2.1 College & Advanced Algebra

PRC Electronics Engineering Licensure Examination Study Guide Exam

Track your progress as you read

Finished sections get a check mark.

1 Chapter 1: Introduction & Board Exam Strategy

1.1 PRC ECE Licensure Exam Overview & Regulation 1.2 Table of Specifications & Exam Strategy

2 Chapter 2: Mathematics Foundations

2.1 College & Advanced Algebra 2.2 Trigonometry & Solid Mensuration 2.3 Analytic Geometry & Differential Calculus 2.4 Integral Calculus & Applications

3 Chapter 3: Advanced Mathematics for ECE

3.1 Differential Equations & Circuit Transients 3.2 Complex Numbers & Laplace Transforms 3.3 Fourier Series, Fourier Transforms & Z-Transforms 3.4 Vector Analysis, Matrices & Probability

4 Chapter 4: GEAS - Engineering Sciences & Economics

4.1 Engineering Physics & Materials Chemistry 4.2 Engineering Mechanics & Strength of Materials 4.3 Engineering Economics & Financial Analysis 4.4 Computer Fundamentals & C/C++ Programming

5 Chapter 5: GEAS - Management, Materials, Environment & CAD

5.1 Engineering Management & Technopreneurship 101 5.2 Materials Science & Engineering 5.3 Environment Science & Engineering 5.4 Computer-Aided Drawing (CAD) & Engineering Graphics

6 Chapter 6: GEAS - ECE Laws, Ethics & Contracts

6.1 Republic Act No. 9292 (Electronics Engineering Law of 2004) 6.2 ECE Code of Ethics & Professional Conduct 6.3 Contracts, Specifications & National Building Code (PD 1096)

7 Chapter 7: Electronics Engineering - Circuits & Semiconductor Diodes

7.1 DC & AC Circuit Analysis & Network Theorems 7.2 Electromagnetics & Passive Components 7.3 Semiconductor Physics & Diode Circuits 7.4 Power Supply Design & Voltage Regulators

8 Chapter 8: Electronics Engineering - Transistors & Analog Circuits

8.1 BJT Operation, Biasing & Small-Signal Amplifiers 8.2 FET Characteristics, Biasing & Amplifier Circuits 8.3 Multistage Amplifiers, Frequency Response & Power Amplifiers 8.4 Operational Amplifiers & Active Filter Design

9 Chapter 9: Electronics Engineering - Digital Electronics & Control Systems

9.1 Feedback Topologies & Oscillator Circuits 9.2 Boolean Algebra, Logic Gates & Combinational Circuits 9.3 Sequential Logic, Flip-Flops & Microprocessor Architecture 9.4 Industrial Power Electronics & Control System Theory

10 Chapter 10: EST - Communications Fundamentals & Transmission Lines

10.1 Signal Fundamentals, Decibel Calculations & Noise Analysis 10.2 Amplitude Modulation (AM) Principles & Receivers 10.3 Frequency Modulation (FM), Phase Modulation & Discriminators 10.4 Transmission Line Parameters, SWR & Smith Chart

11 Chapter 11: EST - Digital Communications, Wireless & Optical Systems

11.1 Pulse & Digital Modulation Systems 11.2 Data Communications, OSI Model & Protocols 11.3 Antenna Principles & Electromagnetic Wave Propagation 11.4 Optical Fiber, Microwave, Satellite & Philippine NTC Regulations

2.1 College & Advanced Algebra

Key Takeaways Explain This Topic AI

Master quadratic equations, discriminant analysis, Vieta's formulas, and their application to RLC circuit characteristic equations.

Apply logarithmic and exponential laws to decibel (dB) power/voltage gain calculations and RC/RL transient response equations.

Perform partial fraction decomposition for linear, repeated, and quadratic factors to solve inverse Laplace transforms in ECE control systems.

Utilize arithmetic and geometric progressions to model cascaded amplifier stage gains, signal attenuation, and infinite series convergence.

Last updated: July 2026

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.

