? s + + j ? ?? 2 2 ?? 2 2 ? ? ?

A0 = sF ( s ) s =0 = 1 F
( s + s + 1) F ( s )
2 1 3 s=? ? j 2 2

= ( A1s + A2 ) s = ? 1 ? j
2

3 2

1 ? 1 ?? 2 ( A1 + A2 ) = 2 ? 即：? ? A1 = ?1, ? 3 (A ? A ) = ? 3 1 2 ? 2 2 ?

A2 = 0

s 1 所以： F ( s ) = ? 2 s s + s +1
1 s = ? 2 2 s ? 1? ? 3? ? ?s + ? +? ? 2? ? 2 ? ? ?

1 1 s+ 1 2 2 = ? + 2 2 2 2 s ? 1? ? 3? 1? ? 3? ? ? ? ?s + ? +? ?s + ? +? ? 2? ? 2 ? 2? ? 2 ? ? ? ? ? ?
1 3 s+ 1 1 2 2 = ? + 2 2 2 2 s ? 3? 1? ? 3? 1? ? 3? ? ? ? ?s + ? +? ?s + ? +? ? 2? ? 2 ? 2? ? 2 ? ? ? ? ?

f (t ) = 1 ? e

?t 2

3 1 ?t 2 3 t+ e sin t cos 2 2 3

2 ?t 2 ? 3 3 1 3 ? cos = 1? e ? t + sin t? ? 2 2 2 2 ? 3 ? ? ? 2 ?t 2 ? 3 ? = 1? e sin ? t + 60 ? , t ≥ 0 ? 2 3 ? ?

s +1 s3 + s 2 + s
a3 + + s 1 3 1 3 s+ + j s+ ? j 2 2 2 2 a1 a2

s +1 X (s ) = 3 = 2 s +s +s

? s +1 ? 1 3 ?? ?? a1 = ? 3 ?? s + + j 2 ? s +s +s ? 2 2 ?? 1 ? ?? s=? ? j ?
2

3 2

1 3 =? + j 2 6

1 3 a2 = ? ? j 2 6

? s +1 ? a3 = ? 3 2 ? s? = 1 ? s + s + s ? s =0
1 3 1 3 ? +j ? ?j s +1 2 6 + 2 6 +1 X (s ) = 3 2 = s +s +s 1 3 1 3 s s+ + j s+ ? j 2 2 2 2

?1 ?1 3? 3? ?? 1 ? ?? 2 + j 2 ?t ? 1 ? ?? 2 ? j 2 ?t ? ? ? ? ? 3 3 ? ? x ( t ) = ?? ? + j e? +?? ? j e? + 1? ?1( t ) ? ? ? 2 ? ? 2 ? 6 ? 6 ? ?? ? ? ? ? ? ?1t ? 3 3 3 ? ? 2 = ?e ? ? 3 sin 2 t ? cos 2 t ? + 1? ?1( t ) ? ? ? ? ? ? ?

There are repeated poles in F(s)

B(s) b0sm + b1sm?1 +L+ bm?1s + bm F (s) = = A(s) (s + p0 )r (s + pr +1 )L(s + pn )
A01 A02 A0r An Ar +1 = + +L+ + +L+ r r ?1 (s + p0 ) (s + p0 ) (s + p0 ) (s + pr +1) (s + pn )

A01 = [ F ( s)(s + p0 ) ]s=? p0
r

?d r ? A02 = ? [ F ( s )( s + p0 ) ]? ? ds ? s =? p0

1 ? d2 r ? A03 = ? 2 [ F ( s)(s + p0 ) ]? 2! ? ds ? s =? p
……

0

1 ? d r ?1 r ? A0 r = ? r ?1 [ F ( s)( s + p0 ) ]? (r ? 1)! ? ds ? s =? p

0

? ? t n ?1 ? p0 t 1 L?1 ? = e n? ? ( s + p0 ) ? ( n ? 1)!

f (t ) = L?1[ F ( s )] ? A01 r ?1 ? ? p0 t A02 r ? 2 =? t + L + A0 r ? e t + (r ? 2)! ? (r ? 1)! ? ? pn t ? p r +1t + Ar +1e + L + An e (t ≥ 0)

s+3 F ( s) = ( s + 2) 2 ( s + 1)

A01 A02 A3 + + 解： F ( s ) = 2 s + 2 s +1 ( s + 2)
? s + 3? A01 = F ( s )( s + 2) 2 s = ?2 = ? = ?1 ? ? s + 1 ? s = ?2

