In a bode magnitude plot, which one of the following slopes would be exhibited at high frequencies by a 4^th order system having two zeros in left half of s-plane?
Answer & step-by-step solution
Each first-order pole contributes −20 dB/decade and each first-order zero contributes +20 dB/decade after its corner frequency. With 4 poles and 2 zeros, the high-frequency slope is 20(2−4)=−40 dB/decade.
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.
बोड मैग्नीट्यूड प्लॉट में, s-प्लेन के बाएँ अर्ध भाग में दो ज़ीरो वाले चतुर्थ कोटि (4th order) सिस्टम द्वारा उच्च आवृत्तियों पर निम्नलिखित में से कौन-सा ढलान दिखाया जाएगा?
उत्तर और चरणबद्ध हल
प्रत्येक प्रथम-कोटि पोल अपनी कॉर्नर आवृत्ति के बाद −20 dB/decade और प्रत्येक ज़ीरो +20 dB/decade देता है। 4 पोल और 2 ज़ीरो के साथ उच्च-आवृत्ति ढलान 20(2−4) = −40 dB/decade है।
विस्तृत हल (मूल भाषा में)
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.
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
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.
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