Calculating controller gain in a feedback loop

[Keeeen]

Homework Statement

FIGURE 3 shows the block biagram of the control of an electric heating system. The heater is driven from a voltage-controlled power supple, the voltage V1 being derived from a potentiometer. The output temperature, θo is subject to distrubances, θd, because of changes in ambient temperature. It is proposed to apply 'disturbance feedback control' to the system by the inclusion of a transducer that measures the external temperature and feeds a signal back to the input via a proportional controller of gain H.

Determine the required value of H to eleminate the effect of the disturbance.

See figure 3 attached.

Homework Equations

The Attempt at a Solution

So the learning material I've been given does not give any help with trying to find the gain of a controller in a loop.
I'm thinking I need to determine the transfer function of the loop and work from there? Or is it a lot easier than that.
If anyone could give me some guidance as to where to start that would be appreciated.

 

[BvU]

Hi,
why not write down a transfer function for $\theta_D$ and see what it brings ?

We can't really help: see guidelines

 

[Keeeen]

Hi,
why not write down a transfer function for $\theta_D$ and see what it brings ?

We can't really help: see guidelines

A transfer function for the disturbance? I thought θD would just = the change in ambient temperature?

 

[BvU]

Yes, it is. But $\theta_D$ is not a transfer function

 

[Keeeen]

Okay I'm not sure if I'm on the right tracks with this, I've got so far -

θo = Vi - H * 0.05 * θD (2*0.4) + θD

Am i going in the right direction? Not sure how to get the transfer function for θD now

I got to this point by -

E = Vi - Mv
Mv = H * 0.05 * θD

E = Vi - H * 0.05 * θD

θo = E (PSU * Heater) + θD

 

[Keeeen]

I have no idea how you would get a transfer function for θD, as it just a disturbance. There's no input to create an output for θD?

 

[Merlin3189]

Use θ_D as an input and calculate the output it produces when other inputs are constant.
(Edit 2: I see you got θo = Vi - H * 0.05 * θD (2*0.4) + θD , which seems ok.)

Then ask yourself, what output do you want to get from variations in θ_D.

Edit 1:

The output temperature, θo is subject to distrubances, θd, because of changes in ambient temperature. It is proposed to apply 'disturbance feedback control' to the system ... Determine the required value of H to eleminate the effect of the disturbance.

 

[Keeeen]

Use θ_D as an input and calculate the output it produces when other inputs are constant.
(Edit 2: I see you got θo = Vi - H * 0.05 * θD (2*0.4) + θD , which seems ok.)

Then ask yourself, what output do you want to get from variations in θ_D.

Edit 1:

So potentially from variations in θD the output you would want is the set-point - variations in disturbance.

 

[Merlin3189]

If there were no disturbances, θ_D = 0,
then your θo = Vi - H * 0.05 * θD (2*0.4) + θD
becomes θ_O = V_i
That is what you want even when θ_D ≠ 0

 

[Keeeen]

If there were no disturbances, θ_D = 0,
then your θo = Vi - H * 0.05 * θD (2*0.4) + θD
becomes θ_O = V_i
That is what you want even when θ_D ≠ 0

Ah yes, so if the disturbance was 2 for example, it would become -

θo = Vi - H * 0.05 * θD(2*0.4) + 2
θo = Vi - H * 2.08
θo = Vi - 1 * 2.08 = Vi

 

[Merlin3189]

Ah yes, so if the disturbance was 2 for example, it would become -
θo = Vi - H * 0.05 * θD (2*0.4) + θD
θo = Vi - H * 0.05 * θD(2*0.4) + 2 Here you've substituted one θ_D, but not the other?
θo = Vi - H * 2.08 I don't quite see this
θo = Vi - 1 * 2.08 = Vi Here you seem to say H=1, and then, 2.08 =0

I agree your expression, θo = Vi - H * 0.05 * θD (2*0.4) + θD
which means, the output = the demanded input and a function of the disturbance
Since you want the output to equal the demanded input, then the function of the disturbance must be zero.

Another way of looking at it would be to rearrange your expression,
θo = Vi - H * 0.05 * θD (2*0.4) + θD
θo = Vi - θD( H * 0.05 * (2*0.4) - 1)
But you want θ_O = V_i , so what does - θD( H * 0.05 * (2*0.4) - 1) equal?

 

[Keeeen]

I agree your expression, θo = Vi - H * 0.05 * θD (2*0.4) + θD
which means, the output = the demanded input and a function of the disturbance
Since you want the output to equal the demanded input, then the function of the disturbance must be zero.

Another way of looking at it would be to rearrange your expression,
θo = Vi - H * 0.05 * θD (2*0.4) + θD
θo = Vi - θD( H * 0.05 * (2*0.4) - 1)
But you want θ_O = V_i , so what does - θD( H * 0.05 * (2*0.4) - 1) equal?

It must equal 0?

Sorry I'm struggling ot understand how I can get to the final gain of H

so rearranging the expression,

θo + θD (H * 0.05 * 0.8 - 1) = Vi

EDIT : H = 25

I only got this by trial and error by inputting numbers. I can't see or picture how this works though, could you explain for my piece of mind how i'd get to this point.
Thanks so much for the help btw

EDIT 2: Also how did you go from
this : θo = Vi - H * 0.05 * θD (2*0.4) + θD
to this : θo = Vi - θD( H * 0.05 * (2*0.4) - 1)

 

[neilparker62]

Considering $θ_d$ at the output summer you have: $$θ_d-θ_d\times0,05H\times2\times0,4=θ_d(1-H\times0,04) $$

 

[Merlin3189]

It must equal 0?EDIT 2: Also how did you go from
this : θo = Vi - H * 0.05 * θD (2*0.4) + θD
to this : θo = Vi - θD( H * 0.05 * (2*0.4) - 1)

I prefer not to go through the formalities of transfer functions here, because it is a simple system.
If θ_D is to have no effect, then the two signal paths must produce equal and opposite outputs.
So straightaway 0.05 x H x (-1) x 2 x 0.4 = - (+1)
so 0.04H = 1

 

[Ray Vickson]

I have no idea how you would get a transfer function for θD, as it just a disturbance. There's no input to create an output for θD?

See, eg., https://en.wikipedia.org/wiki/Control_theory
for appropriate formulas and explanations.