[

]

?d A02 = ? F ( s )( s + 2) 2 ? ds

[

]

? d ? s + 3?? ? =? ? ? ?? ? s =?2 ? ds ? s + 1 ? ? s =?2

? ( s + 3)′( s + 1) ? ( s + 3)( s + 1)′ ? =? ? ( s + 1) 2 ? ? s =?2 = ?2

A3 = [F ( s )( s + 1)]s = ?1 = 2
F ( s) = ?1 2 2 ? + ( s + 2) 2 s + 2 s + 1

f ( t ) = L [ F ( s )] = ? (t + 2 ) e ? 2 t + 2 e ? t (t ≥ 0 )

?1

Solve di S l ordinary differential equation by Laplace transform diff ti l ti b L l t f

d 2 xo (t ) dx d o (t ) +5 + 6 xo (t ) = xi (t ) 2 dt dt

? d 2 xo (t ) ? 2 ′ L? = s X o ( s ) ? sxo (0) ? xo (0) 2 ? ? dt ?

dx ? d o (t ) ? L ?5 ? = 5sX o ( s ) ? 5 xo (0) ? dt ?

L[6 xo (t )] = 6 X o ( s )
? d 2 xo (t ) ? dxo (t ) 即： L ? +5 + 6 xo (t )? 2 dt ? dt ? ′ = ( s 2 + 5s + 6) X o ( s ) ? ( s + 5) xo (0) ? xo (0)

1 L[xi (t )] = X i ( s ) = L[1(t )] = s

1 ′ ( s + 5s + 6) X o ( s ) ? [( s + 5) xo (0) + xo (0)] = s
2

′ 1 ( s + 5) xo (0) + xo (0) + X o ( s) = 2 s 2 + 5s + 6 s( s + 5s + 6) A1 A2 A3 B1 B2 = + + + + s s+2 s+3 s+2 s+3

1 1 ? ? A1 = ? 2 = ? ? s + 5s + 6 ? s = 0 6 1 ? 1 ? A2 = ? =? ? 2 ? s( s + 3) ? s = ?2
1 ? 1 ? = A3 = ? ? ? s( s + 2) ? s = ?3 3 ′ ? ( s + 5) xo (0) + xo (0) ? ′ B1 = ? = 3xo (0) + xo (0) ? s+3 ? ? s = ?2 ′ ? ( s + 5) xo (0) + xo (0) ? ′ B2 = ? = ?2 xo (0) ? xo (0) ? s+2 ? ? s = ?3

′ ′ 2 + 3 + 3 xo ( 0 ) + xo ( 0 ) + ? 2 xo ( 0 ) ? xo ( 0 ) X o (s) = 6 + s s+2 s+3 s+2 s+3 1 ?1 1

1 1 ?2t 1 ?3t 零输入响应 xo (t ) = ? e + e 6 2 3 ′ ′ + [3xo (0) + xo (0)]e?2t ? [2xo (0) + xo (0)]e?3t

(t ≥ 0)

1 1 ? 2t 1 ? 3t xo (t ) = ? e + e 3 6 2 (t ≥ 0)

X o ( s) G( s) = ? X i ( s) (s

( n & a 0 xon ) (t ) + a1 xo ?1 (t ) + L + a n ?1 xo (t ) + a n x0 (t )

& = b0 xi( m ) (t ) + b1 xim ?1 (t ) + L + bm ?1 xi (t ) + a m xi (t )

X o ( s ) b0 s m + b1 s m ?1 + L + bm ?1 s + bm G (s) = = X i ( s ) a 0 s n + a1 s n ?1 + L + a n ?1 s + a n 1

Table 2.2 equivalent spring constant
Dynamic model Equation Laplace transform

Equivalent spring constant

Spring

k

x(t)

f (t ) = kx (t )

F (s ) = kX (s )

k

Dashpot

D

x(t)

& f (t ) = D x (t )

F (s ) = DsX (s )

Ds

Mass

M

x(t)

f (t ) = M&&(t ) x

F (s ) = Ms 2 X (s )

Ms 2

Table 2.3 Equivalent complex impedance

d2 d m 2 xo (t ) + D xo (t ) + kxo (t ) = fi (t ) dt dt

ms 2 X o ( s ) + DsX o ( s ) + kX o ( s ) = Fi ( s )

G (s) = X o (s) 1 = 2 Fi ( s ) ms + Ds + k D

R-L-C无源电路网络的传递函数
d2 d LC 2 uo (t ) + RC uo (t ) + uo (t ) = ui (t ) dt dt

LCs 2U o ( s ) + RCsU o ( s ) + U o ( s ) = U i ( s )

