Calculate the response of the first-order difference equation 17-26 for a1 =−0.368, b1 =1.264, and y(0)=0 using z-transforms and long division for k=0,1,…5. Compare the result with the unit step response for a first-order continuous-time system (K=20,τ=1),where a1 = −e−Δt/τ, b1 =K(1−e−Δt/τ), and

Question

Calculate the response of the first-order difference equation
[latex] y(k)+a_{1}y(k−1)=b_{1}u(k−1) [/latex]
for [latex]a_{1} =−0.368, \space b_{1} =1.264, [/latex] and y(0)=0 using z-transforms and long division for k=0,1,…5. Compare the result with the unit step response for a first-order continuous-time system (K=20,τ=1),where [latex] a_{1} = −e^{−Δt/τ},\space b_{1} =K(1−e^{−Δt/τ}), [/latex] and Δt=1, as discussed in Section 7.4 and Equation
[latex] Z(y(t-i\Delta t)) = z^{−i}\sum\limits_{j=0}^{\infty }{y(jΔt)z^{-j}}= z^{−i}Y(z) [/latex]

Accepted Answer

Using Equation
[latex] Y(z)= \frac{b_{1}z^{-1}}{1+a_{1}z^{-1}}U(z)=G(z)U(z) [/latex] ,
we find that the response for a unit step input [latex] (U(z)=1/(1−z^{−1})) [/latex] is
[latex] Y(z) = \frac{1.264}{1-0.368z^{-1}} .\frac{1}{1-z^{-1}}=\frac{1.264z^{-1}}{1-1.368z^{-1}+0.368z^{-2}} [/latex]
Next long division is used to divide the denominator into the numerator. The order of the numerator and denominator polynomials starts with the lowest powers of [latex]z^{−k} [/latex] for the division operation (shown on the right).
Because of space limitations, only the first three terms are shown above:
y(1)=1.264, y(2)=1.729, and y(3)=1.900. Continuing on, we calculate y(4)=1.963 and y(5)=1.986. Ultimately, y(k) reaches its steady-state value of 2.0 (k large), which agrees with the fact that the process gain K is 2 and the input is a unit step change. The step response in continuous time is [latex] y(t)=2(1−e^{−t}), [/latex] and the sampled values of the discrete-time response for Δt=1 are the same (k=0,1,2,3 …). Thus, the discretization is exact; that is, it is based on the analytical solution for a piecewise constant input.

The same answer could be obtained from simulating the first-order difference equation (1) , with u(k)=1 for k≥0; that is,

[latex]y(k)+a_{1} y(k-1)=b_{1} u(k-1)[/latex]
y(k)=0.368 y(k−1)+1.264(1)
Starting with y(0)=0, it is easy to generate recursively the values of y(1)=1.264, y(2)=1.729, and so on. Note that the steady-state value can be obtained in the above equation by setting [latex]y(k)=y(k−1) = y_{ss} [/latex] and solving for [latex] y_{ss} [/latex] . In this case [latex]y_{ss} =2.0, [/latex] as expected.

Question 17.2: Calculate the response of the first-order difference equatio...

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