Question

Sketch the asymptotes of the Bode plot magnitude and phase for the open-loop transfer ()=100(S+1)/((S+10)(S+100)) Use MATLAB to verify your solution.

Answer 1

The asymptotes of the open loop transfer are:

• Horizontal: y = 0
• Vertical: x = -10 and x = -100

How to plot the asymptotes?

The open loop transfer function is given as:

f(s) = 100(s + 1)/((s + 10)(s + 100))

Set the numerator of the function to 0.

So, we have:

f(s) = 0/((s + 10)(s + 100))

Evaluate

f(s) = 0

This means that, the vertical asymptote is y = 0

Set the denominator of the function to 0.

(s + 10)(s + 100)  0

Split

s + 10 = 0 and s + 100 = 0

Solve for s

s = -10 and s = -100

This means that, the horizontal asymptotes are s = -10 and s = -100

See attachment for the graph of the asymptotes

Read more about asymptotes at:

brainly.com/question/4084552

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