Uo ( s) 1 = G( s) = U i ( s ) LCs 2 + RCs + 1

dn d n ?1 d xo (t ) + a1 n ?1 xo (t ) + L + an ?1 xo (t ) + an xo (t ) dt dt n dt dm d m ?1 d = b0 m xi (t ) + b1 m ?1 xi (t ) + L + bm ?1 xi (t ) + bm xi (t ) ( n ≥ m) dt dt dt

X o ( s ) b0 s m + b1s m ?1 + L + bm ?1s + bm G( s) = = X i ( s ) a0 s n + a1s n ?1 + L + an ?1s + an (n ≥ m)

N ( s ) = a0 s n + a1s n ?1 + L + an ?1s + an
X o ( s) M ( s) 则： G ( s ) = = X i ( s) N ( s)

N(s)=0称为系统的特征方程，其根称为系统 N(s)=0称为系统的特征方程 其根称为系统 的特征根。特征方程决定着系统的动态特性。 N(s)中s的最高阶次等于系统的阶次。 中 的最高阶次等于系统的阶次

X o ( s ) b0 ( s ? z1 )( s ? z2 ) L( s ? zm ) G( s) = = X i ( s ) a0 ( s ? p1 )( s ? p2 ) L( s ? pn )

jω 2 1 -3 -2 -1 -1 -2 G(s)= S+2 (s+3)(s2+2s+2) 的零极点分布图 0 1 2 3 σ

Y (s) = G(s) X (s) = G (s)

y (t ) = g (t ) ? x (t ) = ∫0 g (τ ) x (t ? τ ) dτ
t

= ∫ x (τ ) g (t ? τ ) dτ 式中，当t < 0时，g(t) = x(t) = 0。
t 0

X o ( s ) b0 ( s ? z1 )( s ? z2 ) L( s ? zm ) G( s) = = X i ( s ) a0 ( s ? p1 )( s ? p2 ) L( s ? pn )

b+2c = m v+d+2e = n

s ? zi = s + α i =

1

τi

(τ i s + 1), τ i =

1

αi

1 1 s ? p j = s + β j = (T j s + 1), T j = Tj βj

( s ? zl )( s ? zl +1 ) = ( s + α l ? jωl )( s + α l + jωl ) = s 2 + 2α l s + α l2 + ωl2 = 1 (τ l2 s 2 + 2? lτ l s + 1)
, ?l =

τ l2
1
2 l

τl =

αl α l2 + ωl2

α +ω

2 l

( s ? pk )( s ? pk +1 ) = ( s + β k ? jωk )( s + β k + jωk ) = s 2 + 2β k s + β k2 + ωk2 1 2 2 = 2 (Tk s + 2? k Tk s + 1) Tk

Tk =

1

β +ω
2 k

2 k

, ?k =

βk β k2 + ωk2

K ∏ (τ i s + 1)∏ (τ l2 s 2 + 2? lτ l s + 1) s v ∏ (T j s + 1)∏ (Tk2 s 2 + 2? k Tk s + 1)
j =1 k =1 i =1 d l =1 e b c

G(s) =

e b0 b 1 c 1 d 2 式中， K = ? ∏ ? ∏ 2 ? ∏ T j ? ∏ Tk a0 i =1 τ i l =1 τ l j =1 k =1

1 1 1 K , τ s + 1, τ s + 2ξτ s + 1, , , 2 2 s Ts + 1 T s + 2? Ts + 1
2 2

K

τs+1
τ 2 s 2 + 2?τ s + 1

1 s 1 Ts + 1
1 T 2 s 2 + 2? Ts + 1

X o ( s ) = e ?τs X i ( s )

e 。

?τs

X o ( s) G( s) = =K X i ( s)
z1 ni(t) no(t) z2 ui(t) 运算放大器 R1 uo(t) ()

R2

N o ( s ) z1 G(s) = = =K N i (s) z2

U o ( s) R2 G ( s) = =? =K U i ( s) R1

d T xo (t ) + xo (t ) = Kxi (t ) dt

X o ( s) K G( s) = = X i ( s ) Ts + 1

xi(t) xo(t)

K

D

dxo (t ) + Kxo (t ) = Kxi (t ) D dt d
G(s) = K 1 D = , T= Ds + k Ts + 1 K

dxi (t ) 运动方程为： xo (t ) = τ dt
X o ( s) 传递函数为： G ( s ) = =τ s X i ( s)

dθ i (t ) uo (t ) = K t dt
uo(t)

