Ch2 LTI系统的时域分析
(Time-Domain Representations for
LTI System)
Ch2.1 引言
（Introduction）
本章重点:
• The Convolution Sum and The Convolution Integral
卷积和与卷积积分
• Interconnections of LTI system 系统的互联
• Relations between LTI system properties and the impulse
response
LTI系统的特性与冲激响应的关系
• Solving the response of the LTI system
LTI系统响应的求解
• Block Diagram Representations LTI系统的方框图表示
• State-Variable Descriptions of LTI system 系统的状态变量描述
Impulse response
冲激响应
Definition：
Impulse Response: The Output of a LTI system due to a
unit impulse signal input applied at time t=0或 n=0.
LTI系统对 n=0或t=0时刻的输入单位冲激信号或单位脉冲
信号的响应。记为h[n]或h(t)。
Ch2.2 The Convolution Sum
卷积和
任意离散信号x(n)可以表示为时移的脉冲序列的加权叠加。
x(n)
x(0) (n)
x(1) (n  1)
-2
-1
x(2) (n  2)
x(1) (n  1)
0
1
2
3
n
x(2) (n  2)
x(n)    x(1) (n  1)  x(0) (n)  x(1) (n  1)    x(k ) (n  k )  


 x(k ) (n  k )
k  
The Convolution Sum
卷积和
Illustration of the convolution sum.
(a) LTI system with impulse response h[n] and input x[n].
(b) The decomposition
of the input x[n] into a
weighted sum of timeshifted impulses results
in an output y[n] given
by a weighted sum of
time-shifted impulse
responses.
The Convolution Sum
卷积和
输入为时移脉冲序列的加权和：
x[n] 

 x[k ] [n  k ]
k  
则输出信号为时移脉冲响应的加权和：
y[n] 

 x[k ]h[n  k ]
k  
称为卷积和，记为：
x[n]  x[n] * [n]
（信号与的卷积就是其本身）
y[n]  x[n] * h[n]
（系统响应=输入信号*单位脉冲响应）
Ch2.3 Convolution Sum Evaluation Procedure
离散卷积和的计算过程
卷积和的定义：
y[n] 

 x[k ]h[n  k ]
k  
定义中间信号：
wn [k ]  x[k ]h[n  k ]
则系统的输出：
y[n] 

 w [k ]
k  
n
Convolution Sum Evaluation Procedure
离散卷积和的计算过程
Convolution Sum Evaluation Procedure
离散卷积和的计算过程
Example 2.2. (a) The input signal x[k] above the reflected and time-shifted impulse
response h[n – k], depicted as a function of k. (b) The product signal w5[k] used to
evaluate y [–5]. (c) The product signal w5[k] used to evaluate y[5]. (d) The product
signal w10[k] used to evaluate y[10].
Convolution Sum Evaluation Procedure
离散卷积和的计算过程
Convolution Sum Evaluation Procedure
离散卷积和的计算过程
Example 2.3.
(a) The system impulse response
h[n].
(b) The input signal x[n].
(c) The input above the reflected
and time-shifted impulse
response h[n – k], depicted as a
function of k.
(d) The product signal wn[k] for
the interval of shifts 0  n  3.
(e) The product signal wn[k] for
the interval of shifts 3 < n  9.
(f) The product signal wn[k] for
the interval of shifts 9 < n  12.
(g) The output y[n].
Example 2.4. (a) The input signal x[k] depicted above the reflected and time-shifted
impulse response h[n – k].
(b) The product signal wn[k] for –4  n.
(c) The output y[n] assuming that p = 0.9 and b = 0.8.
Cash flow into an investment. Deposits of $1000 are made at the start of
each of the first 10 years, while withdrawals of $1500 are made at the start
of each of the second 10 years.
Example 2.5. (a) The input signal x[k] depicted above the reflected and time-shifted
impulse response h(n – k). (b The product signal wn[k] for 0  n  9. (c) The product
signal wn[k] for 10  n  16. (d) The product signal wn[k] for 17  n.
(e) The output y[n] representing the value of the investment immediately
after the deposit or withdrawal at the start of year n.
Problem
Evaluate the following discrete-time convolution sums:
( a)
y[n]  u[n] * u[n  3]
(b)
y[n]  (u[n]  2u[n  3]  u[n  6]) * (u[n  1]  u[n  10])
(c)
y[n]  x[n] * h[n], where x[n] and h[n] are
Ch2.4 卷积积分
(Convolution Integral)
Impulse Response（冲激响应）：
The Output of a continuous-time LTI system due to a unit
impulse signal input applied at time t=0.
连续时间LTI系统对 t=0时刻的输入单位冲激信号的响应，
记为h(t)。
N 阶连续时间LTI系统的冲激响应h(t)满足
h (t )  an1h
(n)
 bm
( m)
( n 1)
(t )  bm1
(t )    a1h' (t )  a0 h(t )
( m 1)
(t )    b1 ' (t )  b0 (t )
Convolution Integral(卷积积分)
连续信号可以表示为一系列位移的冲激信号的加权和
f (t)
f ( kD )
 D 0 D 2D
kD (k 1)D
t
连续信号表示为冲激信号的迭加
f (t )    f (0)[u(t )  u(t  D)]  f (D)[u(t  D)  u(t  2D)]  
 f (kD)[u(t  kD)  u(t  kD  D)]  
[u (t )  u (t  D)]
[u (t  D)  u (t  2D)]
f (t )    f (0)
D  f (D)
D 
D
D
 f (kD)
f (t ) 


