1 Time-Domain Representations of LTI Systems CHAPTER 2.11 Characteristics of Systems Described by Differential and Difference Equations and Difference Equations Complete solution: y = y (n) + y (f) y (n) = natural response,y (f) = forced response The Natural Response Example 2.24 Example 2.24 RC Circuit (continued): Natural Response Find the natural response of the this system, assuming that y(0) = 2 V, R = 1  and C = 1 F. The system In Example 2.17 is described by the differential equation <Sol.> 1. Homogeneous sol.: 2. I.C.: y(0) = 2 V y (n) (0) = 2 Vc 1 = 2 3. Natural Response:

2 Time-Domain Representations of LTI Systems CHAPTER Example 2.25 Example 2.25 First-Order Recursive System (Continued): Natural Response Example 2.21 The system in Example 2.21 is described by the difference equation Find the natural response of this system. <Sol.> 1. Homogeneous sol.: 2. I.C.: y[  1] = 8 c 1 = 2 3. Natural Response: The Forced Response The forced response is the system output due to the input signal assuming zero initial conditions. The forced response is valid only for t  0 or n  0

3 Time-Domain Representations of LTI Systems CHAPTER  The at-rest conditions for a discrete-time system, y[  N] = 0, …, y[  1] = 0, must be translated forward to times n = 0, 1, …, N  1 before solving for the undetermined coefficients, such as when one is determining the complete solution. Example 2.26 Example 2.26 First-Order Recursive System (Continued): Forced Response Example 2.21 The system in Example 2.21 is described by the difference equation Find the forced response of this system if the input is x[n] = (1/2) n u[n]. <Sol.> 1. Complete solution: 2. I.C.: Translate the at-rest condition y[  1] to time n = 0 y[0] = 1 + (1/4)  0 =1 3. Finding c 1 : c 1 =  1

4 Time-Domain Representations of LTI Systems CHAPTER 4. Forced response: Example 2.27 Example 2.27 RC Circuit (continued): Forced Response Find the forced response of the this system, assuming that x(t) = cos(t)u(t) V, R = 1  and C = 1 F. The system In Example 2.17 is described by the differential equation <Sol.> 1. Complete solution: From Example I.C.: y(0  ) = y(0 + ) = 0 c =  1/2 3. Forced response:

5 Time-Domain Representations of LTI Systems CHAPTER The Impulse Response  Relation between step response and impulse response 1. Continuous-time case: 2. Discrete-time case: Linearity and Time Invariance InputForced response x1x1 y1(f)y1(f) x2x2 y2(f)y2(f)  x 1 +  x 2  y 1 (f) +  y 2 (f)  Forced response  Linearity Initial Cond.Natural response I1I1 y1(n)y1(n) I2I2 y2(n)y2(n)  I 1 +  I 2  y 1 (n) +  y 2 (n)  Natural response  Linearity not  The complete response of an LTI system is not time invariant. Response due to initial condition will not shift with a time shift of the input Roots of the Characteristic Equation

6 Time-Domain Representations of LTI Systems CHAPTER  Roots of characteristic equation Forced response, natural response, stability, and response time. ★ BIBO Stable: 1. Discrete-time case: 2. Continuous-time case: and  The system is said to be on the verge of instability Block Diagram Representations  A block diagram is an interconnection of the elementary operations that act on the input signal.  Three elementary operations for block diagram: 1.Scalar multiplication: y(t) = cx(t) or y[n] = cx[n], where c is a scalar. 2.Addition: y(t) = x(t) + w(t) or y[n] = x[n] + w[n]. 3.Integration for continuous-time LTI system: ; and a time shift for discrete-time LTI system: y[n] = x[n  1]. Fig

7 (a) (b) (c) Figure 2.32 (p. 162) Symbols for elementary operations in block diagram descriptions of systems. (a) Scalar multiplication. (b) Addition. (c) Integration for continuous-time systems and time shifting for discrete-time systems. Time-Domain Representations of LTI Systems CHAPTER Fig Ex. A discrete-time LTI system: Fig In dashed box: (2.49) 2. y[n] in terms of w[n]:  Direct Form I:

8 Time-Domain Representations of LTI Systems CHAPTER (2.50) 3. System output y[n] in terms of input x[n]: (2.51) Figure 2.33 (p. 162) Block diagram representation of a discrete-time LTI system described by a second-order difference equation. Cascade Form (Direct Form I)

9 Time-Domain Representations of LTI Systems CHAPTER  Direct Form II: 1. Interchange the order of Direct Form I. 2. Denote the output of the new first system as f[n]. (2.52) Input: x[n] 3. The signal is also the input to the second system. The output of the second system is (2.53) Fig Figure 2.35 (p. 164) Direct form II representation of an LTI system described by a second-order difference equation.

