1

Audio &

DS

P Lab.

Audio & DSP Lab.

Lecture 2-3

Linear Time-Invariant System(LTI System)線性非時變系統

2

Audio &

DS

P Lab.

Audio & DSP Lab.

Interconnections of LTI SystemsInterconnection of two LTI systems. (a) Parallel connection of two systems. (b) Equivalent system.

τττ dthxthtx )()()(*)( 11 −= ∫

τττ dthxthtx )()()(*)( 22 −= ∫

ττττττ

τττττ

ττττ

dthxdthx

dthxthx

dththxththtx

)()()()(

))()()()((

))()(()())()((*)(

21

21

2121

−+−=

−+−=

−+−=+

∫∫

∫

∫

3

Audio &

DS

P Lab.

Audio & DSP Lab.

Interconnection of two LTI systems. (a) Cascade connection of two systems. (b) Equivalent system. (c) Equivalent system: Interchange system order.

)()()()()()()()(),()(

212

1

thththtzthtythtzttxif

all ∗=∗==∴== δ

4

Audio &

DS

P Lab.

Audio & DSP Lab.

Given impulse responses of each subsystem for the interconnection of systems for Example 2.11, find the total system output =?

5

Audio &

DS

P Lab.

Audio & DSP Lab.

(a) Reduction of parallel combination of LTI systems in upper branch of Fig. 2.20. (b) Reduction of cascade of systems in upper branch of Fig. 2.21(a).

(c) Reduction of parallel combination of systems in Fig. 2.21(b) to obtain an equivalent system for Fig. 2.20.

6

Audio &

DS

P Lab.

Audio & DSP Lab.

Interconnection of LTI systems for Problem 2.8.

( )( ))()()()()()( 32415 ththththththall +−∗∗=

7

Audio &

DS

P Lab.

Audio & DSP Lab.

Interconnection of LTI systems for Problem 2.9.

8

Audio &

DS

P Lab.

Audio & DSP Lab.

LTI System Properties

•Memory-less LTI System

•Causal LTI System

•Stable LTI System

•Invertible LTI System and De-convolution

9

Audio &

DS

P Lab.

Audio & DSP Lab.

Memory-less LTI System: 輸出只與現在輸入有關，輸出為輸入乘以一個純量。

).()(

),()0()()()(

].[][

],[]0[][][

][][][][][

τδτ

τττ

δ

chifonlyandif

txhdtxhty

kckhifonlyandif

nxhknxkh

nxnhnhnxny

k

=

=−=

=

=−=

∗=∗=

∫

∑

∞+

∞−

∞+

−∞=

10

Audio &

DS

P Lab.

Audio & DSP Lab.

Discrete-Time Causal LTI System:

輸出只與現在與過去的輸入有關，系統在脈衝輸入之前不能有輸出。

orknxkhnxnhnyk∑+∞

−∞=

−=∗= ][][][][][

L

L

+−+−+++−++−+=

]2[]2[]1[]1[][]0[]1[]1[]2[]2[][

nxhnxhnxhnxhnxhny

因為輸出只與現在與過去的輸入有關，令

0,0][ <= kforkh

11

Audio &

DS

P Lab.

Audio & DSP Lab.

Discrete-Time Causal LTI System:

∑+∞

=

−=∗=0

][][][][][k

knxkhnxnhny

12

Audio &

DS

P Lab.

Audio & DSP Lab.

Continuous-Time Causal LTI System:

0,0)( <= ττ forh

τττ dtxhtxthty )()()()()(0

−=∗= ∫∞

13

Audio &

DS

P Lab.

Audio & DSP Lab.

Discrete-Time Stable LTI System:

∑∑+∞

−∞=

+∞

−∞=

−≤−=kk

knxkhknxkhny ][][][][][Q

,][ ∞<≤ xMnxIf the input is bounded (BI),

.][ ∞<≤− xMknxit implies that

∑+∞

−∞=

≤k

x khMny ][][Hence,

14

Audio &

DS

P Lab.

Audio & DSP Lab.

If the impulse response is absolutely summable,

∞<∑+∞

−∞=kkh ][

∞<≤ ∑+∞

−∞=kx khMny ][][Hence,

15

Audio &

DS

P Lab.

Audio & DSP Lab.

Continuous-Time Stable LTI System:

連續時間 LTI 系統是穩定(BIB0)，若且唯若其脈衝響應是絕對可積分的。 (if and only if)

Iff:∞<∫

+∞

∞−ττ dh )(

16

Audio &

DS

P Lab.

Audio & DSP Lab.

Invertible System and Deconvolution

444 3444 21

)()()( tthth inv δ=∗

回復為 x(t) 的過程稱為：Deconvolution

17

Audio &

DS

P Lab.

Audio & DSP Lab.

Invertible System and Deconvolution

Continuous-Time:

{ })()()(,

,)()()()(tththimplies

ththtxtxinv

inv

δ=∗

∗∗=

Discrete-Time:

{ }][][][,

,][][][][nnhnhimplies

nhnhnxnxinv

inv

δ=∗

∗∗=

18

Audio &

DS

P Lab.

Audio & DSP Lab.

]1[][][ −+= nxanxnyEX 2.13

Impulse Response h[n] = ?

]1[][][ −+= nannh δδ

How to find the impulse response from the inverse system, hinv[n] = ?

][][][ nnhnh inv δ=∗

19

Audio &

DS

P Lab.

Audio & DSP Lab.

][][][ nnhnh inv δ=∗Q

][]1[][][]}1[][{][][

nnhanhnhnannhnh

invinv

invinv

δ

δδ

=−+=

∗−+=∗∴

1]1[]0[,0 =−+= invinv hahnwhen

If the hinv[n] is causal, h ,0]1[ =−inv

1]0[ =∴ invh

20

Audio &

DS

P Lab.

