BYST CPE200 - W2003: LTI System 79 CPE200 Signals and Systems Chapter 2: Linear Time-Invariant Systems

BYST CPE200 - W2003: LTI System 80 Introduction 2. 1 linear time-invariant 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. impulse response, h(t) or h[n] 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

BYST CPE200 - W2003: LTI System 81 convolution sum convolution integral The impulse response is an output of the LTI system when the input is an impulse (unit sample) signal  (t) or  [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. linear constant- coefficient differentialdifference equation 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. 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

BYST CPE200 - W2003: LTI System 82 solving differential and difference equations and we will discuss how to represent LTI systems using block diagram. D-T LTI Systems: The 2. 2 Convolution Sum The Representation of Discrete- Time Signals in Terms of Impulses 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.

BYST CPE200 - W2003: LTI System 83 i.e. This idea is fairly obvious to understand by visualizing the graphical representation of d-t signal x[n] as depicted in Fig 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, As briefly mentioned in Ch. 1, the d-t unit impulse can be used to construct any d-t signal (See Eq. 1.47). Any D-T signal is the sum of scaled and shifted unit impulses. (1.47)

BYST CPE200 - W2003: LTI System 84 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]  [n+1]+x[0]  [n]+x[1]  [n-1] +x[2]  [n-2] (2.1)

BYST CPE200 - W2003: LTI System Figure 2.1 Decomposition of a discrete-time signal into a weighted sum of shifted impulses. sifting property Eq is called the sifting property of the d-t unit impulse since only the value of x[k] corresponding to k=n is preserved. 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 = X[n] 2

BYST CPE200 - W2003: LTI System The D-T Unit Impulse Response Unit Impulse Response linear impulse The response of a linear system when the input (excitation) signal is the impulse signal. = unit sample unit sample response 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:

BYST CPE200 - W2003: LTI System 87 Since Linear d-t System,  x[n]y[n] and  Because the system is linear, we can applied the operation  to the shifted unit sample signal  [n-k] before performing the summation operation. Hence,.

BYST CPE200 - W2003: LTI System 88 (2.2) Let  {  [n-k]} = h [n,k]. Hence h k [n] is the response of the linear system when the input is equal to  [n-k]. i.e. Linear d-t System,   [n-k] h [n,k] h [n,k]unit impulse response h [n,k] is known as the “unit impulse response” of a linear d-t system.

BYST CPE200 - W2003: LTI System 89 h [n,k] 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. nk time invariantan issue In general, the response h[n,k] is a function of n and the time k which is a time when the unit sample  [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,

BYST CPE200 - W2003: LTI System 90 h[n,k] = h[n-k](2.4) h[n] unit sample response That is, the response of the LTI system when the input is  [n] is defined as h[n] which is called the “unit sample response”. Then for an LTI system, Eq. 2.3 becomes (2.5) convolution sum convolution 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: The output is the sum of scaled and shifted unit sample response.

BYST CPE200 - W2003: LTI System 91 y[n] = x[n] * h[n](2.6) The convolution process defined by Eq. 2.6 involves these following steps: FLIP 1. FLIP h[k] about k=0 which is h[-k] SHIFT 2. SHIFT h[-k] to the right by n which is MULTIPLY 3. MULTIPLY x[k] by h[n-k] which is the h[n-k] flipped and shifted version of h[k]. ADD 4. ADD across all values of k to obtain the value of the output at one value of n Repeat 5. Repeat step 2-4 for all possible value of n

BYST CPE200 - W2003: LTI System 92 Note: Useful Summation Formulas Finite Summation Formulas

BYST CPE200 - W2003: LTI System 93 Infinite Summation Formulas C-T LTI Systems: The 2. 3 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

BYST CPE200 - W2003: LTI System 94 superpositionintegration summation 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(  ) d  is calculated from the value of x(t) at the time at which each impulse occurs, . 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)} as the output of the system in response to an

BYST CPE200 - W2003: LTI System 95 impulse input. Thus the response of the linear c-t system to any arbitrary input can be evaluated as: If the linear system is time invariant, h(t,  ) in Eq. 2.8 will become h(t-  ). Hence, for an LTI c-t system, the response of the system to x(t) is defined as: (2.8) (2.9)

BYST CPE200 - W2003: LTI System 96 convolution integral 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) Properties of Linear Time Invariant Systems The Commutative Properties x(t) * h(t) = h(t) * x(t) x[n] * h[n] = h[n] * x[n] (2.11)

BYST CPE200 - W2003: LTI System 97 The Distributive Properties x[n] * {h 1 [n]+h 2 [n]} = x[n] * h 1 [n] + x[n] * h 2 [n] (2.12) x(t) * {h 1 (t)+h 2 (t)} = x(t) * h 1 (t) + x(t) * h 2 (t) The Associative Properties x[n] * {h 1 [n] * h 2 [n]} = {x[n] * h 1 [n]} * (2.13) x(t) * {h 1 (t) * h 2 (t)} = {x(t) * h 1 (t)} * {x(t) * h 2 (t)} Parallel Connection of Systems {x[n] * h 2 [n]} Cascade Connection of Systems

BYST CPE200 - W2003: LTI System 98 The Shifting Properties Ify[n] = x[n] * h[n], then (2.14) Convolution with the unit impulse x[n] *  [n] = x[n] (2.15) y[n-k] = x[n-k] * h[n] = x[n] * h[n-k] Ifh[n] =  [n], then x[n] *  [n-k] = x[n-k] (2.16) and Invertibility of LTI System exists inverse system If a system is invertible, there exists an inverse system such that when cascaded with the original system, yields an output

BYST CPE200 - W2003: LTI System 99 equal to the original input (see Sec ). LTI System h(t) Inverse System h -1 (t) x(t)y(t) Figure 2.2 Cascade of an LTI system with impulse response w(t) = x(t) 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)

