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Bode Plots Questions and Answers

11.

The term (1/jw) on the logmagnitude plot has a slope of

Exam Year(s)

UGC NET - 2013,

Answer: A

The term \(1/j\omega\) has a slope of \(-20\) dB/decade, the first option. Its magnitude is \(1/\omega\), so the log magnitude is \(-20 \log \omega\) decibels, which falls by 20 dB for each tenfold rise in frequency.

The slope of +20 dB/decade belongs to a zero at the origin, \(j\omega\), and the slopes of \(\pm 40\) dB/decade belong to double poles or zeros at the origin, \(1/(j\omega)^2\) or \((j\omega)^2\). So the other options are wrong.

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12.

  
(a) Bode's-Plot representation(i) not a frequency domain technique.
(b) Nyquist Diagram(ii) low frequency and high frequency characteristics of the transfer function can be determined
(c) Nichols Charts(iii) decides the Stability criteria
(d) Root-locus Method(iv) also known as Polar plots

Codes :
(a) (b) (c) (d)

Exam Year(s)

UGC NET - 2013,

Answer: D

The correct match is a-ii, b-iii, c-iv, d-i, the fourth option. The Bode plot lets the low and high frequency characteristics of the transfer function be read, the Nyquist diagram decides the stability criterion, the Nichols chart is paired by the key with the polar plot description, and the root locus method is not a frequency domain technique.

The only option with the Bode plot matched to (ii) is the fourth, so the other options, which begin with (iv) or (iii), are wrong.

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