This question was asked inGPSC Deputy Executive Engineer (Electrical) Paper 2024Question 12 of 33 published · Gujarat Public Service CommissionOpen paper
Q13ModerateFrequency Response

બોડ મેગ્નિટ્યુડ પ્લોટમાં, s-પ્લેનના ડાબા અર્ધ ભાગમાં બે ઝીરો ધરાવતી ચોથા ક્રમ (4th order) ની સિસ્ટમ ઉચ્ચ આવૃત્તિઓ પર નીચેનામાંથી કયો ઢાળ દર્શાવશે?

જવાબ અને પગલાવાર ઉકેલ
સાચો જવાબ: B. −40 dB/decade
ટૂંકી સમજૂતી

દરેક પ્રથમ-ક્રમનો પોલ તેની કોર્નર આવૃત્તિ પછી −20 dB/decade અને દરેક ઝીરો +20 dB/decade આપે છે. 4 પોલ અને 2 ઝીરો સાથે ઉચ્ચ-આવૃત્તિ ઢાળ 20(2−4) = −40 dB/decade છે.

વિગતવાર ઉકેલ (મૂળ ભાષામાં)

What This Question Is Really Testing

At frequencies well above all finite corner frequencies, every pole and zero has contributed its full asymptotic slope. A first-order pole contributes −20 dB/decade; a first-order zero contributes +20 dB/decade.

Concept Foundation

The high-frequency slope can be obtained from the pole-zero count:

Final slope = 20(N_z − N_p) dB/decade

Step 1 — Count the Poles

A 4th-order denominator has four poles in the pole count.

Pole contribution = 4(−20) = −80 dB/decade

Step 2 — Count the Zeros

The system has two zeros.

Zero contribution = 2(+20) = +40 dB/decade

Step 3 — Add the Contributions

Net slope = −80 + 40

Net slope = −40 dB/decade

Therefore option B is correct.

Why This Method Works

In the Bode magnitude, a numerator factor grows with frequency and contributes positive slope; a denominator factor reduces the magnitude and contributes negative slope. The net slope therefore depends on the relative degree .

Independent Verification / Cross-Check

Relative degree = N_p − N_z = 4 − 2 = 2

Final slope = −20(2) = −40 dB/decade

The relative-degree method independently gives the same result.

Concept Extension

  • LHP versus RHP location affects phase behavior.
  • For magnitude asymptotic slope, a first-order zero still contributes +20 dB/decade.

Exam Strategy

At high frequency, use the one-line check −20 × (poles − zeros) dB/decade.

Quick Trick

4 poles − 2 zeros = 2 excess poles → −40 dB/decade.

Why Other Options Are Wrong

A counts the four poles but ignores the two zeros. C and D have the wrong positive sign because poles outnumber zeros by two.

Common Mistake

Using only the 4th-order denominator and writing −80 dB/decade without adding the +40 dB/decade contribution of the two zeros.

Solution diagram – બોડ મેગ્નિટ્યુડ પ્લોટમાં, s-પ્લેનના ડાબા અર્ધ ભાગમાં બે ઝીરો ધરાવતી ચોથા ક્રમ…
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