Chap5Solutions - M. J. Roberts - 10/7/06 Chapter 5 -...

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Unformatted text preview: M. J. Roberts - 10/7/06 Chapter 5 - Discrete-Time System Properties Solutions Modeling Discrete-Time Systems 1. Bernoulli’s method can be used to numerically find the dominant root of a polynomial equation (if it exists). It is an example of a discrete-time system. If the equation is of the form aN q n + aN 1q n 1 + + a1q + a0 = 0 , the method consists of solving the difference equation aN q n + aN 1q n 1 + with the initial values q 1 =q + a1q n 2 = N + 1 + a0 q n =q N +1 N =0 0 and q 0 = 1. The dominant root is the limit approached by q n + 1 / q n . Draw a DT system to find the dominant root of a fourth-degree polynomial equation. Find the dominant root of 2q 4 + 3q 3 8q 2 + q 3 = 0 . a3 a4 a2 a4 a1 a4 a0 a4 % % D D D D Program to find the dominant root of a fourth-degree equation using Bernoulli's method q = [1 0.01 0.01 0.01] ; a = [2 3 -8 1 -3] ; while abs(q(1)/q(2) - q(2)/q(3)) > abs(q(1)/q(2))/10000, qn = -a(2:end)*q'/a(1) ; q = [qn [q(1:end-1)]] ; q(1)/q(2) end The dominant root is -2.964. Solutions 5-1 M. J. Roberts - 10/7/06 Properties of DT Systems 2. Show that the system of Figure E-1 is non-linear, BIBO stable, static and noninvertible. (The response of an analog multiplier is the product of its two excitations.) Analog Multiplier x[n] y[n] 2 Figure E-1 A DT system Homogeneity: Let x1 n = g n . Then y1 n = 2 g 2 n ( Let x 2 n = K g n . Then y2 n = 2 K g n Multiplying the first equation by K, ) 2 = 2K 2 g 2 n . K y1 n = 2K g 2 n . Then y2 n . K y1 n Not homogeneous. Let x1 n = g n . Then y1 n = 2 g 2 n Let x 2 n = h n . Then y2 n = 2 h 2 n Let x 3 n = g n + h n . ( Then y3 n = 2 g n + h n ) 2 ( =2 g n 2 +h n 2 + 2g n h n ). Adding the first two equations, y1 n + y2 n = 2 g 2 n + 2 h 2 n y3 n . Not additive. Since the system is not homogeneous and not additive it is also not linear. The system is also not incrementally linear. The system is statically non-linear because the non-linearity is inherent in the squared relationship between excitation and response. Time Invariance: Let x1 n = g n . Then y1 n = 2 g 2 n . Solutions 5-2 M. J. Roberts - 10/7/06 Let x 2 n = g n n0 . Then y2 n = 2 g 2 n n0 . The first equation can be rewritten as y1 n n0 = 2 g 2 n no = y2 n Time Invariant Stability: If the excitation is bounded, the response is bounded. Stable Causality: At any discrete time, n = n0 , the response depends only on the excitation at that same time. Causal. Memory: At any discrete time, n = n0 , the response depends only on the excitation at that same time. System has no memory. Invertibility: Inverting the functional relationship, x n = y n 1 2 2 The square root is multiple valued. Two different excitations can cause the same response. Not Invertible. 3. Show that a system with excitation x n and response y n described by y n = nx n , is linear, time variant and static. Homogeneity: Let x1 n = g n . Then y1 n = n g n Let x 2 n = K g n . Then y2 n = nK g n . Multiplying the first equation by K, K y1 n = nK g n . Then K y1 n = y2 n . Homogeneous. Solutions 5-3 M. J. Roberts - 10/7/06 Let x1 n = g n . Then y1 n = n g n Let x 2 n = h n . Then y2 n = n h n L e t ( x3 n = g n + h n . y3 n = n g n + h n Additive. ) = ng T h e n n + n h n = y1 n + y2 n . Since the system is homogeneous and additive it is also linear. The system is also incrementally linear. The system is not statically non-linear