This is testing Algorithm Writing For this assessment we will be c.pdf

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This is testing Algorithm Writing For this assessment we will be combining while and if control structures together. (A):Write a pseudocode (pen and paper) algorithm that solves the ODE y\'= f(t, y) using Euler’s method. (B):Convert this algorithm into MATLAB syntax, as the function euler method . B1:Your inputs should be the start time t0, the end time tend, the function f, an initial condition y0 and an initial step size h. B2:If h × f(t, y) at any point exceeds 1, your function should halve the step size h. B3:Equally, if h × f(t, y) at any point drops below 0.01, your function should double the step size h. B4:You should also make sure that the final solution step doesn’t ‘overshoot’ tend. (i.e. change h during the final step to exactly reach tend). B5:Your MATLAB code should use the function header below, and be wellcommented with a sensible layout. (C)Include your pseudocode algorithm, as well as the MATLAB code in your Portfolio submission. Also include a discussion of when the code could break, or give incorrect outputs (you DO NOT need to design your code to avoid these). Solution Answer:) A) The derivative term in the first order ivp: y\' = f(t, y), y(t0) = y0 is approtimated by making use of Taylor series approtimation of the dependent variable y(t) at the point ti+1. That is y(ti+1) = y(ti+ Dt) = y(ti) + Dty\'(ti) + (Dt2 / 2)y\'\'(ti) + . . . = y(ti) + Dtf(ti, yi) + (Dt2 / 2)y\'\'(ti) + . . . (... y\'(ti) = f(ti, yi)) if the infinite series is truncated from the term Dt2 onwards, then y(ti+1) = y(ti) + Dt y\'(ti) (or) yi+1 = yi + Dt fi for all i That is, for i = 0, y1 = y0 + Dt f0 i = 1, y2 = y1 + Dt f1 ! i = n-1, yn = yn-1 + Dt fn-1 Since y0 and hence f0 are known (from initial condition) in the equation corresponding to i = 0, all the terms on the r.h.s are known. So y1 that is, y at t1 is calculated easily from this equation. Similarly once y1 is known, r.h.s of the equation corresponding to i = 1 is also known so y2 can be computed. As we proceed in the same way until i = n-1, yn can be obtained. This is an etplicit method because in any equation there is only one unknown which can be separated to the left side of the equation. Local truncation error : The error in the approtimation, that is the difference between the exact solution at ti+1 and the numerical solution yi+1 is called the local truncation error (assumed that yi+1 is calculated with exact arithmetic with out any round off error). Ti+1 = y(ti+1) - yi+1 = y(ti+1) - yi - Dtfi = h2/2 y\'(t) (by Taylor series & remainder theorem) where ti < t < ti+1. Hence the order of the local truncation error for Euler scheme is O(Dt2) as Dt -> 0.

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i = 1, y2 = y1 + Dt f1
!
i = n-1, yn = yn-1 + Dt fn-1
Since y0 and hence f0 are known (from initial condition) in the equation corresponding to i = 0,
all the terms on the r.h.s are known. So y1 that is, y at t1 is calculated easily from this equation.
Similarly once y1 is known, r.h.s of the equation corresponding to i = 1 is also known so y2 can
be computed. As we proceed in the same way until i = n-1, yn can be obtained. This is an etplicit
method because in any equation there is only one unknown which can be separated to the left
side of the equation.
Local truncation error :
The error in the approtimation, that is the difference between the exact solution at ti+1 and the
numerical solution yi+1 is called the local truncation error (assumed that yi+1 is calculated with
exact arithmetic with out any round off error).
Ti+1 = y(ti+1) - yi+1
= y(ti+1) - yi - Dtfi
= h2/2 y'(t) (by Taylor series & remainder theorem)
where ti < t < ti+1. Hence the order of the local truncation error for Euler scheme is
O(Dt2) as Dt -> 0

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I am Bing Jr. I am a Signal Processing Assignment Expert at matlabassignmentexperts.com. I hold a Master's in Matlab Deakin University, Australia. I have been helping students with their assignments for the past 9 years. I solve assignments related to Signal Processing. Visit matlabassignmentexperts.com or email info@matlabassignmentexperts.com. You can also call on +1 678 648 4277 for any assistance with Signal Processing Assignments.