k  
[u (t  kD)  u (t  kD  D)]
D 
D
[u (t  kD)  u (t  kD  D)]
f (kD)
D
D
当D0时，kD，Dd，且
[u (t  kD)  u (t  kD  D)]
  (t   )
D


f (t )  
f ( ) (t   )d
Derivation of Convolution Integral(卷积积分的推导)
x(t )
LTI
 (t )
Time Invariance
Scaling
y (t )
h(t )
 (t   )  h(t   )
x( ) (t   )  x( )h(t   )
Superposition

x(t )   x( )   (t   )d

 x(t )   (t )

y(t )   x( )  h(t   )d

 x(t )  h(t )
Ex: Evaluate the convolution integral for a continuous-time
LTI system with impulse response h(t)=2e-3tu(t) and input
x(t)=3u(t)。已知LTI系统的冲激响应和输入，求系统的响应。

Solution: yzs (t )  x(t )  h(t )   x( )  h(t   )d

=  3u( )  2e3(t  )u(t   )d

t

  3  2e- 3(t - ) d
= 0

0
t
3 t

6e   e3 d
= 
0

0
2(1  e 3t )
=
0
= 2(1  e3t )u(t )
t  0
t  0
t  0
t  0
t0
t0
Ch2.5 Convolution Integral Evaluation Procedure
连续卷积积分的计算过程
卷积积分的定义：
y(t )  x(t )  h(t )  


x( )h(t   )d
定义中间信号：
wt ( )  x( )h(t  )
则系统的输出：
y(t )  


