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
1 answer:
The asymptotes of the open loop transfer are:
- Horizontal: y = 0
- Vertical: x = -10 and x = -100
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:
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Answer:
minimum length of a surface crack is 15.043 mm
Explanation:
given data
strain fracture toughness K = 78 MPa
tensile stress = 345 MPa
Y = 1.04
to find out
minimum length of a surface crack
solution
we find here length of critical interior flaw from formula that is
α = ....................1
put here value we get
α =
α = 15.043 mm
so minimum length of a surface crack is 15.043 mm