Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected]

[email protected] • ENGR-25_Plot_Model-4.ppt1

Bruce Mayer, PE Engineering/Math/Physics 25: Computational Methods

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engr/Math/Physics 25

Chp6 MATLAB

Fcn Discovery



Learning Goals Create “Linear-Transform” Math Models

for measured Physical Data• Linear Function → No Xform• Power Function → Log-Log Xform• Exponential Function → SemiLog Xform

Build Math Models for Physical Data using “nth” Degree Polynomials

Use MATLAB’s “Basic Fitting” Utility to find Math models for Plotted Data



Learning Goals Use Regression Analysis as quantified

by the “Least Squares” Method• Calculate

– Sum-of-Squared Errors (SSE or J)The Squared Errors are Called “Residuals”

– “Best Fit” Coefficients– Sum-of-Squares About the Mean (SSM or S)– Coefficient of Determination (r2)

• Scale Data if Needed– Creates more meaningful spacing



Learning Goals cont

Build Math Models for Physical Data using “nth” Degree Polynomials

Use MATLAB’s “Basic Fitting” Utility to find Math models for Plotted Data

Use MATLAB to Produce 3-Dimensional Plots, including• Surface Plots• Contour Plots



Function Discovery Physical Processes for some

Response (OutPut), y, as Resulting from some Excitation (InPut), x, can many times be approximated by 3 Functions1. The LINEAR function: y = mx + b

– Produces a straight line with SLOPE a of m and an INTERCEPT of b when plotted on rectilinear axes

2. The POWER function: y = bxm – gives a straight line when plotted on log-log

axes

3. The exponential function y = b·10mx or y = b·emx – Yields a straight line when plotted on a

semilog plot with logarithmic y-axis



Linear Transformations For a Linear

Function We can Easily Find the Slope, m, and y-Intercept, b

bmxy

b

y

xx

yxym

Transform the Power Function to Line-Like Form

mbxyPwrFcn Take ln of both sides

bxmyxbbxy mm

lnlnlnlnlnlnln

pY pX pB



Power Function Xform Thus the Power

Function Takes the Form:

Yields a Staight Line So if we suspect a

PwrFcn, Plot the Data in Log-Log Form

xy ln vs.ln

Example:

ppp

mm

BmXYbxmyxbbxy

lnlnlnlnln

If The Data Follows the Power Function Then a Plot of

91713 73.1

xxy



Power Function y = 13x1.73

Rectlinear Plot Log-Log Plot

0 10 20 30 40 50 60 70 80 90 1000

0.5

1

1.5

2

2.5

3

3.5x 10

4

x

y

Power Function

1.5 2 2.5 3 3.5 4 4.5 55.5

6

6.5

7

7.5

8

8.5

9

9.5

10

10.5

ln(x)

ln(y

)

Power Function



y = 13x1.73 by MATLAB loglog

>> x = linspace(7, 91, 500);

>> y = 13*x.^1.73;

>> loglog(x,y), xlabel('x'), ylabel('y'), title('y = 13x^1^.^7^3'), grid

100

101

102

102

103

104

105

y = 13x1.73

x

y



Exponential Function Xform Recall the

General form of the Exponential Fcn

In This Case Let

Then the Xformed Exponential Fcn

mxbey Again taking the ln

bmxyemxb

bey mx

lnlnlnln

lnln

e

e

BbYy

lnln

ee BmxY A SemiLog (log-y)

plot Should Show as a Straight Line



Exponential Fcn Plot Plot 115ln61.0ln115 61.0 tvev t

0 1 2 3 4 5 6 7 8 9 100

20

40

60

80

100

120

t

v

Exponential Function

0 1 2 3 4 5 6 7 8 9 10-2

-1

0

1

2

3

4

5

t

ln(v

)

Exponential Function

ln(115) = 4.79

Rectlinear Plot SemiLog Plot



V = 115e-0.61t by MATLAB semilogy

0 1 2 3 4 5 6 7 8 9 1010

-1

100

101

102

103

v = 115e-0.61t

t

v

>> t = linspace(0, 10, 500);

>> v = 115*exp(-0.61*t);

semilogy(t,v), xlabel('t'), ylabel('v'), title('v = 115e^-^0^.^6^1^t'), grid



Steps for Function Discovery1. Examine the data near the origin.

• The linear function can pass through the origin only if b = 0

• The exponential function (y = bemx) can never pass through the origin – as et > 0 (positive) for ALL t; e.g., e−2.7 = 0.0672– unless of course b = 0, which is a trivial case: y

= 0·emx

• The power function (y = bxm) can pass through the origin (e.g.; y = 7x3) but only if m > 0 (positive) as – As y = bx-m = b/xm → Hyperbolic for negative m



“Discoverable” Functions

-5 0 50

50

100

150

x

ye

-5 0 50

5

10

15

20

25

x

ypp

-5 0 50

50

100

150

200

250

300

350

400

x

ypm

xe ey 2xypp

2xy pm

In most applications x is NONnegative



Steps for Function Discovery2. Plot the data using

rectilinear scales. • If it forms a straight line, then it

can be represented by the linear function and you are finished.