BYST CPE200 - W2003: LTI System 100 Compare Eq with Eq. 2.15, it implies that {h(t) * h -1 (t)} =  (t) (2.18) Causal LTI Systems causal 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) Future inputs Pass and present inputs

BYST CPE200 - W2003: LTI System 101 k < 0 The first term in Eq is associated with indices k < 0 and can be expressed as: (2.20)= …+h[-2]x[n+2]+h[-1]x[n+1] k ≥ 0 The second term in Eq is associated with indices k ≥ 0 and can be expressed as: (2.21)= h[0]+h[1]x[n-1]+h[2]x[n-2]+... causalh[k] = 0 for k<0 From Eq 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)

BYST CPE200 - W2003: LTI System 102 causal h(  ) = 0 for  <0 Similarly, a causal c-t system has impulse response that satisfies h(  ) = 0 for  <0. Thus, the output is expressed as the convolution integral A causal LTI c-t system 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]| ≤ M x < ∞ for all n, then the output must satisfy |y[n]| ≤ M y < ∞ for all n. (2.22)

BYST CPE200 - W2003: LTI System 103 Since then Because all the input values are bounded, say by M x, therefore, (2.23) From Eq. 2.23, ifis absolutely summable, the output |y[n]| is bounded. Thus, for a stable LTI system, the impulse response must satisfies the following condition:

BYST CPE200 - W2003: LTI System 104 (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 d-t system (2.25) A stable LTI c-t system Unit Step Response of LTI 2. 5 Systems Unit Step Sudden Change

BYST CPE200 - W2003: LTI System 105 the step responses[n] 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) u[n-k] = 0k > nu[n-k] = 1 k ≤ n Since u[n-k] = 0 for k > n and u[n-k] = 1 for k ≤ n, hence (2.27)

BYST CPE200 - W2003: LTI System 106 Eq 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 h[n] = s[n] - s[n-1] (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

BYST CPE200 - W2003: LTI System 107 (2.30) Causal LTI Systems 2. 6 Described by Differential and Difference Equations linear constant-coefficient difference (or differential) equation 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.

BYST CPE200 - W2003: LTI System 108 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 and a k and b k = the constant coefficients N and M = the highest delayed orders

BYST CPE200 - W2003: LTI System 109 (2.32) 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: We can notice that Eq 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 or

BYST CPE200 - W2003: LTI System , we must specify a set of initial conditions. Generally, the solution of both Eq and 2.32 can be divided into two types of solutions as shown below: y[n] = y c [n] + y p [n] (2.33) complementary solution particular solution The term y c [n] (or y c (t)) is known as the complementary solution, whereas y p [n] (or y p (t)) is called the particular solution. y(t) = y c (t) + y p (t) the input is zero natural response 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.

BYST CPE200 - W2003: LTI System 111 The complementary solution is usually of the form:,for a difference equation, and (2.34),for a differential equation. (2.35) Where C, s, and are constants to be determined. the given input The particular solution, on the other hand, represents any solution to the differential or difference equation for the given input.

BYST CPE200 - W2003: LTI System 112 forced response 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-TD-T InputParticular Sol.InputParticular Sol. 1C1C e -st Ce -st n C n cos(  t +  )C 1 cos(  t) +C 2 sin(  t) cos(  t +  )C 1 cos(  t) +C 2 sin(  t)

BYST CPE200 - W2003: LTI System 113 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 The Complementary Solution of the Difference Equation To find the complementary solution, we begin with writing the homogeneous equation which is Eq 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.

BYST CPE200 - W2003: LTI System 114 Basically, we assume that the solution of the homogeneous equation is of the form: y c [n] = n (2.37) If we substitute Eq into Eq. 2.36, we obtain the polynomial equation: (2.38) or n-N ( N +a 1 N-1 +…+a N-1 +a N ) = 0 characteristic polynomial a k The polynomial “ N +a 1 N-1 +…+a N-1 +a N ” is called the characteristic polynomial of the system. The roots of Eq can be real or complex valued but the coefficients “a k ”, in practice, are usually real.

BYST CPE200 - W2003: LTI System 115 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 C 1, C 2, …, C N 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] When x[n] = 0 and we substitute y c [n] = n in Eq. 2.40, we obtain (2.40)

BYST CPE200 - W2003: LTI System 116 n +a 1 n-1 = 0 n-1 ( +a 1 ) = 0 = -a 1 Therefore, the solution to the homogeneous difference equation is y c [n] = C n = C(-a 1 ) n 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 y [0] = -a 1 y[-1] (2.41) (2.42) (2.43) From Eq. 2.42, we have y c [0] = C Thus, the homogeneous solution of this system is y c [n] = (-a) n+1 y[-1] n ≥ 0 Ans.

BYST CPE200 - W2003: LTI System multiplicity m 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 must be modified. Let assume 1 is a root of multiplicity m, then Eq will be expressed as: (2.39) The Particular Solution of the Difference Equation The particular solution y p [n] is required to satisfy the difference equation for the

BYST CPE200 - W2003: LTI System 118 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] = 2 n, n ≥ 0 and zero elsewhere. To solve this problem, we begin with assuming the particular solution is y p [n] = C2 n n ≥ 0 Substitute yp[n] into the difference equation, we obtain

BYST CPE200 - W2003: LTI System 119 C2 n u[n] = (5/6)C2 n-1 u[n-1]-(1/6)C2 n-2 u[n-2] 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 Solving the above equation, we get C = 8/5. Therefore, the particular solution is y p [n] = (8/5)2 n n ≥ 0 +2 n u[n] 4C = (5/6)2C - (1/6)C + 4 Ans.