Engineering mathematics is fundamental in mechatronics technology, particularly when dealing with signals, control systems, and circuit analysis. Complex numbers simplify the representation of oscillations, impedances, and system behaviors in electrical and mechanical domains. Mastery of indices is essential for manipulating powers and roots of complex numbers efficiently in design and analysis tasks.

1.1 Indices

Indices (or exponents) express repeated multiplication of a number by itself and are crucial when working with powers of complex numbers. In mechatronics, indices appear in calculations involving exponential signals, frequency response, and stability analysis.

1.1.1 Power zero

Any nonzero number raised to the zero power equals one. This property is used to simplify expressions in circuit and signal calculations.

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

Worked Examples

Example 1: Simplify \(5^0\) in a sensor calibration formula.
Given: Base \(a = 5\)$$a^0 = 1$$$$5^0 = 1$$ Answer: 1

Example 2: Calculate \((j)^0\), where \(j\) is the imaginary unit.
Given: Base \(a = j\)$$a^0 = 1$$$$j^0 = 1$$ Answer: 1

Example 3: Evaluate \((\frac{3}{2})^0\) for a gain factor.
Given: Base \(a = \frac{3}{2}\)$$a^0 = 1$$$$(\frac{3}{2})^0 = 1$$ Answer: 1

1.1.2 Negative powers

Negative powers represent the reciprocal of the base raised to the corresponding positive power. This is vital when dealing with inverse relationships in transfer functions and impedance calculations.

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

Worked Examples

Example 1: Compute \(2^{-3}\) for a decay rate calculation.
Given: \(a=2, n=3\)$$a^{-n} = \frac{1}{a^n}$$$$2^{-3} = \frac{1}{2^3}$$$$= \frac{1}{8}$$ Answer: \frac{1}{8}

Example 2: Find \((j)^{-2}\) for an electrical impedance expression.
Given: \(a=j, n=2\)$$a^{-n} = \frac{1}{a^n}$$$$(j)^{-2} = \frac{1}{j^2}$$ Since \(j^2 = -1\),$$(j)^{-2} = \frac{1}{-1} = -1$$ Answer: -1

Example 3: Evaluate \((\frac{4}{5})^{-1}\) in a ratio conversion.
Given: \(a=\frac{4}{5}, n=1\)$$a^{-1} = \frac{1}{a}$$$$(\frac{4}{5})^{-1} = \frac{1}{\frac{4}{5}} = \frac{5}{4}$$ Answer: \frac{5}{4}

1.1.3 Fractional powers

Fractional powers denote roots. The exponent \(\frac{m}{n}\) means the \(n\)th root of the base raised to the \(m\)th power. This is used in signal processing and control systems for root locus and stability margin calculations.

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

Worked Examples

Example 1: Calculate \(9^{\frac{1}{2}}\) to find a magnitude in a sensor reading.
Given: \(a=9, m=1, n=2\)$$a^{\frac{1}{2}} = \sqrt{a}$$$$9^{\frac{1}{2}} = \sqrt{9}$$$$= 3$$ Answer: 3

Example 2: Evaluate \(27^{\frac{2}{3}}\) for a power conversion.
Given: \(a=27, m=2, n=3\)$$a^{\frac{2}{3}} = (\sqrt[3]{a})^2$$$$= (\sqrt[3]{27})^2$$$$= 3^2$$$$= 9$$ Answer: 9

Example 3: Find \((16j)^{\frac{3}{4}}\) for a complex gain calculation.
Given: \(a=16j, m=3, n=4\)
First, find the magnitude and argument of \(16j\):
Magnitude: \(|16j| = 16\), Argument: \(\frac{\pi}{2}\) radians$$a^{\frac{3}{4}} = r^{\frac{3}{4}} (\cos(\frac{3}{4}\theta) + j\sin(\frac{3}{4}\theta))$$ Calculate \(r^{\frac{3}{4}} = 16^{\frac{3}{4}}\):$$16^{\frac{3}{4}} = (16^{\frac{1}{4}})^3 = (2)^3 = 8$$ Calculate angle:$$\frac{3}{4} \times \frac{\pi}{2} = \frac{3\pi}{8}$$$$a^{\frac{3}{4}} = 8 (\cos\frac{3\pi}{8} + j \sin\frac{3\pi}{8})$$ Numerical values:$$\cos\frac{3\pi}{8} \approx 0.383, \quad \sin\frac{3\pi}{8} \approx 0.924$$$$= 8(0.383 + j0.924) = 3.064 + j7.392$$ Answer: 3.064 + j7.392

