Skip to main content

Series programming of a third-order transfer function

Short answer

Series programming splits X(z) into sections: G1(z)=(1+z^-1)/(1-2z^-1) and G2(z)=(1+3z^-1+7z^-2)/(1+z^-1+9z^-2). Each section uses standard programming with one and two delays respectively, total three delays.

Realize the following transfer function by series programming: X(z)=(z+1)(z²+3z+7)/((z-2)(z²+z+9)).

In series programming, X(z)X(z) (the given pulse transfer function) is split into cascaded low-order sections. Each section is realized by standard programming.

Step 1: Sections in powers of z−1z^{-1}

X(z)=1+z−11−2z−1⏟G1(z)⋅1+3z−1+7z−21+z−1+9z−2⏟G2(z)X(z) = \underbrace{\frac{1 + z^{-1}}{1 - 2z^{-1}}}_{G_1(z)}\cdot\underbrace{\frac{1 + 3z^{-1} + 7z^{-2}}{1 + z^{-1} + 9z^{-2}}}_{G_2(z)}

Both quadratics have complex roots, so each is kept as one second-order factor.

Step 2: Difference equations

Let u(k)u(k) be the input and y(k)y(k) the output.

Section 1:

  • w1(k)=u(k)+2w1(k−1)w_1(k) = u(k) + 2w_1(k-1)
  • y1(k)=w1(k)+w1(k−1)y_1(k) = w_1(k) + w_1(k-1)

Section 2:

  • w2(k)=y1(k)−w2(k−1)−9w2(k−2)w_2(k) = y_1(k) - w_2(k-1) - 9w_2(k-2)
  • y(k)=w2(k)+3w2(k−1)+7w2(k−2)y(k) = w_2(k) + 3w_2(k-1) + 7w_2(k-2)

Step 3: Block diagram

Section 1:

u(k)-->(S)------------o--[1]-------->(S)--> y1(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w1(k-1)       |
        +<[2]---------o--[1]----------+

Section 2:

y1(k)-->(S)------------o--[1]-------->(S)--> y(k)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-1)       |
        (S)<[-1]-------o--[3]-------->(S)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-2)       |
         +<[-9]--------o--[7]----------+
u(k) --> [ G1(z) ] --y1(k)--> [ G2(z) ] --> y(k)

Three delay elements are used. Each section can be scaled and checked separately.

Note: The poles z=2z = 2 and z=−0.5±j2.96z = -0.5 \pm j2.96 lie outside the unit circle, so the system is unstable. The series structure is still a valid realization.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Digital Control System chapter 4 questions →