Trigonometry and hyperbolic functions are fundamental mathematical tools in electronics engineering, especially in Kenya's rapidly developing technology sectors. These functions enable engineers to analyze alternating current (AC) circuits, signal waveforms, and electromagnetic fields. Mastery of trigonometric ratios and the unit circle is essential for understanding phase relationships, resonance, and impedance in electronic components.

3.1 Importance of trigonometry in engineering

Trigonometry is crucial in electronics engineering for analyzing waveforms, signals, and circuit behaviors. It allows precise calculation of angles and magnitudes in AC circuits, essential for designing filters, amplifiers, and oscillators used across Kenyan industries such as telecommunications and renewable energy sectors. Understanding trigonometry supports problem-solving in signal processing and control systems, improving device performance and reliability.

3.1.1 Role in AC circuit analysis

Alternating current (AC) circuits involve voltages and currents that vary sinusoidally with time. Trigonometry helps represent these quantities as sinusoidal functions, enabling calculation of phase angles between voltage and current. This is key to determining power factor and reactive power in circuits.

$$V = V_m \sin \theta$$

where \(V_m\) is the maximum voltage and \(\theta\) is the phase angle.

Worked Examples

Example 1: An AC voltage has a maximum amplitude of 170 V. Calculate the instantaneous voltage at \(\theta = 30^\circ\).

Given: \(V_m = 170\, V\), \(\theta = 30^\circ\)

$$V = V_m \sin \theta$$

$$V = 170 \times \sin 30^\circ$$

$$V = 170 \times 0.5$$

Answer: 85\, V

Example 2: In an AC circuit, the current leads the voltage by \(45^\circ\). If the voltage is \(120\, V\), find the current value when \(\theta = 60^\circ\) and the current amplitude is \(10\, A\).

Given: \(I_m = 10\, A\), voltage phase = \(60^\circ\), current leads by \(45^\circ\)

Current phase: \(60^\circ + 45^\circ = 105^\circ\)

$$I = I_m \sin 105^\circ$$

$$I = 10 \times 0.9659$$

Answer: 9.659\, A

Example 3: A voltage waveform is described as \(V = 100 \sin (100 \pi t)\). Find the voltage at \(t = 0.01\, s\).

Given: \(V_m = 100\, V\), \(\omega = 100 \pi\), \(t = 0.01\, s\)

Calculate \(\theta = \omega t = 100 \pi \times 0.01 = \pi\)

$$V = 100 \sin \pi$$

$$V = 100 \times 0 = 0$$

Answer: 0\, V

3.1.2 Application in signal processing

Trigonometric functions model signals in electronics, especially sinusoidal signals fundamental to communication systems. Engineers use these functions to analyze signal amplitude, frequency, and phase, which are critical for modulation, filtering, and noise reduction in Kenyan telecom infrastructure.

$$x(t) = A \cos (2 \pi f t + \phi)$$

where \(A\) is amplitude, \(f\) frequency, \(t\) time, and \(\phi\) phase angle.

Worked Examples

Example 1: A signal is given by \(x(t) = 5 \cos (200 \pi t + 45^\circ)\). Find the signal value at \(t = 0.005\, s\).

Given: \(A=5\), \(f = 100\, Hz\) (since \(\omega = 2 \pi f = 200 \pi\)), \(\phi = 45^\circ\), \(t = 0.005\, s\)

Calculate angle:

$$\theta = 2 \pi f t + \phi = 200 \pi \times 0.005 + 45^\circ$$

Convert \(45^\circ\) to radians: \(45^\circ = \frac{\pi}{4}\)

$$\theta = 200 \pi \times 0.005 + \frac{\pi}{4} = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$$

Calculate cosine:

$$x(t) = 5 \cos \frac{5\pi}{4} = 5 \times (-0.7071)$$

Answer: -3.5355\, V

Example 2: A signal of amplitude \(3\, V\) and frequency \(50\, Hz\) has zero phase angle. Find its value at \(t = 0.01\, s\).

Given: \(A=3\, V\), \(f=50\, Hz\), \(\phi=0\), \(t=0.01\, s\)

Calculate:

$$\theta = 2 \pi \times 50 \times 0.01 = \pi$$

$$x(t) = 3 \cos \pi = 3 \times (-1)$$

Answer: -3\, V

Example 3: Determine the time when the signal \(x(t) = 4 \cos (100 \pi t + 30^\circ)\) first reaches zero after \(t=0\).

Given: \(A=4\), \(\omega = 100 \pi\), \(\phi=30^\circ = \frac{\pi}{6}\)

Set \(x(t) = 0\):

$$\cos (100 \pi t + \frac{\pi}{6}) = 0$$

The cosine is zero at angles \(\frac{\pi}{2}, \frac{3\pi}{2}, ...\)

Set:

$$100 \pi t + \frac{\pi}{6} = \frac{\pi}{2}$$

Solve for \(t\):

$$100 \pi t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$$

$$t = \frac{\pi}{3} \div 100 \pi = \frac{1}{300} = 0.00333\, s$$

Answer: 0.00333\, s

3.1.3 Use in phasor representation

Phasors represent sinusoidal voltages and currents as complex numbers, simplifying AC circuit analysis. Trigonometric functions convert time-domain signals into phasors, which are essential for calculating impedance and power in Kenyan power systems and electronics manufacturing.

$$\tilde{V} = V_m (\cos \theta + j \sin \theta)$$

where \(j\) is the imaginary unit.

Worked Examples

Example 1: Convert a voltage \(V = 120 \sin 60^\circ\) to phasor form.

Given: \(V_m = 120\, V\), \(\theta = 60^\circ\)

Convert sine to cosine form:

$$V = V_m \sin \theta = V_m \cos (90^\circ - \theta) = 120 \cos 30^\circ$$

Phasor:

$$\tilde{V} = 120 (\cos 30^\circ + j \sin 30^\circ)$$

Calculate:

$$\cos 30^\circ = 0.8660, \quad \sin 30^\circ = 0.5$$

$$\tilde{V} = 120 (0.8660 + j 0.5) = 103.92 + j 60\, V$$

Answer: \(\tilde{V} = 103.92 + j 60\, V\)

Example 2: A current is \(I = 10 \cos 45^\circ\). Express as a phasor.

Given: \(I_m = 10\, A\), \(\theta = 45^\circ\)

$$\tilde{I} = 10 (\cos 45^\circ + j \sin 45^\circ)$$

Calculate:

$$\cos 45^\circ = \sin 45^\circ = 0.7071$$

$$\tilde{I} = 10 (0.7071 + j 0.7071) = 7.071 + j 7.071\, A$$

Answer: \(\tilde{I} = 7.071 + j 7.071\, A\)

Example 3: Find the magnitude and phase of the phasor \(\tilde{Z} = 3 + j 4\, \Omega\).

Given: \(\tilde{Z} = 3 + j 4\, \Omega\)

Magnitude:

$$|\tilde{Z}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, \Omega$$

Phase angle:

$$\theta = \tan^{-1} (\frac{4}{3}) = 53.13^\circ$$

Answer: Magnitude = \(5\ \Omega\), Phase = \(53.13^\circ\)

Chapter Summary