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CPE200 Signals and Systems

Chapter 2: Linear Time-Invariant Systems. CPE200 Signals and Systems. Introduction. 2.1.

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CPE200 Signals and Systems

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  1. Chapter 2: Linear Time-Invariant Systems CPE200 Signals and Systems

  2. Introduction 2.1 In this chapter, we will consider a linear time-invariant (LTI) system which is a system satisfying both linearity and the time-invariance properties. Such systems play a fundamental role in signal and system analysis since highly useful tools and concepts associated with LTI system analysis offer the most insight into system behavior. Although, only a small amount of systems in the world are truly LTI, nonlinear systems can still be approximated as being linear within a small enough input range. An LTI system can be characterized in terms of its impulse response, h(t) or h[n] as a consequence of linear and time-invariance properties. The behavior of an LTI system

  3. can also be described by a linear constant- coefficient differential or difference equation. Differential equations are used to represent c-t systems, while difference equations represent d-t systems. The impulse response is anoutput of the LTI system when the input is an impulse (unit sample) signal d(t) or d[n]. Knowing the impulse response, we can determine the output of the system to any arbitrary input by a weighted sum of time-shifted impulse responses. This operation is called the “convolution sum” for d-t systems and the “convolution integral” for c-t systems. In this chapter, we will define the impulse response and derives the convolution operation. Then properties of liner time- invariant systems will be discussed. Finally, we will briefly review a method for

  4. D-T LTI Systems: The 2.2 solving differential and difference equations and we will discuss how to represent LTI systems using block diagram. Convolution Sum As mentioned in the previous section, the output of an LTI system to any arbitrary input can be determined by the convolution process. We will discuss the convolution process for d-t systems in this section first, since it is much easier to understand than one for c-t systems. 2.2.1 The Representation of Discrete- Time Signals in Terms of Impulses

  5. As briefly mentioned in Ch. 1, the d-t unit impulse can be used to construct any d-t signal (See Eq. 1.47). i.e. Any D-T signal is the sum of scaled and shifted unit impulses. (1.47) This idea is fairly obvious to understand by visualizing the graphical representation of d-t signal x[n] as depicted in Fig. 2.1. From Fig. 2.1, the d-t signal x[n] is decomposed into four time-shifted, scaled unit impulse signals where the scaling on each impulse equals the value of x[n] at the particular instant the unit sample occurs. For example,

  6. Hence, the sum of the four signals in Fig. 2.1 equals x[n] for -1 ≤ n ≤ 2 and we can represent x[n] as follows: x[n] = x[-1]d[n+1]+x[0]d[n]+x[1]d[n-1] +x[2]d[n-2] (2.1)

  7. 3 2 2 2 = - 1 0 1 X[n] - 2 At any time n, only one of the terms on the right-hand side of Eq. 2.1 is nonzero. Similarly, for any d-t signals, we can represent them by Eq. 1.47. + + + Figure 2.1 Decomposition of a discrete-time signal into a weighted sum of shifted impulses. Eq. 1.47 is called the sifting property of the d-t unit impulse since only the value of x[k] corresponding to k=n is preserved.

  8. Unit Impulse Response 2.2.2 The D-T Unit Impulse Response The response of a linear system when the input (excitation) signal is the impulse signal. = Since, in the case of d-t systems, the impulse signal is normally called the “unit sample” signal, the unit impulse response for a linear d-t system is widely called the “unit sample response”. We can derive the mathematical representation of the unit sample response by starting with an arbitrary linear d-t system defined as follows:

  9. x[n] y[n] Linear d-t System, t Since and . Because the system is linear, we can applied the operation t to the shifted unit sample signal d[n-k] before performing the summation operation. Hence,

  10. d[n-k] h[n,k] (2.2) Let t{d[n-k]} = h[n,k]. Hence hk[n] is the response of the linear system when the input is equal to d[n-k]. i.e. Linear d-t System, t h[n,k] is known as the “unit impulse response” of a linear d-t system.