because it is linear. Time Invariance: Let x1 n = g n . Then y1 n = n g n . Let x 2 n = g n n0 . Then y2 n = n g n n0 . The first equation can be rewritten as ( ) y1 n n0 = n n0 g n n0 y2 n Time Variant Stability: If the excitation is a constant, the response is unbounded as n approaches infinity. Unstable Causality: At any discrete time, n = n0 , the response depends only on the excitation at that same time. Causal. Memory: At any discrete time, n = n0 , the response depends only on the excitation at that same time. System has no memory. Invertibility: Inverting the functional relationship, x n =y n /n . Invertible. 4. Show that the system of Figure E-4 is linear, time-invariant, BIBO unstable and dynamic. Solutions 5-4 M. J. Roberts - 10/7/06 x[n] y[n] + + D Figure E-4 A DT system y n =x n +y n 1 y n 1 =x n 1 +y n 2 y n =x n +x n 1 +y n 2 Then, by induction, +x n k + y n =x n +x n 1 + = x n k k =0 Let m = n k . Then y n = n x m = m= n x m m= Homogeneity: n Let x1 n = g n . Then y1 n = g m m= Let x 2 n n = K g n . Then y2 n = Kg m = K m= n g m = K y1 n . m= Homogeneous. n Let x1 n = g n . Then y1 n = g m m= Let x 2 n n = h n . Then y2 n = h m m= Let x 3 n = g n + h n . n Then y3 n = m= (g m +h m )= n m= g m + n h m = y1 n + y2 n . m= Additive. Since the system is homogeneous and additive it is also linear. The system is also incrementally linear because it is linear. The system is not statically non-linear because it is linear. Time Invariance: Solutions 5-5 M. J. Roberts - 10/7/06 n Let x1 n = g n . Then y1 n = g m . m= Let x 2 n = g n n0 . Then y2 n = n g m n0 . m= The first equation can be rewritten as y1 n n0 = n n0 g m m= Let m = q n0 . Then y1 n n0 = n g q q= n0 = y2 n Time invariant Stability: If the excitation is a constant, the response increases without bound. () n Also the solution of the homogeneous difference equation is y h n = K 1 = K . Therefore the eigenvalue is 1 whose magnitude is not less than 1 and the system must be BIBO unstable. Unstable Causality: At any discrete time, n = n0 , the response depends only on the excitation at that discrete time and previous discrete times. Causal. Memory: At any discrete time, n = n0 , the response depends on the excitation at that discrete time and previous discrete times. System has memory. Invertibility: Taking the first backward difference of both sides of the original system equation, y n y n 1 = n n 1 x m m= x n =y n x m m= y n 1 The excitation is uniquely determined by the response. Invertible. 5. Show that a system with excitation x n and response y n described by Solutions 5-6 M. J. Roberts - 10/7/06 ( ), y n = rect x n is non-linear, time invariant and non-invertible. Homogeneity: Let x1 n = g n . Then y1 n = rect g n ( ) ( Let x 2 n = K g n . Then y2 n = rect K g n Not homogeneous. ) K y1 n . ( ) = rect ( h n ) Let x1 n = g n . Then y1 n = rect g n Let x 2 n = h n . Then y2 n Let x 3 n = g n + h n . ( Then y3 n = rect g n + h n Not additive. ) y1 n + y2 n . Since the system is not homogeneous and not additive it is also not linear. The system is also not incrementally linear. The system is statically non-linear because the non-linearity arises from the nonlinearity of the rectangle relation between excitation and response. Time Invariance: Let x1 n = g n . Then y1 n = rect g n . ( ( ) Let x 2 n = g n n0 . Then y2 n = rect g n n0 The first equation can be rewritten as ( y1 n no = rect g n no ). )= y 2 n Time invariant Stability: No matter what values the excitation may have the response can only have the values, zero or one. Stable Causality: At any discrete time, n = n0 , the response depends only on the excitation at that discrete time and not on any future excitation. Causal. Memory: At any discrete time, n = n0 , the response depends only on the excitation at that discrete time and not on any past excitation. Solutions 5-7 M. J. Roberts - 10/7/06 System has no memory. Invertibility: Many different excitations can cause the same response. Not invertible. 