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CALIFORNIA STATE UNIVERSITY, NORTHRIDGEMECHANICAL ENGINEERIN.docx

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CALIFORNIA STATE UNIVERSITY, NORTHRIDGE MECHANICAL ENGINEERING DEPARTMENT MARCH 30 2015 ME 309 HOMEWORK #3 Ahmed Mohammed Problem statement 1. Write a general computer code to solve system of up to five couples first order initial value problems using Heun and Newton iteration trapezoidal methods. These are combined in the same algor4thim in such way that Heun’s method is automatically employed to provide the initial guess at each time step for the Newton iteration required by fully implicit Heun’s method, and no iterations are needed. The complete pseudo-language algorithm is attached. Test this code by solving the following problem. Use step size h= 0.1, 0.05 and 0.025. Solve the problem first with explicit Hun’s method, and then with implicit Newton iteration integration for each value of h. employ a convergence tolerance E= 0.000001 for the Newton iteration of the trapezoidal method. The exact solution to this problem is, Make a table of results of convergence tests (based on the exact solution) at t=1, 2, 3, 4 & 6. This table should include the exact value, the computed solution, the error and the error ratios from successive step size for each of the required values of t. discuss how these results compare with theory. Also discuss what factors should influence the choice of convergence tolerance E for the Newton iterations, and whether the specified value given above is appropriate. 2. Solve the following problem using only the newton iteration trapezoidal method. Employ step size h=0.1, 0.05, 0.01 and iteration convergence tolerance E= 0.000001. Consider the h=0.01 solution to be the “exact” in order to carry out convergence testes between the h= 0.1 and h=0.05 solutions at t=0.5, 1, 1.5, 2. Make a table, similar to the table in problem number 1. Mathematical Description HEUN’S METHOD Heun’s Method, explained in a short manner, uses the line tangent to the function at the beginning of an interval. Now if a small step is applied to it, the error with the function result will be small. Heun’s method can be explained in more detailed in the following way: 2.- To obtain solution point (t1,y1) we can use the fundamental theorem of calculus and integrate y’(t) over [t0,t1] to get 3.- Solving for y(t1) we find, 4.- We can use a numerical integration to approximate definite integral. If we use trapezoidal rule with step size h = t1 – t0, then we get 5.- We still need to find, y(t1) but an estimation for this value will work. After this we get the following, which is the Heun’s method. 6.- When this process is repeated it generates a sequence of points that approximate the solution curve y =y(t). At each step, Euler’s method is used as a prediction then the trapezoidal rule helps to make the correction to obtain the final value. [1] Newton’s Method It is way to approximate the roots of an equation by taking out the curve in the equation and then replace with a tangent line. Then, it find ...

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Find the compact trigonometric Fourier series for the periodic signal.pdf