  11. Therefore, once h[n,k] of the linear d-t system is determined, y[n] of the system for any arbitrary x[n] can be evaluated by this following Eq.: (2.3) Eq. 2.3 indicates that the response of a linear d-t system to the input x[n] is a linear combination of the responses to the individual scaled and shifted impulses. In general, the response h[n,k] is a function of n and the time k which is a time when the unit sample d[n] is applied to the system. However, if the linear system is also time invariant, then the time-shifted k is not an issue. Thus, for an LTI d-t system,

  12. h[n,k] = h[n-k] (2.4) That is, the response of the LTI system when the input is d[n] is defined as h[n] which is called the “unit sample response”. Then for an LTI system, Eq. 2.3 becomes (2.5) The output is the sum of scaled and shifted unit sample response. This result is referred to as the convolution sum or the superposition sum. The operation on the right-hand side of Eq. 2.5 is known as the convolution of the sequence x[n] and h[n] which can be denoted as:

  13. y[n] = x[n]*h[n] (2.6) The convolution process defined by Eq. 2.6 involves these following steps: 1. FLIP h[k] about k=0 which is h[-k] 2. SHIFT h[-k] to the right by n which is h[n-k] 3. MULTIPLY x[k] by h[n-k] which is the flipped and shifted version of h[k]. 4. ADD across all values of k to obtain the value of the output at one value of n 5. Repeat step 2-4 for all possible value of n

  14. Note: Useful Summation Formulas Finite Summation Formulas

  15. C-T LTI Systems: The 2.3 Infinite Summation Formulas Convolution Integral The output of a c-t LTI system can be determined from knowledge of the input and the impulse response of the system. The approach and result are analogous to the d-t case. For c-t systems, the

  16. superposition is evaluated by an integration instead of a summation because of the continuous nature of the input. Similarly, any c-t signal x(t) may express as the superposition of scaled and shifted impulses: (2.7) Here the scaled x(t) dt is calculated from the value of x(t) at the time at which each impulse occurs, t. Eq. 2.7 is also called the sifting property of the c-t impulses. Now, for any linear c-t system, let define the impulse response h(t) = t{d(t)} as the output of the system in response to an

  17. impulse input. Thus the response of the linear c-t system to any arbitrary input can be evaluated as: (2.8) If the linear system is time invariant, h(t,t) in Eq. 2.8 will become h(t-t). Hence, for an LTI c-t system, the response of the system to x(t) is defined as: (2.9)

  18. Properties of Linear Time- 2.4 This result is referred to as the convolution integral or the superposition integral. As before, this operation is denoted by the symbol “*”; that is y(t) = x(t)*h(t) (2.10) Invariant Systems The Commutative Properties x[n]*h[n] = h[n]*x[n] (2.11) x(t)*h(t) = h(t)*x(t)

  19. The Distributive Properties x[n]*{h1[n]+h2[n]} = x[n]*h1[n] + x[n]*h2[n] (2.12) x(t)*{h1(t)+h2(t)} = x(t)*h1(t) + x(t)*h2(t) Parallel Connection of Systems The Associative Properties x[n]*{h1[n]*h2[n]} = {x[n]*h1[n]} * {x[n]*h2[n]} (2.13) x(t)*{h1(t)*h2(t)} = {x(t)*h1(t)}*{x(t)*h2(t)} Cascade Connection of Systems

  20. The Shifting Properties If y[n] = x[n]*h[n], then y[n-k] = x[n-k]*h[n] = x[n]*h[n-k] (2.14) Convolution with the unit impulse If h[n] = d[n], then x[n]*d[n] = x[n] (2.15) and x[n]*d[n-k] = x[n-k] (2.16) Invertibility of LTI System If a system is invertible, there exists an inverse system such that when cascaded with the original system, yields an output

  21. equal to the original input (see Sec. 1.6.2). x(t) y(t) w(t) = x(t) Inverse System h-1(t) LTI System h(t) Figure 2.2 Cascade of an LTI system with impulse response h(t) and the inverse system with impulse response h-1(t). The relationship between the impulse response of a system, h(t), and the corresponding inverse system, h-1(t), is easily derived. From Fig. 2.2, the impulse response of the cascade connection is the convolution of h(t) and h-1(t). Hence, x(t)*{h(t)*h-1(t)} = x(t) (2.17)