10 Time-Domain Representations of LTI Systems CHAPTER  Block diagram representation for continuous-time LTI system: 1. Differential Eq.: (2.54) 2. Let v (0) (t) = v(t) be an arbitrary signal, and set v (n) (t) is the n-fold integral of v(t) with respect to time 3. Integrator with initial condition: (2.55) Integrate N times to eq. (2.54)

11 Time-Domain Representations of LTI Systems CHAPTER Figure 2.37 (p. 166) Block diagram representations of a continuous-time LTI system described by a second-order integral equation. (a) Direct form I. (b) Direct form II. (a) (b) Ex. Second-order system: (2.56)

12 Time-Domain Representations of LTI Systems CHAPTER 2.13 State-Variable Description of LTI Systems  The state of a system may be defined as a minimal set of signals that represent the system’s entire memory of the past. Given initial point n i (or t i ) and the input for time n  n i (or t  t i ), we can determine the output for all times n  n i (or t  t i ) The State-Variable Description Fig Direct form II of a second-order LTI system: Fig Choose state variables: q 1 [n] and q 2 [n]. 3. State equation: (2.57) (2.58) 4. Output equation: (2.59) (2.60) 5. Matrix Form of state equation:

13 Time-Domain Representations of LTI Systems CHAPTER Figure 2.39 (p. 167) Direct form II representation of a second-order discrete-time LTI system depicting state variables q 1 [n] and q 2 [n].

14 Time-Domain Representations of LTI Systems CHAPTER 6. Matrix form of output equation: (2.61) Define state vector as the column vector We can rewrite Eqs. (2.60) and (2.61) as (2.62) (2.63) where matrix A, vectors b and c, and scalar D are given by Example 2.28 Example 2.28 State-Variable Description of a Second-Order System Fig Find the state-variable description corresponding to the system depicted in Fig by choosing the state variable to be the outputs of the unit delays. <Sol.>

15 Time-Domain Representations of LTI Systems CHAPTER Figure 2.40 (p. 169) Block diagram of LTI system for Example State equation:

16 Time-Domain Representations of LTI Systems CHAPTER 2. Output equation: 3. Define state vector as In standard form of dynamic equation: (2.63) (2.62)  State-variable description for continuous-time systems: (2.64) (2.65)

17 Time-Domain Representations of LTI Systems CHAPTER Example 2.29 Example 2.29 State-Variable Description of an Electrical Circuit Fig Consider the electrical circuit depicted in Fig Derive a state-variable description of this system if the input is the applied voltage x(t) and the output is the current y(t) through the resistor. <Sol.> Figure 2.42 (p. 171) Circuit diagram of LTI system for Example State variables: The voltage across each capacitor. 2. KVL Eq. for the loop involving x(t), R 1, and C 1 : (2.66) Output equation 3. KVL Eq. for the loop involving C 1, R 2, and C 2 :

18 Time-Domain Representations of LTI Systems CHAPTER (2.67) 4. The current i 2 (t) through R 2 : (2.68) Use Eq. (2.67) to eliminate i 2 (t) 5. KCL Eq. between R 1 and R 2 : Current through C 1 = i 1 (t) where (2.69) ◆ Eqs. (2.66), (2.68), and (2.69) = State-Variable Description. (2.64) (2.65)

19 Time-Domain Representations of LTI Systems CHAPTER and Example 2.30 Example 2.30 State-Variable Description from a Block Diagram Fig Determine the state-variable description corresponding to the block diagram in Fig The choice of the state variables is indicated on the diagram. Figure 2.44 (p. 172) Block diagram of LTI system for Example <Sol.>

