Question

+ a D b 1. Consider the signal flow diagram representing a r/n)- y[n] discrete-time LTI system as shown, with a = 0.9 and b= -0.2. Assume initial rest conditions. In each case, explain your work. D a. (1 mark) Show that the difference equation for this system is y[n] – a •y[n – 1] – b•y[n – 2] = x[n]. b. (2 marks) Calculate the first 50 terms for the impulse response h[n] of this system. C. (2 marks) Calculate the first 50 terms of the output of this system for an input of x[n] = u[n], making use of the impulse response you determined in step b. d. (5 marks) Determine the natural, forced, and total response of this system for an input of x[n] = 1.u[n]. Calculate the first 50 terms of the total response. e. (1 mark) What is the transfer function of this system? f. (1 mark) Using the transfer function, determine the output of this system for an input of x[n] = 1.u[n]. Note that x[n] = (1)” • u[n]. g. (1 mark) If you've done your work correctly, neglecting rounding errors, your output for step c. and your total response from step d. should be the same, and your forced response from step d. and the output from step f. should be the same. Why is this? h. (5 marks) Determine the natural, forced and total response for of this system for an input of x[n] = (0.7)” • u[n]. Calculate the first 50 elements of the total response i. (2 marks) Using the difference equation itself, calculate the first 50 elements of the system output for an input of x[n] = (0.7)” • u[n].

Please do h) and i). The answers have to match (show 3 or 4 answers for each to show that they match I do not need 50).

Answer 1

x [n] y En D] < Delay ayin-1) ly(n-1) 9 bly D b y (0-2) According to system diagram, output com be written as, y (m) = 2(M) + ay (nt) + by (0-2) yin) – ay (n=1) – by(n-2) = n(n) given, a = 0.9, b= -0.2 y (m) - 0.9 y(n-1) + 0.2 y la - 2y (m-2) = 2 (m2 This in the difference equation of the system. solution (ch) Natural Response - We have to find homogeneous for Natural Response. gling cirim + Ca 8 CS Scanned with CamScanner

Now we will find r, and r2 , for xrm) = 0 ( Natural Response) yim) – 0.9 y(n-1) +0.27 (0-2) = 0 Y(z) - 0.927 Y(2) +0.2 Z? Y(2) =0 (1-0.927 +0.222) Y(z) =0 1-09 21 +0.222=0 z? =0 0.92 +0.2 Z=0.455, 0.455, 0.445 These are the values of ri and rz. ri = 0.455 and 0,445 so, yhem = C (0.4559" + C2 (0.445) It is given that system in initially in rest condition, so, output yine will be zero for n=0,-1,-2, -- for nao, => Gt Cz (1) 0 a tq for n n=1, 그 0 G (0.455)' + C2 C (0.4455 CS Scanned with CamScanner

2.1984 t 2.247 C2 = 0 (2) By solving equ. (1) and (2), G=0 and C₂ = 0 So, Natural Response is, ghimo = It is obvious because initial conditions of system Zuo, so, Natural Response must be zao. in CS Scanned with CamScanner

2 (m) .24 (1-2) Forced Response - Forced Response in due to the input. We will w find forced Response by Z- Taars form. yon) 0.9 y (n-1) +0.2 given, am) = (0.77% 41m2, Taking 2-toam.form, 412) – 0.9 zły(z) +0.222Y(z) = X(2) 22 Y(2) = z²_0192 +0.2 (2)X (1-019.27 +0.222) 2 X(z) X19) = (0-7)(n) (2-0.7) 3 z SO, Y(2) (22-0.92 +0.2)(2-0.7) z EZ Y(27 (2-0.495) (2-0-445) (2-07) 3 Z Y(2) = (Z-0.455) (Z-0.445)(2-0.7) 407.994 392,156 2.55 + + = (2) h Z (2-0.455) (2-0.445) (7-07) ( By partial foaction) CS Scanned with CamScanner

4/2) = 2.5573 - 407.996.73 2-01455 ) + 392.562 (2-0-445 ) + (2 (2-0.7 Taking inverse 2 - Trans form, 7+2 n-+2 Y(n) = - 407.996 (0.455) 4 (n+2). + 392.156 (0.445) uin+2) + 2.55 (0.7)1* 4 (M+2) So, Total Response is, yo com (m) + yon) Natural forced yln) y (n= 1 nt2 n+2 01+2 y (m) = - 407.996 (0.455) urn+2) + 392.156 (0.445)** *4(+2) + 2.5 (0.5M (P2) We can calculate the elements By putting n=0,1,2,3, of yin). CS Scanned with CamScanner

C) Difference equation wi, yim) – 019 y(n-1) +0.2 y (M-2) = a(n) given 71m) = (0.77"urn) for the input we have calculated the output in th) pants ym=[-407.996 (0.4532 + 392.158 (0.4459827 2.556.77727 (2) By putting n=0, 1, 2, 3, We can calulate any number of elements of the Response CS Scanned with CamScanner