Linear Circuit Analysis

In an oscilloscope a timing signal called a horizontal sweep acts as a time base, which allows one to view measured input signals as a function of time.

In practice, the linear increase in voltage is approximated by the “linear” part of an exponential response of an RC circuit.

First-Order RL and RC Circuits 1. What is a first-order circuit? A first-order circuit is characterized by a first-order differential equation. It consists of resistors and the equivalent of one energy storage element. Typical examples of first-order circuits: (a) First-Order RL circuit(b) First-Order RC circuit USUS L R + _ C R USUS + _

Interconnections of sources, resistors, capacitors, and inductors lead to new and fascinating circuit behavior. C R USUS + _ + _ First-Order RC circuit A loop equation leads to since or, equivalently, This equation is called Constant-coefficient first-order linear differential equation Apply duality principle,

2. Some mathematical preliminaries The first-order RL and RC circuits have differential equation models of the form RC circuit first-order differential equation RL circuit first-order differential equation or, equivalently, valid for t≥t 0, where x(t 0 )=x 0 is the initial condition. The term f(t) denotes a forcing function. Usually, f(t) is a linear function of the input excitations to the circuit. The parameter λdenotes a natural frequency of the circuit.

The main purpose of this chapter is to find a solution to the differential equation. or The solution to the equation for t≥t 0 has the form (1) Satisfies the differential equation (2) Satisfies the correct initial condition, x(t 0 )=x 0

Integrating factor method First step, multiply both sides of the equation by a so-called integrating factor e － λt. By the product rule for differentiation, Integrate both sides of the equation from t 0 to t as follows

3. Source-free or zero-input response L R R C A parallel connection of a resistor with an inductor or capacitor without a source. In these circuits, one assumes the presence of an initial inductor current or initial capacitor voltage. (a) KCL implies(b) KVL implies + _ + _ + _ Both differential equation models have the same general form,

The solution to the equation for t≥t 0 has the form Where τ is a special constant called the time constant of the circuit. The response for t≥t 0 of the undriven RL and RC circuit are, respectively, given by RC circuit RL circuit The time constant of the circuit is the time it takes for the source-free circuit response to drop to e － 1 =0.368 of its initial value.

For more general circuits, those containing multiple resistors and dependent sources, it is necessary to use the Thevenin equivalent resistance seen by the inductor or capacitor in place of the R. Linear Resistive circuit No independent sources L Linear Resistive circuit No independent sources C + _ Thevenin equivalent L + _ C Thevenin equivalent

Example 1. For the circuit of the figure, find i L (t) and u L (t) for t ≥0 given that i L (0 － )=10A and the switch S closes at t=0.4s. Then compute the energy dissipated in the 5Ω resistor over the time interval [0.4, ∞). + _

USUS C + _ u(t) + _ R K USUS 过渡期为零 USUS R + _ u(t) K + _ （ 1 ）问题的提出 USUS 第一个稳定状态第二个稳定状态第三个稳定状态 过渡状态 瞬态（ transient state ） 稳态（ steady state ） 换路 代数方程描述 微分方程描述 一阶电路的初始条件

（ 2 ）电路的初始条件 ① t ＝ 0 － 和 t ＝ 0 ＋ 的概念 认为换路在 t ＝ 0 时刻进行 换路前一瞬间 换路后一瞬间 t ＝ 0 ＋ t ＝ 0 － 初始条件为 t ＝ 0 ＋ 时 u 、 i 及其各阶导数的值 ② 电容的初始条件 C + _ t＝0＋ 时t＝0＋ 时 当为有限值 电荷守恒 换路瞬间，若电容电流保持为有限值，则电容电压（电荷）换路前后保持不变。

③ 电感的初始条件 L + _ t＝0＋ 时t＝0＋ 时 当为有限值 磁链守恒 换路瞬间，若电感电压保持为有限值，则电感电流（磁链）换路前后保持不变。 ④ 换路定律 换路瞬间，若电容电流保持为有限值， 则电容电压（电荷）换路前后保持不变。 换路瞬间，若电感电压保持为有限值， 则电感电流（磁链）换路前后保持不变。 反映了能量不能跃变

例 1 电路如图所示，试求 。 解： + _ + _ + _ + _ + _ + _ ② 由换路定律 ① 画出 电路，求出 电路 ③ 画出 电路，求出 可见， 电容 开路 电容用 电压源 替代

