2. Time-Domain Analysis of Continuous- Time Signals and Systems
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1 2. Time-Domain Analysis of Continuous- Time Signals and Systems 2.1. Continuous-Time Impulse Function (1.4.2) 2.2. Convolution Integral (2.2) 2.3. Continuous-Time Impulse Response (2.2) 2.4. Classification of a Linear Time-Invariant Continuous- Time System by its Impulse Response (2.3) 2.5. Linear Constant-Coefficient Differential Equations (2.4.1)
2 2.1. Continuous-Time Impulse Function Continuous-Time Impulse Function The continuous-time impulse function, (t), is defined by (t)dt 1 (t) 0, t 0. (2.1) The illustration of the continuous-time impulse function is given in figure 2.1. (t) 1 t A(tt 0 ) Figure 2.1. Illustration of Continuous-Time Impulse Function. t 0 A t
3 (t) can be considered as (t) where lim 0 (t), 1/, t / 2 (t) 0, t / 2 (figure 2.2). (t) 1/ (2.2) (2.3) /2 /2 Figure 2.2. (t). t (t) has a sampling property, i.e.,
4 where t 0 is an arbitrary real number Continuous-Time Step Function x(t)(tt 0 )=x(t 0 )(tt 0 ), (2.4) The continuous-time step function is defined as 0, u(t) 1, (figure 2.3). At t=0, u(t) is undefined but should be considered to be finite. t t 0 0 (2.5) 1 u(t) t Figure 2.3. Continuous-Time Step Function.
5 u(t) can be expressed as the running integral of (t), i.e., t u(t) ( )d. (2.6) It follows from (2.6) that (t) can be considered as the derivative of u(t), i.e., 2.2. Convolution Integral ( t) du (t) dt The convolution integral of x 1 (t) and x 2 (t) is defined as x (2.7) (t) x (t) x ( )x (t )d. (2.8) Note that the integration is carried out with respect to an introduced variable,, and the final result is a function of t. The convolution integral satisfies the commutative property
6 the associative property and the distributive property x 1 (t)x 2 (t)=x 2 (t)x 1 (t), (2.9) [x 1 (t)x 2 (t)]x 3 (t)=x 1 (t)[x 2 (t)x 3 (t)], (2.10) x 1 (t)[x 2 (t)+x 3 (t)]=x 1 (t)x 2 (t)+ x 1 (t)x 3 (t). (2.11) The convolution integral can be calculated in the following steps. (1) Reflect x 2 () about the origin to obtain x 2 (). (2) Shift x 2 () by t to obtain x 2 (t). (3) Calculate the convolution integral at t. Steps (2) and (3) often need to be carried out in different ways for different intervals of t. Example. Find x 1 (t)x 2 (t), where
7 (1) x 1 (t)=e at u(t) and x 2 (t)=u(t). 1, 0 t T (2) x1(t) and x 2(t) 0, otherwise (3) x 1 (t)=e 2t u(t) and x 2 (t)=u(t3). t, 0, 2.3. Continuous-Time Impulse Response 0 t 2T. otherwise Definition of Continuous-Time Impulse Response A linear time-invariant continuous-time system can be completely characterized by the continuous-time impulse response, which is defined as the response of the system to the continuous-time impulse function. In addition, a linear time-invariant continuous-time system can be completely characterized by the continuous-time step response, which is defined as the response of the system to the continuous-time
8 step function I/O Relation by Continuous-Time Impulse Response The I/O relation of a linear time-invariant continuous-time system can be expressed by its impulse response. Let x(t) and h(t) be the input and the impulse response of a linear time-invariant continuoustime system, respectively. Then, the output of the system will be y(t)=x(t)h(t). (2.12) Proof. Since x(t) x( ) (t )d, (2.13) then y(t) T[x(t)] T Since the system is linear, then x( ) (t )d. (2.14)
9 y(t) (t ) d. x( )T (2.15) Since T[(t)]=h(t) and the system is time-invariant, then T[(t)]=h(t). (2.16) Substituting (2.16) into (2.15), one obtains (2.12) Classification of a Linear Time-Invariant Continuous-Time System by its Impulse Response Memoryless Systems versus Systems with Memory Assume that h(t) is the impulse response of a linear time-invariant continuous-time system. The system is memoryless if and only if h(t)=0 for t0. (2.17) Causal Systems versus Noncausal Systems Assume that h(t) is the impulse response of a linear time-invariant
