The Z-transform of the following function f[k] =22*+/- (-5)*(-4)* is So the constants A and B are........ O a. None of them O b. 4, 2 O C. -2,4 O d. 1, 2 O e. -1,3 Z(A+BZ²) (Z-2)(2+4)(Z+5)
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![Z(A+BZ²)
(Z-2)(2+4)(Z+5)
The Z-transform of the following function f[k] =22*+. (-5)* − 3 (-4)* is ·
42
So the constants A and B are........
O a. None of them
O b. 4, 2
O C. -2,4
O d. 1, 2
O e. -1,3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc995f9ff-c4fe-4e28-91a4-b909fde27f1e%2F98690e92-5ca9-427f-867c-4f53d18f0ea7%2Fwpnkxl_processed.jpeg&w=3840&q=75)
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- using mathematical definition of z transform derive [rn sinw0n]u[n] = [rsinw0]z-1 /1-[2rcoswn]z-1+r2 z-2 |z|>r And, {an , 0<=n<=N-1 and 0, otherwise} = 1-aN z-N /1-az-1 |z|>0Solution to IVP using laplace and inverse laplace transform. Show the form of rational function--->form of partial fraction y''+7y'+12y=0, y(0)=1, y'(0)=221.Consider the equation (35) ay′′+by′+cy=0,ay″+by′+cy=0, where a, b, and c are constants with a ≠ 0. In Chapter 3 it was shown that the general solution depended on the roots of the characteristic equation (36) ar2+br+c=0.ar2+br+c=0. a.Transform equation (35) into a system of first-order equations by letting x1 = y, x2 = y′. Find the system of equations x′ = Ax satisfied by x=(x1x2)x=x1x2.
- 29.In this problem we show how a general partial fraction expansion can be used to calculate many inverse Laplace transforms. Suppose that F(s)=P(s)Q(s),F(s)=P(s)Q(s), where Q(s) is a polynomial of degree n with n distinct zeros r1, …, rn, and P(s) is a polynomial of degree less than n. In this case it is possible to show that P(s)/Q(s) has a partial fraction expansion of the form (36) P(s)Q(s)=A1s−r1+⋯+Ans−rn,P(s)Q(s)=A1s−r1+⋯+Ans−rn, where the coefficients A1, …, An must be determined. a.Show that Ak=P(rk)Q′(rk),k=1,…,n. Hint:One way to do this is to multiply equation (36) by s − rk and then to take the limit as s → rk. Note that limits are used because it is not appropriate to simply evaluate equation (36) multiplied by s − rk because equation (36) is not defined at each root of Q(s).Z- Transform 1. x(n) = 2n u(n) + 3 (1/2)n u(n) - 4n u(-n-1) 2. cos ( nβ + θ ) u(n)Example:- Find the Taylor polynomials Pn(x) generated by f(x) = ex at x = 0 Solution:- The given function and its derivative are. f(x) = ex, f'(x) = ex,...., f(n)(x) = ex
- x(0)=4, x(1)=3, x(2)=2, x(3)=1, x(4)=0, x(5)=0 give as, x(n) series ( 0≥ ? ≥ 5 ). a)calculate x(k) with discrete fourier transform coefficients. (step-by-step should be written)Answer: Basis for W: ((1, 0, 0), (1,0,1)). Characteristic polynomial for Tw: g(t) = t2 - 2.(a) Find the third-order Taylor polynomial generatedby y(x) = e−x about x = 1.(b) State the third-order error term.(c) Find an upper bound for the error term given|x| < 1