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Find the compact trigonometric Fourier series for the periodic signal x(t) and sketch the amplitude and phase spectrum for first 4 frequency components. By inspection of the spectra, sketch the exponential Fourier spectra. By inspection of spectra in part b), write the exponential Fourier series for x(t) Solution ECE 3640 Lecture 4 – Fourier series: expansions of periodic functions. Objective: To build upon the ideas from the previous lecture to learn about Fourier series, which are series representations of periodic functions. Periodic signals and representations From the last lecture we learned how functions can be represented as a series of other functions: f(t) = Xn k=1 ckik(t). We discussed how certain classes of things can be built using certain kinds of basis functions. In this lecture we will consider specifically functions that are periodic, and basic functions which are trigonometric. Then the series is said to be a Fourier series. A signal f(t) is said to be periodic with period T0 if f(t) = f(t + T0) for all t. Diagram on board. Note that this must be an everlasting signal. Also note that, if we know one period of the signal we can find the rest of it by periodic extension. The integral over a single period of the function is denoted by Z T0 f(t)dt. When integrating over one period of a periodic function, it does not matter when we start. Usually it is convenient to start at the beginning of a period. The building block functions that can be used to build up periodic functions are themselves periodic: we will use the set of sinusoids. If the period of f(t) is T0, let 0 = 2/T0. The frequency 0 is said to be the fundamental frequency; the fundamental frequency is related to the period of the function. Furthermore, let F0 = 1/T0. We will represent the function f(t) using the set of sinusoids i0(t) = cos(0t) = 1 i1(t) = cos(0t + 1) i2(t) = cos(20t + 2) . . . Then, f(t) = C0 + X n=1 Cn cos(n0t + n) The frequency n0 is said to be the nth harmonic of 0. Note that for each basis function associated with f(t) there are actually two parameters: the amplitude Cn and the phase n. It will often turn out to be more useful to represent the function using both sines and cosines. Note that we can write Cn cos(n0t + n) = Cn cos(n) cos(n0t) Cn sin(n)sin(n0t). ECE 3640: Lecture 4 – Fourier series: expansions of periodic functions. 2 Now let an = Cn cos n bn = Cn sin n Then Cn cos(n0t + n) = an cos(n0t) + bn sin(n0t) Then the series representation can be f(t) = C0 + X n=1 Cn cos(n0t + n) = a0 + X n=1 an cos(n0t) + bn sin(n0t) The first of these is the compact trigonometric Fourier series. The second is the trigonometric Fourier series.. To go from one to the other use C0 = a0 Cn = p a 2 n + b 2 n n = tan1 (bn/an). To complete the representation we must be able to compute the coefficients. But this is the same sort of thing we did before. If we can show that the set of functions {cos(n0t),sin(n0t)} is in fact an orthogonal set, then we can use the same.

MATLAB sessions Laboratory 3MAT 275 Laboratory 3Numeric.docx

andreecapon

MATLAB sessions: Laboratory 3 MAT 275 Laboratory 3 Numerical Solution s by Euler and Improved Euler Methods (scalar equations) In this session we look at basic numerical methods to help us understand the fundamentals of numerical approximations. Our objective is as follows. 1. Implement Euler’s method as well as an improved version to numerically solve an IVP. 2. Compare the accuracy and efficiency of the methods with methods readily available in MATLAB. 3. Apply the methods to specific problems and investigate potential pitfalls of the methods. Instructions: For your lab write-up follow the instructions of LAB 1. Euler’s Method To derive Euler’s method start from y(t0) = y0 and consider a Taylor expansion at t1 = t0 + h: y(t1) = y(t0) + y ′(t0)(t1 − t0) + . . . = y0 + hf(t0, y(t0)) + . . . = y0 + hf(t0, y0) + . . . For small enough h we get an approximation y1 for y(t1) by suppressing the . . ., namely y1 = y0 + hf(t0, y0) (L3.1) The iteration (L3.1) is repeated to obtain y2 ≃ y(t2), . . . such that yn+1 = yn + hf(tn, yn) tn+1 = tn + h Geometrically, the approximation made is equivalent to replacing the solution curve by the tangent line at (t0, y0). From the figure we have f(t0, y0) = f(t0, y(t0)) = y ′(t0) = tan θ = y1 − y0 h , from which (L3.1) follows. . ........... ........... ........... ..........s s y0 y1 y(t1) t0 t1 θ h � � � � � � As an example consider the IVP y′ = 2y = f(t, y) with y(0) = 3. Note that here f does not explicitly depend on t (the ODE is called autonomous), but does implicitly through y = y(t). To apply Euler’s method we start with the initial condition and select a step size h. Since we are constructing arrays t and y without dimensionalizing them first it is best to clear these names in case they have been used already in the same MATLAB work session. >> clear t y % no comma between t and y! type help clear for more info >> y(1)=3; t(1)=0; h=0.1; c⃝2011 Stefania Tracogna, SoMSS, ASU hamadalmershed Typewritten Text %%%%%%%%%%%% hamadalmershed Typewritten Text hamadalmershed Typewritten Text hamadalmershed Typewritten Text hamadalmershed Typewritten Text xxxxxxxxxxxxxxxxxxxx hamadalmershed Typewritten Text sssssssssssssssssssss hamadalmershed Typewritten Text hamadalmershed Typewritten Text eeeeeeeeeeeeeeeeeee hamadalmershed Typewritten Text hamadalmershed Typewritten Text wwwwwwwwwwwwwww hamadalmershed Typewritten Text IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII hamadalmershed Typewritten Text XXXXXXXXXXXXX hamadalmershed Typewritten Text MATLAB sessions: Laboratory 3 Since f is simple enough we may use the inline syntax: >> f=inline(’2*y’,’t’,’y’) f = Inline function: f(t,y) = 2*y Note that the initialization y(1)=3 should not be interpreted as “the value of y at 1 is 3”, but rather “the first value in the array y is 3”. In other words the 1 in y(1) is an index, not a time value! Unfortunately, MATLAB indices in arrays must be positive (a legacy from Fortran...). The ...

