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### ece306-9

Course: ECE 306, Winter 2008
School: Cal Poly Pomona
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Word Count: 1369

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of Solution Linear Constant-Coefficient Difference Equations Solution of Linear Constant-Coefficient Difference Equations Example: Determine the response y ( n), n 0 of the system described by the second-order difference equation y ( n) = 0.7 y ( n - 1) - 0.1 y ( n - 2) + 2 x ( n) - x ( n - 2) to the input x ( n ) = 4 n u ( n) The homogenous solution is y ( n) - 0.7 y ( n - 1) + 0.1 y ( n - 2) = 0 n - 0.7...

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of Solution Linear Constant-Coefficient Difference Equations Solution of Linear Constant-Coefficient Difference Equations Example: Determine the response y ( n), n 0 of the system described by the second-order difference equation y ( n) = 0.7 y ( n - 1) - 0.1 y ( n - 2) + 2 x ( n) - x ( n - 2) to the input x ( n ) = 4 n u ( n) The homogenous solution is y ( n) - 0.7 y ( n - 1) + 0.1 y ( n - 2) = 0 n - 0.7 n -1 + 0.1 n - 2 = 0 n + 2 ( 2 - 0.7 + 0.1) = 0 1 = 0.5 and 2 = 0.2 yh (n) = c1 0.5n + c2 0.2n 1 Solution of Linear Constant-Coefficient Difference Equations Particular Solution: y p ( n) = K 4 n u ( n ) K 4n u (n) - 0.7 K 4 n -1 u (n - 1) + 0.1K 4n - 2 u (n - 2) = (2)4n u (n) - 4n - 2 u (n - 2) n=2 K 42 - 0.7 K 41 + 0.1K 40 = 2(4)2 - 40 16 K - 2.8K + 0.1K = 32 - 1 y p (n) = 2.33(4) n u (n) The total solution K= 31 = 2.33 13.3 y (n) = c1 0.5n + c2 0.2n + 2.33(4) n u (n) Solution of Linear Constant-Coefficient Difference Equations To find c1 and c2 For n=0: From difference equation, y (0) = 0.7 y (0 - 1) - 0.1 y (0 - 2) + 2 x (0) - x (0 - 2) y (0) = 2 From the total solution, , y (0) = c1 + c2 + 2.33 For n=1: From difference equation, y (1) = 0.7 y (1 - 1) - 0.1 y (1 - 2) + 2 x (1) - x (1 - 2) y (1) = 1.4 + 8 = 9.4 From difference equation, y (1) = 0.5c1 + 0.2c2 + 9.32 c1 = 0.466 Therefore, 2 = c1 + c2 + 2.33 9.4 = 0.5c1 + 0.2c2 + 9.32 c2 = -0.807 Total Solution y ( n) = 0.466(0.5) n - 0.807(0.2) n + 2.33(4) n u ( n) 2 The Impulse Response of a LTI recursive system y zs (n) = n k =0 h(k ) x(n - k ), n 0 x ( n) = ( n) yzs (n) = h(n) The impulse response can be obtained from the linear constantcoefficient difference equation. That is the solution of homogeneous equation and particular solution to the excitation function. In the case where the excitation function is an impulse function. The particular solution is zero y p (n) = 0 , since x ( n) = 0 for n>0. The Impulse Response of a LTI recursive system Example: Find the impulse response form the following equation, y ( n) - 3 y (n - 1) - 4 y (n - 2) = x (n) + 2 x( n - 1) The homogenous solution is yh (n) = C1 (-1) n + C2 4n The particular solution is zero when x ( n) = ( n ) To find C1and C2, we evaluate difference equation and homogenous solution for n=0 and n=1.