1. Quadratic Equations & Roots Analysis

A quadratic equation in standard polynomial form is expressed as: a x 2 + b x + c = 0 ( a ≠ 0 ) ax^2 + bx + c = 0 \quad (a \neq 0) a x 2 + b x + c = 0 ( a  = 0 )

Its solutions (roots) are given by the quadratic formula: x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c ​ ​

The Discriminant and Damping Behavior

The expression under the radical, Δ = b 2 − 4 a c \Delta = b^2 - 4ac Δ = b 2 − 4 a c , 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):

• Discriminant Value Mathematical Nature of Roots Physical ECE System Damping State
• Δ > 0 \Delta > 0 Δ > 0 Two distinct real roots Overdamped System : Slow exponential decay without oscillation
• Δ = 0 \Delta = 0 Δ = 0 Real repeated (double) root Critically Damped System : Fastest decay without overshoot

Δ < 0 \Delta < 0 Δ < 0 Complex conjugate roots ( x = α ± j β x = \alpha \pm j\beta x = α ± j β ) Underdamped System : Oscillatory decay at damped natural frequency

Vieta's Formulas

For a quadratic equation with roots x 1 x_1 x 1 ​ and x 2 x_2 x 2 ​ :

• Sum of roots : x 1 + x 2 = − b a x_1 + x_2 = -\frac{b}{a} x 1 ​ + x 2 ​ = − a b ​
• Product of roots : x 1 x 2 = c a x_1 x_2 = \frac{c}{a} x 1 ​ x 2 ​ = a c ​
• Worked ECE Board Example 1: RLC Characteristic Equation

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

Find the value of resistance R R R that causes the system to be critically damped .

Determine the roots of the characteristic equation when R = 100 Ω R = 100\ \Omega R = 100 Ω .

Solution:

Step 1: For critical damping, set the discriminant Δ = b 2 − 4 a c = 0 \Delta = b^2 - 4ac = 0 Δ = b 2 − 4 a c = 0 , where a = 1 a = 1 a = 1 , b = R L b = \frac{R}{L} b = L R ​ , and c = 1 L C c = \frac{1}{LC} c = L C 1 ​ . ( R L ) 2 − 4 ( 1 ) ( 1 L C ) = 0 ⟹ R 2 = 4 L 2 L C = 4 L C \left(\frac{R}{L}\right)^2 - 4(1)\left(\frac{1}{LC}\right) = 0 \implies R^2 = \frac{4L^2}{LC} = \frac{4L}{C} ( L R ​ ) 2 − 4 ( 1 ) ( L C 1 ​ ) = 0 ⟹ R 2 = L C 4 L 2 ​ = C 4 L ​ R = 2 L C = 2 1 × 10 − 3 H 0.1 × 10 − 6 F = 2 10000 = 2 ( 100 ) = 200 Ω 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 R = 2 C L ​ ​ = 2 0.1 × 1 0 − 6 F 1 × 1 0 − 3 H ​ ​ = 2 10000 ​ = 2 ( 100 ) = 200 Ω

Step 2: When R = 100 Ω R = 100\ \Omega R = 100 Ω , b = 100 1 × 10 − 3 = 10 5 b = \frac{100}{1 \times 10^{-3}} = 10^5 b = 1 × 1 0 − 3 100 ​ = 1 0 5 , c = 1 ( 1 × 10 − 3 ) ( 0.1 × 10 − 6 ) = 10 10 c = \frac{1}{(1 \times 10^{-3})(0.1 \times 10^{-6})} = 10^{10} c = ( 1 × 1 0 − 3 ) ( 0.1 × 1 0 − 6 ) 1 ​ = 1 0 10 . Δ = ( 10 5 ) 2 − 4 ( 1 ) ( 10 10 ) = 10 10 − 4 × 10 10 = − 3 × 10 10 < 0 \Delta = (10^5)^2 - 4(1)(10^{10}) = 10^{10} - 4 \times 10^{10} = -3 \times 10^{10} < 0 Δ = ( 1 0 5 ) 2 − 4 ( 1 ) ( 1 0 10 ) = 1 0 10 − 4 × 1 0 10 = − 3 × 1 0 10 < 0 s = − 10 5 ± − 3 × 10 10 2 = − 50 , 000 ± j 86 , 602.54 rad/s s = \frac{-10^5 \pm \sqrt{-3 \times 10^{10}}}{2} = -50,000 \pm j 86,602.54\text{ rad/s} s = 2 − 1 0 5 ± − 3 × 1 0 10 ​ ​ = − 50 , 000 ± j 86 , 602.54 rad/s The circuit is underdamped and oscillatory.