MATLAB sessions: Laboratory 4 MAT 275 Laboratory 4 MATLAB solvers for First-Order IVP In this laboratory session we will learn how to 1. Use MATLAB solvers for solving scalar IVP 2. Use MATLAB solvers for solving higher order ODEs and systems of ODES. First-Order Scalar IVP Consider the IVP { y′ = t − y, y(0) = 1. (L4.1) The exact solution is y(t) = t − 1 + 2e−t. A numerical solution can be obtained using various MATLAB solvers. The standard MATLAB ODE solver is ode45. The function ode45 implements 4/5th order Runge-Kutta method. Type help ode45 to learn more about it. Basic ode45 Usage The basic usage of ode45 requires a function (the right-hand side of the ODE), a time interval on which to solve the IVP, and an initial condition. For scalar first-order ODEs the function may often be specified using the inline MATLAB command. A complete MATLAB solution would read: 1 f = inline(’t-y’,’t’,’y’); 2 [t,y] = ode45(f,[0,3],1); 3 plot(t,y) (line numbers are not part of the commands!) • Line 1 defines the function f as a function of t and y, i.e., f(t, y) = t − y. This is the right-hand side of the ODE (L4.1). • Line 2 solves the IVP numerically using the ode45 solver. The first argument is the function f, the second one determines the time interval on which to solve the IVP in the form [initial time, final time], and the last one specifies the initial value of y. The output of ode45 consists of two arrays: an array t of discrete times at which the solution has been approximated, and an array y with the corresponding values of y. These values can be listed in the Command Window as [t,y] ans = 0 1.0000 0.0502 0.9522 0.1005 0.9093 0.1507 0.8709 0.2010 0.8369 ........ 2.9010 2.0109 2.9257 2.0330 2.9505 2.0551 2.9752 2.0773 3.0000 2.0996 c⃝2011 Stefania Tracogna, SoMSS, ASU MATLAB sessions: Laboratory 4 Since the output is quite long we printed only some selected values. For example the approximate solution at t ≃ 2.9257 is y ≃ 2.0330. Unless specific values of y are needed it is better in practice to simply plot the solution to get a sense of the behavior of the solution. • Line 3 thus plots y as a function of t in a figure window. The plot is shown in Figure L4a. 0 0.5 1 1.5 2 2.5 3 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 Figure L4a: Solution of (L4.1). Error Plot, Improving the Accuracy Error plots are commonly used to show the accuracy in the numerical solution. Here the error is the difference between the exact solution y(t) = t−1+2e−t and the numerical approximation obtained from ode45. Since this approximation is only given at specified time values (contained in the array t) we only evaluate this error at these values of t: err = t-1+2*exp(-t)-y err = 1.0e-005 * 0 0.0278 0.0407 0.0162 -0.0042 ...... -0.0329 -0.0321 -0.0313 -0.0305 -0.0298 (in practice the exact solution is unknown and this error is estimated, for example by comparing the solutions obtained by different methods). Again, since th ...

