- Asked 2/22
- 2068 Jestha · 8 marks
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, (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
Both quadratics have complex roots, so each is kept as one second-order factor.
Step 2: Difference equations
Let be the input and the output.
Section 1:
Section 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 and 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)
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