2. Logarithms, Exponentials & Decibel Systems

Logarithmic functions are inverse operations of exponential functions: log ⁡ b ( x ) = y ⟺ b y = x \log_b(x) = y \iff b^y = x lo g b ​ ( x ) = y ⟺ b y = x .

Core Properties of Logarithms

Product Rule : log ⁡ b ( x y ) = log ⁡ b ( x ) + log ⁡ b ( y ) \log_b(xy) = \log_b(x) + \log_b(y) lo g b ​ ( x y ) = lo g b ​ ( x ) + lo g b ​ ( y )

Quotient Rule : log ⁡ b ( x y ) = log ⁡ b ( x ) − log ⁡ b ( y ) \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) lo g b ​ ( y x ​ ) = lo g b ​ ( x ) − lo g b ​ ( y )

Power Rule : log ⁡ b ( x k ) = k log ⁡ b ( x ) \log_b(x^k) = k \log_b(x) lo g b ​ ( x k ) = k lo g b ​ ( x )

Change of Base : log ⁡ b ( x ) = ln ⁡ ( x ) ln ⁡ ( b ) = log ⁡ 10 ( x ) log ⁡ 10 ( b ) \log_b(x) = \frac{\ln(x)}{\ln(b)} = \frac{\log_{10}(x)}{\log_{10}(b)} lo g b ​ ( x ) = l n ( b ) l n ( x ) ​ = l o g 10 ​ ( b ) l o g 10 ​ ( x ) ​

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 dB = 10 log ⁡ 10 ( P out P in ) G_{\text{dB}} = 10 \log_{10}\left(\frac{P_{\text{out}}}{P_{\text{in}}}\right) G dB ​ = 10 lo g 10 ​ ( P in ​ P out ​ ​ )

Voltage Gain/Loss (in dB) : A v , dB = 20 log ⁡ 10 ( V out V in ) A_{v,\text{dB}} = 20 \log_{10}\left(\frac{V_{\text{out}}}{V_{\text{in}}}\right) A v , dB ​ = 20 lo g 10 ​ ( V in ​ V out ​ ​ ) (assuming equal input/output impedance)

Absolute Power in dBm : P dBm = 10 log ⁡ 10 ( P 1 mW ) P_{\text{dBm}} = 10 \log_{10}\left(\frac{P}{1\text{ mW}}\right) P dBm ​ = 10 lo g 10 ​ ( 1 mW P ​ )

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

Calculate the total system power gain in dB.

Determine the final output power in milliwatts (mW) and dBm.

Solution:

Step 1: For cascaded systems, decibel gains and losses add algebraically: G total, dB = + 18 dB − 4 dB + 22 dB = + 36 dB G_{\text{total, dB}} = +18\text{ dB} - 4\text{ dB} + 22\text{ dB} = +36\text{ dB} G total, dB ​ = + 18 dB − 4 dB + 22 dB = + 36 dB

Step 2: Convert input power P in = 2 μ W = 0.002 mW P_{\text{in}} = 2\text{ }\mu\text{W} = 0.002\text{ mW} P in ​ = 2 μ W = 0.002 mW into dBm: P in, dBm = 10 log ⁡ 10 ( 0.002 mW 1 mW ) = 10 log ⁡ 10 ( 2 × 10 − 3 ) = 10 ( − 2.69897 ) = − 26.99 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} P in, dBm ​ = 10 lo g 10 ​ ( 1 mW 0.002 mW ​ ) = 10 lo g 10 ​ ( 2 × 1 0 − 3 ) = 10 ( − 2.69897 ) = − 26.99 dBm