wt ( )d
Convolution Integral Evaluation Procedure
连续卷积积分的计算过程
Example 2.6: Evaluate the convolution integral for a system
with Input signal x(t) and impulse response h(t), respectively.
Convolution Integral Evaluation Procedure
连续卷积积分的计算过程
Evaluation of the convolution integral for Example 2.6. (a) The input x() depicted
above the reflected and time-shifted impulse response. (b) The product signal wt()
for 1  t < 3. (c) The product signal wt() for 3  t < 5. (d) The system output y(t).
Example2.7. Consider the RC circuit system and assume that the circuit’s
time is RC =1s. The impulse response of the system is
h(t)=e-tu(t), Use convolution to determine the voltage across the capacitor,
y(t), resulting from an input voltage x(t) = u(t) - u(t-2), as output.
RC circuit system with the voltage source x(t) as input and the voltage
measured across the capacitor y(t), as output.
Convolution Integral Evaluation Procedure
连续卷积积分的计算过程
Evaluation of the convolution integral for Example 2.7. (a) The input x()
superimposed over the reflected and time-shifted impulse response h(t – ),
depicted as a function of . (b) The product signal wt() for 0 t < 2. (c) The
product signal wt() for t  2. (d) The system output y(t).
Signals for Problem 2.5.
Example 2.10: Radar range measurement. (a) Transmitted RF pulse. (b) The
received echo is an attenuated and delayed version of the transmitted pulse.
(a) Impulse response of the matched filter for processing the received signal.
(b) The received signal r()
superimposed on the reflected
and time-shifted matched filter
impulse response hm(t – ),
depicted as functions of . (c)
Matched filter output x(t).
Ch2.6 LTI系统的互联
(Interconnection of LTI systems )
1. parallel connection of LTI systems(并联)：
h1(t)
x(t )
y (t )
x (t )
h1(t)+h2(t)
h2(t)
两个系统并联
等效于
一个总系统
Distributive property(分配律):
x(t)  h1(t) + x(t)  h2(t) = x(t)  [ h1(t) + h2(t) ]
y (t )
2. cascade connection of systems (级联)：
x (t )
h1(t)
h2(t)
两个系统级联
h1(t)*h2(t)
x (t )
y (t )
等效于
一个总系统
Associative Property(结合律):
[ x(t)  h1(t) ]  h2(t) = x(t)  [ h1(t)  h2(t) ]
两个级联系统的先后次序可交换：
x(t )
h2(t)
h1(t)
y (t )
Communicative Property(交换律)：
h1(t)  h2(t) = h2(t)  h1(t)
y (t )
Ex: Determine the impulse response of the system, where
h1(t) = e-3t u(t)，h2(t) =δ(t -1) ，h3(t) = u(t)。
h1(t)
h2(t)
f(t)
+
h3(t)
solution：
h(t )  h1 (t ) * h2 (t )  h3 (t )
  (t  1) * e 3t u(t )  u(t )
 e 3(t 1) u(t  1)  u(t )
y(t)
Ex: Determine the impulse response of the system, where
h1[k] =2ku[k]，h2[k] = d[k-1] ，h3[k] = 3ku[k]， h4[k] = u[k]。
h1[k]
x[k]
+
h2[k]
h4[k]
h3[k]
solution：
h[k ]  h1[k ]   [k ]  h2[k ]  h3[k ] h4[k ]
 2(2)k u[k ]  [1.5(3)k 1  0.5]u[k 1]
y[k]
Ch2.7 用冲激响应表示LTI系统的特性
(Properties of the Impulse Response
Representations for LTI systems)
• Causal System(因果系统)
• Stable System(稳定系统)
Causal System(因果系统)
Definition:
The output of a causal LTI system depends only on
past or present values of the input .因果系统的输出只取
决于过去时刻及当前时刻的输入。
因果连续时间LTI系统 ( Continuous-time causal LTI system)的
单位冲激响应必须满足:
h(t )  0, t  0
因果离散时间LTI系统( Discrete-time causal LTI systems)的单
位脉冲响应必须满足:
h[n]  0, n  0
Stable System(稳定系统)
Definition: A system is bounded input-bounded output
(BIBO) stable if the output is guaranteed to be bounded
for every bounded input. 对任意的有界输入, 系统的输出也
有界，则称该系统是稳定系统。
稳定的连续时间LTI系统(Stable Continuous-time LTI systems)
的单位冲激响应必须满足:



h( ) d  
稳定的离散时间LTI系统(Stable Discrete-time LTI systems )的
单位脉冲响应必须满足:


n  
h[n]  
Ex: Moving Average System(滑动平均系统).
y(n)=[x(n+1)+x(n)+x(n-1)]/3. Is this system causal?
Stable?是否因果?是否稳定?
solution： The Impulse Response(单位脉冲响应) is
h(n)=[(n+1)+ (n)+ (n-1)]/3
So, This system is not causal.(系统非因果)
对h[k]求和，可得

1
k  
k  1
 h[k ]  
1
1
3
So, This system is stable.(系统稳定)
Ex: A causal LTI system with impulse response h(t)
= eat u(t). is the system stable? 已知因果LTI系
统的单位冲激响应，判断系统是否稳定。
solution:



h( ) d  

0
1 a
e d  e
a
a

0
当 a<0 时，



1
h( ) d 
a
stable (稳定)
h( ) d  
not stable(不稳定)
当 a0 时，



Ch2.8 Step Response(阶跃响应)
Definitions:
The Step response is defined as the output due to a unit
step input signal.单位阶跃信号作用在LTI系统上产生的响应
称为单位阶跃响应，用符号s[k]或s(t)表示。
s ( n ) (t )  an 1s ( n 1) (t )    a1s' (t )  a0 s(t )
 bmu ( m ) (t )  bm1u ( m1) (t )    b1u ' (t )  b0u (t )
Step Response(阶跃响应)
阶跃响应的求解方法:
利用单位阶跃响应与单位冲激（脉冲）响应的关系
s[n] 
n
 h[k ]
k  
t
s(t )   h( )d

