Control Systems Engineering : Masons Gain

[BartlebyS]

Homework Statement

Here is the system I am looking at

Homework Equations

So I worked out a formula for the steady state error and put it in a form similar to masons gain formula

The Attempt at a Solution

But when I try and draw a block diagram and work out e I get confused by forward path and loop gain. It's not a question I have been asked, I am just playing, perhaps I cannot apply masons gain to my drawing? Or my drawing is wrong?

 

[rude man]

Your equations are not correct. For one thing, they are not dimensionally consistent: dh/dt = qin - qout has different dimensions on each side. q has dimension L^3 whereas dh/dt has dimension LT^(-1). L = length, T = time.

Your block diagram is good up to the point qin but then falls apart. Correct your equation for dh/dt: h cannot = qin - qout, you've already written dh/dt = qin - qout (which was incorrect also).

What is the correct relationship between dh/dt and qin - qout? (Hint: area of the bottom of the tank is a factor). And, once you get dh/dt right, how does one go from dh/dt to h?

 

[BartlebyS]

Thanks for pointing that out, so I came to the idea that

a*dh/dt = dv/dt = q_in(t) - q_out(t) = 0 for steady state

which would make the rest of the derivation ok I hope?

Which then gave me a different drawing for the block diagram.

Which gives me a loop gain of (c_1 - c_2)/as

Instead of the c_1 / c_2 i got when I used the assumption that dv/dt = 0 for steady state. Am I still wrong with my equations or drawing?

I thought I had it for a minute there! Thanks for your help. I don't need to do it, it's just bugging me as I cannot relate it to the rest of my notes.

 

[rude man]

Your block diagram is now correct.

Loop gain = H(s)/H₀(s).

 

[BartlebyS]

Thanks again mate, I played with it some more, got the answers I wanted and have a much better understanding of signal flow graphs and masons formula.