U o ( s) = Kt s G( s) = Θi ( s )

θi (t)

1 ui (t ) = ∫ i (t )dt + i (t ) R C uo (t ) = i (t ) R
G( s) =

C ui(t) i(t) R uo(t)

RCs Ts = , T = RC RCs + 1 Ts + 1

G (s) = X o (s ) (s = K (τ s + 1) X i (s)

Integral link I t l li k

1 t 运动方程为： xo (t ) = ∫ xi (t )dt T 0
X o ( s) 1 = 传递函数为： G ( s) = X i ( s ) Ts

1 t 1 xo (t ) = ∫0 Adt = At T T

R a ? +

i2(t)
C

ui(t)

uo(t)

duo (t ) RC = ?ui (t ) (t dt 1 1 G( s) = ? = ? , T = RC RCs Ts

d2 d 2 T x (t ) + 2? T xo (t ) + xo (t ) = Kxi (t ), 0 < ? < 1 x 2 o dt dt

X o (s) K G (s) = = 2 2 X i ( s ) T s + 2? Ts + 1

ωn2 1 , ωn = G (s) = 2 2 s + 2? ωn s + ωn T

ωn称为无阻尼固有角频率。

d2 d m 2 xo (t ) + D xo (t ) + Kxo (t ) = f i (t ) dt dt

1 1/ K = 2 2 G (s) = 2 ms + Ds + K T s + 2? Ts + 1

T=

m D ,? = K 2 mK

D < 2 mk k

?τ s

the Signal flow graphs

The block diagram

X(s), x(t) 信号线

X(s) () X(s) X( ) X(s) 引出线 X(s) X(s) X(s) X( )

?

X1(s)±X2(s)

± X2(s)

B A

?

A-B

?

A-B+C

C B A+C-B B A-B+C

A

?
C

A+C

?

A

?

C

?

1 R

1 Cs C

Ri (t ) = ui (t ) ? uo (t )
ui(t) i(t) 无源RC电路网络 R

C

1 uo (t ) = ∫ i (t )dt C

RI ( s ) = U i ( s ) ? U o ( s ) Uo ( s) = 1 I ( s) Cs

uo(t)

1 I ( s ) = [U i ( s ) ? U o ( s )] R 1 Uo ( s) = I ( s) Cs

Ui(s)

?

Ui-Uo

1 R

I(s)

I(s)

1 Cs

Uo(s)

Uo(s)

1 (a) I ( s ) = [U i ( s ) ? U o ( s )] R

1 (b) U o ( s ) = I ( s) Cs C

Ui( ) (s)

?

U(s) U( )

1 R

I(s) I( )

1 Cs

Uo ( ) (s)

fi(t) () fi(t) ()

m1
K1

0 D x(t)

m1
fK1 fD

fm1

m2
K2

0 xo(t)

m2
fK2

fm2

fi(t)

m1
fK1

0 fD x(t)

fm1

m1 &&(t ) = f i (t ) ? f D (t ) ? f K1 (t ) x f K1 (t ) = K1 [x(t ) ? xo (t )]

m2
fK2

0 xo(t)

fm2

? dx(t ) dxo (t ) ? ? f D (t ) = D ? ? dt dt ? ?
m2 &&o (t ) = f K1 (t ) + f D (t ) ? f K 2 (t ) x f K 2 (t ) = K 2 xo (t )

FK1 ( s ) = K1 [ X ( s ) ? X o ( s ) ] FD ( s ) = Ds [ X ( s ) ? X o ( s ) ]

1 ? F ( s ) ? FD ( s ) ? FK1 ( s ) ? ? X ( s) = 2 ? i m1s

1 ? F ( s ) + FD ( s ) ? FK2 ( s ) ? X o (s) = 2 ? K1 ? m2 s FK2 ( s ) = K 2 X o ( s )

FK1(s) Fi(s)

?
FD(s)

1 X(s) m1 s 2

X(s) ()

?

