2.1 College & Advanced Algebra

2.1 College & Advanced Algebra

Algebraic concepts form the foundational toolkit for electronics engineers. In the PRC Electronics Engineering Licensure Examination, algebraic methods are heavily tested both as standalone mathematical problems and as embedded steps within circuit analysis, signal processing, and control systems.

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1. Quadratic Equations & Roots Analysis

A quadratic equation in standard polynomial form is expressed as:

$$ax^2 + bx + c = 0 \quad (a \neq 0)$$

Its solutions (roots) are given by the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The Discriminant and Damping Behavior

The expression under the radical, $\Delta = b^2 - 4ac$, is the discriminant. In electronics engineering, the discriminant directly determines the transient response regime of second-order dynamic systems (such as series or parallel RLC circuits):

Vieta's Formulas

For a quadratic equation with roots $x_1$ and $x_2$:

• Sum of roots: $x_1 + x_2 = -\frac{b}{a}$
• Product of roots: $x_1 x_2 = \frac{c}{a}$

Worked ECE Board Example 1: RLC Characteristic Equation

Problem: A series RLC circuit has an inductor $L = 1\text{ mH}$ and a capacitor $C = 0.1\text{ }\mu\text{F}$. The characteristic equation governing the circuit's transient current is given by $s^2 + \left(\frac{R}{L}\right)s + \frac{1}{LC} = 0$.

1. Find the value of resistance $R$ that causes the system to be critically damped.
2. Determine the roots of the characteristic equation when $R = 100\ \Omega$.

Solution:

• Step 1: For critical damping, set the discriminant $\Delta = b^2 - 4ac = 0$, where $a = 1$, $b = \frac{R}{L}$, and $c = \frac{1}{LC}$. $$\left(\frac{R}{L}\right)^2 - 4(1)\left(\frac{1}{LC}\right) = 0 \implies R^2 = \frac{4L^2}{LC} = \frac{4L}{C}$$ $$R = 2 \sqrt{\frac{L}{C}} = 2 \sqrt{\frac{1 \times 10^{-3}\text{ H}}{0.1 \times 10^{-6}\text{ F}}} = 2 \sqrt{10000} = 2(100) = 200\ \Omega$$
• Step 2: When $R = 100\ \Omega$, $b = \frac{100}{1 \times 10^{-3}} = 10^5$, $c = \frac{1}{(1 \times 10^{-3})(0.1 \times 10^{-6})} = 10^{10}$. $$\Delta = (10^5)^2 - 4(1)(10^{10}) = 10^{10} - 4 \times 10^{10} = -3 \times 10^{10} < 0$$ $$s = \frac{-10^5 \pm \sqrt{-3 \times 10^{10}}}{2} = -50,000 \pm j 86,602.54\text{ rad/s}$$ The circuit is underdamped and oscillatory.

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2. Logarithms, Exponentials & Decibel Systems

Logarithmic functions are inverse operations of exponential functions: $\log_b(x) = y \iff b^y = x$.

Core Properties of Logarithms

1. Product Rule: $\log_b(xy) = \log_b(x) + \log_b(y)$
2. Quotient Rule: $\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)$
3. Power Rule: $\log_b(x^k) = k \log_b(x)$
4. Change of Base: $\log_b(x) = \frac{\ln(x)}{\ln(b)} = \frac{\log_{10}(x)}{\log_{10}(b)}$

The Decibel (dB) System in Telecommunications

The decibel is a logarithmic unit used to express power and voltage ratios in communication systems:

• Power Gain/Loss (in dB): $G_{\text{dB}} = 10 \log_{10}\left(\frac{P_{\text{out}}}{P_{\text{in}}}\right)$
• Voltage Gain/Loss (in dB): $A_{v,\text{dB}} = 20 \log_{10}\left(\frac{V_{\text{out}}}{V_{\text{in}}}\right)$ (assuming equal input/output impedance)
• Absolute Power in dBm: $P_{\text{dBm}} = 10 \log_{10}\left(\frac{P}{1\text{ mW}}\right)$

Worked ECE Board Example 2: Cascaded Amplifier Gain & Power Calculation

Problem: A communication receiver consists of a low-noise pre-amplifier with a gain of $+18\text{ dB}$, a coaxial transmission line with an attenuation (loss) of $-4\text{ dB}$, and a main amplifier stage with a gain of $+22\text{ dB}$. If the input signal power to the pre-amplifier is $2\text{ }\mu\text{W}$:

1. Calculate the total system power gain in dB.
2. Determine the final output power in milliwatts (mW) and dBm.

Solution:

