Matlab and Python programming: how to get started

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1 Malab and Pyhon programming: how o ge sared Equipping readers he skills o wrie programs o explore complex sysems and discover ineresing paerns from big daa is one of he main goals of his book. In his chaper, we explain how o do Malab and Pyhon programming. While Malab is imporan for sudying boh complex sysems and big daa, Pyhon is paricularly convenien for dealing wih big daa. The maerials presened here are chosen such ha ) afer sudying his chaper, laymen will no longer be laymen, and 2) seasoned programmers will also feel inspired. We organize he maerials hrough five examples. Example Wrie a malab code o generae a sine wave signal from = sin() () To generae a curve, we have o decide which sampling ime o use as well as how long he curve should be ploed. To his end, we firs noe ha sin() is a periodic funcion wih period 2π. To generae a smooh curve, we have o choose such ha each period is sampled a few ens of imes. To make he curve look like a sine wave, we have o plo ou a few periods of he signal. Based on hese consideraions, he complee malab code, saved as generae sinewave.m, is he following. Noe lines 2-4 are he header. I is beer o include hem in any of your malab codes. The main commands for our purpose are he lines 6, 7, and. In paricular, line 6 says ha sars from and ends a = 6π. The sampling ime inerval is =.. While line 7 is almos self-explanaory, we noe ha in malab, when evaluaing rianglular funcions, radians insead of angles are used. Therefore, if you wan o compue cos(8 ), you wrie cos(pi), where pi denoes π. The linewidh in line specifies he hickness of he curve. Lines 2-5 specify how he curve is ploed. They should be self-explanaory. The curve is ploed in Fig. (a). Wha happens if a differen sampling ime inerval is used? When a smaller is used, visually he curve remains he same. However, if a much larger is used, he curve could look very differen. To see his, you may choose =.6. If you uncommen lines 8 and 9, execue hem, and plo ou x2 vs. 2, you ge he curve shown in Fig. (b). I is quie differen from he sine wave everybody knows. This is because is oo big! So far, all is simple and almos rivial. Now le us play wih somehing more fun. Specifically, we ask: how may we generae oher ypes of plos from he sine wave signal? One ineresing soluion is provided by he ime delay embedding echnique of chaos heory, which we will sudy in deph laer. As a 2-dimensional (2-D) graphical represenaion, he echnique ells us o consruc vecors of he form: V (i) = [x(i),x(i + L)] (2) Associaing he 2 componens wih he x and y axes, he above equaion amouns o ploing x(i+l) vs. x(i). When L = 6, he plo is shown in Fig. (c) (blue curve). I is generaed by he codes given a lines 7 and 8. I is almos a perfec circle. Why is i so? The reason is L =.6. I is very close o π/2. If we wrie x(i) as sin(), hen x(i+l) is close o cos(). Now recall ha in polar coordinaes, a circle is represened by [cosθ,sinθ]. Tha is why he shape of he curve in Fig. (c) is almos a circle. The above argumen makes i clear ha L < 6 will When you save your malab file, you mus save i as.m!

2 make he circle-like curve in Fig. (c) an ellipse along he main diagonal, while 6 < L < 32 an ellipse along he 35 diagonal. In erms of he erminology of nonlinear dynamics, Fig. (c) is a phase space represenaion. The shape of he curve will no change wih he oal ime inerval sin() is evaluaed, so long as he oal ime inerval is slighly longer han one period. Tha is he beauy of phase space represenaion i is more able o capure he invarians of a dynamical process. To beer appreciae he above saemens, le us re-consider he case of =.6. Now we should choose L = so ha he acual ime delay L is he same as before. Based on he curve shown in Fig. (b), we ge Fig. (d). I is no he same as Fig. (c). Why? The reason is successive poins in Fig. (d) are conneced. Since is so big, successive poins on he circle are acually quie far apar. When conneced, somehing like a circular disk emerges in Fig. (d). If we do no connec he poins, we ge he discree poins designaed by red diamonds in Fig. (c). This is obained by un-commening line 9 in he code and execuing i. They all perfecly fall on he uni circle. This is he beauy of invarian dynamics! % g e n e r a e s i n e w a v e.m 2 c l c ; %removes a l l l i n e s i n h e command window 3 c l e a r a l l ; %d e l e e s a l l s o r e d v a r i a b l e s i n h e workspace 4 c l o s e a l l ; %c l o s e s a l l of h e f i g u r e s g e n e r a e d i n h e program 5 6 = :. : 3 p i 2 ; 7 x = s i n ( ) ; 8 % 2 = :. 6 : 2 p i 2 ; 9 %x2 = s i n ( 2 ) ; f i g u r e ; p l o (, x, l i n e w i d h, 2 ) ; 2 a x i s i g h ; 3 s e ( gca, f o n s i z e, 6 ) ; % s p e c i f i e s s i z e of h e numbers on x and y a x i s 4 x l a b e l (, f o n s i z e, 7 ) ; 5 y l a b e l ( s i n ( ), f o n s i z e, 7 ) ; 6 7 L=6; 8 f i g u r e ; p l o ( x ( : end L ), x (+L : end ), l i n e w i d h, 2 ) ; hold on ; 9 %p l o ( x2 ( : end ), x2 ( + : end ), dr, l i n e w i d h, 2, m a r k e r s i z e, 2 ) ; 2 s e ( gca, f o n s i z e, 6 ) ; 2 x l a b e l ( x ( ), f o n s i z e, 7 ) ; 22 y l a b e l ( x ( +L ), f o n s i z e, 7 ) ; 2

