M-3

M.A.E 318 Bonus Assignment J. Murray

Problem 1: This questions asks that the model composed of the equations (1) and (2) be linearized. Using the linear equations the transfer function of this model is to be found. (1)

μ &r& − θ& 2 r = − 2 + u r

(2)

rθ&& + 2r&θ& = 0

To linearize this let:

r = R + ∆r simplify by letting ∆r =α θ& = Ω + Δθ& simplify by letting ∆θ΄=β u = 0 + ∆u simplify by letting ∆u=γ To linearize the problem these expressions must be substituted for r, θ, and u. These are then substituted into equations (1) and (2). Then retain only the terms which are linear in α, β and γ. First the value of μ must be found. This is done by plugging in r=R, θ’=Ω, and u=0 into equation (1). Since R is constant R’’=0, and μ=Ω2R3. Substitutions are now made into (1) and (2). (3) (4)

Ω2R3 +γ (R + α ) 2 ( R + α )(Ω + β )′ + 2( R + α )′(Ω + β ) = 0 ( R + α )′′ − (Ω + β ) 2 ( R + α ) =

Linearizing the left side of (3): ( R + α )′′ − (Ω + β ) 2 ( R + α ) = α ′′ − (Ω 2 + 2Ωβ + β 2 )( R + α ) = α ′′ − (2 RΩβ + 2Ωαβ + RΩ 2 + Rβ 2 + Ω 2α + αβ 2 ) Neglecting nonlinear terms the equation becomes: (5)

α ′′ − Ω 2α − 2 RΩβ − RΩ 2 Linearizing the right side of (3): Ω2 R3 R3 2 = Ω ⋅ (R + α ) 2 R 2 + 2 Rα + α 2

= Ω 2 ( R − 2α ) (this contains only the linear terms of the division)

Plugging this back into the right-hand side of (3), we get: (6)

Ω2 R3 − + γ = − RΩ 2 + 2Ω 2α + γ 2 (R + α )

Substitute (5) and (6) into (3):

α ′′ − Ω 2α − 2 RΩβ − RΩ 2 = − RΩ 2 + 2Ω 2α + γ , which can be simplified to: (1′) . This is the linearization of equation (1). To linearize equation (2) we will start with (4) (as it already has the variables substituted): ( R + α )(Ω + β )′ + 2( R + α )′(Ω + β ) = ( R + α ) β ′ + 2α ′(Ω + β ) = Rβ ′ + αβ ′ + 2α ′Ω + 2α ′β = Rβ ′ + 2α ′Ω (neglecting nonlinear terms)

(2′) . Rβ ′ + 2Ωα ′ = 0 This is the linearization of equation (2). To find the transfer function for the linearized system ( 1′ ), ( 2′ ), take the derivative of ( 1′ ): (7)

α ′′′ − 3Ω 2α ′ − 2Ω( Rβ ′) = γ ′ Then from ( 2′ ),

(8)

Rβ ′ = −2Ωα ′

Next substitute (8) into (7): (9)

α ′′′ − 3Ω 2α ′ − 2Ω(−2Ωα ′) = γ ′ simplify this:

(10)

α ′′ + Ω 2α = γ

To get the linearized transfer function, take the Laplace transform of both sides. This gives the transfer function of: T (s) =

1 s + Ω2 2

This tells us that the linearized solution is an undamped oscillation about the operating point (R,Ω,0), at a radian frequency of Ω.

Problem 2:

To determine the L and C values of circuits (a) and (b), the first step is to find the Transfer functions of each circuit. For the low pass circuit (a), this was found by:

Let Z RC

R ( jω )C = 1 R+ ( jω )C

Next

H ( jω ) =

This equals Z rc =

Z RC ( jω ) L + Z RC

This gives a transfer function of: 1 (1) H ( s ) = L s 2 LC + s + 1 R For the high pass circuit (b), this was found by: Let Z RL =

( jω ) RL R + ( jω ) L Z RL

Next H ( jω ) =

1 ( jω )C

+ Z RL

This gives a transfer function of: (2) H ( s ) =

s2 s 1 + s2 + RC LC

( jω ) RC ( jω ) RC + ( jω )C 2

Solving L and C for the low pass filter. Because (1) is a standard form second-order system, and ζ=0.707 L = 2ζω n and R 1 (4) LC = 2

(3)

ω

Solving (3) for L when ωn=100Hz, L=0.11312 Henrys Solving (4) for C when ωn=100Hz, C=0.000884 Farads This can then be put into the transfer function to give: (5) H ( s ) =

1 0.0001s + 0.01414s + 1 2

Using MATLab to create a Bode plot of (5) gives the plot in Figure 1.