K1 Ds D

FK1(s) FD(s)

Xo(s)

X ( s) =

1 ? F (s) ? FD (s) ? FK1 (s)? 2 ? i ? m1s

FK1 ( s ) = K1 [X ( s ) ? X o ( s )]

(a)

FD ( s ) = Ds [ X ( s ) ? X o ( s ) ]
(b)

FK2(s) FK1(s)

? ?
+
FD(s)

1 m2 s 2

Xo(s) Xo(s) K2 FK2(s)

1 ? F ( s ) + FD ( s ) ? FK 2 ( s ) ? ? X o (s) = 2 ? K1 m2 s (c)

FK 2 ( s ) = K 2 X o ( s )
(d)

FK1(s) Fi(s)

FK2(s)

K2

?
FC(s)

1 X(s) m1s 2

?
Xo(s) ()

FK1(s) K1 Ds FD(s)

? ?
+

1 m2 s 2

Xo(s)

Block diagram reduction

Series connection Xi( ) (s) G1(s) X1(s) G2(s) X2(s) ... Xn-1(s) Gn(s) Xo(s)

Xi(s)

G(s)=G1(s) G2(s) ··· Gn(s)

Xo(s)

Parallel connection

G1(s) + G2(s) . . . Gn(s) +

Xi(s)

+

?

Xo(s)

Xi(s)

G1(s)+ G2(s)+? ? ? + Gn(s)

Xo(s)

Feedback Xi(s) () E(s) Xo(s)

B(s)

± Xi(s)

?

G(s) H(s)

X o ( s) = G( s) E ( s) E ( s) = X i ( s) m B( s) B( s) = H ( s) X o ( s)
X o (s) Φ( s) = X i ( s) G (s) = 1 ± G ( s) H (s)

G(s) 1± G(s)H (s)

Xo(s)

The rules of the block diagram transforming g g

A

? ±
B A

G(s)

C

A

G(s) G( )

?

C

± B C

G(s) G(s) 求和点后移

?

C

A

?

G(s)
1 G (s)

±

± B

B

A G(s) C C A G(s) C A

A

G(s) G(s) 引出点前移

C C

A

G(s)

C A 1 G (s) 引出点后移

H2(s) Xi(s)

? ?

B

+

G1(s)

?
H1(s)

G2(s)

G3(s)

() A Xo(s)

H3(s)

H2(s)G3(s) Xi(s)

? ?

+

G1(s)

?
H1(s)

G2(s)

G3(s)

Xo(s)

H3(s)

2、消去H (s)G (s)反馈回路 2 消去H2( )G3( )反馈回路
Xi(s)

? ?

+

G1(s)

G2 (s) 1+ G2 (s)G3 (s)H2 (s)

G3(s)

Xo(s)

H1(s) H3(s)

3、消去H (s) 3 消去H1( ) 反馈回路
Xi(s) ()

?

G1 ( s)G2 ( s )G3 ( s) 1 ? G1 ( s)G2 ( s) H1 ( s) + G2 ( s)G3 ( s) H 2 ( s)

Xo(s) ()

H3(s)

4、消去H3(s) 反馈回路
Xi(s) Xo(s) G (s)G2(s)G3(s) 1 1?G (s)G2(s)H1(s) +G2(s)G3(s)H2(s) +G (s)G2(s)G3(s)H3(s) 1 1

Signal flow graphs of systems and Mason’s formula

x 2 = x1 + ex 3 x 3 = ax 2 + fx 4 x1 f x 4 = bx 3 x 5 = d 2 + cx 4 + gx 5 dx
1
x2

d a x3 b x4 e f c x5 1

g

x5

d

x1

g 1 x2 e f a x3 b x4 c x5 1 x5

g 1 x1 x2 e f a x3 b x4 c x5 1 x5

d g 1 x1 x2 e f a x3 b x4 c x5 1 x5

g 1 x1 x2 e f a x3 b x4 c x5 1 x5

R1 ui(t) i1(t) C1 uA(t) R2 i2(t) C2 uo(t)

ui (t ) ? u A (t ) i1 (t ) = R1 u A (t ) = 1 ∫ [i1 (t ) ? i2 (t )]dt C1

u A (t ) ? uo (t ) i2 (t ) = R2 1 uo (t ) = ∫ i2 (t )dt C2

U i ( s) ? U A ( s) I1 ( s ) = R1 1 [I1 ( s ) ? I 2 ( s )] U A ( s) = C1s U A ( s) ? Uo ( s) I 2 ( s) = R2 U o ( s) = 1 I 2 ( s) C2 s

a)

U i (s) ? U A (s) I1 ( s ) = R1
Ui(s) 1

1 R1
I1(s) -1 UA(s)

1 b) U A ( s ) = ) [I1 (s) ? I 2 ( s)] C1s
-1 1 I1(s)
1 C1s

UA(s) ()

I2(s) ()

c)

