Matlab Experiments Analysis - Assignment Example

MATLAB Affiliation MATLAB Experiment 1.3. Solution The first experiment employed the use of the two systems provided in the lab manual as shown below. They were incorporated into a Matlab script with the help of the ‘tf’ inbuilt transfer function. Outlined below are the System Functions calculations that were used to facilitate the input of the MATLAB code function ‘tf’. The MATLAB was then used in the calculation of the Transfer functions T(s). The resultant simplified system transfer functions are illustrated in Figure 1.3.1 below Figure1.3.1 1.4. Test Results Table 1: Experiment 1 results 1.5. Discussion & Conclusions Q1: Analyse each system using the values of Kpos, Kv, or Ka that have been provided. It is evident from Table 1 that the two systems analysed in the experiment are under the category of system type 0. This is primarily because in the step input, they both displayed a real value as an error and in the ramp and parabolic input; they displayed an infinite value error. Figure 1.1.3 illustrated the mode of determination of system types. Q2: What is the significance of the positive/negative sign of the steady-state error? The results indicated show that system (a) had a significantly larger absolute steady state error as compared to system (b). It is equally clear that system (B) displayed a greater accuracy with less absolute steady state error as compared to that of system (a). This leads to a conclusion that the positive error value of system (b) established the system output’s stability. Prior to the introduction and calculation of the error formulas, it is possible to determine the system’s performance by the use of its constant (K) value. A perfect system with no steady state error, ideally, Kp would be equal to zero, Kv would be equal to one and Ka would be equal to one. It is however extremely challenging to attain this kind of a system. If Kv or Ka would be equal to zero, then the steady state error would be infinite as illustrated in the results in the table. If Kpos would be equal to negative one, then the steady state error would become infinite. The results from the table show that the Kpos is very close to negative one. This is basically the reason as to why it has a large value of the steady state error. The sign recorded by the steady state error gives an indication of whether the output is greater than the input or vice versa. For instance, the table shows that system (a) displays a negative error value making the output of the system to be greater than the input. This designates some degree of amplification. On the other hand, system (b) displays a positive error making the output smaller than the input. 2.3. Solution Description The box diagram below is a closed loop system with proportion control. It has been simplified using the following values: Km= 19.513 Kθ=0.796 τ=0.1 The method shown above is the one used for purposes of simplifying the servo mechanism initial response. The use of simple mathematics makes the equation to appear simpler and easier to comprehend especially for people who have no background knowledge of the block diagrams. The values below have been extracted from the appendix: “Kp=2; -Setting variable Kp to 2 Km=19.513; -Setting variable Km to 19.513 Ko=0.796; -Setting variable Kp to 0.796 t=0.1; -Setting variable Kp to 0.1 var = 10*Kp*Km*Ko; -Setting variable var to multiply 10*Kp*Km*Ko num = var; -Setting variable num to var den = [1,10,var]; -Setting variable Kp to 1 10 and variable var G = tf(num,den) -Transfer function num divided by den “ Before proceeding to the second task, it was important to set the necessary values in the formula to a variable. This is for the purposes of simplifying the making of the final formula. Failure to carry out this step would lead to a very long and complicated final equation which would definitely be prone to a number of mistakes. It is equally of importance to set some values in place of the full formula. This ensures that no mistakes are realized in the code. The variables that can be used are numerator (num) and denominator (den). The last line of the code illustrated above show the letters ‘tf’ in brackets. This creates a transfer function model of any values placed immediately after the brackets. 2.4. Test Results Servomechanism with Gp = 2 Servomechanism with Gp=1 Results where Kp=2, and an input of 1V applied : Transfer function Geq(s) System type 0 1 Kpos 1 Inf Kv 0 31.0647 Ka 0 0 ess (r = u(t)) 0.5V 0 ess (r = tu(t)) Inf 0.0322V ess (r = ½ t2u(t)) Inf Inf Results where Kp = 2, and input increased to 2V : ess (r = u(t)) 1V 0.5V – No change Results where an input of 1V applied , and Kp is increased to _____V : ess (r = Ku(t)) 0.5V – No change 0V – No change Table 2: Results for experiment 2 The table above displays all the values that were given from Matlab and the ESVL system. Figure 2.1: ESVL at 1V input The image above shows the voltages from each reader obtained from the virtual mechanism and labelled on the diagram on the bottom right corner. Number (5) demonstrates the input voltage which is around 1.014V. This value is above the input value of 1. Number (4) shows the steady state error which is around 0.485V and the same is consistent with the Matlab result of 0.5V. It is therefore clear that these values are in line with the results given in Table 2 above. Figure 2.2: ESVL at 2V input The image above equally displays the same format. The voltages for each voltage reader from the virtual mechanism are shown above and labelled on the bottom right corner of the diagram. Number [5] shows the input voltage which is 2.017V. This value is close to the input value of 2V. Number [4] on the other hand shows the steady state error which is 1.019V. This is consistent with the Matlab result of 1V. 2.5. Discussion & Conclusions Experiment 2A Q1: Demonstrate your understanding of the system using the given the values of Kpos, Kv, or Ka. If the Kpos, Kv or Ka values were given, it would be possible to determine the type of the system. For instance, if the Kpos result was a constant value and not zero, then the system type would be 0. If the Kpos value was infinite and