Bode Diagram 20

Magnitude (dB)

0 -20 -40 -60 -80 0

Phase (deg)

-45 -90 -135 -180 0

10

1

10

2

10

3

10

Frequency (rad/sec)

Figure 1: Bode Plot for Low Pass Filter

4

10

Solving L and C for the high pass filter. Because (2) is also a standard form second-order system it can be solved the same way. (6) 2ζω n =

1 and CR

1 CL

(7) ω n =

Solving (6) for C when ωn=1000Hz, C=0.000088 Farads Solving (7) for L when ωn=1000Hz, L=.011364 Henrys This can be put into the transfer function to give: (8) H ( s ) =

s2 s 2 + 1420.45s + 999968

Bode Diagram 0

Magnitude (dB)

-20 -40 -60 -80 -100 -120 180

Phase (deg)

135 90 45 0 1

10

2

10

3

10

4

10

Frequency (rad/sec)

Figure 2: Bode Plot for High Pass Filter

5

10

Problem 3:

The problem is to find the transfer function of the step response data that is provided. The first step is to plot the data and decide what type of least squares will solve the problem. The first data plot is Figure 3. Data Plot From response.txt 3.5

3

y(t)

2.5

2

1.5

1 response.txt data point 0.5

0

0.5

1

1.5

2

2.5 t

3

3.5

4

4.5

5

Figure 3: Plot of response.txt Data Points

Using this data plot there can possibly be two types of linearization. The first is a saturated growth rate. The second is a linear model with nonpolynomial terms. The saturated growth model does not model this data well. To find the linear model with nonpolynomial terms the assumption was used that the model would use the standard formula of: y = a0 + a1e − t + a 2 te −t

Using MATLab the unknown coefficients could be found using matrix manipulation. This was done by creating a matrix X=[ones(size(t)) exp(-t) t.exp(-t)]. This

matrix was then multiplied by the inverse of matrix b. Doing this provided the coefficients of: 2.9808, -2.7968, 0.3085. This provides an equation of y = 2.9808 − 2.7968e − t + 0.3085 ⋅ t ⋅ e −t

This function was then plotted over the response.txt data, and is shown in Figure 4. Linear Model With Nonpolynomial terms Over response.txt Data 3.5

3

2.5

y(t)

2

1.5

1

0.5

0 0

Liner Model with Nonpolynomial Terms Respone.txt 0.5

1

1.5

2

2.5 t

3

3.5

4

4.5

5

Figure 4: Plot of response.txt with Transfer Function Overlaid

This model accurately represents the data points. The transfer function is then found by taking the Laplace transform of the equation. This gives: T ( s) =

2.9808 2.7968 .9519 − + s s +1 ( s + 1) 2

The Code for Figure 1: >> n=[ 0 0 1 ] >> d=[ 0.0001 0.01414 1 ] >> bode(n,d) The Code for Figure 2: >> n=[ 1 0 0 ] >> d=[ 1 1420.45 999968] >> bode(n,d) The code for problem 3: function problem3 load response.txt mu=mean(response) sigma=std(response) [n,p] =size(response); MeanMat=repmat(mu,n,1); SigmaMat=repmat(sigma,n,1); outliers=abs(response-MeanMat)>3*SigmaMat; nout=sum(outliers) response(any(outliers,2),:)=[]; corrcoef(response) t=response(:,1); y=response(:,2); %q=plot(t,y,'og'); X=[ones(size(t)) exp(-t) t.*exp(-t)]; a=X\y T=(0:0.1:5)'; Y=[ones(size(T)) exp(-T) T.*exp(-T)]*a; %Y = X*a; %MaxErr = max(abs(Y - y)) plot(T,Y,'-',t,y,'ro'); syms t; f=inline('y=2.9808-2.7968*exp(-t)+.9519*t*exp(-t)'); laplace(2.9808-2.7968*exp(-t)+.9519*t*exp(-t)) % Laplace transform comes back as % 1863/625/s-1748/625/(1+s)+9519/10000/(1+s)^2