20 Time-Domain Representations of LTI Systems CHAPTER 1. State equation: 2. Output equation: 3. State-variable description: Transformations of The State The transformation is accomplished by defining a new set of state variables that are a weighted sum of the original ones. The input-output characteristic of the system is not changed. 1. Original state-variable description: (2.70) (2.71) 2. Transformation:q’ = Tq T = state-transformation matrix q = T  1 q’

21 Time-Domain Representations of LTI Systems CHAPTER 3. New state-variable description: 1) State equation: q = T  1 q’ 2) Output equation: 3) If we set then and Example 2.30 Ex. Consider Example 2.30 again. Let us define new states Find the state-variable description. <Sol.> 1. State equation:

22 Time-Domain Representations of LTI Systems CHAPTER 2. Output equation: 3. State-variable description: Example 2.31 Example 2.31 Transforming The State A discrete-time system has the state-variable description and Find the state-variable description A, b, c, D corresponding to the new states and <Sol.> 1. Transformation: q = Tq, where

23 Time-Domain Representations of LTI Systems CHAPTER 2. New state-variable description: and  This choice for T results in A being a diagonal matrix and thus separates the state update into the two decoupled first-order difference equations and 2.14 Exploring Concepts with MATLAB  Two limitations: 1. MATLAB is not easily used in the continuous-time case. 2. Finite memory or storage capacity and nonzero computation times.  Both the MATLAB Signal Processing Toolbox and Control System Toolbox are use in this section.

24 Time-Domain Representations of LTI Systems CHAPTER Convolution 1. MATLAB command:y = conv(x, h) x and h are signal vectors. 2. The number of elements in y is given by the sum of the number of elements in x and h, minus one. <pf.> 1) Elements in vector x: from time n = k x to n = l x 2) Elements in vector h: from time n = k h to n = l h 3) Elements in vector y: from time n = k y = k x + k h to n = l y = l x + l y 4) The length of x[n] and h[n] are L x = l x  k x + 1 and L h = l h  k h +1 5) The length of y[n] is L y = L x + L h  1 Ex. Repeat Example 2.1 Impulse and Input : From time n = k h = k x = 0 to n = l h = 1 and n = l x =2 Convolution sum: From time n = k y = k x + k h = 0 to n = l y = l x + l h = 3 The length of convolution sum: L y = l y – k y + 1 = 4 MATLAB Program: >> h = [1, 0.5]; >> x = [2, 4, -2]; >> y = conv(x,h) y =

25 Time-Domain Representations of LTI Systems CHAPTER Repeat Example 2.3 Givenand Impulse responseInput Find the convolution sum x[n]  h[n]. 1. In this case, k h = 0, l h = 3, k x = 0 and l x = 9 <Sol.> 2. y starts at time n = ky = 0, ends at time n = l y =12, and has length L y = Generation for vector h with MATLAB: >> h = 0.25*ones(1, 4); >> x = ones(1, 10); 4. Output and its plot: >> n = 0:12; >> y = conv(x, h); >> stem(n, y); xlabel('n'); ylabel('y[n]') Fig

26 Time-Domain Representations of LTI Systems CHAPTER Figure 2.45 (p. 177) Convolution sum computed using MATLAB.

27 Time-Domain Representations of LTI Systems CHAPTER The Step Response 1. Step response = the output of a system in response to a step input 2. In general, step response is infinite in duration. 3. We can evaluate the first p values of the step response using the conv function if h[n] = 0 for n < k h by convolving the first p values of h[n] with a finite-duration step of length p. 1) Vector h = the first p nonzero values of the impulse response. 2) Define step: u = ones(1, p). 3) convolution: s = conv(u, h). Ex. Repeat Problem 2.12 Determine the first 50 values of the step response of the system with impulse response given by with  =  0.9, by using MATLAB program. <Sol.> 1. MATLAB Commands:

28 Time-Domain Representations of LTI Systems CHAPTER >> h = (-0.9).^[0:49]; >> u = ones(1, 50); >> s = conv(u, h); >> stem([0:49], s(1:50)) Fig Step response: Fig Figure 2.47 (p. 178) Step response computed using MATLAB Simulating Difference equations 1. Difference equation: (2.36) Command: Command: filter