• Otherwise, if you have data at x = 0, thena. If y(0) = 0, then try the power

function.b. If y(0) 0, then try the exponential

functionc. If data is not given for x = 0,

then proceed to step 3.

mx

m

bey

bxy

bmxy



Steps for Function Discovery3. If you suspect a power

function, then plot the data using log-log scales. • Only a power function will form a

straight line on a log-log plot.

4. If you suspect an exponential function, then plot the data using the SemiLogy scale. • Only an exponential function will

form a straight line on a SemiLog plot.

mx

m

bey

bxy

bmxy



SemiLog and LogLog Scales

-5 0 510

-3

10-2

10-1

100

101

102

103

x

ye

10-2

100

102

10-3

10-2

10-1

100

101

102

10-2

100

102

10-2

10-1

100

101

102

103

x

ypm

• Note Change in power Function x-axis scales– In this case x MUST be POSITIVE

xe ey 2xypp

2xy pm



Steps for Function Discovery5. In function discovery

applications, use the log-log and semilog plots only to identify the function type, but not to find the coefficients b and m. • The reason is that it is difficult to

interpolate on log scales• To Determine Quantities for m &

b Perform the Appropriate Linearization Transform to plot one of– ln(y) vs. ln(x) → Power Fcn– ln(y) vs x → Exponential Fcn

mx

m

bey

bxy

bmxy



The polyfit Function The Command → p =

polyfit(x,y,n)• This Function Fits a polynomial of

degree n to data described by the vectors x and y, where x is the independent variable.

• polyfit Returns a row vector p of length n + 1 that contains the polynomial coefficients in order of descending powers

Note That a FIRST Degree Polynomial Describes the Eqn of a LINE• If w =p1z +p2 then polyfit on data Vectors

W & Z returns: p = ployfit(Z,W,1) = [p1, p2] → [m, b]



Using polyfit For Discovery polyfit of degree-1 (n = 1) returns the

parameters of a Line• p1 → m (slope)• p2 → b (intercept)

Thus polyfit can provide m & b for any of the previously noted functions AFTER the appropriate Linearization Transform



m & b by polyfit(x,y,1) We need to find

an Eqn for the Vapor pressure of Ethanol, C2H5OH, as a fcn of Temperature

Find Pv vs T Data by Consulting• P. E. Liley and W.

R. Gambill, Chemical Engineers’ HandBk, New York, McGraw-Hill Inc., 1973, p. 3-34 & 3-54

Pv for Ethanol (Torr)

Temperature(Kelvins)

760 351.4400 336.5200 321.4100 307.960 29940 29220 28110 270.75 2611 241.7



m & b by polyfit(x,y,1)cont

Since this a Vapor Pressure we Suspect an Antoine or Clapeyron Relation Pv ~ em/T

As a Starting Point Make a Rectilinear Plot

>> Pv = [1, 5, 10, 20, 40, 60, 100, 200, 400, 760];>> T = [241.7, 261, 270.7, 281, 292, 299, 307.9, 321.4, 336.5, 351.4];>> plot(T,Pv,'x', T,Pv,':'), xlabel('T (K)'), ylabel('Pv (Torr)'),...title('Ethanol Vapor Pressure'), grid

240 260 280 300 320 340 3600

100

200

300

400

500

600

700

800

T (K)

Pv

(Tor

r)

Ethanol Vapor Pressure



Compare LogLog vs SemiLogY Log Log Plot

100.5

100.6

100

101

102

103


1000/T (1/K)

Pv

(Tor

r)

logY vs linX

2.8 3 3.2 3.4 3.6 3.8 4 4.210

0

101

102

103


1000/T (1/K)

Pv

(Tor

r)

Looks like an Exponential in 1/T • e.g., the Clapeyron eqn




The Plots Looks Pretty Well Exponential For a 1st-Cut Assume the Clapeyron

form

Tmv

TmBBTmv

v

beP

eeeP

BT

mP

1

11

or1ln

Now Xform

BTmPv 1ln

x eBeY Thus Plot

ln(Pv) vs 1/T




The Command Session

Nicely Linear → Clapeyron is OK

>> Ye = log(Pv);>> x = 1./T>> plot(x,Ye,'d', x,Ye,':'), xlabel('1/T (1/K)'), ylabel('ln(Pv) (ln(Torr))'),...title('Ethanol Vapor Pressure - Clapeyron Plot'), grid

2.8 3 3.2 3.4 3.6 3.8 4 4.2

x 10-3

0

1

2

3

4

5

6

7

1/T (1/K)ln

(Pv)

(ln(

Torr)

)