( y ( -1) = 0, y ( -2) = 0, since the system must be relaxed) y (0) = 1 y (1) = 3 y (0) + 2 = 5 The impulse response is y (0) = C1 + C2 y (1) = -C1 + 4C2 C2 = 6 5 C1 = - 1 5 1 6 h(n) = - ( -1) n + 4n u ( n) 5 5 3 Implementation of Discrete-Time Systems A system can be described by a linear constant-coefficient difference equation. Let's consider the first order system y (n) = -a1 y (n - 1) + b0 x(n) + b1 x(n - 1) The system can be described by two systems in cascade. The first is a nonrecursive system described by the equation v(n) = b0 x(n) + b1 x(n - 1) The second part is recursive system y (n) = -a1 y (n - 1) + v(n) x(n) z-1 b0 b1 + v(n) + -a1 z-1 y(n) This is called a direct form I structure. Implementation of Discrete-Time Systems Using convolution properties, we can interchange the order of the recursive and nonrecursive system x(n) + -a1 z-1 w(n) z-1 b0 b1 + y(n) w(n-1) w(n-1) w(n) = -a0 w(n - 1) + x(n) x(n) + -a1 w(n) z-1 w(n-1) b0 b1 y (n) = b0 w(n) + b1w(n - 1) + y(n) Using only one delay. It is more efficient in terms of memory requirement. It is called the direct form II structure ! 4 Implementation of Discrete-Time Systems In general form, The difference equation is given by y (n) = - N k =1 N k =0 ak y (n - k ) + bk x (n - k ) N k =0 The nonrecursive system is The recursive system is y ( n) = - N k =1 v(n) = bk x(n - k ) ak y ( n - k ) + v ( n ) Direct form II structure , the recursive system is w(n) = - N k =1 ak w(n - k ) + x(n) y ( n) = N k =0 The nonrecursive system is bk w(n - k ) Implementation of Discrete-Time Systems x(n) z-1 b0 b1 v(n) + + z-1 y(n) x(n) + w(n) z-1 b0 b1 + y(n) + + -a1 z-1 + -a1 + z-1 b2 + + -a2 + -a2 z-1 b2 + z-1 bM -aN z-1 + -aN-1 z-1 bM z-1 -aN w(n-N) Step-1 Step-3 " 5 Implementation Discrete-Time of Systems A special case of general case N k =0 ak = 0, k = 1,2,..., N y ( n) = bk x ( n - k ) which is a nonrecursive LTI system. Such a system coefficients bk h( k ) = bk 0 k M 0 otherwise The second part, set M=0 to obtain the general case difference equation y ( n) = - N k =1 ak y (n - k ) + b0 x(n) " " The crosscorelation of x(n) and y(n) is s sequence rxy (l ) is defined as rxy (l ) = n =- x( n) y ( n - l ), l = 0, 1, 2,... Or rxy (l ) = n =- x(n + l ) y (n), l = 0, 1, 2,... The reverse crosscorrelation is ryx (l ) = n =- n =- y ( n) x( n - l ), l = 0, 1, 2,... y (n + l ) x( n), l = 0, 1, 2,... Or ryx (l ) = rxy (l ) = ry x ( -l ) " 6 In special case , we have autocorrelation, which is defined as So, that rxx (l ) = n =- x ( n) x ( n - l ) Or rxx (l ) = n =- x(n + 1) x(n) Example: x ( n ) = {...0,0, 2, -1,3,7,1, 2, -3,0,0,...} y ( n ) = {...0,0,1, -1, 2, -2, 4,1, -2,5,0,0,...} rxy (0) = rxy (1) = n=- x(n) y (n) = 0 + 0 + 2 + 6 - 14 + 4 + 2 + 6 + 0 + 0 = 7 x(n) y ( n - 1) = 0 + 0 + -1 - 3 + 14 - 2 + 8 - 3 + 0 + 0 = 13 n =- rxy (l ) = {10, -9,19,36, -14,33,0,7,13, -18,16, -7,5, -3} " # Problem 43.a \$% Determine the direct form II realization for the following LTI system 2 y (n) + y (n - 1) - 4 y (n - 3) = x(n) + 3x(n - 5) y ( n) + 1 1 3 y (n - 1) - 2 y (n - 3) = x(n) + x(n - 5) 2 2 2 x(n) + z-1 + -1/2 z-1 z-1 2 z-1 z-1 3/2 1/2 + y(n) " 7 # Problem 44.a x(n) + \$% + Compute the first samples of its impulse response y(n) z-1 1/2 x( n) = ( n) x( n) = {1, 0, 0,...} y ( n) = 1 y (n - 1) + x(n) + x(n - 1) 2 y (1) = 1 3 y (0) + x(1) + x(0) = 2 2 y (0) = x(0) = 1 1 3 y (1) + x(2) + x(1) = 2 4 1 3 y (3) = y (2) + x(3) + x(2) = 2 8 y (2) = 3 3 3 3 3 3 y ( n ) = 1, , , , , , ,... 2 4 8 16 32 64 " # Problem 44.b y ( n) = \$% Find the input output relation 1 y (n - 1) + x(n) + x(n - 1) 2 Problem 44.c The input x( n) = {1,1,1,...