6. Show that the system of Figure E-6 is non-linear, time-invariant, static and invertible. 5 x[n] 10 + y[n] Figure E-6 A DT system y n = 10 x n Homogeneity: Let x1 n = g n . Then y1 n = 10 g n 5, 5 Let x 2 n = K g n . Then y2 n = 10K g n Not homogeneous. Let x1 n = g n . Then y1 n = 10 g n K y1 n . 5 Let x 2 n = h n . Then y1 n = 10 h n 5 5 Let x 3 n = g n + h n . ( Then y3 n = 10 g n + h n Not additive. ) 5 y1 n + y2 n . Since the system is not homogeneous and not additive it is also not linear. The system is incrementally linear because the only deviation from linearity is caused by the presence of the non-zero, zero-excitation response. The system is not statically non-linear because it is incrementally linear. Time Invariance: Let x1 n = g n . Then y1 n = 10 g n 5. Let x 2 n = g n n0 . Then y2 n = 10 g n n0 The first equation can be rewritten as y1 n n0 = 10 g n n0 5. 5 = y2 n Time invariant Solutions 5-8 M. J. Roberts - 10/7/06 Stability: If the excitation is bounded, the response is bounded. Stable Causality: At any discrete time, n = n0 , the response depends only on the excitation at that discrete time and not on any future excitation. Causal. Memory: At any discrete time, n = n0 , the response depends only on the excitation at that discrete time and not on any past excitation. System has no memory. Invertibility: Solving the system equation for the excitation as a function of the response, x n = y n +5 10 Invertible. 7. Show that the system of Figure E-7 is time-invariant, BIBO stable, and causal. x[n] 0.25 y[n] + + + D -1 D 2 Figure E-7 A DT system Homogeneity: Let x1 n = g n . Then 4 y1 n Let x 2 n = K g n . Then 4 y2 n y1 n 1 + 2 y1 n 2 = g n y2 n 1 + 2 y2 n 2 = K g n Multiply the first equation by K. 4K y1 n Then, equating results, 4 y2 n K y1 n 1 + 2K y1 n 2 = K g n y2 n 1 + 2 y2 n 2 = 4K y1 n If this equation is to be satisfied for all n, y2 n = K y1 n . Homogeneous. Solutions 5-9 K y1 n 1 + 2K y1 n 2 M. J. Roberts - 10/7/06 Additivity: Let x1 n = g n . Then 4 y1 n y1 n 1 + 2 y1 n 2 = g n Let x 2 n = h n . Then 4 y2 n y2 n 1 + 2 y2 n 2 = h n Let x 3 n = g n + h n . Then 4 y3 n y3 n 1 + 2 y3 n 2 = g n + h n Add the two first two equations. ( 4 y1 n + y2 n ) (y 1 ) ( n 1 + y2 n 1 + 2 y1 n 2 + y2 n 2 )=g n +h n Then, equating results, 4 y3 n ( y3 n 1 + 2 y3 n 2 = 4 y1 n + y2 n ) (y 1 ) ( n 1 + y2 n 1 + 2 y1 n 2 + y2 n 2 ) If this equation is to be satisfied for all n, y3 n = y1 n + y2 n . Additive. Since the system is both homogeneous and additive, it is linear. Since the system is linear it is also incrementally linear. Since the system is linear, it is not statically non-linear. Time Invariance: Let x1 n = g n . Then 4 y1 n y1 n 1 + 2 y1 n 2 = g n Let x 2 n = g n n0 . Then 4 y2 n We can re-write the first equation as 4 y1 n n0 y2 n 1 + 2 y2 n 2 = g n n0 y1 n n0 1 + 2 y1 n n0 2 = g n n0 Then, equating results, 4 y1 n n0 y1 n n0 1 + 2 y1 n n0 2 = 4 y2 n If this equation is to be satisfied for all n, then Solutions 5-10 y2 n 1 + 2 y2 n 2 