the Kv value was a constant, then the system would be type 1. If none of the criteria above were correct, the system would be a type 2. The table below demonstrates the same. Figure 3: Table showing the relationship between input and errors. It was extracted from Control systems engineering textbook online. Known values of Kpos, Kv or Ka would also give an indication of the shape of the input signal. This would be a step function, ramp function or a parabolic function. Q2: Give an explanation of any differences that were observed between the Matlab and ESVL simulated results. The main difference that was observed between Matlab and the ESVL program was that the results presented from Matlab would either be rounded off or were slightly less accurate. This is clear from the steady state error for ramp function with Gp=2 and input voltage of 1V gave a presentation of the Matlab value as 0.5V. For the same experiment using ESVL, the steady state error for ramp function with Gp=2 was 0.485V. This is shown in figure 2, 5 above. Q3: What is the effect of increasing the input signal magnitude on the steady-state error? The images in figure 2.6 show that increasing the input voltage to 2V and keeping the pre-set value for Gp being 2, there would be no steady state error. The Error is half that of the input which is 1V. This is the same for the previous voltage of 1V and the steady state error being half that of 0.5V. Q4: What is the effect of increasing gain Kp on the output signal? Three tests were carried out: kp= 2, kp = 4 and kp = 10. The voltage was set back and remained at 1V to ensure that there were no other variances. These tests indicated that no change was occurring with the output response from the ESVL program. Experiment 2B Q5: Demonstrate your understanding of the system using the values of Kpos , Kv, or Ka. As earlier mentioned, knowledge of the Kpos, Kv or Ka values would help one to know the system type. For example, if the Kpos result was a constant value and not zero, then the system type would be 0. If the Kpos value was infinite and the Kv value was a constant then this would be a type 1 system. If none of the criteria above were correct, the system would be a type 2. This is clearly shown in the table. Likewise, if the Kpos, Kv or Ka values were given, one would be better placed to identify the input signal shape as a step function, ramp function or a parabolic function. Figure 2.7 gives a perfect demonstration of this. Q6: What is the effect of increasing the input signal magnitude on the steady-state error? A look at figure 2.6 shows that increasing the input voltage to 2V and keeping the pre-set value for Gp being 2, a steady state error would not arise. The Error is half that of the input which is 1V. This is the same for the previous voltage of 1V and the steady state error being half that of 0.5V. Q7: What is the effect of increasing gain Kp on the output signal? There is no effect whatsoever on increasing the Kp value on the output signal. Q8: Summarize the effect of reducing the feedback path gain GP to 1 on the transfer function, system type and steady state error. Reducing the feedback path gain Gp from 2 to1 has a major impact on the results. A change of the feedback can easily result to a change in the whole system. Table 2 demonstrates the possible effects that can take place. The transfer function changes, the constant value in function Gp=1 falls off and this leads to an automatic change of the poles. Subsequently, the Kpos, Kv and the Ka values are all altered and the system type changes as well. For the servomechanism with Gp=2, the Kpos is 1, the Kv is 0 and the Ka is 0. Figure 2.7 shows us that the system is one of a type 0. For the servomechanism with Gp=1, the Kpos is infinite, the Kv is 31.07 and the Ka is 0. Figure 2.7 equally shows us that this is a type 1 system. It is therefore evident that if the Kpos, Kv or Ka values were given, we would know the system type, which includes type 0, type 1 or type 2. In the same breath, if the Kpos, Kv or Ka values were given, we would identify the shape of the input signal as a step function, ramp function or a parabolic function. The main difference that was noticed between Matlab and the ESVL program was that the results presented from Matlab were less accurate than ESVL. For example, the steady state error for ramp function with Gp=2 and input voltage of 1V presented a Matlab value of 0.5V. For the same experiment with the use of ESVL, the steady state error for ramp function with Gp=2 was 0.485V. This is shown in figure 2.5 above. Figure 2.6 shows that increasing the input voltage to 2V and keeping the other values constant results in a no chance in the steady error. To determine whether or not kp plays a major role in the results, a simple test was constructed which involved changing the kp to three different values and viewing the results for any inconsistencies. An alteration of the feedback gain Gp can result to a change in the whole system. Table 2 demonstrates the exact impacts that have taken place. Basically all the values are different as a result of changing the feedback gain. All K values with the exception of Ka were altered. This therefore leads to a change in the system type. On occasions whereby the gain is 1, the Kpos value is a constant. However, on occasions when the gain is 2, the Kv value is a constant. This is a clear demonstration of how the system changes. In conclusion, it is evident that the Kpos value plays no significant role in the steady state error. This is simply due to the fact that a change on it will have no impact whatsoever on the entire system. This is clear from the foregoing analysis. Prove was given that a variance in the Kpos from 2 to 10 does not change the error. It was also evident that a change in the feedback gain Gp has the potential to alter the entire system such that Kpos , Kv and the Ka values will change. This means that it will equally lead to a change in the system type. It is also clear that the use of Matlab results in accurate results. Despite the fact that they seem to be less accurate than ESVL, an alteration of the code would improve the accuracy of the Matlab. Read More