Design of System - Lab Report Example

Running Head: Design of system Design of system [Student Name] [Course Title] [Instructor Name] [Date] Part 1: Analyzing a simple system Section 1A: Mathematical Analysis The aim here is to carry out differentiation of the equation to obtain an equation that can be solved using Matlab. The differentiation of the linear constant-coefficient differential equation is as shown below The quadratic equation formed above can be used to solve various parameters relating to the system. The Lagrange method uses the fact that any relative extremum of a function subject to a constraint must occur at a critical point. The equation characteristic and eigenvalues For the second equation, it shows the relationship between mass and stiffness. That is, mas and stiffness is influenced by length . x(t)= The frequency response of this system x(t) = y(t) = S (j) one assumes that x(t )and yt are integrated processes, and there is a linear combination integrated then both variables are said to co-integrated. That is, if the variables of a dynamic linear model are co-integrated, fluctuations are considered stationary = k[h (= k ( + ma k[h (= k ( + k[h] = k ( + k [h]= (k - Co-integration can easily be tested by examining the ranks of a matrices obtained by getting their eigen-values (characteristic roots) that are not zero. When it is zero then all variables are considered non-stationary hence no co-integrating factors and If >1, there must be more than 1 co-integrating vectors. Generally Johansen’s co-integration test as shown above indicates that only linear combinations of a number co-integrating factors may be stationary and that this behavior of co-integration is expected to exist system Section 1B: Analysis using Matlab Introducing figures like mass, k, and l as follows M =0. 2, k = 1800 N/m and /0 = 0.25 m 4. the frequency response of the equation. This is done by converting frequency into time using Frequency = 1/time the scripts is as follows S= 0.1; s=0:S:1/20; c=[0.2,0.25,800]; d=[0.1, 800]; [y,z]=freqs(d,c,200); figure(1) subplot(211), plot(z, abs(y)); title('magnitude'),xlabel('frequency,rad/s'), ylabel('phase response') subplot(212), plot(z, 180*angle(y)/pi); title('Phase response'),xlabel('frequency,rad/s'), ylabel('phase response') the screenshot is as follows 5. The impulse response and step response since time must never be negative then the subsidy for the first scenario must be 0 < s < 5 . S= 0.1; s=0:T:5; c=[200,0.25,800]; d=[100, 800]; [y,z]=freqs(d,c,200); figure(1) subplot(211), plot(z, abs(y)); title('magnitude'),xlabel('frequency,rad/s'), ylabel('phase response') subplot(212), plot(z, 180*angle(y)/pi); title('Phase response'),xlabel('frequency,rad/s'), ylabel('phase response') p=roots(c); sys=ss(c,d,0); impulse(sys); step(d,c,s); subplot(211), plot(s,h) subplot(212), plot(s,y) S= 0.1; s=0:S:1/20; c=[200,0.25,800]; d=[100, 800]; [y,z]=freqs(d,c,200); figure(2) p=roots(c); sys=ss(c,d,0); impulse(sys); step(d,c,s); subplot(211), plot(s,h) subplot(212), plot(s,y) 6. Frequency response of the system shows that it is similar to values of Part 1A meaning that the setup can perform the required function . The frequency response is due to vibration present immediately after the input is applied. As time goes on, the frequency response diminishes, and the only response present is due to the forced response, or the step input. The frequency response depends on frequency, so frequency, so the amplitude depends on frequency. Part 2: Improving the system New component The new component which is a shock absorber to the simple shock absorber will lead to changing the equation. This will also require the differentiation of the new equation to accommodate the new component part of the equation. The new component will add a value of c ( and the equation will be 8. plot the frequency response using matlab c: s = 0:.1:20; tr = 2; po = 0.2; bv = tr^2; de = [1 2*po*tr tr^2]; y1 = step(bv,de,t); tr = 2; po = 0.1; bv = tr^2; de = [1 2*po*tr tr^2]; y2 = step(bv,de,t); tr = 3; po =0.1; bv = tr^2; de = [1 2*po*tr tr^2]; y3 = step(bv,de,t); figure(2) subplot(221), plot(s, y1); title('c=200') subplot(222), plot(s, y2); title('c=400') subplot(223), plot(s, y3); title('c=1800') subplot(224), plot(s, y1); title('c=0') 9. From the behavior of the plots above it shows that increasing value of C will lead to increasing of damping ratio which leads to reduction of discomfort due to reduced vibration changes. The changes of vibration relates to changes of the frequencies of the system. When decreasing the damping ratio there will be reduced uncomfortable oscillatory response leading to increased comfort. This gives the relationship of damping coefficient and the stiffness as inverse. Damping ratio affects the time‐domain of system are again important thus affecting smooth ride of a car. The road surface provides friction that needs to be reduced in order to increase conformability. The road surface is always in contact with wheels and vibration will be higher when the value of C is lower that means a lower C value will have higher uncomfortability. 10. The optimal value of c is 800 where a natural frequency and cutoff frequency match making all car users comfortable. At this point it can be noted that the parameters can be changed in damping coefficient which is in form of shock absorbers. The differences between cutoff and natural frequency did not provide a lot of difference although it can be noted that the a C with the lower range showed a slightly lower conformability. The best settings for the parameters used in this model are 5seconds of time and C values of 800. Part 3: Analyzing system response for various conditions 11. The following shows model in matlab to ‘‘high’ speed and low’ speed. In the model the assumption is that the road section is irregular. The results clearly indicated that the road conditions are much persistent in establishing the models over speed of vehicle. A consequence of such estimations relates to a time and phase that forecasts in advance about the speed. A major problem encountered modeling relates to the speed being naturally unobservable. 12. Matlab plot off different road conditions The response the plots for low, high speed and smoothing the high speed values. If a car hits kerb it will be very uncomfortable therefore the use of a shock absorber will reduce vibration of the vehicle thus leading to conformability of passengers. The plots are shown below shows the modeling where the aim is to make to improve car users conformability. The phase diagram below have an unusual feature; the slope of the curve is positive. From graph above it can be noted that stability is achieved at low speed because when the car hits road kerbs, the oscillation increases. The road kerb affects the velocity of the vehicle by reducing acceleration. This results indicate that the road kerb affects the velocity of the veicle due changes to acceleration reduction. From the results it can be noted that when the cutoff frequency is of the same value with the natural frequency then movement or velocity because lower due to effects of kerbs. The screenshots shows oscillations that are uniform when the speed is low . The main disadvantage of the model is the use of partial differentiation to estimate the values of the parameters. Partial differentiation fails to provide correct numerical approximation is for system stability. It produces magnitudes using timestep as well as step. References French, A.P, 2003. Vibrations and Waves. New York: CBS Publishers & Distributors Kleppner, D. & Kolenkon, R. 2006. An Introduction to Mechanics. New York: McGraw-Hill Morin D., 2008. Introduction to Classical Mechanics: With Problems and Solutions. London: Cambridge University Press Read More