+ _ + _ + _ + _ + _ + _ 例 2 电路如图所示，试求 。 解： ② 由换路定律 ① 画出 电路，求出 电路 ③ 画出 电路，求出 可见， 电感 短路 电感用 电流源 替代

求解初始条件的步骤 ① 画 0 － 等效电路，即换路前电路（稳定状态），求 u C ( 0 － ) 和 i L ( 0 － ) 。 ② 由换路定律得 u C ( 0 + ) 和 i L ( 0 + ) 。 ③ 画 0 + 等效电路，即换路后的电路。 ④ 由 0 + 电路求所需各变量的 0 + 值。 电容用电压源来替代，大小为 u C ( 0 ＋ ) 电感用电流源来替代，大小为 i L ( 0 ＋ ) 。 电压源和电流源的方向均与原来的电压、电流方向一致。 其中 电容相当于开路 电感相当于短路 其中

+ _ + _ + _ + _ + _ 例 3 电路如图所示，试求 ， 。 解： ② 由换路定律 ① 画出 电路，求出 ， 电路 ③ 画出 电路，求出 ，

+ _ + _ + _ 例 4 电路如图所示，试求开关 K 闭合瞬间各支路的电流和电感上的电压。 解： 电路 + _ + _ + _ + _ + _

+ _ + _ + _ + _ + _ 例 5 电路如图所示，试求开关 K 闭合瞬间流过它的电流值。 解： 电路 + _ + _

单位阶跃函数 延时（ delayed ）单位阶跃函数 分段常量信号（ piecewise-constant signal ） （矩形）脉冲（ pulse ）脉冲串（ pulse train ） 单位阶跃函数及分段常量信号

运用阶跃函数和延时阶跃函数，分段常量信号可以表示为一系列阶跃信号之和。

Solution Example 8.2. For the circuit of the figure, find i L (t) and u L (t) for t ≥0 given that i L (0 － )=10A and the switch S closes at t=0.4s. Then compute the energy dissipated in the 5Ω resistor over the time interval [0.4, ∞). + _ Step 1. With switch S open, compute the response for 0≤t ≤0.4s. From the continuity property of the inductor current, Step 2. With switch S closed, compute the response for t ≥0.4s.

Step 3. Write the complete response as a single expression. Step 4. Plot the complete response. The 0.4s time constant has a much faster rate of decay than the lengthy 2s time constant.

Step 5. Compute u L (t). + _ for 0≤t ≤0.4s, in particular, hence, for t ≥0.4s, in particular, Step 6. Compute energy dissipated in the 5Ω resistor over the time interval [0.4, ∞).

Example 8.3. Find u C (t) for t ≥0 for the circuit of the figure given that u C (0)=9V. Solution + _ Step 1. Compute the response for 0≤t ≤1s. By the continuity of the capacitor voltage, hence, Step 2. Compute the response for t ≥ 1s. Step 3. Use step functions to specify the complete response.

Step 4. Obtain a plot of the response. Here the part of the response with the 0.3s time constant shows a greater rate of decay than the longer 0.8s time constant.

Exercise. Suppose that in example 2 the switch moves to the 4.5Ω resistor at t=0.5s instead of 1s. Compute the value u C (t) at t=1.2s. + _

Example 8.4. Find u C (t) for the circuit of the figure, assuming that g m =0.75S and u C (0 － )=10V. Solution It is straightforward to show that the Thevenin equivalent seen by the capacitor is a negative resistance, + _ + _ Thevenin equivalent + _ hence,

Negative resistance causes capacitor voltage to increase without bound. Because of the negative resistance, this response grows exponentially, as shown in the figure. A circuit having a response that increases without bound is said to be unstable.

Exercise. In example 3, let g m =0.125S. Find the equivalent resistance seen by the capacitor and u C (t), t ≥0. + _ + _

+ _ + _ 示例 已知图示电路中的电容原本充有 24V 电压，求 K 闭合后，电容电压和 各支路电流随时间变化的规律。 解： 本题为求解一阶 RC 电路零输入响应问题 等效电路 t > 0 则有 又由已知条件 利用并联分流，得

