Skip to main content

Calculator Techniques for the Casio FX-991ES and FX-991EX Unraveled


      In solving engineering problems, one may not have the luxury of time. Most situations demand immediate results. The price of falling behind schedule is costly and demeaning to one's reputation. Therefore, every bit of precaution must be taken to expedite calculations. The following introduces methods to tackle these problems speedily using a Casio calculator FX-991ES and FX-991EX.


►For algebraic problems where you need to find the exact value of a dependent or independent variable, just use the CALC or [ES] Mode 5 functions or [EX] MENU A functions.


►For definite differentiation and integration problems, simply use the d/dx and integral operators in the COMP mode.


►For models that follow the differential equation: dP/dx=kt and models that follow a geometric function(i.e. A*B^x).

[ES]
-Simply go to Mode 3 (STAT) (5)      e^x
-For geometric functions Mode 3 (STAT) 6 A*B^x
-(Why? Because the solution to the D.E. dP/dx=kt is an exponential function e^x.
When we know the boundary conditions, we can input them for regression.)

[EX]
-Simply go to MENU 6 (STAT) (DOWN) (1)     a*e^(b*x)
-For geometric functions MENU 6 (STAT) (DOWN) (2)     a*b^x
-(Why? Because the solution to the D.E. dP/dx=kt is an exponential function e^x.
When we know the boundary conditions, we can input them for regression.)



Examples:

     I arrive at an urban area where I plan to set up cell sites that could satisfy the locale's needs for 10 years. The population is 200. 2 years later, the population has increased to 1000. I assume a basic exponential model. How many cells should I plan for if I alot a Grade of Service of 0.02 corresponding to 1000 people per cell?

[ES]

Mode 3 5

X      Y

0      200

2      1000

10 - Shift - 1 - 5 - 5 = 625000/1000 = 625

Therefore, I have to plan for 625 cells.


[EX]

MENU (6) (DOWN) (1)

X      Y

0      200

2      1000

10 - OPTN (DOWN) (4) (5) = 625000/1000 = 625

Therefore, I have to plan for 625 cells.





►For models following an arithmetic progression or linear relationship,

[ES]
-Simply go to Mode 3 (STAT)      2 A+B*X
-(Again, this is because the nature of arithmetic progression is repeated addition. With the STAT function, we can do regression given sufficient data.)

[EX]
-Simply go to MENU (6) (STAT) (2)      a+b*x
-(Again, this is because the nature of arithmetic progression is repeated addition. With the STAT function, we can do regression given sufficient data.)




Example:

     I have a 105 kohm resistor at 30 degrees Celsius in an airconditioned room. Its temperature increases to 47 degrees Celsius after I take it out on a hot summers day, thereby increasing its resistance by 700 ohms (positive temperature coefficient). What would be its resistance when I take it to the living room where there is no airconditioning at 37 degrees Celsius?

[ES]

Mode 3 2

X           Y

30      105000

47      105700

37 - Shift - 1 - 5 - 5 = 105288 ohms or 105.3 kohms.


[EX]

MENU (6) (2)

X           Y

30      105000

47      105700

37 - (OPTN) (DOWN) (4) (5) = 105288 ohms or 105.3 kohms.







►For approximations and functions where the limit is to be taken, the TABLE function can be used.


►For finding the angle between 2 vectors, we can use [ES] Mode 8 (VECTOR) or [EX] MENU (5) (VECTOR). This is a bit conspicuous but is worth mentioning here:

In our vector analysis course we remember that:
dotproduct(A,B)=abs(A)*abs(B)cos(theta)
therefore:
theta=arccos(dotproduct(A,B)/(abs(A)*abs(B)));
We simply just have to provide the values for vectors A and B.


►For Laplace Transforms, we can solve for a specific value using the integral function of the fx-991ES/EX.


Example:

Find the Laplace Transform of 3*cos(t)*t^4. (remember to put the calc. in radian mode)

the answer:
(360*s)/(s^2 + 1)^3 - (1440*s^3)/(s^2 + 1)^4 + (1152*s^5)/(s^2 + 1)^5

Let s=2.5
(true answer:  -0.1657)

-integrate from 0.1 to 5 with the function multiplied by exp(-(2.5)x)

      = -0.168
   (close enough)

Note: There is a Matlab function that will get the Laplace of any function.
Declare variables as string first. >>syms t
Find the Laplace.                        >>laplace(3*cos(t)*t^4)



►For Fourier Transforms, we can solve for a specific value using Simpson's 1/3 rule (since we can't use the integral function of the calculator) and by using the CMPLX mode.

Comments

  1. Sir, paturo naman po ng Simpson's 1/3 rule using CMPLX mode. Please po. Salamat.

    ReplyDelete
  2. www.facebook.com/CALculatorTECHniques
    para sayo yan.. hehehe
    puntahan mu page ni ENGINEER ROMEO Q. TOLENTINO
    sya mismo sasagot sayo.. he is the father of CALTECH

    ReplyDelete
  3. Sir paturo naman po ng computation for z transform sa cal.tech

    ReplyDelete
  4. Okay, try ko i-update ang post na ito.
    Sorry sa late reply.
    Justin

    ReplyDelete

Post a Comment

Popular posts from this blog

What Ifs? Sigmoid Function vs. Error Function in Machine Learning through Logistic Regression

While brushing up on some study materials in Mathematics, a familiar function piqued my interest. It was the error function, the solution to the non-elementary integral exp(-x^2) and whose complement is used in determining the conditional probability of bit error due to noise:






Or quite simply, the probability of error due to noise.




But the real point of interest was the nature of the curve of the function shown below.



Now, why be so interested in such a function? When I compared erf(x) with the sigmoid function commonly used in defining the decision boundary in machine learning algorithms, it returned a steeper slope. Then the thought came to me. What would be the differences of using the error function instead of the sigmoid function? Would the cost improve? Would the training accuracy improve?

And so my curiosity got the better of me and I played around with both of the functions to see what would happen.

Sigmoid vs. Error
First of all, replacing the sigmoid function with the error…

What are all these Nanogenerator stuff, Anyway?

Was there a time when you were introduced to the piezoelectric effect in one of your Physics classes and wondered, “If piezoelectric crystals generate voltage when subjected to vibration, can’t we harness this voltage to power our electronics?” It was a pretty interesting afterthought. What about the voltage developed from the Seebeck effect? There are a lot of naturally occurring temperature gradients in our environment such as the thermal gradient between our body, the engines we use, even our gadgets and ambient temperature. It would feel wasteful to watch all the energy from these potential sources dissipate to the empty void. Apparently, such sources would only yield power just enough for the mobile phone of an ant. But recent developments in materials science as well as improvements in power consumption of modern electronics have aroused interest anew. Thus, in 2006 the first nanogenerator emerged drawing energy through the piezoelectric and semiconductor characteristics of a …