M. J. Roberts - 10/7/06 y2 n = y1 n n0 . Time Invariant. Stability: The eigenvalues of the system homogeneous solution are found from the characteristic equation, 4 2 +2=0 . They are 1,2 = 0.125 ± j0.696 or 1,2 = 0.7071e± j1.3931 Therefore the homogeneous solution is of the form, ( ) ( n ) n y h n = K h1 0.7071 e+ j1.3931n + K h2 0.7071 e j1.3931n and, as n approaches infinity the homogeneous solution approaches zero and the total solution approaches the particular solution. The particular solution is bounded because it consists of functions of the same form as the excitation and all its unique differences and the excitation is bounded in the BIBO stability test. Therefore if x is bounded, so is y. Stable. Causality: We can rearrange the system equation into y n = ( 1 x n + y n 1 + 2y n 2 4 ) showing that the response at time, n, depends on the excitation at time, n, and the response at previous times. It does not depend on any future values of the excitation. Causal. Memory: The response depends on past values of the response. The system has memory. Invertibility: The original system equation, 4 y n excitation in terms of the response. Invertible. y n 1 + 2 y n 2 = x n , expresses the Modeling Discrete-Time Systems 8. At the beginning of the year 2000, the country, Freedonia, had a population, p, of 100 million people. The birth rate is 4% per annum and the death rate is 2% per annum, compounded daily. That is, the births and deaths occur every day at a uniform fraction of the current population and the next day the number of births and deaths changes because the population changed the previous day. For example, every day Solutions 5-11 M. J. Roberts - 10/7/06 0.02 , of the total population at the end 365 of the previous day (neglect leap-year effects). Every day 275 immigrants enter Freedonia. the number of people who die is the fraction, (a) Write a difference equation for the population at the beginning of the nth day after January 1, 2000 with the immigration rate as the excitation of the system. (b) By finding the zero-exctiation and zero-state responses of the system determine the population of Freedonia be at the beginning of the year 2050. The difference equation is p n +1 = p n + 0.04 p n 365 (1 + 5.48 p n +1 10 0.02 p n + 275 365 5 )p n = 275 The eigenvalue is 1 + 5.48 10 5 = 1.0000548 and the zero-excitation response is ( p n = 108 1.0000548 ( ) ) n n The impulse response is h n = 1.0000548 u n . The response to the immigration rate is the convolution of the impulse response with the immigration rate, 275u n , or ( ) n p n = 1.0000548 u n 273.9 u n = 273.9 n 1 m=0 (1.0000548) m Using the summation formular for a geometric series, p n = 275 ( ) 1 1.0000548n 1.0000548n 1 u n = 275 u n = 5018248.2 1.0000548n 1 u n 1 1.0000548 0.0000548 The total solution is ( ) ( n ) p n = 108 1.0000548 + 5018248.2 1.0000548n 1 u n ( p n = 1.05 108 1.0000548 ) n 5018248.2 , n The beginning of the year 2050 is the 18250th day. Solutions 5-12 0 , n 0 M. J. Roberts - 10/7/06 ( p 18250 = 1.05 108 1.0000548 9. ) 18250 5018248.2 = 280,420,000 In Figure E-9 is a MATLAB program simulating a DT system. (a) Without actually running the program, find the value of x when solving the difference equation for the system in closed form. (b) n = 10 Run the program and check the answer in part (a). by x = 1 ; y = 3 ; z = 0 ; n = 0 ; while n <= 10, z = y ; y = x ; x = 2*n + 0.9*y - 0.6*z ; n = n + 1 ; end Figure E-9 This system is a second-order DT system. It can be