In Mechatronics Technology, algebraic manipulation is foundational for solving complex engineering problems involving sensors, actuators, control systems, and robotics. Mastery of indices enables efficient handling of exponential relationships common in signal processing, motor control, and system modeling. This chapter focuses on applying the laws of indices to simplify expressions and solve equations critical for mechatronics design and analysis.

1.1 Indices

1.1.1 Power zero

The power zero rule states that any non-zero number raised to the power of zero equals one. This principle is essential in simplifying expressions in mechatronics calculations, such as normalizing sensor outputs or scaling control parameters. The rule is expressed as:

$$ a^0 = 1 \quad \text{where} \quad a \neq 0 $$

Worked Examples

Example 1:
A sensor output voltage is modeled as \( V = 5^0 \). Calculate \( V \).

Given: \( a = 5 \), \( n = 0 \)$$ V = 5^0 $$$$ V = 1 $$ Answer: 1 V

Example 2:
Calculate \( (12)^0 \) for a control signal scaling factor.

Given: \( a = 12 \), \( n = 0 \)$$ (12)^0 = 1 $$ Answer: 1

Example 3:
If a digital encoder count is expressed as \( N = (x)^0 \) where \( x \neq 0 \), find \( N \).

Given: \( x \neq 0 \), \( n = 0 \)$$ N = x^0 = 1 $$ Answer: 1 count

1.1.2 Negative powers

Negative powers represent the reciprocal of the base raised to the corresponding positive power. This concept is vital in mechatronics for interpreting inverse relationships such as gain factors and decaying signals. The rule is:

$$ a^{-n} = \frac{1}{a^n} \quad \text{where} \quad a \neq 0 $$

Worked Examples

Example 1:
Calculate \( 4^{-2} \) representing an inverse squared distance sensor attenuation.

Given: \( a = 4 \), \( n = 2 \)$$ 4^{-2} = \frac{1}{4^2} $$$$ = \frac{1}{16} $$ Answer: \frac{1}{16}

Example 2:
A motor speed gain is modeled as \( G = (3)^{-3} \). Find \( G \).

Given: \( a = 3 \), \( n = 3 \)$$ G = 3^{-3} = \frac{1}{3^3} $$$$ = \frac{1}{27} $$ Answer: \frac{1}{27}

Example 3:
Determine the value of \( (0.5)^{-4} \) used in an inverse fourth power law in sensor calibration.

Given: \( a = 0.5 \), \( n = 4 \)$$ (0.5)^{-4} = \frac{1}{(0.5)^4} $$$$ = \frac{1}{0.0625} $$$$ = 16 $$ Answer: 16

1.1.3 Fractional powers

Fractional powers denote roots, with the denominator indicating the root type. This is critical in mechatronics for calculations involving square roots in signal RMS values or cube roots in volumetric flow rates. The general formula is:

$$ a^{\frac{m}{n}} = \sqrt[n]{a^m} $$

Worked Examples

Example 1:
Calculate \( 9^{\frac{1}{2}} \), the square root of 9, representing RMS voltage.

Given: \( a = 9 \), \( m = 1 \), \( n = 2 \)$$ 9^{\frac{1}{2}} = \sqrt{9^1} = \sqrt{9} $$$$ = 3 $$ Answer: 3

Example 2:
Find \( 27^{\frac{1}{3}} \), the cube root of 27, used in volume flow rate calculations.

Given: \( a = 27 \), \( m = 1 \), \( n = 3 \)$$ 27^{\frac{1}{3}} = \sqrt[3]{27^1} = \sqrt[3]{27} $$$$ = 3 $$ Answer: 3

Example 3:
Evaluate \( 16^{\frac{3}{4}} \), used in power scaling in control systems.

Given: \( a = 16 \), \( m = 3 \), \( n = 4 \)$$ 16^{\frac{3}{4}} = \sqrt[4]{16^3} $$$$ = \sqrt[4]{4096} $$ Since \( 16 = 2^4 \),$$ 16^{\frac{3}{4}} = (2^4)^{\frac{3}{4}} = 2^{4 \times \frac{3}{4}} = 2^3 = 8 $$ Answer: 8

1.1.4 Laws of indices

The laws of indices allow simplification of expressions involving powers, critical for solving equations in mechatronics control algorithms and signal processing. The main laws are:

• Multiplication: \( a^m \times a^n = a^{m+n} \)
• Division: \( \frac{a^m}{a^n} = a^{m-n} \)
• Power of a power: \( (a^m)^n = a^{mn} \)
• Power of a product: \( (ab)^n = a^n b^n \)
• Power of a quotient: \( (\frac{a}{b})^n = \frac{a^n}{b^n} \)

Worked Examples

Example 1:
Simplify \( 2^3 \times 2^4 \) representing combined sensor gains.

Given: \( a = 2 \), \( m = 3 \), \( n = 4 \)$$ 2^3 \times 2^4 = 2^{3+4} = 2^7 $$$$ = 128 $$ Answer: 128

Example 2:
Simplify \( \frac{5^6}{5^2} \) used in signal attenuation calculations.

Given: \( a = 5 \), \( m = 6 \), \( n = 2 \)$$ \frac{5^6}{5^2} = 5^{6-2} = 5^4 $$$$ = 625 $$ Answer: 625

Example 3:
Calculate \( (3^2)^4 \) representing repeated power in control loops.

Given: \( a = 3 \), \( m = 2 \), \( n = 4 \)$$ (3^2)^4 = 3^{2 \times 4} = 3^8 $$$$ = 6561 $$ Answer: 6561

Example 4:
Simplify \( (2 \times 5)^3 \) used in combined actuator response.

Given: \( a = 2 \), \( b = 5 \), \( n = 3 \)$$ (2 \times 5)^3 = 2^3 \times 5^3 $$$$ = 8 \times 125 $$$$ = 1000 $$ Answer: 1000

Example 5:
Simplify \( (\frac{4}{2})^2 \) representing ratio squared in system calibration.

Given: \( a = 4 \), \( b = 2 \), \( n = 2 \)$$ (\frac{4}{2})^2 = \frac{4^2}{2^2} $$$$ = \frac{16}{4} $$$$ = 4 $$ Answer: 4

Chapter Summary

This chapter covered the application of algebra beginning with indices, explaining zero, negative, and fractional powers, as well as the fundamental laws of indices involving addition, subtraction, division, and multiplication of powers. The order of operations was clarified through BODMAS, ensuring correct calculation sequences. Roots were examined, including square roots, cube roots, and nth roots, providing a foundation for understanding radical expressions. Logarithms were introduced with their essential laws of product, quotient, and power, which simplify complex multiplicative and exponential relationships. The effective use of scientific calculators was detailed, including powering on and off, selecting modes such as degree and radian, and using functions like clear, save, and shift. Methods for solving simultaneous equations with up to three variables were explored, including elimination, substitution, reduction, and graphical approaches. Finally, quadratic equations were addressed through factorization, the quadratic formula, completing the square, and graphical solutions, equipping learners with multiple strategies for solving polynomial equations.