• Step 1: For cascaded systems, decibel gains and losses add algebraically: $$G_{\text{total, dB}} = +18\text{ dB} - 4\text{ dB} + 22\text{ dB} = +36\text{ dB}$$
• Step 2: Convert input power $P_{\text{in}} = 2\text{ }\mu\text{W} = 0.002\text{ mW}$ into dBm: $$P_{\text{in, dBm}} = 10 \log_{10}\left(\frac{0.002\text{ mW}}{1\text{ mW}}\right) = 10 \log_{10}(2 \times 10^{-3}) = 10(-2.69897) = -26.99\text{ dBm}$$
• Step 3: Compute output power in dBm and mW: $$P_{\text{out, dBm}} = P_{\text{in, dBm}} + G_{\text{total, dB}} = -26.99\text{ dBm} + 36\text{ dB} = +9.01\text{ dBm}$$ $$P_{\text{out}} = 1\text{ mW} \times 10^{\frac{9.01}{10}} = 1\text{ mW} \times 10^{0.901} = 7.96\text{ mW}$$

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3. Partial Fraction Decomposition (PFD)

Partial Fraction Decomposition expands complex rational algebraic expressions $\frac{P(s)}{Q(s)}$ into simpler terms. In ECE, PFD is essential for taking the Inverse Laplace Transform to find time-domain responses $h(t)$ from s-domain transfer functions $H(s)$.

Standard Factor Expansions

1. Distinct Linear Factors: $\frac{P(s)}{(s-a)(s-b)} = \frac{A}{s-a} + \frac{B}{s-b}$
2. Repeated Linear Factors: $\frac{P(s)}{(s-a)^2} = \frac{A}{s-a} + \frac{B}{(s-a)^2}$
3. Irreducible Quadratic Factors: $\frac{P(s)}{(s-a)(s^2 + bs + c)} = \frac{A}{s-a} + \frac{Bs + C}{s^2 + bs + c}$

Worked ECE Board Example 3: Laplace Transfer Function Decomposition

Problem: Decompose $H(s) = \frac{7s + 18}{s^2 + 5s + 6}$ using partial fractions to obtain the impulse response $h(t) = \mathcal{L}^{-1}\{H(s)\}$.

Solution:

• Step 1: Factor denominator: $s^2 + 5s + 6 = (s + 2)(s + 3)$.
• Step 2: Set up decomposition: $\frac{7s + 18}{(s + 2)(s + 3)} = \frac{A}{s + 2} + \frac{B}{s + 3}$.
• Step 3: Use Heaviside Cover-Up Method: $$A = \left. \frac{7s + 18}{s + 3} \right|_{s = -2} = \frac{7(-2) + 18}{-2 + 3} = \frac{4}{1} = 4$$ $$B = \left. \frac{7s + 18}{s + 2} \right|_{s = -3} = \frac{7(-3) + 18}{-3 + 2} = \frac{-3}{-1} = 3$$
• Step 4: Time-domain impulse response: $$H(s) = \frac{4}{s+2} + \frac{3}{s+3} \implies h(t) = (4e^{-2t} + 3e^{-3t})u(t)$$

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4. Arithmetic & Geometric Progressions

Arithmetic Progressions (AP)

An AP is a sequence where consecutive terms differ by a constant $d$: $a_n = a_1 + (n-1)d$.

• Sum of first $n$ terms: $S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}\left[2a_1 + (n-1)d\right]$

Geometric Progressions (GP)

A GP is a sequence where each term after the first is found by multiplying the previous term by a common ratio $r$: $a_n = a_1 r^{n-1}$.

• Sum of finite $n$ terms: $S_n = \frac{a_1(1 - r^n)}{1 - r} \quad (r \neq 1)$
• Sum of an infinite geometric series ($|r| < 1$): $S_\infty = \frac{a_1}{1 - r}$

Worked ECE Board Example 4: Cascaded Signal Attenuation Series

Problem: An optical fiber network link consists of an infinite series of repetitive passive optical splitters. The first splitter dissipates $50\text{ W}$ of optical power ($a_1 = 50\text{ W}$). Each subsequent splitter dissipates $15\%$ less power than the preceding one ($r = 0.85$). Compute:

1. The power dissipated by the 6th splitter.
2. The theoretical total power dissipated by the entire infinite sequence of splitters.

Solution:

• Step 1: Power dissipated by 6th splitter ($n = 6$): $$a_6 = a_1 r^{6-1} = 50 \times (0.85)^5 = 50 \times 0.4437 = 22.19\text{ W}$$
• Step 2: Total power dissipated by infinite splitters ($S_\infty$): $$S_\infty = \frac{a_1}{1 - r} = \frac{50}{1 - 0.85} = \frac{50}{0.15} = 333.33\text{ W}$$

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5. Summary of Formulae & Board Exam Tips