modeled by the difference equation x n 0.9 x n 1 + 0.6 x n 2 = 2n . The homogeneous solution is ( ) ( n x h n = K h1 0.45 + j0.63 + K h2 0.45 or ( x h n = K h1 0.7746e j0.9509 ) n ( + K h2 0.7746e ) j0.63 j0.9509 n ) n . The form of the particular solution is a linear combination of functions of the same form of the forcing function 2n and all its unique differences. A forward difference of 2n is 2 n + 1 2n = 2 , a constant and a second forward difference is zero. Therefore the form of the particular solution is ( ) x p n = K p1n + K p2 . Substituting that into the difference equation and solving, K p1n + K p2 K p1n + K p2 ( ( ) ) ( ( ) ) 0.9 K p1 n 1 + K p2 + 0.6 K p1 n 2 + K p2 = 2n ( ) 0.9K p1 n 1 ( ) 0.9K p2 + 0.6K p1 n 2 + 0.6K p2 = 2n 0.7K p1 = 2 0.3K p1 + 0.7K p2 = 0 Solutions 5-13 M. J. Roberts - 10/7/06 K p1 = 2.857 , K p2 = 1.224 So the total solution is ( x n = 2.857n + 1.224 + K h1 0.7746e j0.9509 ) n ( + K h2 0.7746e j0.9509 ) n . From the recursion in the MATLAB program, x 0 = 0.9 1 0.6 3 = 0.9 x 1 = 2 + 0.9 ( 0.9 ) 0.6 1 = 0.59 . Therefore 1.224 + K h1 + K h2 = 0.9 ( 2.857 + 1.224 + K h1 0.7746e j0.9509 ) n ( + K h2 0.7746e j0.9509 ) n = 0.59 or 1 0.7746e j0.9509 K h1 1 0.7746e j0.9509 K h2 = 2.124 3.491 and K h1 = 1.062 + j2.012 = 2.2751e j 2.054 , K h2 = 1.062 j2.012 = 2.2751e j 2.054 . So the final total numerical solution is ( x n = 2.857n + 1.224 + 2.2751e j 2.054 0.7746e j0.9509 ) n + 2.2751e j 2.054 (0.7746e j0.9509 and x 10 = 29.99 . Properties of Discrete-Time Systems 10. n+1 A DT system is described by y n = m= x m . Classify this system as to time invariance, BIBO stability and invertibility. Homogeneity: Let x1 n = g n . Then y1 n = n+1 g m m= Let x 2 n = K g n . Then y2 n = n+1 Kg m = K m= Homogeneous. Solutions 5-14 n+1 m= g m = K y1 n . ). n M. J. Roberts - 10/7/06 n+1 Let x1 n = g n . Then y1 n = g m m= n+1 Let x 2 n = h n . Then y2 n = h m m= Let x 3 n = g n + h n . Then y3 n = n+1 m= ( g m +h m )= n+1 g m + m= n+1 h m = y1 n + y2 n . m= Additive. Since the system is homogeneous and additive it is also linear. The system is also incrementally linear because it is linear. The system is not statically non-linear because it is linear. Time Invariance: n+1 Let x1 n = g n . Then y1 n = g m . m= Let x 2 n = g n n0 . Then y2 n = n+1 g m n0 . m= The first equation can be rewritten as y1 n n0 = n n0 +1 g m = n+1 g q q= m= n0 = y2 n Time invariant Stability: If the excitation is a constant, the response increases without bound. Unstable Causality: At any discrete time, n = n0 , the response depends on the excitation at the next discrete time in the future. Not causal. Memory: At any discrete time, n = n0 , the response depends on the excitation at that discrete time and previous discrete times. System has memory. Invertibility: Inverting the functional relationship, y n = n+1 x m . m= Solutions 5-15 M. J. Roberts - 10/7/06 Invertible. Taking the first backward difference of both sides of the original system equation, n+1 y n 1 = y n n+1 1 x m m= x n +1 = y n x m m= y n 1 The excitation is uniquely determined by the response. Invertible. 