h[n]=s[n]s[n1]
ds (t )
h(t ) 
dt
Ex: A discrete-time LTI system with the impulse response
h[n] = [ -(-1)n + 2(-2)n ] u[n]. Find the step response of
this system.已知离散时间LTI系统的单位脉冲响应，求系
统的单位阶跃响应 s[n]。
solution：
利用h[n]与s[n] 的关系，可得
n
n
k 0
k 0
s[n]   (1) k  2 (2) k
1
4
1
n
n
 [ (1)  (2)  ]u[n]
2
3
6
Ex: A continuous-time LTI system with the impulse
response h(t) = 2e3t u(t) , Find the step response of
this system.已知连续时间LTI系统的单位冲激响应，求系
统的单位阶跃响应 s(t)。
solution：
利用冲激响应与阶跃响应的关系，可得
t
s(t )   2e
0
3
d
2
 (1  e 3t )u(t )
3
Ch2.9 Differential and Difference Equation Representations of LTI Systems
连续LTI系统的微分方程描述
离散LTI系统的差分方程描述
• 连续时间系统用N阶常系数微分方程描述
y ( n ) (t )  an 1 y ( n1) (t )    a1 y (t )  a0 y (t )
 bm x ( m ) (t )  bm1 x ( m1) (t )    b1 x (t )  b0 x(t )
ai 、 bi为常数。
• 离散时间系统用N阶常系数差分方程描述
n

i 0
ai 、 bi为常数。
m
ai y[k  i]   b j x[k  j ]
j 0
Example of an RLC circuit described by a differential equation.
The voltage drops around the loop gives
dy (t ) 1 t
dx (t )
Ry (t )  L
  y ( )d 
dt
C 
dt
Differentiating both sides of this equation with respect to
results is
1
dy(t )
d 2 y (t ) dx(t )
y (t )  R
L

2
C
dt
dt
dt
Example 2.16. A system is described by the difference equation
y(n)  1.143y(n  1)  0.4128y(n  2)  0.0675x(n)  0.1349x(n  1)  0.0675x(n  2)
Write the recursive formula (递归方程)that computes the present output
from the past outputs and the current inputs. Use a computer to
determine the step response of the system, the system when the input
is zero and the initial conditions(初始条件) are y[-1]=1 and y[-2]=2,

and the output in response to sinusoidal inputs x1 (n)  cos( n)u (n) ,

7
10
and x3 (n)  cos( 10 n)u (n) , assuming zero initial
5
conditions. Last find the output of the system if the input is the weekly
closing price of Intel stock, assuming zero initial conditions.
x2 (n)  cos(
n)u (n)
Weekly closing price of Intel stock.
Illustration of the solution to Example 2.16.
(a) Step response of system.
(b) Output due to nonzero initial conditions
with zero input.
(c) Output due to x1[n] = cos (1/10n).
(d) Output due to x2[n] = cos (1/5n).
(e) Output due to x3[n] = cos(7/10n).
Ch2.10 Solving Differential and Difference Equation
微分和差分方程求解
x(t )
连续LTI
y (t )
x(n)
y (n)
离散LTI
输出 = 齐次解(homogeneous solution) + 特解(particular solution )
y(t )  yh (t )  y p (t )
y(n)  yh (n)  y p (n)
The homogeneous solution is obtained by setting all
terms involving the input to zero. It is depends on the
roots of the system’s characteristic equation.齐次解yh(t)
的形式由齐次方程的特征根确定。
homogeneous solution(齐次解)
齐次解yh(t)的形式:
(1) 特征根是不等实根s1, s2, , sn
yh (t )  K1es1t  K2es2t   Knesnt
(2) 特征根是等实根s1=s2==sn
yh (t )  K1es t  K2tes t   Knt n1es t
(3) 特征根是成对共轭复根 si   i  ji , i  n / 2
yh (t )  e1t (K1 cos1t  K1 sin 1t )   e it (Ki cosit  Ki sin it )
particular solution(特解)
The particular solution is obtained by assuming an
output of the same form as the input.特解yp(t)的形式由
方程右边激励信号的形式确定
常用激励信号对应的特解形式
Ex: Find the complete response of a second-order LTI system described
by the differential equation
y" (t )  6 y' (t )  8 y(t )  f (t ), t  0
if the initial condition is y(0)=1, y’(0)=2, and the input is x(t)=e-t u(t).
已知二阶LTI系统的动态方程、初始条件和输入，求系统的完全响应y(t)。
solution:
(1)求齐次方程y''(t)+6y'(t)+8y(t) = 0的齐次解yh(t)
特征方程为
特征根为
s 2  6s  8  0
s1  2，s2  4
齐次解yh(t)
yh (t )  K1e—2t  K2e—3t
(2) 求非齐次方程y‘’(t)+6y‘(t)+8y(t) = x(t)的特解yp(t)
由输入x (t)的形式，设方程的特解为
yp(t)=Ce-t
将特解带入原微分方程即可求得常数C=1/3。
(3) Complete Solution(全解)
y (t )  yh (t )  y p (t )  Ae
 2t
 Be
 4t
1 t
 e
3
1
y (0)  A  B   1
3
解得 A=5/2，B= 11/6
1
y ' (0)  2 A  4 B   2
3
5  2t 11  4t 1 t
y (t )  e  e  e , t  0
2
6
3
Ch2.12 LTI系统的方框图表示
(Block Diagram Representations of LTI system)
Elementary operations(基本运算单元)：
1、标量乘法( scalar multiplication)
x(n)
y (n)  cx(n)
c
y (t )  cx(t )
x(t )
2、加法（addition）
x1 (n)
x1 (t )
+
y (n)  x1 (n)  x2 (n)

