控制工程基礎(chǔ)第二版課后答案

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機(jī)械掌握工程基礎(chǔ)答案提示

第二章系統(tǒng)的數(shù)學(xué)模型

2-1試求如圖2-35所示機(jī)械系統(tǒng)的作用力F(t)與位移y(t)之間微分方程和傳遞函數(shù)。

F(t)

圖2-35題2-1圖

解：依題意：m?

dy?t?

2

dt

2

2

?

ab

?F?t??f?

dy?t?dt

?ky?t?

故m

dy?t?dt

2

?f?

dy?t?dta

?ky?t??

ab

?F?t?

傳遞函數(shù):G?s??

Y?s?F?s?

?

ms

2

b

?fs?k

2-2對(duì)于如圖2-36所示系統(tǒng)，試求出作用力F1(t)到位移x2(t)的傳遞函數(shù)。其中，f為粘性阻尼系數(shù)。F2(t)到位移x1(t)的傳遞函數(shù)又是什么？

圖2-36題2-2圖

解：依題意：

dx1?t??dx1?t?dx2?t??

??F?kxt?f??mm對(duì)1：11112?dtdt?dt??

2

對(duì)兩邊拉氏變換：F1?s??k1x1?f?sX

1

?s??sX2?s???2

m1sX1?s?①

dx2?t??dx?t?dx2?t??

對(duì)m2：F2?t??f?1????kxt?m2222?dtdtdt??

2

對(duì)兩邊拉氏變換：F2?s??f?sx1?s??sx2?s???k2x2?s??m2s2X2?s?②?m1s2?fs?k1x1?s??fsx2?s??F1?s?故：?2

??fsx1?s??m2s?fs?k2x2?s??F2?S?

??

??

2

?F1?s??m2s?fs?k2?fsF2?s??x1?s??222?m1s?fs?k1m2s?fs?k2??fs?故得：?2

fsF1?s??F2?s?m1s?fs?k2

?x?s??2222???ms?fs?kms?fs?k?fs1122?

??

?

?

故求F1?t?到x2?t?的傳遞函數(shù)令：F2?s??0

G1?s??

x2?s?F1?s?

4

?

fs

?

?ms

1

2

?fs?k1??m2s?fs?k2???fs?

2

2

fs

m1m2s?f?m1?m2?s??m1k2?m2k1?s?f?k1?k2?s?k1k2

3

2

求F2?t?到x1?t?的傳遞函數(shù)令：F1?s??0

G1?s??

x1?s?F2?s?

4

?

fs

?ms

1

2

?fs?k1??m2s?fs?k2???fs?

2

2

?

fs

m1m2s?f?m1?m2?s??m1k2?m2k1?s?f?k1?k2?s?k1k2

3

2

2-3試求圖2-37所示無源網(wǎng)絡(luò)傳遞函數(shù)。

o

圖2-37題2-3圖

解（a）系統(tǒng)微分方程為

1C

?i?t?dt?i?t?R

1

2

1

ui?i2?t?R1?i?t?R2u0?i?t?R2

i?t??i1?t??i2?t?

拉氏變換得1Cs

I1?s??R1I2?s?

Ui?s??I2?s?R1?I1?s?R2U0?s??I1?s?R2I?s??I1?s??I2?s?

R2

消去中間變量I1?s?,I2?s?,I?s?得：G?s??

U0?s?Ui?s?

?

R2?CsR1?1?R1?R2?CsR1?1?

?

R1?R2

?R1Cs?1?

Cs?1

R1R2R1?R2

（b）設(shè)各支路電流如圖所示。

系統(tǒng)微分方程為

ui?t??R1i3?t??uR1i3?t??Lu0?t??

1C2

di2?t?

1

?t??1??2??3??4??5??6?

dt

4

?i?t?dt

di5?t?dt

u0?t??L2

u0?t??R2i6?t?

i2?t??i3?t??i4?t??i5?t??i6?t?

由（1）得：Ui?s??R1I3?s??Uo?s?由（2）得：R1I3?s??L1sI2?s?由（3）得：Uo?s??

1C2s

i4?s?

由（4）得：Uo?s??L2sI5?s?由（5）得：Uo?s??R2I6?s?

由（6）得：I2?s??I3?s??I4?s??I5?s??I6?s?

故消去中間變量I1?s?,I2?s?,I3?s?,I4?s?,I5?s?,I6?s?得：L2

?Us??L1s?1?

U?

L?L?

o?12?R1

?

i?s?L1L2LL?R?R?LC2?1212

2ss?1

1?L2?L1?L2?R1R2

2-4證明L?cos?t??

ss2

??

2

證明：設(shè)f?t??cos?t

由微分定理有L?d2f??t????s2

F?s??sf?0??(1)

?0??dt2

f?

2

由于f?0??cos0?1，f

?1?

?0????sin0?0，

df?t?2

dt

2

???cos?t將式（2）各項(xiàng)帶入式（1）中得

L?2

???cos?t??

?s2F?s??s即??2

F?s??s2

F?s??s

整理得F?s??ss2

??

2

2-5求f(t)?

12

t2的拉氏變換。

解：F?s??L?12?

?2t?????

?12

?st

2

t2e

?st

dt?

1

?12

?0

s

3

?st?e

d?st?

令st?x，得

F?s??

12s

3

?

?2?x

xe

dx

由于伽馬函數(shù)??n?1???

?

xne?x

dx?n!，在此n?2

所以F?s??

12s

3

2!?

1s

3

2-6求下列象函數(shù)的拉氏反變換。（１）X(s)?

5s?3

(s?1)(s?2)(s?3)

1）2）（

（

（２）X(s)?

s?2s?3(s?1)

3

2

（３）X(s)?解：（1）X(s)?

1

s(s?1)(s?2)

3

5s?3

(s?1)(s?2)(s?3)

?

A1s?1

?

A2s?2

?

A3s?3

A1??X(s)(s?1)?s??1?

5??1??3(?1?2)(?1?3)

??1

同理A2?7，A3??6

X(s)??

1s?1

?

7s?2

?

6s?3

拉式反變換得

x(t)??e

?t

?7e

?2t

?6e

?3t

2

（2）X(s)?

s?2s?3(s?1)

3

2

?

?s?1?

?2

3

(s?1)

?

2(s?1)

3

?

1s?1

拉式反變換得

x(t)?te

2

?t

?e

?t

A1(s?1)

3

（3）X(s)?

1

s(s?1)(s?2)

3

??

A2(s?1)

2

?

A3(s?1)

?

A4s

?

A5s?2

A1?

1s(s?2)

s??1

?

1?1(?1?2)

??1

A2?

?d?1

??ds?s(s?2)?1d

2

?

s??1

??2s?2?s(s?2)

2

2s??1

?0

??1

A3??2?

2ds?s(s?2)?

s??1

1d???2s?2????22?2ds?s(s?2)?

?

s??1

s(s?2)?(s?1)