JNTU B.Tech III Year LINEAR SYSTEMS ANALYSIS Previous Question Paper (November/December - 2020 - R16)

To determine the state-space model for the given differential equation, we first rewrite the system in the form of a transfer function or by using the method of direct decomposition. Given the equation:

We define the system using the phase-variable canonical form. First, we express the input-output relationship as a ratio of polynomials in the Laplace domain:

We can decompose this into a direct form by introducing an intermediate variable Z(s) such that:

Defining state variables as x₁ = z, x₂ = z', and x₃ = z'', we obtain the state equations:

• x₁' = x₂
• x₂' = x₃
• x₃' = -0.5x₁ + 2x₂ - 3x₃ + u(t)

The output y(t) is given by y = 4z''' + 2.5z'' + 1.5z' + 2z. Substituting the expressions for the derivatives of z:

The resulting state-space representation is:

The state-space representation for a discrete-time system is given by the state equation x(n+1) = Ax(n) + Bf(n) and the output equation y(n) = Cx(n) + Df(n). Given A = [[2, 0], [1, 0]], B = [[0], [1]], C = [0, 1], D = [1], x(0) = [[2], [1]], and f(n) = u(n), we solve for the state response and output.

First, we find the state transition matrix Φ(n) = Aⁿ. Since A = [[2, 0], [1, 0]], we calculate powers of A:

The state response x(n) is given by the formula:

For n=1: x(1) = Ax(0) + Bf(0) = [[2, 0], [1, 0]][[2], [1]] + [[0], [1]][1] = [[4], [2]] + [[0], [1]] = [[4], [3]].

For n=2: x(2) = Ax(1) + Bf(1) = [[2, 0], [1, 0]][[4], [3]] + [[0], [1]][1] = [[8], [4]] + [[0], [1]] = [[8], [5]].

The output y(n) is calculated using y(n) = Cx(n) + Df(n):

For n=0: y(0) = Cx(0) + Df(0) = [0, 1][[2], [1]] + 1(1) = 1 + 1 = 2.

For n=1: y(1) = Cx(1) + Df(1) = [0, 1][[4], [3]] + 1(1) = 3 + 1 = 4.

For n=2: y(2) = Cx(2) + Df(2) = [0, 1][[8], [5]] + 1(1) = 5 + 1 = 6.

By induction, the state vector at step n is x(n) = [[2ⁿ⁺¹], [2ⁿ+1]] and the output is y(n) = 2ⁿ + 2.