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:
point B where has the largest Q value at section a–a
Explanation:
The missing diagram that is suppose to be attached to this question can be found in the attached file below.
So from the given information ;we are to determine the point that has the largest Q value at section a–a
In order to do that; we will work hand in hand with the image attached below.
From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.
We also have block partitioned into different point segments . i,e A,B,C, D
For point A ;
Let Q be the moment of the Area A;
SO ;
where ;
For point B ;
Let Q be the moment of the Area B;
SO ;
where ;
For point C ;
Let Q be the moment of the Area C;
SO ;
where ;
For point D ;
Let Q be the moment of the Area D;
SO ;
since there is no area about point D
Area = 0
Thus; from the foregoing ; point B where has the largest Q value at section a–a