+ _ + _ + _ 示例 t = 0 时 ，开关 K 由 1→2 ，求电感电压和电流。 解： RL 电路零输入响应问题

① 一阶电路的零输入响应是由储能元件的初值引起的响应, 都是由初始值 衰减为零的指数衰减函数。 ② 衰减快慢取决于时间常数  。 ③ 同一电路中所有响应具有相同的时间常数。 ④ 一阶电路的零输入响应和初始值成正比，称为零输入线性。 小结 RC 电路 RL 电路 RC 电路 RL 电路 R 为与动态元件相连的一端口电路的等效电阻。

4. DC or step response of first-order circuits This section takes up the calculation of voltage and current responses when constant voltage or constant current sources are present. Linear Resistive circuit With independent sources L Linear Resistive circuit With independent sources C + _ Thevenin equivalent Thevenin equivalent C + _ + _ L + _ + _

C + _ + _ L + _ + _ Deriving the differential equation models characterizing each circuit’s voltage and current responses. By KVL and Ohm’ law,By KCL and Ohm’ law,

Exercise. Constant differential equation models for the parallel RL and RC circuits of the figure. Note that these circuits are Norton equivalents of those in the figure. Again choose i L (t) as the response for the RL circuit and u C (t) as the response for the RC circuit. L + _ + _ C Answers:

Observe that four differential equations have the same structure: where And the general formula for solving such a differential equation: for RC case for RL case

whereas long as x(t) is a capacitor voltage or inductor current, and f(τ)=F is a constant (nonimpulsive) forcing function. Which is valid for t≥t 0. After some interpretation, this formula will serve as a basis for computing the response to RL and RC circuits driven by constant sources.

if then for RC case for RL case Hence, the solution of the differential equation given constant or dc excitation becomes for RC case for RL case

Example 8.5. For the circuit of the figure, suppose a 10V unit step excitation is applied at t=1 when it is found that the inductor current is i L (1 － )=1A. The 10V excitation is represented mathematically as u in (t)=10u(t － 1)V for t≥1. Find i L (t) and u L (t) for t≥1. + _ + _ Solution Step 1. Determine the circuit’s differential equation model. where the time constant Step 2. Determine the form of the response.

+ _ + _ Step 3. Compute i L (∞) and set forth the final expression for i L (t). replace the inductor by a short circuit, It follows that, and since Step 4. Plot i L (t).

Step 5. Compute u L (t). + _ + _ Exercise. Verify that in example 4, u L (t) can be obtained without differentiation by Exercise. In example 4, suppose R is changed to 4Ω. Find i L (t) at t=2s. Specifically, we need only compute,, and the time constant or.

Example 8.6. The source in the circuit of the figure furnishes a 12V excitation for t<0 and a 24V excitation for t≥0, denoted by u in (t)=12u( － t)+24u(t)V. The switch in the circuit closes at t=10s. First determine the value of the capacitor voltage at t=0 －, which by continuity equals u C (0 + ). Next determine u C (t) for all t≥0. + _ + _ Solution Step 1. Compute initial capacitor voltage Step 2. Obtain u C (t) for 0≤t ≤ 10s.

Step 3. Compute the initial condition for the interval t>10. + _ + _ Step 4. Find u C (t) for t>10. Thevenin equivalent + _ + _ Step 5. Set forth the complete response using step functions.

Step 6. Plot u C (t). Exercise. Suppose the switch in example 5 opens again at t=20s. Find u C (t) at t=25s.

+ _ + _ 示例 t = 0 时, 开关 K 闭合，已知 u C (0 － ) = 0V ，求（ 1 ）电容电压和电流； （ 2 ） u C ＝ 80V 时的充电时间 t 。 解： （ 1 ） RC 电路零状态响应问题 （ 2 ）设经过 t 秒， u C ＝ 80V

+ _ + _ 例 1 t = 0 时, 开关 K 打开，求 t > 0 后 i L ， u L 的变化规律 。 解： RL 电路零状态响应问题， 先化简电路

+ _ + _ + _ + _ 例 2 t = 0 时, 开关 K 打开，求 t > 0 后 i L ， u L 以及电流源两端的电压 u 。 解： RL 电路零状态响应问题， 先化简电路

全响应的两种分解方式 ① 根据电路的两种工作状态 稳态分量暂态分量 全响应 USUS U0－USU0－US 稳态分量 暂态分量 U0U0 物理概念清晰 ② 根据激励与响应的因果关系 零输入响应零状态响应全响应 便于叠加计算 USUS U0U0 零输入响应 零状态响应 全响应