10 continuous-time system. The system is causal if and only if h(t)=0 for t<0. (2.18) Stable Systems versus Unstable Systems Let us suppose that the impulse response of a linear time-invariant continuous-time system is h(t). The system is stable if and only if h(t) is absolutely integrable, i.e., there is a finite constant B such that h(t)dt B. (2.19) Proof. Consider the sufficiency first. Let x(t) be bounded, i.e., where A is a finite constant. Then, x(t) A, (2.20) y(t) h( )x(t )d h( ) x(t ) d
11 A h( ) d AB. (2.21) That is, y(t) is also bounded. Thus, the system is stable. Consider the necessity next. For the input x(t) * h ( t) / 0, the output of the system at t=0 is y(0) x( )h( )d h( ) 0 h( ) d h( t), h( ) 0 h( t) h( t) x( )h( )d h( ) d 0, 0 h(t) dt. (2.22) (2.23) The system is assumed to be stable. Since x(t) is bounded, y(t) is also bounded. Thus, there exists a finite constant B such that h(t)dt B. (2.24)
12 Example. Determine whether the following systems are stable: (1) h(t)=(tt 0 ). (2) h(t)=u(t). (3) h(t)=e 2t u(t). (4) h(t)=e 2t u(t). (5) h(t)=e 2t u(t) Invertible Systems versus Noninvertible Systems We assume that two linear time-invariant continuous-time systems A and B have the impulse responses g(t) and h(t), respectively. Then, A and B are mutually inverse if and only if g(t)h(t)=(t). (2.25) (2.25) can be used to construct the inverse of a given system.
13 2.5. Linear Constant-Coefficient Differential Equations A continuous-time system may be described by a linear constantcoefficient differential equation. However, it should be mentioned that only a linear constant-coefficient differential equation cannot specify a continuous-time system uniquely. Other conditions, such as the output and its derivatives at a given time under a given input or the statements about linearity, time-invariance, causality and stability, are also required. A continuous-time system is often described by a linear constantcoefficient differential equation, a right-sided input and some initial conditions. We will focus on these cases Homogeneous Solution and Particular Solution A linear constant-coefficient differential equation can be solved in the following steps.
14 (1) Find the homogeneous solution., a characteristic value of order K, corresponds to K terms in the homogeneous solution, i.e., y (t)... K 1 k0 where A k is the coefficient to be determined. h A k t k e t..., (2.26) (2) Find the particular solution. Let the free term be P(t)e t, where P(t) is a polynomial, and is a characteristic value of order L. Then, the particular solution will have the form y p (t)=t L Q(t)e t, (2.27) where Q(t) is a polynomial with the same order as P(t). Substituting (2.27) into the linear constant-coefficient differential equation, one can determine the coefficients in Q(t). In addition, it should be noted that the particular solution is linear with respect to the free term. (3) Combine the homogeneous solution and the particular solution
15 to obtain the complete solution, and determine the coefficients in the homogeneous solution. The homogeneous solution is also called the natural response, and the particular solution is also called the forced response. Example. A continuous-time system is given by dy(t) dt where x(t)=e t u(t) and y(0 )=2. Find y(t). 2y(t) x(t), (2.28) First consider y(t) for t<0. When t<0, (2.28) becomes dy(t) dt The characteristic equation is 2y(t) 0. (2.29) +2=0. (2.30)
