2. 1 College of Communication Engineering Undergraduate Course: Signals and Linear Systems Lecturer: Kunbao CAI

3. 2 Chapter 2 Time-Domain Analysis of Continuous-Time Systems Lecture 5 Main Contents: 2.1 Introduction 2.2 Linear Constant-Coefficient Differential Equations and Their Solutions 1. Mathematical Models 2. Nth-Order Linear Constant-Coefficient Differential Equations 2.2.1 Classical Solution of the Differential Equations 1. Homogeneous Solution 2. Particular Solution 3. Complete Solution 4. An Example

4. 3 2.1 Introduction What is the system analysis? For a given system model with its input, the process to find its response is called the system analysis. When we read scientific literatures, we may meet some terminologies such as the system identification and the system synthesis that are beyond the scope of our course. What is the time-domain analysis of the systems? The so-called time-domain analysis of a continuous-time system is that the solving procedure for its solution is totally confined within the time domain. From this definition, you may ask that if there is the system analysis in other domains. The answer is yes. In this chapter, however, we will concentrate our attention to the time-domain analysis of the systems. Chapter 2 Time-Domain Analysis of Continuous-Time Systems

5. 4 2.1 Introduction (continued) Mathematical Models of the Continuous-Time Systems A large class of continuous-time systems can be represented by three time-domain mathematical models as follows. (1) Linear constant-coefficient differential equations. (2) Impulse response of LTI systems. (3) State-variable formulations.

6. 5 2.2 Linear Constant-Coefficient Differential Equations and Their Solutions 1. Mathematical Models Figure 2.1 Parallel RLC circuit with a current source The mathematical model relating its input and output can be determined as follows. and output can be determined as follows. Step 1. Establish the relation between the current and voltage for every passive circuit element. Resistor: (2-1) Inductor: (2-2) Capacitor: (2-3)

7. 6 1. Mathematical Models Step 2. Using KCL (Kirchhoff’s Current Law), we can obtain a current equilibrium equation as (2-4) Step 3. Substituting Eqs.(2-1), (2-2) and (2-3) into Eq.(2-4) and applying some basic operations, we can obtain a second-order linear differential equation of the given circuit, which is of the form (2-5) In the viewpoint of systems, this is called the single-input single-output equation of the dynamic system described by the circuit in Figure 2.1 (on slide 5). ▲ If a known input is applied to the system at time, then Eq.(2-5) together with two initial conditions and is sufficient to determine the system output for uniquely. ▲ Furthermore, if and, then Eq.(2-5) together with these two zero-initial conditions determines a causal LTI system with order two.

8. 7 1. Mathematical Models (continued) In this chapter, we consider a class of continuous-time dynamic systems whose single-input single-output relation can be characterized by an nth-order linear constant-coefficient differential equation (2-6) where system output or response; system input or excitation; constant coefficients (real). ▲ If a known input is applied to the system at, then Eq.(2-6) together with n independent initial conditions can uniquely determine the system output for. ▲ Furthermore, if all of the n initial conditions are equal to zero, then Eq.(2-6) together with these zero-initial conditions determines an nth-order causal LTI system.

9. 8 2.2.1 Classical Solution of the Differential Equations Assume that a dynamic system is described by (2-7) (back to slide 11 )back to slide 11 where we have assumed, without loss of generality, that. With its n initial conditions given by the complete solution of the system consists of two parts, that is, (2-8) where complete solution; homogeneous solution; particular solution. The procedure to solve this mathematical problem is given as follows.

10. 9 1. Homogeneous Solution 1. Homogeneous Solution The homogeneous equation corresponding to the original equation in Eq.(2-7) is given by (2-9) From this equation we can get the following algebraic equation (2-10) This is called the characteristic equation of the system. Its n roots are called the characteristic roots of the systems, or the natural frequencies of the system. The function form of the homogeneous solution depends on the multiplicity of characteristic roots.

11. 10 1. Homogeneous Solution (continued) (1) Simple Characteristic Roots If all of the characteristic roots are simple, then the homogeneous solution is of the form (2-11) where are n unknown constants which will be determined in a later step. (2) Repeated Characteristic Roots Assuming that is a repeated root with order and the remainder roots are all simple, the homogeneous solution is of the form (2-12) Also, n unknown constants will be determined in a later step.

12. 11 2. Particular Solution 2. Particular Solution The function form of the particular solution which satisfies the original differential equation in Eq.(2-7) (on slide 8)(on slide 8) depends on the excitation. To find the particular solution, we generally need to follow the steps: (1) Substituting the known into Eq.(2-7). (2) By inspecting the function form on the right-hand side of the equation resulted in step (1) and considering the characteristic roots, we can select an appropriate particular solution with unknown constants. This step has been assumed to be familiar with you, while taking the course of advanced mathematics. (3) Substituting the selected particular solution into the equation obtained in the step (1), we can determine the values of the unknown constants. Thus the particular solution can be completely determined.

13. 12 3. Complete Solution 3. Complete Solution The complete solution with n unknown constants is given by (2-13) Here we have assumed that all of the characteristic roots are simple. Clearly, n independent initial conditions (2-14) are needed to determine n constants in Eq.(2-13). There are several methods to find the initial conditions at from the initial conditions at, which will be introduced through several Examples. From Eqs.(2-13) and (2-14), we can obtain a set of Simultaneous linear equations as follows.

14. 13 Solving this set of equations yields n constant values. Thus, we totally determine the complete solution. Example 2.1 Consider a linear dynamic circuit shown in Figure 2.2. The circuit has reached steady state at. If the switch is moves from position 1 to position 2 at, calculate for. Figure 2.2 The linear dynamic circuit for Example 2.1 Solution: Step 1. Modeling the circuit for. Applying two basic laws for the circuit analysis, we can obtain following three equations. (Back to slide 16) (Back to slide 17)

15. 14 Continue Example 2.1 KVL: (2-15a) (2-15b) KCL: (2-15c) From these three equations, we can get a differential equation as follows. Substituting the circuit-element parameters into the above equation, we can obtain (2-16)

16. 15 Continue Example 2.1 Step 2. Find the homogeneous solution with undetermined constants. Characteristic equation: Characteristic roots: and Homogeneous solution: (2-17) Step 3. Determine the particular solution. Substituting into Eq.(2-16), we can get (2-18) According to the knowledge obtained in the course of advanced calculus, the particular solution is of the form Substituting this into Eq.(2-18), the particular solution is given by

17. 16 Continue Example 2.1 Step 4. Find the complete solution. The complete solution: (2-19) We need and to determine its two constants. Since the circuit in Figure 2.2 has reached steady state at, we know that: the inductor: a short circuit; the capacitor: an open circuit. Therefore, we have, and (See slide 13.)

18. 17 Continue Example 2.1 After switching, the designed current satisfies Since the inductor current and the capacitor voltage for the given circuit can not abruptly change at, we have and Thus, and (See slide 13.)

19. 18 Continue Example 2.1 From these two initial conditions and Eq.(2-19) (2-19) we get a set of simultaneous linear equations Solving this set of equations yields and Substituting them into Eq.(2-19), we obtain the designed complete Solution (2-20)

20. 19 The End! Thank you for your attention!