Step 3: Compute output power in dBm and mW: P out, dBm = P in, dBm + G total, dB = − 26.99 dBm + 36 dB = + 9.01 dBm P_{\text{out, dBm}} = P_{\text{in, dBm}} + G_{\text{total, dB}} = -26.99\text{ dBm} + 36\text{ dB} = +9.01\text{ dBm} P out, dBm ​ = P in, dBm ​ + G total, dB ​ = − 26.99 dBm + 36 dB = + 9.01 dBm P out = 1 mW × 10 9.01 10 = 1 mW × 10 0.901 = 7.96 mW P_{\text{out}} = 1\text{ mW} \times 10^{\frac{9.01}{10}} = 1\text{ mW} \times 10^{0.901} = 7.96\text{ mW} P out ​ = 1 mW × 1 0 10 9.01 ​ = 1 mW × 1 0 0.901 = 7.96 mW

3. Partial Fraction Decomposition (PFD)

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

Standard Factor Expansions

Distinct Linear Factors : P ( s ) ( s − a ) ( s − b ) = A s − a + B s − b \frac{P(s)}{(s-a)(s-b)} = \frac{A}{s-a} + \frac{B}{s-b} ( s − a ) ( s − b ) P ( s ) ​ = s − a A ​ + s − b B ​

Repeated Linear Factors : P ( s ) ( s − a ) 2 = A s − a + B ( s − a ) 2 \frac{P(s)}{(s-a)^2} = \frac{A}{s-a} + \frac{B}{(s-a)^2} ( s − a ) 2 P ( s ) ​ = s − a A ​ + ( s − a ) 2 B ​

Irreducible Quadratic Factors : P ( s ) ( s − a ) ( s 2 + b s + c ) = A s − a + B s + C s 2 + b s + c \frac{P(s)}{(s-a)(s^2 + bs + c)} = \frac{A}{s-a} + \frac{Bs + C}{s^2 + bs + c} ( s − a ) ( s 2 + b s + c ) P ( s ) ​ = s − a A ​ + s 2 + b s + c B s + C ​

Worked ECE Board Example 3: Laplace Transfer Function Decomposition

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

Solution:

Step 1: Factor denominator: s 2 + 5 s + 6 = ( s + 2 ) ( s + 3 ) s^2 + 5s + 6 = (s + 2)(s + 3) s 2 + 5 s + 6 = ( s + 2 ) ( s + 3 ) .

Step 2: Set up decomposition: 7 s + 18 ( s + 2 ) ( s + 3 ) = A s + 2 + B s + 3 \frac{7s + 18}{(s + 2)(s + 3)} = \frac{A}{s + 2} + \frac{B}{s + 3} ( s + 2 ) ( s + 3 ) 7 s + 18 ​ = s + 2 A ​ + s + 3 B ​ .

Step 3: Use Heaviside Cover-Up Method: A = 7 s + 18 s + 3 ∣ s = − 2 = 7 ( − 2 ) + 18 − 2 + 3 = 4 1 = 4 A = \left. \frac{7s + 18}{s + 3} \right|_{s = -2} = \frac{7(-2) + 18}{-2 + 3} = \frac{4}{1} = 4 A = s + 3 7 s + 18 ​ ​ s = − 2 ​ = − 2 + 3 7 ( − 2 ) + 18 ​ = 1 4 ​ = 4 B = 7 s + 18 s + 2 ∣ s = − 3 = 7 ( − 3 ) + 18 − 3 + 2 = − 3 − 1 = 3 B = \left. \frac{7s + 18}{s + 2} \right|_{s = -3} = \frac{7(-3) + 18}{-3 + 2} = \frac{-3}{-1} = 3 B = s + 2 7 s + 18 ​ ​ s = − 3 ​ = − 3 + 2 7 ( − 3 ) + 18 ​ = − 1 − 3 ​ = 3