16 The characteristic value, i.e., the solution to (2.30), is =2. Thus, the homogeneous solution has the form y h (t)=ae 2t. (2.31) Here, the homogeneous solution is also the complete solution. Thus, Using y(0 )=2, we obtain A=2. Thus, y(t)=ae 2t. (2.32) y(t)=2e 2t. (2.33) Then consider y(t) for t>0. When t>0, (2.28) becomes dy(t) dt The homogeneous solution has the form t 2y(t) e. (2.34) y h (t)=ae 2t. (2.35)
17 The particular solution has the form y p (t)=be t. (2.36) Substituting (2.36) into (2.34), we obtain B=1. Thus, y p (t)=e t. (2.37) Adding (2.35) and (2.37), we obtain the complete solution, i.e., y(t)=ae 2t +e t. (2.38) From (2.28) and x(t)=e t u(t), we know that y(t) is continuous at t=0. So, y(0 + )=y(0 )=2. Applying this condition to (2.38), we obtain A=1. Then, substituting A=1 into (2.38), we obtain y(t)=e 2t +e t. (2.39) Finally, combining (2.33) and (2.39), we obtain y(t)=2e 2t u(t)+(e 2t +e t )u(t). (2.40)
18 Zero-Input Response and Zero-State Response The solution to a linear constant-coefficient differential equation can be decomposed into the inherent response and the response due to the input. The former is called the zero-input response because it equals the response when the input is zero. The latter is called the zero-state response because it equals the response when the state (the inherent response) is zero. Example. A causal continuous-time system is given by dy(t) dt 2y(t) x(t), (2.41) where x(t)=e t u(t) and y(0 )=2. Find the zero-input response, the zero-state response and the complete response. The input is right-sided. Since the system is causal, the zero-state response is also right-sided. This means that y(0 ) has nothing to do the zero-state response and is only related to the zero-input response.
19 Thus, the zero-input response is the solution to dy zi(t) dt 2y (t) 0, (2.42) where y zi (0 )=2, and the zero-state response is the solution to dy zs(t) dt zi t 2y (t) e u(t), (2.43) zs where y zs (0 )=0. Using the method in section 2.5.1, we obtain the zero-input response and the zero-state response y zi (t)=2e 2t (2.44) y zs (t)=(e 2t +e t )u(t). (2.45) The complete response is the sum of the zero-input response and the zero-state response, i.e.,
20 y(t)= 2e 2t +(e 2t +e t )u(t). (2.46) Figure 2.7 is used to clarify the homogeneous solution (the natural response), the particular solution (the forced response), the zero-input response, and the zero-state response in this example. Zero-Input Response 2e 2t Homogeneous Solution (Natural Response) Zero-State Response e t u(t) e 2t u(t) t Particular Solution (Forced Response) Figure 2.7. Clarification of Several Concepts. The zero-state response is linear and time-invariant with respect to the input. This means that the relation between the input and the zero-state response can be characterized by an impulse response. The
21 impulse response is the zero-state response when the input is the continuous-time impulse function. The zero-state response is the convolution integral of the input and the impulse response. Example. A causal continuous-time system is given by dy(t) dt Find the impulse response h(t). The zero-state response satisfies dy zs(t) dt When x(t)=(t), y zs (t)=h(t). Thus, 2y(t) x(t). (2.47) 2y (t) x(t). (2.48) zs dh (t) 2h(t) (t). (2.49) dt
22 Since the system is causal, When t>0, (2.49) becomes h(t)=0 for t<0. (2.50) dh (t) dt Using the method in section 2.5.1, we obtain 2h(t) 0. (2.51) h(t)=ae 2t. (2.52) From (2.49), we conclude that h(t) has a step of 1 from 0 to 0 +. Thus, h(0 + )=h(0 )+1=1. From this condition, we obtain A=1. Thus, Combining (2.50) and (2.53), we obtain h(t)=e 2t. (2.53) h(t)=e 2t u(t). (2.54)
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