U A ( s) ? U o ( s) I 2 (s) = R2
1 UA(s)
1 R2

Uo(s) I2(s) -1

d)

1 U o (s) = I 2 ( s) C2 s

1 C2 s
I2(s) Uo(s)

-1 1 Ui(s) 1

1 R1

1 I1(s) -1

1 C1s

1 UA(s)

1 R2

1 C2 s
I2(s) -1

1 Uo(s)

Ui(s) 1

1 R1

1 C1s

-1 1 UA(s)
1 R2

1 C2 s

1 Uo(s) ()

I1(s)–I2(s) -1

I2(s) -1

Xi( ) (s) 1

? H(s) E(s) E( ) G(s) Xo ( ) (s) 1 -H(s) Xo ( ) (s)

?

E(s)

G(s)

Xo(s)

※ 线与节点对应关系：
G4(s) G1(s) E1 E3 G1 -G2 G G4(s) G1(s) E1 E3 G4 G3 G4 G1 E1 1 E3 G3 -G2

?

G3(s) G2(s)

E2

?

E3

G3(s) G2(s) ()

E2

1 P = ∑ Pk Δ k Δ k

Δ = 1 ? ∑ La + ∑ Lb Lc ?
a b, c

d , e, f

∑ Ld Le L f

+L

∑ La
a
b, bc

—所有不同回路的传递函数之和；

∑ Lb Lc —每两个互不接触回路传递函数

d , e, f

Δk— 第 条前向 路特 式的余 第k条前向通路特征式的余因子，即对于 流图的特征式Δ，将与第k 条前向通路相 接触的回路传递函数代以零值，余下的Δ 即为Δk。

-1 Ui(s) 1

1 R1
I1(s)–I2(s) (s) -1

1 C1s

1 UA(s)

1 R2

1 C2 s
I2(s) -1

1 Uo(s)

1 1 1 1 P = ? ? ? 1 R1 C1s R2 C2 s

L3 Ui(s) 1

-1
1 C1s

1 R1
I1(s)–I2(s) L1 -1

1 UA(s) L2

1 R2

1 C2 s
I2(s) -1

1 Uo(s)

1 1 L1 = ? ? R1 C1s
1 1 L2 = ? ? R2 C2 s

1 1 L3 = ? ? R2 C1s

Δ = 1 ? ( L1 + L2 + L3 ) + L1L2 1 1 1 1 1 = 1+ + + + ? R1C1s R2C2 s R2C1s R1C1s R2C2 s

1 1 ∑ Pk Δ k = Δ P1Δ1 Δ k

1 = R1R2C1C2 s 2 + ( R1C1 + R2C2 + R1C2 ) s + 1

Xo(s) () = Xi(s) G1G 2G3G 4G5+ G1G6G 4G5 + G1G 2G7(1+ G 4H1) 1+ G 4H1+ G 2G7 H2 + G6G 4G5H2 + G 2G3G 4G5H2 + G 4H1G 2G7H2

N(s) Xi(s)

?

ε (s)

G1(s)

+ +

?

G2(s)

Xo(s) ()

B(s) H(s)

Xi(s)到Xo(s)的信号传递通路称为前向通道； Xo( )到B( )的信号传递通路称为反馈通道 (s)到B(s)的信号传递通路称为反馈通道；

B( s ) GK ( s ) = = G1 ( s )G2 ( s ) H ( s ) ε ( s)

xi(t)作用下系统的闭环传递函数
Xi(s)

?

ε (s)

G1( ) (s) H(s)

G2( ) (s)

Xo1(s)

B(s) xi(t)作用下的闭环系统

X 01 ( s ) G1 ( s )G2 ( s ) Φ i ( s) = = X i ( s ) 1 + G1 ( s )G2 ( s ) H ( s )

Xi(s)

?
H(s)

1 G2(s) G1(s)

ε (s)

1 Φε i ( s) = = X i ( s ) 1 + G1 ( s )G2 ( s ) H ( s )

ε i ( s)

n(t)作用下系统的闭环传递函数
N(s)

?
G1(s)

G2(s) H(s)

Xo2(s)

n(t)作用下的闭环系统 ()

X 02 ( s ) G2 ( s ) Φ N ( s) = = N ( s ) 1 + G1 ( s )G2 ( s ) H ( s )

N(s)

?