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This problem illustrates one of the possible pitfalls of blindly appl.pdf

arihantmum

This problem illustrates one of the possible pitfalls of blindly applying numerical methods without paying attention to the theoretical aspects of the differential equation itself. Consider the equation ty\' + 3y - 9t^2 = 0. Use the MATLAB program myeuler.m from Chapter 8 to compute the Euler Method approximation to the solution with initial condition y(-0.5) = 3.15, using step size h = 0.2 and n = 10 steps The program will generate a lot of ordered pairs (x_i, .y_i). Use plot to graph the piecewise linear function connecting the points (x_i, y_i). Now modify the program to implement the Improved Euler Method. Can you make sense of your answers? Next, use ode45 to find an approximate solution on the interval (-0.5, 0.5), and plot it with plot. Print out the values of the solution at the points -0.06: 0.02: 0.06. What is the interval on which the approximate solution is defined? Solve the equation explicitly and graph the solutions for the initial conditions y(0) = 0, y(-0.5) = 3.15, y(0, 5) = 3.15, y(-0.5) = -3.45, and y(0.5) = -3.45. Now explain your results in (a)-(c). Could we have known, without solving the equation, whether to expect meaningful results in parts (a) and (b)? Why? Can you explain how ode45 avoids making the same mistake? Solution clc clear all close all x=input(\'Enter the first sequence: \'); l1=input(\'Enter the lower limit: \'); u1=input(\'Enter the upper limit: \'); x1=l1:1:u1; h=input(\'Enter the second sequence: \'); l2=input(\'Enter the lower limit: \'); u2=input(\'Enter the upper limit: \'); h1=l2:1:u2; l=l1+l2; u=u1+u2; n=l:1:u; s=numel(n); i=1; for i=1:s y(i)=0; for k=1:numel(x) if (i+1-k)<=0 y(i)=y(i)+(x(k)*0); else if (i+1-k)>numel(h) y(i)=y(i)+(x(k)*0); else y(i)=y(i)+(x(k)*h(i+1-k)); k=k+1; end end end i=i+1; end disp(y); subplot(2,2,1);stem(x1,x); title(\'First sequence\');xlabel(\'n\');ylabel(\'x(n)\'); subplot(2,2,2);stem(h1,h); title(\'Second Sequence\');xlabel(\'n\');ylabel(\'h(n)\'); subplot(2,2,[3 4]);stem(n,y); title(\'Convoluted sequence\');xlabel(\'n\');ylabel(\'y(n)\'); Comment only X = input(\'Enter x: \'); %input vector X and H H = input(\'Enter h: \') ; LenX = length(X); %defining their lenghts LenH = length(H); y = zeros(1,LenX+LenH); %defing vector y of zeroes and of size % lenth of X + length of H t = zeros(1,LenH); % definign a vector t of same length as H for i = 1:LenH+LenX-1 % Running a for loop from 1 to length of Y -1 if i<=LenX % till I IS Lesser then length of X i.e overlap about to begin t(1)= X(i); % put x(i) on t(1) later it is shifted forwards in the vector t i.e. later t(2)=t(1) for j = 1:LenH % in the if condition a for loop from 1 to length of H y(i) = y(i) + H(j)*t(j); % summing for all H(j)*t(j) and putting it at y(1) or y(2) or Y(i) in first iteration % i.e. for i=1 only firt multiplication would % be non zero rest all zeroes. end for k = LenH:-1:2 % shifting old value of t(i) to t(i+1) now there would me 1+ non zeroes values in t % this cycle would continue until i is lesse.

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