{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions_hw6

# Solutions_hw6 - Homework ECE147A Problem 1 1.1 Your...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework ECE147A November 14, 2009 Problem 1 1.1 Your Controller Nothing much to do for me here. You should show simulation results, tell me whether or not it shows a stable response and why or why not it is stable, then you get 4 points. 1.2 Controller from the Midterm Solutions The proposed controller is C ( s ) = k ( s + 1)( s + 2)( s + 3) s ( s 2 + 4) . (1) For now, the control gain is set to k = 2, like in the Midterm Solutions. The response to a reference of r ( t ) = 4 and a load disturbance d ( t ) = 3cos(2 t ) is shown in Figure 2 and the response is obviously converging to the reference signal. 2 points for plot and observation. The transfer functions from r and d to the tracking error e := r- y are easily computed as T r → e ( s ) = 1 1 + C ( s ) P ( s ) = s ( s 2 + 4) s 3 + 12 s 2 + 52 s + 3 (2) T d → e ( s ) = P ( s ) 1 + C ( s ) P ( s ) = s ( s 2 + 4)( s 2 + 4 s + 5) s 6 + 18 s 5 + 135 s 4 + 510 s 3 + 1004 s 2 + 972 s + 360 . (3) The frequencies of reference and disturbance are ω r = 0 and ω d = 2, so the long-term response if either of the two is present is given by e ( t ) = | T r → e ( jω r ) | cos( ω r t + 6 T r → e ( jω r )) = 0 (4) e ( t ) = | T d → e ( jω d ) | cos( ω d t + 6 T d → e ( jω d )) = 0 . (5) 1 To Workspace y Step Sine Wave Plant P Controller C e u d r y y Figure 1: Problem 1.2. Simulink Model 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Output t in s Figure 2: Problem 1.2: Response of Plant + Controller to d and r Note that those responses are zero because zeros of the closed-loop transfer functions to the tracking error e are exactly at the frequencies of reference and disturbance, respectively. 3 points for the TFs, 3 more if you made use of them to explain the response. This is a general result, sometimes known as the Inner Model Principle , which states that in order to fully reject a signal, the feedback path must contain the poles of the signal. To see why, write P ( s ) = N P ( s ) D P ( s ) and C ( s ) = N C ( s ) D C ( s ) and form the transfer functions from Equation ( 3 ), this time generally: T r → e ( s ) = 1 1 + C ( s ) P ( s ) = D P ( s ) D C ( s ) D P ( s ) D C ( s ) + N P ( s ) N C ( s ) (6) T d → e ( s ) = P ( s ) 1 + C ( s ) P ( s ) = N P ( s ) D C ( s ) D P ( s ) D C ( s ) + N P ( s ) N C ( s ) . (7) In both cases, the poles of the Controller end up as the zeros of the relevant transfer function, which means that signals at the frequencies corresponding to that zeros are going to be fully rejected. 1.3 Routh criterion Now we want to check stability w.r.t. the controller gain k in Equation ( 1 ). The closed loop denominator is D ( s ) = s 3 + 6 ks 2 + (24 k + 4) s + 30 k (8) and your Routh Array should resemble 3 : 1 24 k + 4 2 : 6 k 30 k 1 : 24 k- 1 0 : 30 k ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 8

Solutions_hw6 - Homework ECE147A Problem 1 1.1 Your...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online