1、公差累積分析1第一頁，共16頁。1. Worst Case Analysis Worst case is used to determine what happens when all the parameters are at their limits Set every parameter to its spec limit Measure the response This is the same as adding tolerances If all the parameters represent 3 sigma limits then The number of parts bad

2、 for one parameter is 0.0026 Four parameters (0.0026) 4 = 0.00000000000286 (3 in a trillion) For 20 parameters we get 2 out of 10 58 bad! Analysis should limit probabilities to 6 sigma This is about 3 parameters all at their limits2第二頁，共16頁。Tolerance Stack UpWhat is the tolerance stack up from the t

3、op to bottom?Will they line up?3第三頁，共16頁。Worst Case Analysis Tough!Shaft 1 +-0.002Hole 1 +.003, -.001Shaft 2 +-0.002Hole 2 +-.003 -.001Shaft 3 +-.002Hole 3 +-.003 -.001Total Worst case is .002+.003+.002+.003+.002+.003=+0.015, -.009The SHAFT WONT FIT IN THELAST HOLE UNLESS ITS .009SMALLER!4第四頁，共16頁。W

4、orst CaseFor the first shaft and hole the shaft can be +-.002 and the hole +.003, -.001.If the hole is small then -.001 and the shaft is large then +.002. The difference between the two must be 0.003 to make sure it fits.Add another set and the worst case says the second hole and shaft must be 0.006

5、 different in sizeA third set would have to 0.009 etc. The problem with this is it makes for a too conservative design when you stack up tolerancesPerfectWorst case5第五頁，共16頁。2. Sensitivity Analysisa) Monte Carlo SimulationSimulate variation in the resultsb) Taylor Series ExpansionMathematical evalua

6、tion of variationsc) Design of ExperimentsCareful experimentation (testing) approach6第六頁，共16頁。2a) Monte Carlo Analysis Monte Carlo is the name of a gambling city on the Mediterranean Sea. With MC analysis you “roll the dice” and estimate the outcome many, many times If you randomize the result often

7、 enough you get a good idea of the outcome of a system You need to know the following An equation for what you are simulating The distribution of the parameters so they can vary over time7第七頁，共16頁。Circuit ExampleIVRfL222() Mean Std Dev V = voltage 100 5 r = resistance 10 1 f = capacitance 50 5 L= in

8、ductance .004 .0008The equation for the current through the circuit is shown.The voltage of the power supply is usually at 100 Volts but it varies +-5 Volts 1 sigma1 Sigma means 68% of the time, 2 Sigma is 95% and 3 Sigma is 99%, so the voltage could be 100-3*5= 85 Volts some times.If the package sa

9、ys a 1% tolerance on 10 Ohm resistors, that means the resistors can be between 9 or 11 ohms 99% of the time (or 3 sigma).8第八頁，共16頁。Monte Carlo AnalysisCircuit Example Mean Std Dev V = voltage 100 5 r = resistance 10 1 f = capacitance 50 5 L= inductance .004 .0008(1) V = 5 * Z + 100(2) R = 1 * Z + 10

10、(3) f = 5 * Z + 50(4) L = 0.0008 * Z + .004(5) Calculate I and plot Z is a random number(6) Repeat 1000s of times Normally distributedIVRfL222()9第九頁，共16頁。2b) Taylor Series Expansion The Taylor Series calculates the system variance based on the derivatives of the function with respect to each paramet

11、er Give the equation g(x1, x2, x3) thenggxxgxxgxx211222233210第十頁，共16頁。Sensitivity Analysis The sensitivity of the output to changes in the input can be calculated from the derivatives. The sensitivity of I to changes in V This means that when V increases from its mean by 1 sigma, I increases by 0.5I

12、VV 05 .11第十一頁，共16頁。Taylor Series ExampleIVRfL222()IVRfL22122()IRVRfLR 05222232.()IfVRfLL 05242232.()ILVRfLf 05242232.()12第十二頁，共16頁。Example Evaluate the expression for I using the mean and standard deviations IV 01 .IR 10 .If 00005.IL 313 .I212506 .I110.IVRfLIVIRIfIL2222213第十三頁，共16頁。Root Sum Squared

13、(RSS) Root Sum Squared is simply the Taylor Series expansion for a linear systemg = x1 + x2 + x3 This only work for linear system (mechanical tolerances) This is VERY popular in industry!2322212321xxxgxgxgxg2322212xxxg14第十四頁，共16頁。Revisiting Tolerance Stack upShaft 1 +-0.002 implies that all the unit

14、s vary by this much. So 3 Sigma is 0.002 and Sigma = 0.00067Hole 1 +.003, -.001 implies the 3 Sigma is .001/3 = .00033RSS for 2 parts is SQRT (0.000672 + .000332 ) = 0.00075 for 1 Sigma of the system. 3 Sigma of the system is then .00225 not 0.003.For 6 parts in the system you get SQRT (0.000672 + .

15、000332 + 0.000672 + .000332 + 0.000672 + .000332 ) = .00129 for 1 Sigma. So 3 Sigma is 0.004 not 0.009 shown by worst case.15第十五頁，共16頁。2c) Design Of Experiments DOE is a statistical tool that allows the most information to be extracted from an experiment Normal experiments hold every variable constant and change one. In the previous example hold all fixe