+
x2 (t ) x2 (n)
y (t )  x1 (t )  x2 (t )
LTI系统的方框图表示
(Block Diagram Representations of LTI system )
3.积分（integration）---对于连续系统
x(t )

y (t )   x( )d
t

延迟（time shift）---对于离散系统
x(n)
s
y (n)  x(n  1)
Ex:Draw direct form I and direct form II implementations of the LTI
system described by thr differential equation
y’’(t)+a1y’(t)+a2y(t)= b0x(t) + b1x’(t) + b2x’’(t)
已知2阶连续LTI系统的微分方程，画出其方框图。
(a) Direct form I.
(b) Direct form II.
Ex: Draw direct form I and direct form II implementations of the LTI
system described by thr differential equation
y(n) - 3/4 y(n -1) +1/8y(n -2) = x(n) +1/5x(n -1)
已知2阶离散LTI系统的差分方程， 画出其方框图。
solution: the direct form II implementation is
1
x(n)

3/4

s
s
-1/8
1/5
y (n)
Ch2.13 LTI系统的状态变量描述
(state-variable description for LTI systems)
1.状态方程的普遍形式
2.状态方程的建立
(1)连续系统状态方程的建立
(2)离散系统状态方程的建立
Definitions of state
The state(状态) of a system may be defined as a
minimal set of signals that represent the system’s
entire memory of the past.
That is, given only the value of the state at an initial
point in time, ni (or ti ), and the input for times n> ni
(or t>ti ), we can determined the output for all times
n> ni (or t>ti ).
1. 状态方程的普遍形式
对一个具有m个输入p个输出的n阶连续时间LTI系统，可用n个
一阶微分方程组表示：

q1  a11q1  a12 q2    a1n qn  b11 x1  b12 x2    b1m xm

q2  a21q1  a22 q2    a2n qn  b21 x1  b22 x2    b2m xm


qn  an1q1  an 2 q2    ann qn  bn1 x1  bn 2 x2    bnm xm
输出方程
y1  c11q1  c12 q2   c1n qn  d11 x1  d12 x2   d1m xm
y2  c21q1  c22 q2   c2n qn  d21 x1  d22 x2   d2m xm

y p  c p1q1  c p 2q2    c pn qn  d p1 x1  d p 2 x2    d pm xm
  
 q1 
  
q 
 2
    
 
qn 
(n维)
 a11
a
 21
=  

an1
A
a12
a22

an 2
   a1n 
   a2 n 
  

   ann 
(n*n阶)
C
 
 y1 
 
y 
 2
  
 
 yn 
(n维)
 c11 c12
c
=  21 c22
  

c p1 c p 2
 
 q1 
 
q 
 2
  
 
qn 
B
b11 b12
b
 21 b22
+   

bn1 bn 2
(n维)
   b1m 
   b2 m 
  

   bnm 
(n*n阶)


 x1 


x 
 2
  


 xm 
(m维)
D
   c1n 
   c2 n 
  

   c pn 
(n*n阶)
 
 q1 
 
q 
 2
  
 
qn 
 d11 d12
d
+  21 d 22
  

d p1 d p 2
(n维)
   d1n 
   d 2 n 
  

   d pn 
(p*m阶)


 x1 


x 
 2
  


 xm 
(m维)
连续时间系统的状态变量方程为一阶微分方程组，可表示为
矩阵形式：

q(t)=Aq(t)+Bx(t)
y(t)=Cq(t)+Dx(t)
类似地，离散时间系统的状态变量方程为一阶差分方程组：
q[n + 1]=Aq[n]+Bx[n]
y[n]=Cq[n]+Dx[n]
2. 离散时间LTI系统状态方程的建立
(1) 由方框图建立状态方程
(2) 由差分方程建立状态方程
Ex: Find the state-variable description corresponding to
the LTI system. 写出LTI系统的状态方程和输出方程。
解：
该系统有两个延
时器，选取其输出
q1[n] 及 q2[n] 作 为 状 态
变 量。由左端加法 器
列出状态方程为
x1[n]