3 (a) =..5 sin() (b) =.6.5 sin() (c) (d) x(+l) Figure : Example 2 Explore he dynamics described by he following equaion: = sin() + asin( 5), a =.3 (3) This signal is called a quasi-periodic signal. I describes an imporan ype of moion. Since is a superposiion of wo sinusoidal signals, is i a periodic signal? The answer is no. To see his, we noe ha if is periodic, hen here exiss a period T such ha x( + T ) =. Now, sin() is periodic wih T = 2πn. Similarly, here exiss a T 2 such ha sin( 5( + T 2 )) = sin( 5). This yields T 2 = 2πn 2 / 5. For o be 3

4 periodic, among T and T 2, here mus exis a common T. Consequenly, n = n 2 / 5. Bu his equaion canno hold, since n, n 2 are inegers and n /n 2 is a raional number. Therefore, he signal is aperiodic. To explore he dynamics described by Eq. (3), we may plo he 2-D phase space diagrams. The basic code is given by lines 4-7 and below. The plo is shown in Fig. 2(a). I looks like he ire of a car. In fac, in nonlinear dynamics, i is called a orus. To beer indicae how he signal differs from a rue periodic signal, we have used lines and 2 o indicae he sar and end poins of he phase space diagram by a red circle and a red diamond, respecively. Boh are open ended. This is he essence of aperiodic signals. To beer undersand he difference beween quasi-periodic and periodic signals, le us examine wha happens if we replace 5 by is very crude approximaion, 2.2. Wih he codes a lines 9-23, we obain Fig. 2(b). This is clearly a periodic signal, albei wih a very long period. If we replace 5 by a beer approximaion, hen he period will become longer. These consideraions should help one undersand our earlier argumens beer. % g e n e r a e o r u s.m 2 c l c ; c l e a r a l l ; c l o s e a l l ; 3 4 a =. 3 ; 5 = :. 5 : 5 p i 2 ; 6 x = s i n ( ) +a s i n ( s q r ( 5 ) ) ; 7 L = 2 4 ; 8 f i g u r e ; 9 s u b p l o ( 2 ) ; p l o ( x ( : end L ), x (+L : end ), l i n e w i d h,. 5 ) ; hold on ; p l o ( x ( ), x (+L ), or, m a r k e r s i z e, ) ; 2 p l o ( x ( end L ), x ( end ), dr, m a r k e r s i z e, ) ; 3 s e ( gca, f o n s i z e, 6 ) ; 4 x l a b e l ( x ( ), f o n s i z e, 7 ) ; 5 y l a b e l ( x ( +L ), f o n s i z e, 7 ) ; 6 e x (. 3,. 3, ( a ), f o n s i z e, 6 ) ; 7 xlim ( [. 5,. 5 ] ) ; ylim ( [. 5,. 5 ] ) ; 8 9 x = s i n ( ) +a s i n ( 2. 2 ) ; 2 s u b p l o ( 2 2 ) ; 2 p l o ( x ( : end L ), x (+L : end ), l i n e w i d h,. 5 ) ; hold on ; 22 p l o ( x ( ), x (+L ), or, m a r k e r s i z e, ) ; 23 p l o ( x ( end L ), x ( end ), dr, m a r k e r s i z e, ) ; 24 xlim ( [. 5,. 5 ] ) ; ylim ( [. 5,. 5 ] ) ; 25 s e ( gca, f o n s i z e, 6 ) ; 26 x l a b e l ( x ( ), f o n s i z e, 7 ) ; 27 y l a b e l ( x ( +L ), f o n s i z e, 7 ) ; 28 e x (. 3,. 3, ( b ), f o n s i z e, 6 ) ; 4

