FE Electrical & Computer Practice Question Solution included after investigation
CASE E-032 · FE

The Gain-10 Loop That Only Reached 0.909

The forward gain was 10. The closed-loop output did not become 10 times the reference.

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Case E-032 engineering failure visual
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Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

A unity negative-feedback control loop has a constant forward-path gain G = 10. A review note says a 1.0 reference should produce an output of 10 because the controller gain is 10. In test, the output settles near 0.909. Diagnose the block-diagram error, calculate the closed-loop gain and steady-state error, and choose the change that reduces the error below 5%.

Control Systems Electrical & Computer Design Verification Root Cause Analysis Safe Correction
E-032
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CASE CLOSED · ENGINEERING VERDICT

What actually failed — and what should change.

What failed

The output was predicted from forward gain instead of closed-loop gain.

What was assumed

Feedback was treated as if it did not change input-output gain.

Corrected model

Use T = G/(1+GH) before evaluating closed-loop response.

QUICK ANSWER

For unity negative feedback, the closed-loop transfer is T = G/(1+G). With G=10, T=10/11≈0.909. A unit-step input therefore produces about 0.909 output and 9.1% steady-state error.

ENGINEERING TAKEAWAY

Reduce the feedback loop before interpreting gain. For unity negative feedback, T(s)=G(s)/(1+G(s)) and E/R=1/(1+G).

CASE QUESTIONS

Questions this case should settle.

What is the closed-loop gain for unity feedback with G=10?

10/(1+10)=10/11≈0.909.

What is the unit-step steady-state error for G=10?

1/(1+10)=1/11≈9.1%.

NEXT FE PRACTICE QUESTION

The Higher-Gain Loop That Crossed Into Instability

Control Systems · 5–10 min

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