29 Time-Domain Representations of LTI Systems CHAPTER 2. Procedure: 1) Define vectors a = [a 0, a 1, …, a N ] and b =[b 0, b 1, …, b M ] representing the coefficients of Eq. (2.36). 2) Input vector: x 3) y = filter(b, a, x) results in a vector y representing the output of the system for zero initial conditions. 4) y = filter(b, a, x, zi) results in a vector y representing the output of the system for nonzero initial conditions zi.  The initial conditions used by filter are not the past values of the output.  Command zi = filtic(b, a, yi), where yi is a vector containing the initial conditions in the order [y[  1], y[  2], …, y[  N]], generates the initial conditions obtained from the knowledge of the past outputs. Ex. Repeat Example 2.16 The system of interest is described by the difference equation (2.73)Determine the output in response to zero input and initial condition y[  1] = 1 and y[  2] = 2. <Sol.>

30 Time-Domain Representations of LTI Systems CHAPTER 1. MATLAB Program: >> a = [1, , ]; b = [0.0675, , 0.675]; >> x = zeros(1, 50); >> zi = filtic(b, a, [1, 2]); >> y = filter(b, a, x, zi); >> stem(y) Fig. 2.28(b). 2. Output: Fig. 2.28(b). 3. System response to an input consisting of the Intel stock price data Intc: >> load Intc; >> filtintc = filter(b, a, Intc);  We have assume that the Intel stock price data are in the file Intc.mat.  The command [h, t] = impz(b, a, n) evaluates n values of the impulse response of a system described by a different equation.

31 Time-Domain Representations of LTI Systems CHAPTER State-Variable Descriptions  MATLAB command: ss 1. Input MATLAB arrays: a, b, c, d Representing the matices A,b,c, and D. 2. Command: sys = ss(a, b, c, d, -1) produces an LTI object sys that represents the discrete-time system in state-variable form. ★ Continuous-time case: sys = ss(a, b, c, d) No  1  System manipulation: 1. sys = sys1 + sys2Parallel combination of sys1 and sys2. 2. sys = sys1  sys2 Cascade combination of sys1 and sys2.  MATLAB command: lsim 1. Command form: y = lsim(sys, x) 2. Output = y, input = x.  MATLAB command: impulse 2. This command places the first N values of the impulse response in h. 1. Command form: h = impulse(sys, N)  MATLAB routine: ss2ss Perform the state transformation 1. Command form: sysT = ss2ss(sys, T), where T = Transformation matrix

32 Time-Domain Representations of LTI Systems CHAPTER Ex. Repeat Example Original state-variable description: and 2. State-transformation matrix: 3. MATLAB Program: >> a = [-0.1, 0.4; 0.4, -0.1]; b = [2; 4]; >> c = [0.5, 0.5]; d = 2; >> sys = ss(a, b, c, d, -1); % define the state-space object sys >> T = 0.5*[-1, 1; 1, 1]; >> sysT = ss2ss(sys, T) 4. Result:

33 Time-Domain Representations of LTI Systems CHAPTER a = x1 x2 x x b = u1 x1 1 x2 3 c = x1 x2 y1 0 1 d = u1 y1 2 Sampling time: unspecified Discrete-time model. Ex. Verify that the two systems represented by sys and sysT have identical input-output characteristic by comparing their impulse responses. <Sol.> 1. MATLAB Program: >> h = impulse(sys, 10); hT = impulse(sysT, 10); >> subplot(2, 1, 1) >> stem([0:9], h) >> title ('Original System Impulse Response'); >> xlabel('Time'); ylabel('Amplitude') >> subplot(2, 1, 2) >> stem([0:9], hT) >> title('Transformed System Impulse Response'); >> xlabel('Time'); ylabel('Amplitude')

34 Time-Domain Representations of LTI Systems CHAPTER 2. Simulation results: Fig Figure 2.48 (p. 181) Impulse responses associated with the original and transformed state-variable descriptions computer using MATLAB.  We may verify that the original and transformed systems have the (numerically) identical impulse response by computing the error, err = h – hT.

35 Time-Domain Representations of LTI Systems CHAPTER Plot for err = h  hT