  22. Compare Eq. 2.17 with Eq. 2.15, it implies that {h(t)*h-1(t)} = d(t) (2.18) Causal LTI Systems An LTI system is said to be causal if and only if its impulse response is zero for negative values of n (or t). Let consider the convolution sum which is: (2.19) Pass and present inputs Future inputs

  23. The first term in Eq. 2.19 is associated with indices k < 0 and can be expressed as: = …+h[-2]x[n+2]+h[-1]x[n+1] (2.20) The second term in Eq. 2.19 is associated with indices k ≥ 0 and can be expressed as: = h[0]+h[1]x[n-1]+h[2]x[n-2]+... (2.21) From Eq. 2.20 and 2.21, we can noticed that future values of the input are associated with indices k < 0 while present and past values of the input are associated with indices k ≥ 0 in the convolution sum. Hence, for a causal system, h[k] = 0 for k<0, and the convolution sum is reduced to A causal LTI d-t system (2.22)

  24. Similarly, a causal c-t system has impulse response that satisfies h(t) = 0 for t<0. Thus, the output is expressed as the convolution integral A causal LTI c-t system (2.22) Stable LTI Systems Recall from Ch. 1 that a system is bounded input-bounded output (BIBO) stable if the output is guaranteed to be bounded for every bounded input. I.e. , for a stable d-t system, if |x[n]| ≤ Mx < ∞ for all n, then the output must satisfy |y[n]| ≤ My < ∞ for all n.

  25. Since then Because all the input values are bounded, say by Mx, therefore, (2.23) From Eq. 2.23, if is absolutely summable, the output |y[n]| is bounded. Thus, for a stable LTI system, the impulse response must satisfies the following condition:

  26. Unit Step Response of LTI 2.5 A stable LTI d-t system (2.24) Similarly, a c-t LTI system is BIBO stable if and only if the impulse response is absolutely integrable, that is, A stable LTI c-t system (2.25) Systems Unit Step Sudden Change

  27. The unit step response of an LTI system describes how the system responds to sudden changes in the input. Let consider a d-t LTI system having the impulse response h[n] and denote the step response as s[n]. Thus, the step response s[n] can be determined by the following equation: (2.26) Since u[n-k] = 0 for k > n and u[n-k] = 1 for k ≤ n, hence (2.27)

  28. h[n] = s[n] - s[n-1] Eq. 2.27 indicates that the step response is the running sum of the impulse response and h[n] can be recovered from s[n] using the relation (2.28) Similarly, in c-t system, the step response of an LTI system with impulse response h(t) is the running integral of h(t), or (2.29) From Eq. 2.29, the impulse response will be the first derivative of the unit step response, or

  29. Causal LTI Systems 2.6 (2.30) Described by Differential and Difference Equations An extremely important characteristic of d-t (or c-t) systems is that for which the input and output are related through a linear constant-coefficient difference (or differential) equation. That is, linear constant-coefficient difference and differential equations provide another representation for the input-output characteristics of LTI systems.

  30. Difference equations are used to represent d-t systems, while differential equations represent c-t system. The general form of a linear constant-coefficient difference equation is: (2.31) where y[n] = the output x[n] = the input N and M = the highest delayed orders and ak and bk = the constant coefficients

  31. A linear constant-coefficient differential equation has a similar form, with the delayed values replaced by the derivative values of the input x(t) and output y(t), as shown in the following equation: (2.32) We can notice that Eq. 2.31 and 2.32 provide an implicit specification of the system. That is, they describe a relationship between the input and the output, rather than an explicit expression for the system output as a function of the input. To determine an explicit expression, we must solve the difference or differential equation. In general, to solve Eq. 2.31 or

  32. 2.32, we must specify a set of initial conditions. Generally, the solution of both Eq. 2.31 and 2.32 can be divided into two types of solutions as shown below: y[n] = yc[n] + yp[n] (2.33) y(t) = yc(t) + yp(t) The term yc[n] (or yc(t)) is known as the complementary solution, whereas yp[n] (or yp(t)) is called theparticular solution. Generally, the complementary solution will describe the response of a system when the input is zero. Such response is usually called the “natural response” of a system.