Ethanol Vapor Pressure - Clapeyron Plot




Apply PolyFit to Find m and B

>> p = polyfit(x,Ye,1)p = 1.0e+003 * -5.1304 0.0213

or

TP

mxBY

v

ee

14.51303.21ln

To Increase the Sig Figs displayed for B these plots are typically plotted with x1 = 1000/T

>> x1 = 1000./T>> p1 = polyfit(x1,Ye,1)p1 = -5.1304 21.2512

TPv 10001304.52512.21ln




2.8 3 3.2 3.4 3.6 3.8 4 4.20

1

2

3

4

5

6

7

1000/T (1000/K)

ln(P

v) (l

n(To

rr))

Ethanol Vapor Pressure - Clapeyron Plot

ln(Pv) =21.2512-5130.4(1/T)

Tv eP 14.51302512.21



All Done for Today

PowerandExps



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engr/Math/Physics 25

Appendix 6972 23 xxxxf



Power Fcn Plot

>> x =[7:91];>> y = 13*x.^1.73;>> Xp = log(x); >> Yp = log(y);>> plot(x,y), xlabel('x'), ylabel('y'),...grid, title('Power Function')>> plot(Xp,Yp), xlabel('ln(x)'), ylabel('ln(y)'),...grid, title('Power Function')

91713 73.1

xxy



Exponential Fcn Plot

>> t = [0:0.05:10];>> v = 120*exp(-0.61*t);>> plot(t,v), xlabel('t'), ylabel('v'),...grid, title('Exponential Function')>> Ve = log(v);>> plot(t,Ve), xlabel('t'), ylabel('ln(v)'),...grid, title('Exponential Function')

115ln61.0ln115 61.0 tvev t



Altitude of Right Triangle The Area of RIGHT

Triangle

y

x

L

h

yxA 21 The Area of an ARBITRARY

Triangle

hLA 21 By Pythagoras for

Rt-Triangle

22 yxL



Altitude of Right Triangle cont

y

x

L h

hyxyx 222121

Solving for h

22 yx

yxh

Lx 21 Base&Base Equating the A=½·Base·Hgt noting that



Normalized Plot>> T = [69.4, 69.7, 71.6, 75.2, 76.3, 78.6, 80.6, 80.6, 82, 82.6, 83.3, 83.5, 84.3, 88.6, 93.3];>> CPH = [15.4, 14.7, 16, 15.5, 14.1, 15, 17.1, 16, 17.1, 17.2, 16.2, 17, 18.4, 20, 19.8];>> Tmax = max(T);>> Tmin = min(T);>> CPHmax = max(CPH);>> CPHmin = min(CPH);>> Rtemp = Tmax - Tmin;>> Rcroak = CPHmax - CPHmin;>> DelT = T - Tmin;>> DelCPH = CPH - CPHmin;>> Theta = DelT/Rtemp;DelCPH = CPH - CPHmin;>> Omega = DelCPH/Rcroak;>> plot(T, CPH,), grid>> plot(Theta,Omega), grid



Start Basic Fitting Interface 1 FIRST →

Plot the DATA



Start Basic Fitting Interface 2

Expand Dialog Box

Goodness of Fit; smaller is

Better



Start Basic Fitting Interface 3 Result Chk by

polyfit

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Theta

Om

ega

Frog Croaking in the Evening - 2045hrs

y = 0.8737*x + 0.04286

Croak Data linear Fit

>> p = polyfit(Theta,Omega,1)p = 0.8737 0.0429



Caveat



Greek Letters in Plots

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Frog Croaking Frequency

Croak Data Linear Fit

= 0.8737 + 0.04286



Plot “Discoverables”% "Discoverable" Functions Displayed% Bruce Mayer, PE • ENGR25 • 15Jul09%x = linspace(-5, 5);ye = exp(x);ypp = x.^2;ypm = x.^(-2);% plot all 3 on a single grapheplot(x,ye, x,ypp, x,ypm),grid,legend('ye', 'ypp', 'ypm')disp('Showing MultiGraph Plot - Hit ANY KEY to continue')pause%% PLot Side-by-Sidesubplot(1,3,1)plot(x,ye), gridsubplot(1,3,2)plot(x,ypp), gridsubplot(1,3,3)plot(x,ypm), grid



% "Discoverable" Functions Displayed% Bruce Mayer, PE • ENGR25 • 15Jul09%x = linspace(-5, 5);ye = exp(x);ypp = x.^2;ypm = x.^(-2);% plot all 3 on a single grapheplot(x,ye, x,ypp, x,ypm),grid,legend('ye', 'ypp', 'ypm')disp('Showing MultiGraph Plot - Hit ANY KEY to continue')pause%% PLot Side-by-Sidesubplot(1,3,1)plot(x,ye), gridsubplot(1,3,2)plot(x,ypp), gridsubplot(1,3,3)plot(x,ypm), grid

Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected]

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Transcript of Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected]