A common rule of thumb for controller discretization is to have "6 samples per rise time" in order to achieve a reasonable appro
ximation of the CT controller, without wasting computing resources. Derive the effect on PM that this choice will have. Also derive the relationship of this choice to the effective Nyquist rate for the closed-loop system.
1 answer:
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Answer:
Vab ≈ 3.426 V
Explanation:
First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...
R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600
Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:
Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1
The voltage at Vb is also a divider from V1:
Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1
The parallel branches containing Va and Vb have an effective resistance of ...
(1030)(1710)/(1030+1710) = 642.81
That forms a divider with R1 to give V1:
V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V
The difference Va-Vb is ...
Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V
_____
We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...
(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0
(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0
(V1 -Va)/R3 -Va/R4 = 0
The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.
Answer:
Load carried by shaft=9.92 ft-lb
Explanation:
Given: Power P=4.4 HP
P=3281.08 W
<u><em>Power: </em></u>Rate of change of work with respect to time is called power.
We know that P=
rad/sec
So that P=
So 3281.08=
T=13.45 N-m (1 N-m=0.737 ft-lb)
So T=9.92 ft-lb.
Load carried by shaft=9.92 ft-lb