  33. The complementary solution is usually of the form: (2.34) ,for a difference equation, and (2.35) ,for a differential equation. Where C, s, and l are constants to be determined. The particular solution, on the other hand, represents any solution to the differential or difference equation for the given input.

  34. Such response is usually called the “forced response” of a system. The particular solution is usually obtained by assuming the system output has the same general form as the input. Table 2.1 provides the general form of the particular solution for common input signals. Table 2.1 Form of a particular solution corresponding to several types of common inputs. C-T D-T Input Particular Sol. Input Particular Sol. 1 C 1 C ln Cln e-st Ce-st cos(Wt + f) C1cos(Wt) cos(wt + f) C1cos(wt) +C2sin(Wt) +C2sin(wt)

  35. For a convenience, we will discussed only how to solve a difference equation. However, solving a differential equation can be perform in the same manner. 2.6.1 The Complementary Solution of the Difference Equation To find the complementary solution, we begin with writing the homogeneous equation which is Eq. 2.31 with the left side set equal to zero, that is, (2.36) In other words, the complementary solution will describe the response of a system when the input is zero.

  36. Basically, we assume that the solution of the homogeneous equation is of the form: yc[n] = ln (2.37) If we substitute Eq. 2.37 into Eq. 2.36, we obtain the polynomial equation: or ln-N(lN+a1lN-1 +…+aN-1l+aN) = 0 (2.38) The polynomial “lN+a1lN-1 +…+aN-1l+aN” is called the characteristic polynomial of the system. The roots of Eq. 2.38 can be real or complex valued but the coefficients “ak”, in practice, are usually real.

  37. If we assume that the roots are distinct, then, the most general solution to the homogeneous difference equation is in the form described by Eq. 2.34, that is, (2.39) where C1, C2, …, CN are weighting coefficients. These coefficients are determined from the initial conditions specified for the system. Example 2.1 Determine the homogeneous solution of the system described by the first-order difference equation y[n] + a1y[n-1] = x[n] (2.40) When x[n] = 0 and we substitute yc[n] = ln in Eq. 2.40, we obtain

  38. ln+a1ln-1 = 0 ln-1(l+a1) = 0 l = -a1 (2.41) Therefore, the solution to the homogeneous difference equation is yc[n] = Cln = C(-a1)n (2.42) To determine the value of C, some of initial conditions must be provided. From Eq. 2.41, when x[n] = 0 and at n = 0, we obtain (2.43) y[0] = -a1y[-1] From Eq. 2.42, we have yc [0] = C Thus, the homogeneous solution of this system is yc [n] = (-a)n+1y[-1] n ≥ 0 Ans.

  39. Previously, we assumed that the characteristic equation contains distinct root. On the other hand, if the characteristic equation contains multiple roots, the form of the solution given in Eq. 2.39 must be modified. Let assume l1 is a root of multiplicity m, then Eq. 2.39 will be expressed as: (2.39) 2.6.2 The Particular Solution of the Difference Equation The particular solution yp[n] is required to satisfy the difference equation for the

  40. specific input signal x[n], n ≥ 0. It is usually obtained by assuming the system output has the same general form as the input. That is, if x[n] is an exponential, we would assume that the particular solution is also an exponential. Example 2.2 Determine the particular solution of the difference equation y[n]-(5/6)y[n-1]+(1/6)y[n-2] = x[n] when the forcing function x[n] = 2n, n ≥ 0 and zero elsewhere. To solve this problem, we begin with assuming the particular solution is yp[n] = C2n n ≥ 0 Substitute yp[n] into the difference equation, we obtain

  41. C2nu[n] = (5/6)C2n-1u[n-1]-(1/6)C2n-2u[n-2] +2nu[n] To determine the value of K, we can evaluate the above equation for any n ≥ 2, where none of the terms vanish. Thus we obtain 4C = (5/6)2C - (1/6)C + 4 Solving the above equation, we get C = 8/5. Therefore, the particular solution is yp[n] = (8/5)2n n ≥ 0 Ans.

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