} y (1) = 1 5 y (0) + x(1) + x(0) = 2 2 1 y (n) = y (n - 1) + x(n) + x(n - 1) 2 y (2) = 1 13 y (1) + x(2) + x(1) = 2 4 y (0) = x(0) = 1 y (3) = 1 29 y (2) + x(3) + x(2) = 2 8 ...... " 8 # Problem 44.d y (n ) = u(n ) * h(n ) = k =0 \$% u ( k )h ( n - k ) = n k =0 Use convolution h(n - k ) y (0) = h(0) = 1 y (1) = h(0) + h(1) = 5 2 y (2) = h(0) + h(1) + h(2) = 13 4 29 8 y (3) = h(0) + h(1) + h(2) + h(3) = " # Problem 54 \$% Find the y(n) for the following equation y (n) - 4 y (n - 1) + 4 y (n - 2) = x(n) - x(n - 1) x( n) = ( -1) n u ( n) The characteristic equation 2 - 4 + 4 = 0 The homogenous solution is The particular solution is 1 = 2, 2 = 2 yh (n) = c1 2n + c2 n 2n y p (n) = k (-1) n u (n) k (-1) n u (n) - 4k (-1) n -1 u (n - 1) + 4k (-1) n - 2 u (n - 2) = (-1) n u (n) - (-1) n -1 u (n - 1) For n=2 k + 4k + 4k = 1 + 1 k= 2 9 " ! 9 # The total solution is \$% 2 y ( n ) = yh (n ) + y p (n ) = c1 2n + c2n 2n + ( -1) n u(n ) 9 Using the initial condition, y (-1) = y (-2) = 0 we can obtain from difference equation at n=0 y (0) - 4 y (-1) + 4 y (-2) = x(0) - x(-1) y (0) = 1 n=1 y (1) - 4 y (0) + 4 y ( -1) = x(1) - x(0) y (1) = 2 y (1) = c1 2 + c2 2 - 2 =2 9 From the total solution y (0) = c1 + 2 =1 9 c1 = 7 9 y (1) = c1 2 + c2 2 - 2 =2 9 c2 = 1 3 The total solution y ( n) = 7 n 1 n 2 2 + n2 + (-1) n u (n) 9 3 9 " # Problem 55 \$% Find the impulse response h(n) for the following equation y (n) - 4 y (n - 1) + 4 y (n - 2) = x(n) - x(n - 1) The homogenous solution is y h ( n ) = c1 2 n + c2 n 2 n So system response To find the constant y (0) - 4 y ( -1) + 4 y (-2) = (0) - (-1) y (0) = 1 h(n) = c1 2n + c2 n2n u (n) y (1) - 4 y (0) + 4 y (-1) = (1) - (0) y (1) = 3 From the system response h(n) y (0) = c1 = 1 y (1) = c1 2 + c1 2 = 3 c2 = 1 2 1 h( n) = 2 n + n 2 n u ( n) 2 10
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Notes 11/1 Alliances are created- triple alliance turkey, germany, Austria-hungry Triple GB, France, Russia July 1914 This is the breakdown of the early part of the 20th century. It is these alliances that break down peace in Russia. France Ferdinand
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Chapter 25 The soviet union under Lenin and stalin - Whites were the bolshevics opponents (the reds), they were mostly comprised of supporters from the past regime. - The greens joined in with the whites who opposed all central state power, comprised
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Notes 11/13 Two parties are set up in Russia. Mensheviks, and Bolsheviks. What changes the power between the two parties is that the Petrograd Soviets claimed that they would give them food, money, and other social needs but never delivered, Lenin (B
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Composition and Function of the Blood Function: It transports substances between the body cells and the external environment/ Helps maintain a stable cellular environment. Blood and Blood cells 45% blood cells 55% plasma .When if you put your blood i
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Kelsey Jans FRL 201 1/15/07Brief: Case 4.2 pg.85 ( Chapter 4.) Interactive Digital Software Association vs. St. Louis County, MissouriIn the case of Interactive Digital Software Association Vs. St Louis County, Missouri, the court is dealing with
Cal Poly Pomona - FRL - 201
Kelsey Jans FRL 201 1/21/08Brief: Case 5.2 Page 100 In re the Exxon ValdezIn the case In re the Exxon Valdez, the court has to decide whether Exxon was highly reprehensible for their damages as well as determining whether \$5 billion would be too h