In a bode magnitude plot, which one of the following slopes would be exhibited at high frequencies by a 4th order system having two zeros in left half of s-plane?
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Answer & Explanation
Correct AnswerB. −40 dB/decade
Quick Explanation
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.
Formula / Key Relation
High-frequency slope = 20(Nz − Np) dB/decade
Np = 4, Nz = 2
Slope = 20(2 − 4) = −40 dB/decade
Detailed Explanation
Correct Answer
B — −40 dB/decade
The system has two more poles than zeros, so its final Bode magnitude asymptote falls at 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(Nz − Np) 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 = Np − Nz = 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.
Exam Tip
Use relative degree for the final slope: −20(Np−Nz) dB/decade.
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