11. A DT system is described by ny n 8y n 1 = x n . Classify this system as to time invariance, BIBO stability and invertibility. Homogeneity: Let x1 n = g n . Then n y1 n 8 y1 n 1 = g n Let x 2 n = K g n . Then n y2 n 8 y2 n 1 = K g n Multiply the first equation by K. nK y1 n Then, equating results, nK y1 n 8K y1 n 1 = K g n 8K y1 n 1 = n y2 n 8 y2 n 1 If this equation is to be satisfied for all n, y2 n = K y1 n . Homogeneous. Additivity: Let x1 n = g n . Then n y1 n 8 y1 n 1 = g n Let x 2 n = h n . Then n y2 n 8 y2 n 1 = h n Let x 3 n = g n + h n . Then n y3 n 8 y3 n 1 = g n + h n Add the two first two equations. ( n y1 n + y2 n ) 8( y 1 ) n 1 + y2 n 1 = g n + h n Then, equating results, Solutions 5-16 M. J. Roberts - 10/7/06 ( n y1 n + y2 n ) 8( y 1 ) n 1 + y 2 n 1 = n y3 n 8 y3 n 1 If this equation is to be satisfied for all n, y3 n = y1 n + y2 n . Additive. Since the system is both homogeneous and additive, it is linear. Since the system is linear it is also incrementally linear. Since the system is linear, it is not statically non-linear. Time Invariance: Let x1 n = g n . Then n y1 n 8 y1 n 1 = g n Let x 2 n = g n n0 . Then n y2 n We can re-write the first equation as (n ) n0 y1 n n0 8 y2 n 1 = g n n0 8 y1 n n0 1 = g n n0 Then, equating results, (n ) n0 y1 n n0 8 y1 n n0 1 = n y2 n 8 y2 n 1 This equation cannot be satisfied for all n, therefore y1 n n0 . y2 n Time Variant. Stability: The homogeneous equation is ny n = 8y n 1 or y n = 8 y n 1 . n Thus, as n increases without bound, y n must be decreasing because it is its previous value and 8 times n 8 approaches zero. Rearranging the original equation, n Solutions 5-17 M. J. Roberts - 10/7/06 y n = x n n + 8 y n 1 . n For any bounded excitation, x n , as n gets larger, x n / n must be bounded and (8 / n ) y n 1 must be getting smaller because it is a decreasing fraction of its previous value. Therefore for a bounded excitation, the response is bounded. Stable. Causality: We can rearrange the system equation into y1 n = g n + 8 y1 n 1 n showing that the response at time, n, depends on the excitation at time, n, and the response at a previous time. Causal. Memory: The response depends on past values of the response. The system has memory. Invertibility: The original system equation, n y n in terms of the response. Invertible. 12. 8 y n 1 = x n , expresses the excitation A DT system is described by y n = x n . Classify this system as to linearity, BIBO stability, memory and invertibility. Homogeneity: Let x1 n = g n . Then y1 n = g n Let x 2 n = K g n . Then y2 n = Multiplying the first equation by K, Kg n = K y1 n = K g n K g n . y2 n . Not homogeneous. Let x1 n = g n . Then y1 n = g n Let x 2 n = h n . Then y2 n = h n Solutions 5-18 M. J. Roberts - 10/7/06 Let x 3 n = g n + h n . Then y3 n = g n + h n Not additive. y1 n + y2 n . Since the system is not homogeneous and not additive it is also not linear. The system is also not incrementally linear. The system is statically non-linear because of the square-root relationship between excitation and response. Time Invariance: Let x1 n = g n . Then y1 n = g n . Let x 2 n = g n n0 . Then y2 n = g n n0 . The first equation can be rewritten as y1 n n0 = g n n0 = y2 n Time invariant Stability: If the excitation is bounded, the response is bounded. Stable Causality: At any discrete time, n = n0 , the response depends only on the excitation at that same time. Causal. Memory: At any discrete time, n = n0 , the response depends only on the excitation at that same time. System has no memory. Invertibility: Inverting the functional relationship, x n = y2 n . Invertible. Solutions 5-19 ...
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This note was uploaded on 12/10/2010 for the course DCIS 32 taught by Professor Go during the Spring '10 term at College of E&ME, NUST.

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