+ _ + _ 例 1 t = 0 时, 开关 K 打开，求 t > 0 后的 i L 。 解： RL 电路全响应问题 零输入响应 零状态响应 全响应

+ _ + _ + _ 例 2 t = 0 时, 开关 K 闭合，求 t > 0 后的 i C ， u C 以及电流源两端的电 压 u ，已知 u C (0 + ) = 1V 。 解： RC 电路全响应问题 稳态分量 全响应 故

例 3 t = 0 时, 开关闭合，求换路后的 u C (t) 。 + _ 解：

小结 状态变量 uCuC iLiL 零输入响应零状态响应全响应

5. Superposition and linearity Provided one properly accounts for initial conditions, superposition still apply when capacitors and inductors are added to the circuit. supposeand for a capacitor By the same arguments, the current due to the input excitation is.

One the other hand, suppose and Thus linearity and, hence, superposition hold. Arguments analogous to the preceding imply that a relaxed inductor satisfies a linear relationship, and thus superposition is valid, whether the inductor is excited by currents or by voltages.

For a general linear circuit, one can view each initial condition as being set up by an input which shuts off the moment the initial condition is established. This mean that when using superposition on a circuit, one first looks at the effect of each independent source on a circuit having no initial conditions. Then one sets all independent sources to zero and computes the response due to each initial condition with all other initial conditions set to zero. The sum of all the responses to each of the independent sources plus the individual initial condition responses yields the complete circuit response, by the principle of superposition.

Example 8.8. The linear circuit of the figure has two source excitations applied at t=0, as indicated by the presence of the step functions. The initial condition on the inductor current is i L (0 － )= － 1A. Determine the response i L (t) for t≥0 using superposition. Solution Step 1. Compute the part of the circuit response due only to the initial condition, with all independent sources set to zero. + _ + _

Step 2. Determine the response due only to U S =10u(t)V. + _ + _ Step 3. Compute the response due only to the current source I S =2u(t)A.

Step 4. Apply the principle of superposition. due to initial condition due to source U S due to source I S

Question 1. What is the new response if the initial condition is changed to i L (0 － )=5A? Question 2. What is the new response if the voltage source U S is changed to 5u(t)V, with all other parameters held constant at their original values? Question 3. What is the new response if the initial condition is changed to 5A, the voltage source U S is changed to 5u(t)V, and the current source I S is changed to 8u(t)A? The question still remains as to why the superposition principle holds an advantage over the Thevenin equivalent method. This allows one to explore easily a circuit’s behavior over a wide range of excitations and initial conditions.

6. Response classifications Zero-input response The zero-input response of a circuit is the response to the initial conditions when the input is set to zero. Zero-state response The zero-state response of a circuit is the response to a specified input signal or set of input signals given that the initial conditions are all set to zero. Complete response By linearity, the sum of the zero-input and zero-state responses is the complete response of the circuit. Natural response The natural response is that portion of the complete response that has the same exponents as the zero- input response. Forced response The forced response is that portion of the complete response that has the same exponents as the input excitation.

7. Further points of analysis and theory Not only a capacitor voltage or an inductor current, it turns that any voltage or current in an RC or RL first-order linear circuit with constant input has the form by linearity, any voltage or current in the circuit has the form and which implies that

Example 8.9. For the circuit of the figure, let u in (t)= － 18u( － t )+9u(t)V. Find i in (t) for t>0. Solution Step 1. Compute i in (0 + ). + _ + _ 0 + circuit + _ + _ Step 2. Computeτand i in ( ∞ ).

Step 3. Set forth the complete response i in (t). Exercise. In example 7, find i C (0 － ), i C (0 + ), and i C (t) for t>0. + _ + _

例1例1 t = 0 时开关闭合，求 t > 0 后 i L ， i 1 和 i 2 。 解法 1 ： + _ + _ + _ + _

例 1 t = 0 时开关闭合，求 t > 0 后 i L ， i 1 和 i 2 。 解法 2 ： + _ + _

+ _ + _ + _ 例 2 t = 0 时开关由 1→2 ，求换路后的 u C (t) 。 解： 戴维宁等效 + _ + _

例 3 t = 0 时, 开关闭合，求换路后的 i(t) 。 + _ 解：

例 4 已知：电感无初始储能， t = 0 时闭合 K 1 ， t =0.2 s 时闭合 K 2 ， 求两次换路后的电感电流。 解： + _