电路基础英文版CHAPTER 7 .ppt(1)

要求深刻理解与熟练掌握的重点内容有：要求深刻理解与熟练掌握的重点内容有：1.电容元件及电感元件中贮能的计算，会运用换 .电容元件及电感元件中贮能的计算，路定律确定初始状态。路定律确定初始状态。 2.时间常数的概念及计算。 .时间常数的概念及计算。 3.一阶电路方程的建立，全响应的两种分解形式， .一阶电路方程的建立，全响应的两种分解形式，零输入响应与零状态响应，暂态响应与稳态响应。零输入响应与零状态响应，暂态响应与稳态响应。  4.阶跃响应和冲激响应。 .阶跃响应和冲激响应。 5.直流电源作用下一阶电路全响应的三要素法。 .直流电源作用下一阶电路全响应的三要素法。 6.二阶电路方程的建立，固有频率的概念。 .二阶电路方程的建立，固有频率的概念。要求一般理解与掌握的内容有：要求一般理解与掌握的内容有：阶跃函数与阶跃响应。跃响应。 难点：动态电路方程的建立。难点：动态电路方程的建立。

电路基础英文版课件详解

7-1 7-2 7-3 7.4

Introduction The Source-free RC Circuit The Source-free RL Circuit Singularity Functions

7.5 Step Response of an RC Circuit 7-6 Step Response of an RL Circuit 7-7 First-Order Op Amp Circuits 7-10 Summary

电路基础英文版课件详解

Dynamic elementiL + L uL L -

diL uL = L dt 1 t iL (t ) = iL (t0 ) + ∫ u (ξ )dξ L t0

iLiC

ψ L (t ) = ψ L (t0 ) + ∫ u (ξ )dξt0

C uC -

duC iC = C dt 1 t uC (t ) = uC (t0 ) + ∫ i (ξ )dξ C t0 q (t ) = q (t0 ) + ∫ i (ξ )dξt0 t

电路基础英文版课件详解

7.1 Introduction 1. Two Example Ri+uC =US (t≥0)

a S E C

b R

i C

+uC

R Us

+uC

duC dq d (CuC ) i= = =C dt dt dt

duC RC + uC = U S dt

t≥0

电路基础英文版课件详解

S(t=0) + uS -

R + uR

L - + uL - + uC -

i=iL C

KVL: uL+uR+uC=uS VCR: i = C duC dt

(1) (2)

duC u R = Ri = RC (3) dt 2 d uc dψ d (Li ) di uL = = = L = LC 2 dt dt dt dt

(4)

电路基础英文版课件详解

(3)、(4)代入(1)

d 2 uC duC LC + RC + uC = u S 2 dt dt

t≥0

(5)

A first-order circuit is characterized by a firstorder differential equation. A second-order circuit is characterized by a second-order differential equation.

电路基础英文版课件详解

2. Determining The Initial Value First-order circuit:y(0+); Second-order circuit:y(0+),y’(0+); 0-

0 0+

N-order circuit:y(0+),y’(0+) … y(n-1)(0+)。 t=0 denotes the time just before switching event, and t=0+ is the time just after switching event, assuming that the switching event takes place at t=0.

电路基础英文版课件详解

Determining uC (0+) and iL (0+) 设t0=0- t=0+ 1 0+ iL (0+) = iL (0 ) + ∫ u (ξ )dξ L 0

ψ L (0+) = ψ L (0 ) + ∫ u (ξ )dξ0

When the voltage uL is a finitude value(从0-到0+瞬间积分项为0) iL(0+)=iL(0-) ΨL(0+)= ΨL(0-)

电路基础英文版课件详解

1 0+ uC (0+) = uC (0 ) + ∫ iC (ξ )dξ C 0 q(0+) = q(0 ) + ∫ i

C (ξ )dξ0 0+

When the voltage iC is a finitude value uC(0+)=uC(0-) q(0+)=q(0-)

The capacitor voltage is always continuous (cannot change abruptly) so that: uC (0+)=uC (0-). - The inductor current is always continuous (cannot change instantaneously) so that: iL(0+)=iL (0-). -

电路基础英文版课件详解

1). Determine the uC (0-) and iL(0-) in the circuit at t=0-. 2). Replace the inductor with current source and replace the capacitor with voltage source in the circuit at t=0+.C

L iL(0 )

L iL(0-)

+ uC(0 )

uC(0+)=uC(0 )

iL(0+)= iL(0 )

电路基础英文版课件详解

Oringinal circuituS(t) + - iS(t) L

0+ equivalent circuit +uS(0+) + - iS(0+)

iL(0+) L C uC(0+) - + -

电路基础英文版课件详解

Steps to determine the initial values: : 1. By 0- equivalent circuit, - , find uC(0-) or/and iL(0-) - - 2. Obtain uC(0+) or/and iL(0+) + + 3. By 0+ equivalent circuit,find f(0+) ,

电路基础英文版课件详解

Example: The switch in Fig.(a) has been in position A for a long time . At t=0, the switch moves to B .Calculate the values uC(0+) and i(0+).S(t=0) A + 5V - 2 i(t) B 3A 1H 3 0.5F 4 + uC(t) -

(a) Solution: 1. By 0- equivalent circuit as shown in : - Fig.(b),find uC(0-) and iL(0-). , - -

电路基础英文版课件详解

+ 5V -

2 3 iL(0-)

4 + uC(0-) -

5 iL (0 ) = = 1A 2+3 uC (0 ) = 3iL (0 ) = 3V

(b) 2. Calculate uC(0+)、iL(0+). +、 + 根据换路定理 uC(0+)=uC(0-)=3V iL(0+)=iL(0-)=1A

电路基础英文版课件详解

3. By 0+ equivalent circuit as shown in Fig.(c),find , i(0+). i(0+)=3-1=2A ∴uC(0+)=3A i(0+)=2A1A 3A 2 3 4 + uC(0 ) (0+) - i(0+)

(c)

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