S
q1[n]

q2[n]

y1[n]
a1
x2[n]

S
a2
q1[n  1]  a1q1[n]  x1[n]
q2 [n  1]  a2q2[n]  x2 [n]
由右端加法器列写出输出方程为
y1[n]  q1[n]  q2 [n]
y2[n]  q2 [n]  x1[n]
y2[n]
矩阵表示式为
a1
 q1[ n  1] 
 q [ n  1]   0

 2

0
a2 
 q1[ k ]  1 0  x1[k ] 
q [k ]  0 1  x [k ]
 2 
 2  
 y1[ n] 
1 1  q1[n]  0 0  q1[n] 

 y [n]  



q [n]

0
1
q
[
n
]
 2 

  2  1 0  2 
由差分方程建立状态方程
•从差分方程画出相应的方框图
•从方框图建立系统的状态方程。
Ex: Find the state-variable description corresponding to the
LTI system
y(n)  6 y(n  1)  11y(n  2)  6 y(n  3)  30x(n  1)  110x(n  2)  90x(n  3)
3. 连续时间LTI系统状态方程的建立
（1）由方框图建立状态方程
（2）由微分方程建立状态方程
（3）由电路建立状态方程
由方框图建立状态方程
选取积分器的输出作为状态变量；
围绕加法器列写状态方程和输出方程。
Example 2.30 : Determine the state-variable description
corresponding to the block diagram of a LTI system.
已知一个LTI系统的方框图为写出系统结构的状态方程。
solution：选2个积分器输出为系统的状态变量q1，q2，有
d
q1 (t )  2q1 (t )  q2 (t )  x(t )
dt
d
q2 (t )  q1 (t )
dt
输出方程为
y(t )  3q1 (t )  q2 (t )
状态方程的矩阵表示式为
d

q
(
t
)
 dt 1  2
d

 q2 (t ) 1
 dt

 1  q1 (t )  1
  x(t )



0  q2 (t ) 0
 q1 (t ) 
y (t )  3 1 

q
(
t
)
 2 
由微分方程建立状态方程
•从微分方程画出相应的方框图
•从方框图建立系统的状态方程。
由电路建立状态方程
（1）选择电感电流和电容电压作为状态变量；
（2）应用KCL写出电容的电流 c dv c 与其它状态变量和输入量的关系式；
dt
（3）应用KVL写出电感的电压 L
di L
dt
与其它状态变量和输入量的关系式；
（4）对步骤（2）（3）所建立方程，两边分别除以C或L就得到状态方程；
（5）由KCL、KVL写出用状态变量和输入量表示的输出，即得输出方程。
Ex: Consider Derive a state-variable description
of the electrical circuit.
R1
q2 (t)
L
+
+
x (t)
+
i1 (t)
q1 (t)

C
i2 (t)
R2
y (t)


solution：选择电容的电压q1(t)和电感的电流q2(t)作为系统
的状态变量。
回路电流和状态变量的关系为
q2 (t )  i2 (t )
C
dq1 (t )
 i1 (t )  i2 (t )
dt
R1
q2 (t)
L
+
+
x (t)
+
i1 (t)
q1 (t)

C
i2 (t)
R2
y (t)


解：
回路方程为
R1i1 (t )  q1 (t )  x(t )
Li2 (t )  R2i2 (t )  q1 (t )  0
由上面四式可求出状态方程为
d
1
1
1
q1 (t )  
q1 (t ) 
q2 (t ) 
x(t )
dt
R1C
C
R1C
d
1
R2
q2 (t )  q1 (t ) 
q2 (t )
dt
L
L
系统的输出方程为
y(t )  R2 q2 (t )
系统的状态方程和输出方程用矩阵来表示

d

q
(
t
)

 dt 1 
d
  

 q2 (t )

 dt

y (t )  0
1
R1C
1
L
1 
  q (t )
 1 
C  1 

R1C  x (t )



R
q (t )
 2  2   0 
L 
 q1 (t ) 
R2  

q
(
t
)
 2 