Programme - HND Electronic Engineering

Module/Unit - Unit 35 - Further Analytical Methods J/601/1465

Level - QCF Level 5

Assignment Title - Graphical Techniques, Simulation and Iteration

Bob and Carol are researching electrical pulse waveforms used in neural stimulation of brains of Alzheimer patients. Bob says that research shows that optimal stimulation can be achieved using a waveform modeled by the function

i = 0.017e0.24t

where i is the current (amps) and t is pulse time in seconds.

(a) LO2.1 Tabulate values for the function over a suitable short time period, then illustrate the shape of the mathematical curve using techniques with which you are familiar.

(b) LO2.1, LO2.2

Carol says the current in the pulse ramps from 0.017 amps to a maximum of 0.187 amps, but wonders what the mean current might be over this period. Calculate an estimate of the mean current for her, using a technique with which you are familiar.

M1 To achieve M1 for task lb you need to demonstrate a mathematical solution with which you are familiar which produces an exact value. This will require the use of calculus over and above geometrical techniques.

D2 In part (a) you need to show independence of thought in how you illustrate the function including step by step planning and explanation demonstrating organized strategies.

Task 2 Ted and Alice are attempting to work out the flow rate of water in a river. The average velocity of the water is found to be 2.05m/sec according to a flow meter, however, they do not know the cross-section of the river. To find this, they decide to measure the depth of the river from one side of the embankment to the other using intervals of 1.5m. Below are their results. They propose to use a numerical method to determine the approximate cross-sectional area of the river and multiply this by the velocity (above) and therefore obtain the flow rate in m3/sec.

Dist (m)    0.00    1.5    3.0    4.5    6.0    7.5    9.0    10.5    12.0    13.5    15.0
Depth (m)    0.00    1.04    1,65    3.10    4.66    4.12    3.21    2.33    1.78    0.76    0.00

Ted says they can approximate the cross-sectional area of the river using the Trapezium rule. Alice suggests using Simpson rule as she says it is more accurate.

(a) Using the data in the table, use the Trapezium rule to approximate the cross section of the river, and thereby determine the approximate flow rate to 2 d.p).

(b) Repeat your calculation, but this time use Simpsons rule.

Ted believes the cross-section of the river can be modelled by the mathematical function

y = 0.036x2 (where y is the depth (m) at distance x (m))

Alice doesn't think Ted is correct and suggests using the mid-ordinate rule as a rough check to see if it calculates similar values to those in (a) and (b) above.

(d) Using the table below to help you, use a further mathematical technique with 5 strips, each of width 3m, to work out the cross-sectional area of the river based on the model y=0.036x2. Does the answer compare favourably with the approximations obtained in (a) and (b)? Explain your conclusion.

M2 In part (d) You must select and apply a further technique to determine the cross section of the river, with which you are familiar. You must show the formula used and ALL working. You should make a comprehensive comparison of the answer with those obtained earlier, and explain which solution is likely to be the closest to the true cross section.

Bob and Carol are working on a problem in fluid mechanics. From experimentation they find that they can model the velocity flow, v (m/s), of a liquid along a channel using the following mathematical relationship

v3- 6v2 - 348v = -3112

Bob is told that there should be a root of this equation between v=10 and v=11 m/s.

(a)  Re-arrange the equation and, using the interval above, find the root (to 3 decimal places) using the method of bisection.

(b)  Carol knows a quicker way of doing this using the Newton-Raphson method. Using v=10 m/s as a first approximation, show how the Newton-Raphson method can be used to obtain the solution more quickly, and use it to check you answer in (a). Once again, obtain a solution to three decimal places.

(c) Use the Newton-Raphson procedure to find a root of the following equation to two decimal places, using r1 = -2 as a first approximation:

ex - 2x - 5 = 0

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i = 0.017e0.24t

 t 0 1 2 3 4 5 6 7 8 9 10 i 0.017 0.022 0.028 0.035 0.044 0.056 0.072 0.09 0.12 0.147 0.187

A Ramp signal is obtained (shifted upwards)

D2 : We can start from t= 0 and slowly increase the value of t. But this is an increasing function, so plot the values for 10 values of 't'.

Also , ex= 1+x+x2/2! +...

I = 0.017( 1+.24t + (0.24t)2/2! +...)

So, we can see its increasing function.

(b) Mean current = 1/10∫ i(t)dt = Im

i(t)= (.187-.017/10-0)t + .017

= .017(t+1)

Im= .017/10[t2/2 + t]10 =.102

M1 : Another way to find average value = Area under curve/10

=1/2 *10(.187-.017) + .017*10

= .17/2 + .017 = .102

Both results match

Vavg = 2.05 m/s

(a)    Trapezium Rule

A = ∫ f(x) dx = h/2 [(y0 + yn )+ 2(y1 +y2 +...)]

=∫f(x)dx =1.5/2 [(0+0) +2(1.04+1.65+3.10+4.66+4.12+3.21+2.33+1.78+.76)]

(n=10, h=1.5)

=33.975 m2

Flow rate  = A*Vavg= 33.975*2.05 = 69.65 m3/s

(b)   Simpsons Rule

A= h/2 [(y0+yn) +4(y1+y3+....+yn-1) +2(y2+y4+...)]

=1.5/3[0+0+4(1.04+3.1+4.12+2.33+.76)+2(1.65+4.66+3.21+1.78)]

= .5(45.4+22.6) = 34 m2

Flow rate = A*Vavg =34*2.05 = 69.7 m3/s

(c)    The result in (a) and (b) are almost same.

(d)   A = ∫ ydx = h(y1 +y2 +...+yn)     (Mid ordinate rule)

 M 1.5 4.5 7.5 10.5 13.5 Y=.036x2 0.081 0.729 2.025 3.969 6.561

A = 3(.081+.729+2.025+3.969+6.561)

= 40.095

The value of A is very  much different from (a) and (b) parts. It means approximated mathematical function is not correct totally.

M2: Further technique can be integration.

A = ∫y dx (from 0 to 15) = ∫0.036 x2 dx

= 0.036 (x3/3) =  40.5

This result is similar to that obtained in part (d). Hence, if function is modelled properly, the results obtained by mid ordinate rule are almost correct.

(a)    v3-6v-348v+ 3112 = 0 =f(v)

Method of bisection

f(10) = 32, f(11) = -111

f(10.5) = -45.875

f(10.25) = -8.484

The function has a root v= 10.125

10.125

-----]--------]--------]--------]--------------------]---------------------------

10                    10.25   10.5                    11

(+)                      (-)         (-)                       (-)

(b)    Newton - Raphson Method

v1 = 10 m/s

vn+1= vn+ -f(vn)/f(vn)

f'(v) = 3v2-12v-348

v2=v1-f(v1)/f'(v1)

= 10- f(10)/f'(10) = 10-32/(-168)

= 10.191

Approximately (a) and (b) give similar results

(c)    Newton - Raphson Procedure

ex -2x-5 = 0 =f(x)

f'(x) = ex -2

x1 =-2

x2=x1-f(x1)/f'(x1)

= -2-f(-2)/f'(-2)

= -2- [e-2+4-5/e-2-2]

=-2.46