Audio & DSP Lab.

]1[][:,0]1[][,0

−−=

=−+>

nhanhrewriteornhanhnwheninvinv

invinv

ninvinv

invinv

invinv

invinv

inv

anahnh

aahhaahh

aahhh

)(]1[][

]2[]3[]1[]2[]0[]1[

1]0[

3

2

−=−−=

−=−=

=−=

−=−=

=

M

21

Audio &

DS

P Lab.

Audio & DSP Lab.

The impulse response from the inverse system,

][)(][ nuanh ninv −=

Is the inverse system stable?

∑ ∑∞

=

∞

=

=0 0

][k k

kinv anh

.;,1 stablenototherwisestablesystemaif <

22

Audio &

DS

P Lab.

Audio & DSP Lab.

Discrete-Time Step Response

當以單位步階訊號 Unit-Step輸入時，系統輸出以 s[n]表示步階響應：

∑+∞

−∞=

−=∗=k

knukhnhnuns ][][][][][

<<−≥≥−

=−knorknknorkn

knu,0,0,0,1

][

23

Audio &

DS

P Lab.

Audio & DSP Lab.

<≥

=−knkn

knu,0,1

][Q

∑−∞=

=n

kkhns ][][步階響應是脈衝響應的總和：

脈衝響應亦可用步階響應相減獲得：

]1[][][ −−= nsnsnh

24

Audio &

DS

P Lab.

Audio & DSP Lab.

Continuous-Time Step Response

步階響應是脈衝響應的積分：

ττ dhtst

∫ ∞−= )()(

脈衝響應亦可用步階響應微分獲得：

)()()( thdhdtdts

dtd t

== ∫ ∞−ττ

25

Audio &

DS

P Lab.

Audio & DSP Lab.

)(1)( )/( tueRC

th RCt−=EX 2.14 已知脈衝響應，

找出步階響應 s(t) = ?

RC1

0 t

26

Audio &

DS

P Lab.

Audio & DSP Lab.

找出步階響應 = ?

Solution:

RCt

ets−

−=1)(

RC circuit step response for RC = 1 s

27

Audio &

DS

P Lab.

Audio & DSP Lab.

ττ dhtst

∫ ∞−= )()(Q

RCt

RCt

RCt

RC

tRC

e

et

edeRC

dueRC

ts

−

−−−

∞−

−

−=

+−=

−==

=

∫

∫

1

10

1

)(1)(

0

ττ

τ

τ

ττ

28

Audio &

DS

P Lab.

Audio & DSP Lab.

Differential and Difference Equation

微分方程式： 描述連續時間系統

∑ ∑= =

=N

k

M

kk

k

kk

k

k txdtdbty

dtda

0 0)()(

差分方程式： 描述離散時間系統

∑ ∑= =

−=−N

k

M

kkk knxbknya

0 0][][

29

Audio &

DS

P Lab.

Audio & DSP Lab.

Example of an RLC circuit described by a differential equation.

30

Audio &

DS

P Lab.

Audio & DSP Lab.

微分方程式：

輸入為電壓源: 輸出為電流: )(tx )(ty

)()(1)()( txdyC

tRytydtdL

t=++ ∫ ∞−

ττ

對上式等號兩邊對t微分:

)()(1)()(2

2

txdtdty

Cty

dtdRty

dtdL =++

31

Audio &

DS

P Lab.

Audio & DSP Lab.

差分方程式：

∑ ∑= =

−=−N

k

M

kkk knxbknya

0 0][][

∑∑==

−−−=N

kk

M

kk knya

aknxb

any

10 00

][1][1][

32

Audio &

DS

P Lab.

Audio & DSP Lab.

EX: 2.16 Difference Equation:

]2[675.0]1[1349.0][0675.0]2[4128.0]1[143.1][

−+−+=−+−−nxnxnx

nynyny

Initial Condition: x[n] = 0, y[-1] = 1, and y[-2] = 2,Please write a recursive rule to find y[n] = ?

33

Audio &

DS

P Lab.

Audio & DSP Lab.

]2[675.0]1[1349.0][0675.0]2[4128.0]1[143.1][

−+−++−−−=

nxnxnxnynyny

Solution:

]2[4128.0]1[143.1][,0][ −−−=∴= nynynynxQ

M

0199.00207.00406.0)05.0(4128.0)1883.0(143.1]3[1883.01311.00572.0)3174.0(4128.0)05.0(143.1]2[

05.04128.03628.0)1(4128.0)3174.0(143.1]1[3174.0)2(4128.0)1(143.1]0[

1]1[2]2[

−=+−=−−−=−=−−=−−=

−=−=−==−=

=−=−

yyyyyy 已知

34

Audio &

DS

P Lab.

Audio & DSP Lab.

EX: 2.16 : If the input x[n] is described by the following closing price of INTEL stock, please find the output y[n] = ?

Weekly closing price of Intel stock

35

Audio &

DS

P Lab.

Audio & DSP Lab.

Output associated with the weekly closing price of Intel stock. (討論輸出、入差異)

36

Audio &

DS

P Lab.

Audio & DSP Lab.

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 (1/10πn). (d) Output due to x2[n] = cos (1/5πn). (e) Output due to x3[n] =cos(7/10πn).

37

Audio &

DS

P Lab.

Audio & DSP Lab.

EX2.14 請用微分方程式描述下述 RL circuit，其x(t)表輸入電壓、y(t)表輸出電流。

學生試一試 ?