Step 4: Time-domain impulse response: H ( s ) = 4 s + 2 + 3 s + 3 ⟹ h ( t ) = ( 4 e − 2 t + 3 e − 3 t ) u ( t ) H(s) = \frac{4}{s+2} + \frac{3}{s+3} \implies h(t) = (4e^{-2t} + 3e^{-3t})u(t) H ( s ) = s + 2 4 ​ + s + 3 3 ​ ⟹ h ( t ) = ( 4 e − 2 t + 3 e − 3 t ) u ( t )

• 4. Arithmetic & Geometric Progressions
• Arithmetic Progressions (AP)

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

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

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 r r : a n = a 1 r n − 1 a_n = a_1 r^{n-1} a n ​ = a 1 ​ r n − 1 .

Sum of finite n n n terms : S n = a 1 ( 1 − r n ) 1 − r ( r ≠ 1 ) S_n = \frac{a_1(1 - r^n)}{1 - r} \quad (r \neq 1) S n ​ = 1 − r a 1 ​ ( 1 − r n ) ​ ( r  = 1 )

Sum of an infinite geometric series ( ∣ r ∣ < 1 |r| < 1 ∣ r ∣ < 1 ): S ∞ = a 1 1 − r S_\infty = \frac{a_1}{1 - r} S ∞ ​ = 1 − r a 1 ​ ​

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 W 50\text{ W} 50 W of optical power ( a 1 = 50 W a_1 = 50\text{ W} a 1 ​ = 50 W ). Each subsequent splitter dissipates 15 % 15\% 15% less power than the preceding one ( r = 0.85 r = 0.85 r = 0.85 ). Compute:

The power dissipated by the 6th splitter.

The theoretical total power dissipated by the entire infinite sequence of splitters.

Solution:

Step 1: Power dissipated by 6th splitter ( n = 6 n = 6 n = 6 ): a 6 = a 1 r 6 − 1 = 50 × ( 0.85 ) 5 = 50 × 0.4437 = 22.19 W a_6 = a_1 r^{6-1} = 50 \times (0.85)^5 = 50 \times 0.4437 = 22.19\text{ W} a 6 ​ = a 1 ​ r 6 − 1 = 50 × ( 0.85 ) 5 = 50 × 0.4437 = 22.19 W

Step 2: Total power dissipated by infinite splitters ( S ∞ S_\infty S ∞ ​ ): S ∞ = a 1 1 − r = 50 1 − 0.85 = 50 0.15 = 333.33 W S_\infty = \frac{a_1}{1 - r} = \frac{50}{1 - 0.85} = \frac{50}{0.15} = 333.33\text{ W} S ∞ ​ = 1 − r a 1 ​ ​ = 1 − 0.85 50 ​ = 0.15 50 ​ = 333.33 W

5. Summary of Formulae & Board Exam Tips

Concept Core Formula Typical PRC ECE Board Application

Quadratic Discriminant Δ = b 2 − 4 a c \Delta = b^2 - 4ac Δ = b 2 − 4 a c Determining RLC circuit damping state (over, under, or critical)

Decibel Power Gain G dB = 10 log ⁡ 10 ( P 2 P 1 ) G_{\text{dB}} = 10 \log_{10}\left(\frac{P_2}{P_1}\right) G dB ​ = 10 lo g 10 ​ ( P 1 ​ P 2 ​ ​ ) Link budget analysis, amplifier gain, attenuator calculations

Decibel Voltage Gain A v , dB = 20 log ⁡ 10 ( V 2 V 1 ) A_{v,\text{dB}} = 20 \log_{10}\left(\frac{V_2}{V_1}\right) A v , dB ​ = 20 lo g 10 ​ ( V 1 ​ V 2 ​ ​ ) Filter frequency response, Bode plot asymptotes

Infinite Geometric Sum S ∞ = a 1 1 − r S_\infty = \frac{a_1}{1 - r} S ∞ ​ = 1 − r a 1 ​ ​ Infinite resistive ladder networks, repeated signal reflections