I am having an overload trying to figure this problem out. I think my calculus is weak so that is my first problem. I have been working on this one and can't get it. any help or advise would be greatly appreciated


The error voltage input to an integral controller goes from 0 V to 3 V at t=1 second and the remains at 3 V until t=5 seconds. What is the integral controller output voltage at t=5 seconds and what is the slope of the output voltage waveform from t=1 to t=5 assume that the integration constant Ki for the controller is 2.5 /sec.

I am looking for t he integral controller output voltage and the slope.

thanks for all the help you can give me..

 

First, is the integral term the only one we are considering? Is there a proportional term as well?

To "integrate" a piecewise linear function of time with respect to time it is only necessary to graph the function, and using known formulas for polygons like triangles and rectangles compute the "area under the curve". Many integral controllers will have a constant of proportionality which is multiplied by the resulting integral. In this case it is Ki and the units Ibelieve are volts/sec. BTW "/sec" as a unit is a bit ambiguous

Clearly if you integrate an error signal with only positive valuse the result will increase without bound. That is why integral control is used in conjunction with a proportional term. There also has to be some negative feedback or the output will grow without bound -- and we can't have that now can we?

So the output is zero up until t=1. Then we have a step function from 0V to 3V and the integral of a step function (constant with respect to t from t=1 to t=5) is a ramp, or a linear function of t

Vo = 2.5*integral(3dt, t=1, t=5)
Vo = 2.5*3*t] evaluated between t=5 and t=1
Vo = 2.5*((3*5)-(3*1)
Vo = 2.5*12
Vo = 30

Make sense?

 

First, is the integral term the only one we are considering? Is there a proportional term as well?

To "integrate" a piecewise linear function of time with respect to time it is only necessary to graph the function, and using known formulas for polygons like triangles and rectangles compute the "area under the curve". Many integral controllers will have a constant of proportionality which is multiplied by the resulting integral. In this case it is Ki and the units Ibelieve are volts/sec. BTW "/sec" as a unit is a bit ambiguous

Clearly if you integrate an error signal with only positive valuse the result will increase without bound. That is why integral control is used in conjunction with a proportional term. There also has to be some negative feedback or the output will grow without bound -- and we can't have that now can we?

So the output is zero up until t=1. Then we have a step function from 0V to 3V and the integral of a step function (constant with respect to t from t=1 to t=5) is a ramp, or a linear function of t

Vo = 2.5*integral(3dt, t=1, t=5)
Vo = 2.5*3*t] evaluated between t=5 and t=1
Vo = 2.5*((3*5)-(3*1)
Vo = 2.5*12
Vo = 30

Make sense?

yeah that makes sense now. I did that and I wasn't for sure is that was right. I think now I understand it little better. but what is the slope?

 

The slope of a linear function is computed by knowing two (x,y) points on the line. The first point on the line is (t = 1, V = 0), and the second point is (t = 5, V = 30). The slope is ΔV/Δt.

ΔV/Δt = (30 - 0)/(5 - 1)
      = 7.5 Volts/second