For the unity feedback system with feedforward transfer function G(s) a. find the system type b. What steady state error can be expected for a step input of 12?

The correct answer and explanation is:

Sure! To answer your question about the unity feedback system with feedforward transfer function G(s)G(s):

Given:

• Unity feedback system
• Forward transfer function G(s)G(s)
• Input: Step input of magnitude 12

a. Find the system type

System type depends on the number of poles of G(s)G(s) at the origin (i.e., number of integrators).

• Type 0 system: No poles at s=0s=0
• Type 1 system: One pole at s=0s=0
• Type 2 system: Two poles at s=0s=0, and so on…

The system type indicates the system’s ability to track certain inputs with zero steady-state error.

b. What steady-state error can be expected for a step input of magnitude 12?

Step 1: Steady-state error formula for unity feedback system

For a unity feedback system with forward transfer function G(s)G(s), the error transfer function is: E(s)=11+G(s)⋅R(s)E(s) = \frac{1}{1 + G(s)} \cdot R(s)

where R(s)R(s) is the Laplace transform of the input.

Step 2: Use Final Value Theorem

The steady-state error is: ess=lim⁡t→∞e(t)=lim⁡s→0sE(s)=lim⁡s→0s⋅11+G(s)⋅R(s)e_{ss} = \lim_{t \to \infty} e(t) = \lim_{s \to 0} s E(s) = \lim_{s \to 0} s \cdot \frac{1}{1 + G(s)} \cdot R(s)

For a step input of magnitude 12: r(t)=12⋅u(t)⇒R(s)=12sr(t) = 12 \cdot u(t) \Rightarrow R(s) = \frac{12}{s}

Plug into steady-state error: ess=lim⁡s→0s⋅11+G(s)⋅12s=lim⁡s→0121+G(s)=121+Kpe_{ss} = \lim_{s \to 0} s \cdot \frac{1}{1 + G(s)} \cdot \frac{12}{s} = \lim_{s \to 0} \frac{12}{1 + G(s)} = \frac{12}{1 + K_p}

where Kp=lim⁡s→0G(s)K_p = \lim_{s \to 0} G(s)

KpK_p is the position error constant.

Step 3: Interpreting the error based on system type

• Type 0: KpK_p is finite. The steady-state error to step input is finite:

ess=121+Kpe_{ss} = \frac{12}{1 + K_p}

• Type 1 or higher: Since G(s)G(s) has at least one pole at zero, G(0)→∞G(0) \to \infty, so

ess=121+∞=0e_{ss} = \frac{12}{1 + \infty} = 0

Summary

Example

If G(s)=10s+2G(s) = \frac{10}{s+2}

• No pole at zero → Type 0 system
• Kp=lim⁡s→0G(s)=102=5K_p = \lim_{s \to 0} G(s) = \frac{10}{2} = 5
• Steady-state error for step 12:

ess=121+5=126=2e_{ss} = \frac{12}{1 + 5} = \frac{12}{6} = 2

Explanation

In control systems, system type is crucial for predicting steady-state error for standard inputs like step, ramp, or parabolic signals.

• The system type is determined by the number of pure integrators in the forward path. Integrators help eliminate steady-state error for certain inputs.
• For a step input, a Type 0 system can only reduce the error to a finite nonzero value depending on the gain KpK_p.
• A Type 1 or higher system has at least one integrator, making the position error constant infinite, resulting in zero steady-state error for step inputs.
• This means increasing system type improves tracking of lower-order polynomial inputs (step, ramp, parabolic, etc.).
• If the exact G(s)G(s) is known, calculating KpK_p is straightforward and gives the exact steady-state error.
• The magnitude of the step input scales the steady-state error linearly, as seen in the formula.

Understanding system type and error constants allows control engineers to design feedback systems that meet accuracy requirements for specific inputs.