1.1.4 Laws of indices

The laws of indices allow the simplification and manipulation of powers in expressions, essential for solving equations and analyzing systems in mechatronics.

Key laws:

• \(a^m \times a^n = a^{m+n}\)
• \(\frac{a^m}{a^n} = a^{m-n}\)
• \((a^m)^n = a^{mn}\)
• \((ab)^n = a^n b^n\)
• \((\frac{a}{b})^n = \frac{a^n}{b^n}\)

Worked Examples

Example 1: Simplify \(3^4 \times 3^{-2}\) for a signal amplitude calculation.
Given: \(a=3, m=4, n=-2\)$$a^m \times a^n = a^{m+n}$$$$3^4 \times 3^{-2} = 3^{4 + (-2)} = 3^2 = 9$$ Answer: 9

Example 2: Evaluate \(\frac{(5j)^3}{(5j)^1}\) for an impedance ratio.
Given: \(a=5j, m=3, n=1\)$$\frac{a^m}{a^n} = a^{m-n}$$$$\frac{(5j)^3}{(5j)^1} = (5j)^{3-1} = (5j)^2$$ Calculate \((5j)^2\):$$(5j)^2 = 5^2 \times j^2 = 25 \times (-1) = -25$$ Answer: -25

Example 3: Find \(((2j)^2)^3\) for a power amplification factor.
Given: \(a=2j, m=2, n=3\)$$(a^m)^n = a^{mn}$$$$((2j)^2)^3 = (2j)^{2 \times 3} = (2j)^6$$ Calculate \((2j)^6\):$$(2j)^6 = 2^6 \times j^6 = 64 \times j^6$$ Since \(j^2 = -1\), \(j^6 = (j^2)^3 = (-1)^3 = -1\)$$= 64 \times (-1) = -64$$ Answer: -64

Example 4: Simplify \((3 \times 4j)^2\) for a combined system response.
Given: \(a=3, b=4j, n=2\)$$(ab)^n = a^n b^n$$$$(3 \times 4j)^2 = 3^2 \times (4j)^2 = 9 \times (16 \times j^2) = 9 \times 16 \times (-1) = -144$$ Answer: -144

Example 5: Evaluate \((\frac{6j}{2})^3\) for a normalized output.
Given: \(a=6j, b=2, n=3\)$$(\frac{a}{b})^n = \frac{a^n}{b^n}$$$$(\frac{6j}{2})^3 = \frac{(6j)^3}{2^3} = \frac{216 j^3}{8}$$ Calculate \(j^3 = j^2 \times j = (-1) \times j = -j\)$$= \frac{216 \times (-j)}{8} = -27j$$ Answer: -27j

Chapter Summary

This chapter covered the fundamental concepts of indices, including the rules for powers of zero, negative powers, fractional powers, and the laws governing addition, subtraction, division, and multiplication of indices. The importance of the BODMAS rule was emphasized to correctly solve expressions involving multiple operations. Various types of roots such as square roots, cube roots, and nth roots were explored to understand their calculation and applications. The chapter also explained logarithms and their laws, including the product, quotient, and power laws, which simplify complex calculations. The use of scientific calculators was detailed, highlighting essential functions such as power on/off, mode settings for degrees, radians, and gradients, as well as clear, save, and shift operations. Methods for solving simultaneous equations with up to three variables were discussed, including elimination, substitution, reduction, and graphical approaches. Finally, techniques for solving quadratic equations through factorization, the quadratic formula, completing the square, and graphical methods were presented to provide a comprehensive understanding of solving polynomial equations.