Control systems: Simplifying block diagram

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The discussion focuses on simplifying a block diagram to a standard first-order form, specifically K/(1+tD). Participants are attempting to derive the transfer function using Mason's rule for both inner and outer loops of the system. One user expresses uncertainty about how to represent their findings in the required first-order format, while others clarify that the form can be achieved through proper manipulation of the transfer functions. The final equation derived for the outer loop is h(s) = H(s) * Kp * Kb / (2s + 4 + Kp), and the steady-state gain is calculated by evaluating h(0). Overall, the thread emphasizes the application of Mason's rule and the importance of correctly identifying terms in the equations.
MattH150197
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Homework Statement


How to show a block diagram can be simplified to a standard first order form for the diagram shown in the image attached and the question is shown in 1.a

Homework Equations


Standard first order form K/(1+tD) where K gain and t is time constant

The Attempt at a Solution

[/B]
So i got ((Kb*H-h)*Kp-4)*1/2s = h Is this correct? and how do i show it in standard first order form from here? Thanks
 

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Figure 1 shows two closed loops: An inner and an outer loop.

Use Mason's rule to calculate the transfer function as for the inner loop.

Insert this inner transfer function in the outer loop and use Mason again to calculate h(s)/H(s) for the outer loop ( the transfer function for the outer loop ).

The characteristic equation for h(s)/H(s) will be in the form:

as + b = 0 → τ = b / a.
 
Enlarged image of diagram as requested
 

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Hesch sorry I am not quite sure how from what youve said i can show it is a first order system as i need to show it in the form h = K/(1+tD), I am just not sure how to get there from what i worked out
 
MattH150197 said:
i need to show it in the form h = K/(1+tD)

You don't need to show in this form.

Using Mason you should get:

h(s)/H(s) = Kp * Kb / ( 2s + 4 + Kp ) →

h(s) = H(s) * Kp * Kb / ( 2s + 4 + Kp )
( if you wish )
 
MattH150197 said:
So i got ((Kb*H-h)*Kp-4)*1/2s = h Is this correct?
I think it's nearly there; Check what's fed back via the "4" block. That 4 looks mighty lonely in your equation, given that the term its being summed with is bound to have some units associated with it... :smile: .
 
Hesch said:
h(s) = H(s) * Kp * Kb / ( 2s + 4 + Kp )
To determine the steady state gain from the above equation, let s → 0, so

h(0) = H(0) * Kp * Kb / ( 4 + Kp )
 
Ah yeah i understand what you have done now actually Hesch, thanks for the help guys.
 

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