5 .5 (a) x(+l) (b) x(+l) Figure 2: Example 3 Examine an exponenial funcion: = e a, a =.5 (4) Now i should be a simple maer o wrie a few lines of codes o generae he funcion. Please see he codes a lines 3-4 and Fig. 3(a). Now le us ask a quesion: If we observe a curve like ha in Fig. 3(a), how do we deermine wheher i is an exponenial funcion or no, and if yes, how do we esimae is parameer? The answer lies in aking logarihm of. Then he curve becomes a sraigh line he slope of he sraigh line 5

6 should be.5. Please see he code a line 4 and Fig. 3(b). Noe ha exponenial funcions can appear in many differen conexs, and he simple echnique discussed here can be very useful in all hose conexs. c l c ; c l e a r a l l ; c l o s e a l l ; 2 3 a =. 5 ; 4 = :. : ; 5 x = exp( a ) ; 6 f i g u r e ; s u b p l o ( 2 ) ; 7 p l o (, x, l i n e w i d h, 2 ) ; 8 s e ( gca, f o n s i z e, 6 ) ; 9 x l a b e l (, f o n s i z e, 8 ) ; y l a b e l ( x ( ), f o n s i z e, 8 ) ; e x ( 9,. 9, ( a ), f o n s i z e, 6 ) ; 2 3 s u b p l o ( 2 2 ) ; 4 p l o (, l o g ( x ), l i n e w i d h, 2 ) ; 5 s e ( gca, f o n s i z e, 6 ) ; 6 x l a b e l (, f o n s i z e, 8 ) ; 7 y l a b e l ( l n ( x ), f o n s i z e, 8 ) ; 8 e x (9,.5, ( b ), f o n s i z e, 6 ) ;.8 (a) (b) ln(x) Figure 3: 6

7 Example 4 Examine a power-law funcion: =.5, (5) As far as coding is concerned, his example is as simple as he las one. Please see he codes a lines 3-5 and Fig. 4(a). Now our quesion becomes he following: If we have observed Fig. 4(a), how do we deermine wheher i is a power-law or no, and if yes, how do we esimae is parameers? The idea lies in ploing ou he curve in double-logarihmic scale. Then we ge a sraigh line he slope of he sraigh line should be.5. Please see he code a line 4 and Fig. 4(b). c l c ; c l e a r a l l ; c l o s e a l l ; 2 3 a =.5; 4 = :. : ; 5 x =. ˆ a ; 6 f i g u r e ; s u b p l o ( 2 ) ; 7 p l o (, x, l i n e w i d h, 2 ) ; 8 s e ( gca, f o n s i z e, 6 ) ; 9 x l a b e l (, f o n s i z e, 8 ) ; y l a b e l ( x ( ), f o n s i z e, 8 ) ; e x ( 9,. 9, ( a ), f o n s i z e, 6 ) ; 2 3 s u b p l o ( 2 2 ) ; 4 p l o ( l o g ( ), l o g ( x ), l i n e w i d h, 2 ) ; a x i s i g h ; 5 s e ( gca, f o n s i z e, 6 ) ; 6 x l a b e l ( l n ( ), f o n s i z e, 8 ) ; 7 y l a b e l ( l n ( x ), f o n s i z e, 8 ) ; 8 e x ( 4. 2,. 6, ( b ), f o n s i z e, 6 ) ; 7

8 .8 (a) (b) 2 ln(x) ln() Figure 4: Example 5 Wrie a code o generae a plo ha resembles a unnel or a ornado. This exercise is open-ended, and here should exis many soluions. Our soluion involves using sinusoidal signals, exponenial signals, and he ime-delay embedding echnique. The code is given below and he plo is given in Fig. 5. I would be beneficial for you o check how he plo changes when you modify par of he code. c l c ; c l e a r a l l ; c l o s e a l l ; 2 3 = :. : 3 p i 2 ; 4 x = 5 s i n ( ). exp (. 3 ) ; 5 L=6; 6 l e n = l e n g h ( ) ; 7 f i g u r e ; 8 p l o 3 ( x ( : len L ) +5 ( : len L ), x (+L : l e n ), ( : len L ), l i n e w i d h, 2 ) ; 9 a x i s o f f ; 8

9 Figure 5: 9