+

G2(s)

H(s) G1(s)

-1 1

ε (s)

? G2 ( s ) H ( s ) Φε N ( s) = = N ( s ) 1 + G1 ( s )G2 ( s ) H ( s )

ε N ( s)

) 系统的闭环传递函数 Φ i (s )、 Φ ε i (s、 Φ N (s ) 及 Φ ε N (s ) 具有相 的特征多项式 具有相同的特征多项式： 1+G1(s)G2(s)H(s) 其中G1(s)G2(s)H(s)为系统的开环传递函数。 闭环传递函数的极点相同。

X o ( s ) = X 01 ( s ) + X 02 ( s ) = G1 ( s )G2 ( s ) G2 ( s ) X i ( s) + N ( s) 1 + G1 ( s )G2 ( s ) H ( s ) 1 + G1 ( s )G2 ( s ) H ( s )

X o ( s) ≈ 1 X i ( s) H ( s)

? L ? J = m? ? ? 2π ?
2

L — 丝杠螺距，即丝杠每转一周工作台 移动的直线距离。

dv TL ( t ) ? 2 = m ? L 2π dt
ω?L v= 2π

mL d ? ω ? L ? L ? dω L ? d 2θ ? ? ∴ TL ( t ) = = m? ? = m? ? ? ? 2π dt ? 2π ? 2π ? dt 2π ? dt 2 ? ?

2

2

T1、T2：转矩 T 转矩 θ1、θ2：角位移

ω1、ω2：角速度
z2 T2 θ2 z1、z2：齿数 r1、r2：齿轮分度圆半径

T1

T2 θ2

J2

T ：输入转矩 J1、J2 ：齿轮（包括轴）的转动惯量 D1、D2：啮合齿轮、支承粘性阻尼系数

d 2θ1 (t ) dθ1 (t ) 齿轮1： 齿轮1 T (t ) = J1 + D1 + T1 (t ) 2 dt dt d 2θ 2 (t ) dθ 2 (t ) 齿轮2： T2 (t ) = J 2 + D2 2 dt dt

z1 z1 利用： T1 (t ) = T2 (t ), θ 2 (t ) = θ1 (t ) z2 z2

? z1 ? d θ1 (t ) ? z1 ? dθ (t ) ? ? J2 ? ? D2 1 T1 (t ) = ? ? +? ? 2 dt dt ? z2 ? ? z2 ?
2

2

2

d 2θ1 (t ) dθ1 (t ) T (t ) = J I + DI 2 dt dt

DI = D1 + ( z1 z 2 ) D2
2

—— 等效折算到输入端的粘性阻尼系数 其中， (z1 z2 )2 D2 为齿轮2 一侧的粘 性阻尼系数折算到齿轮1一侧的等效 性阻尼系数折算到齿轮1 侧的等效 粘性阻尼系数
T1 θ 2 z1 ω 2 = = = 显然，利用 T2 θ1 z 2 ω1

，齿轮2 一

KI =

1 1 1 + K1 (z1 z 2 )2 K 2

III

xo(t) () Cm J3, K3, D3

z2
II

T
I

Km 丝杠L J2, K2, D2 z3 z1 J1, K1, D1

θi

d 2θ m (t ) dθ m (t ) +D + K [θ m (t ) ? θ i (t )] = 0 J 2 dt dt

θm(t)为工作台位移xo(t)折算到I轴上的等

1 2π θ m (t ) = xo (t ) i1i2 L z3 z1 i1 = , i2 = ，L为丝杠螺距 z2 z4

J、D、K分别为工作台及各轴折算到I轴上的 等效总转动惯量、等效总粘性阻尼系数及等效 总刚度系数。 总刚度系数
J=

)m D = D + i D + (i i ) D + (i i L ) D 2π
2 J1 + i1 J 2
2 1

+ (i1i2 ) J 3 + i1i2 L
2
2 1 2 3 1 2

(

2

2

1

2

m

K=

1 1 1 1 1 + 2 + + 2 2 K1 i1 K 2 (i1i2 ) K 3 i i L 1 2 2π K m

(

)

d 2 xo (t ) dxo (t ) L J +D + Kxo (t ) = i1i2 Kθ i (t ) 2 dt dt 2π

i1i2 K L X o ( s) 2π = = 2 G ( s) = Θ i ( s ) Js + Ds + K J 2π s2 + D s +1 K K i1i2 L

m2 K2 m1 xi(t) K1 D

xo(t)

x(t)

′ ′ m2 xo′(t ) = ? K 2 [xo (t ) ? x(t )] ? D[xo (t ) ? x′(t )] ′ m1 x′′(t ) = K 2 [xo (t ) ? x(t )] + D[xo (t ) ? x′(t )] + K1 [xi (t ) ? x(t )]

(m s
2

2

+ D + K 2 X o ( s ) = (D + K 2 ) X ( s ) Ds Ds + Ds + K1 + K 2 X ( s ) = (Ds + K 2 )X o ( s ) + K1 X i ( s )
1 m1s 2 + Ds + K1 + K2

)

(m s
1

2

)

Xi(s) ()

K1

X(s) ()

Ds+ K2 m2 s 2 + Ds + K2

Xo(s) ()

Ds + K 2
X o (s) K1 (Ds + K 2 ) = X i ( s ) m1m2 s 4 + (m1 + m2 ) Ds 3 + [K1m2 + (m1 + m2 ) K 2 ]s 2 + K1 Ds + K1 K 2

Matlab简介： Matlab简介
? 1980年前后，美国moler博士构思并开 发； ? 最初的matlab版本是用fortran语言编 写，现在的版本用c语言改写； ? 1992年推出了具有重要意义的matlab 版本；并于 年推出了其 4.0版本；并于1993年推出了其 windows平台下的微机版，目前7.0版, 甚至8.0版是比较新的版本。 甚至8 0版是比较新的版本

B(s) b0s + b1s +L+ bm?1s + bm G(s) = = n A(s) a0s + a1sn?1 +L+ an?1s + an
m

m?1

>> p=[1 -12 0 25 126] p= 1 -12 0 25 126

z=[1 2]; p=[-1 2 3] p=[ 1 -2 -3]; k=4; sys=zpk(z,p,k) k( k)

4 ( 1) ( 2) (s-1) (s-2) ----------------(s+1) (s+2) (s+3) ( 1) ( 2) ( 3)

2 3 den = 2 7 3 13 2 14 10 4

B(s) b0sm + b1sm?1 +L+ bm?1s + bm 设： F(s) = = n A(s) a0s + a1sn?1 +L+ an?1s + an

MATLAB提供函数residue用于实现部分分式 提供函数 用 实现部分分式 展开，其句法为：

[r, p, k] = residue(num, den)

F (s) = r (1) r (1) r ( n) + +L + + K ( s) s ? p(1) s ? p(2) s ? p ( n)

r ( j) r ( j + 1) r ( j + q ? 1) + +L + 2 s ? p ( j ) [s ? p ( j ) ] [s ? p ( j ) ]q

s 4 + 11s 3 + 39s 2 + 52s + 26 例：求 F (s) = 4 s + 10s 3 + 35s 2 + 50s + 24

>> num=[1 11 39 52 26]; p= [ ]; >> den=[1 10 35 50 24]; -4.0000 >> [r,p,k]=residue(num,den) -3.0000 r= -2.0000 1.0000 1 0000 -1.0000 1 0000 2.5000 -3.0000 -3 0000 k= 0.5000 1 展开式为： 展开式为 F ( s) = 1 + 2.5 + ? 3 + 0.5 + 1 s + 4 s + 3 s + 2 s +1

s 5 + 10 s 2 + 5s + 6 例：求 F ( s ) = 4 s + 5s 3 + 9 s 2 + 7 s + 2

>> num=[1 0 0 10 5 6]; >> den=[1 5 9 7 2]; >> [ k] [r,p,k]=residue(num,den) id ( d ) r= -4.0000 -4 0000 20.0000 -20.0000 10.0000
s+2

p= -2.0000 -1.0000 1 0000 -1.0000 -1.0000 -1 0000 k= 1 -5
( s + 1)

s + 1 ( s + 1)

[num, den] = residue(r, p, k)

>> [num, den] = residue(r, p, k) num = 10 70 150 96 den = 1 10 35 50 24

10 s 3 + 70 s 2 + 150 s + 96 F ( s) = 4 s + 10 s 3 + 35s 2 + 50 s + 24

num1=[1]; den1=[1,0]; [ , ]; num2=[1]; den2=[1,2]; [numc,denc]=series(num1,den1,num2,den2); [numb,denb]=parallel(num1,den1,num2,den2); [numf,denf]=feedback(num1,den1,num2,den2,-1); [numf denf]=feedback(num1 den1 num2 den2 1); [z,p,k]=tf2zp(numf,denf)

2-1, 2-2, 2-9(b), 2 10(a) 2-11(c) 2 1 2 2 2 9(b) 2-10(a), 2 11(c)

2-6(b), 2-8,2-12(b), 2-19 选做: 2-3, 2-26(b)

THE

END

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