Question

The low-frequency asymptote in the Bode plot of

\(G\left( s \right) = \frac{{6\left( {{s^2} + 10s + 100} \right)}}{{{s^2}\left( {50{s^2} + 15s + 1} \right)}}\) 

Has a slope of

1. -10 dB/dec
2. -20 dB/dec
3. -40 dB/dec
4. -60 dB/dec

Answer 1

Correct Answer - Option 3 : -40 dB/dec

Given:

Concept:

• The low-frequency approximation is called the low-frequency asymptote.
• It is measured at frequencies close to zero (low frequency). ∴ The low-frequency asymptote slope depends upon the poles or zeros at the origin.
• ADD FROM LMS

Analysis:

\(G\left( s \right) = \frac{{6\left( {{s^2} + 10s + 100} \right)}}{{{s^2}\left( {50{s^2} + 15s + 1} \right)}}\) 

As the G(s) function has two poles at the origin:

So, the low freq. asymptote slope will be:

= 2(-20)

= -40 dB/decode