FE Electrical and Computer exam practice: cram pack

Formula sheets, the key method steps for every topic and 40 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Series & parallel resistors

Resistors in series
Req = R1 + R2 + …
Resistors in parallel
1/Req = 1/R1 + 1/R2 + …
Two in parallel
Req = R1R2 / (R1 + R2)
Ohm's law
V = I·R
Power
P = V·I = I²R = V²/R

Method

  1. Start at the end of the network furthest from the terminals.
  2. Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
  3. Redraw the circuit after each step.
  4. Repeat until one resistor is left between the terminals.

Nodal analysis

Kirchhoff’s current law
ΣIin = ΣIout
Branch current
I = (Va − Vb)/R
Reference node
Vground = 0

Method

  1. Choose a reference node (ground) and label the other node voltages.
  2. Write Kirchhoff’s current law at every unknown node: currents leaving through each branch equal the currents entering.
  3. Express every branch current as (Va − Vb)/R.
  4. Solve the system of equations, then find any current you need.

Mesh analysis

Kirchhoff’s voltage law
ΣV = 0 around every loop
Shared resistor
current = I1 − I2
Current source in a mesh
fixes that mesh current

Method

  1. Draw a clockwise current in every mesh (window) of the circuit.
  2. Apply Kirchhoff’s voltage law around each mesh: the sum of voltage drops is zero.
  3. In a resistor shared by two meshes the current is the difference of the two mesh currents.
  4. Solve the equations; a negative result means the current flows the other way.

Superposition

Superposition
x = x1 + x2 + … (one source at a time)
Switched-off voltage source
short circuit (0 V)
Switched-off current source
open circuit (0 A)
Voltage divider
V2 = V·R2/(R1 + R2)
Current divider
I1 = I·R2/(R1 + R2)

Method

  1. Count the independent sources. You will solve one simpler circuit per source.
  2. Keep one source and switch off the others: a voltage source becomes a short circuit (a wire), a current source becomes an open circuit (a gap).
  3. Solve the simpler circuit for the voltage or current you need, keeping track of its direction.
  4. Repeat for every source, then add the contributions with their signs.

Thévenin & Norton

Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)

Method

  1. Remove the load and mark the two terminals a and b.
  2. Find the open-circuit voltage between a and b: this is Vth.
  3. Switch off all independent sources and find the resistance seen from a–b: this is Rth.
  4. Put the load back on the equivalent circuit: Vth in series with Rth.
  5. For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.

RC & RL transients

Time constant
τ = RC | τ = L/R
Response
x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ
Capacitor at DC
open circuit; vC cannot jump
Inductor at DC
short circuit; iL cannot jump
Settled
t ≈ 5τ

Method

  1. Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
  2. Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
  3. Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
  4. Use x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.

AC impedance & phasors

Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z

Method

  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.

AC power & power factor

Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)

Three-phase power

Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η

Transformers

Turns ratio
a = N1/N2 = V1/V2
Current ratio
I1/I2 = N2/N1 = 1/a
Rated current
I = S/V
Referred impedance
Z1′ = a²·ZL
Copper loss at part load
Pcu = x²·Pcu,fl
Efficiency
η = x·S·pf/(x·S·pf + Pcore + x²·Pcu,fl)
Maximum efficiency
x = √(Pcore/Pcu,fl)

Op-amp circuits

Inverting amplifier
Vout = −(Rf/Rin)·Vin
Non-inverting amplifier
Vout = (1 + Rf/R1)·Vin
Summing amplifier
Vout = −Rf(V1/R1 + V2/R2)
Ideal op-amp
V+ = V−, Iin = 0

Diodes & transistors

Diode (constant drop)
VD = 0.7 V when conducting
Active region
VBE = 0.7 V, IC = β·IB, IE = (β + 1)·IB
Fixed bias
IB = (VCC − 0.7)/RB
Divider bias
IB = (VTH − 0.7)/(RTH + (β + 1)RE)
Thévenin of the divider
VTH = VCC·R2/(R1 + R2), RTH = R1∥R2
Collector loop
VCE = VCC − ICRC − IERE
Saturation
IC(sat) = (VCC − VCE(sat))/RC, IB ≥ IC(sat)/β

Number systems

Place values
binary 2k, octal 8k, hex 16k
To decimal
Σ dk·bk
From decimal
divide by the base, read the remainders bottom-up
Hex ↔ binary
one hex digit = 4 bits
Two’s complement
invert the bits and add 1

Digital logic: gates, K-maps & flip-flops

Basic gates
A·B, A + B, A′, (A·B)′, (A + B)′, A ⊕ B
Boolean rules
X + X′ = 1, X·X′ = 0, X + XY = X, XY + XY′ = X
De Morgan
(X·Y)′ = X′ + Y′, (X + Y)′ = X′·Y′
Karnaugh map order
00, 01, 11, 10 (Gray code)
Group of 2ᵏ cells
loses k variables
4-to-1 multiplexer
F = A′B′I0 + A′BI1 + AB′I2 + ABI3
Flip-flops
D: Q⁺ = D, T: Q⁺ = T ⊕ Q, JK: Q⁺ = JQ′ + K′Q
Counters
n-bit: 2ⁿ states; mod-N: smallest n with 2ⁿ ≥ N; fout = fclk/N
Ripple counter
fmax = 1/(n·tpd)
Ring and Johnson counters
n and 2n states

Method

  1. For a circuit, work from the inputs to the output one gate at a time and write the value on every wire, or write its expression and fill in the truth table.
  2. For simplification, put the 1s (and the don’t-cares as X) into a Karnaugh map with Gray-code rows and columns: 00, 01, 11, 10.
  3. Cover the 1s with as few and as large groups of 1, 2, 4, 8 or 16 cells as possible, wrapping round the edges; each group is one product term made of the variables that stay constant. Group the 0s for the POS form.
  4. For flip-flops and counters, read the inputs at each active clock edge, write the state after every edge, and remember that an n-bit counter wraps round after 2ⁿ states.

Stability: Routh–Hurwitz & root locus

Stability
all roots of the characteristic polynomial in the left half-plane
Routh array, s¹ entry of a cubic
b1 = (a2a1 − a3a0)/a2
Right half-plane roots
number of sign changes in the first column
Cubic s³ + a₂s² + a₁s + a₀
stable when all a > 0 and a₂·a₁ > a₀
K/(s(s + a)(s + b))
Kmax = ab(a + b), ω = √(ab)
Characteristic equation
1 + K·G(s) = 0
Asymptotes
σa = (Σp − Σz)/(n − m), θ = (2q + 1)·180°/(n − m)
Break-away and break-in points
dK/ds = 0
Magnitude condition
K = 1/|G(s)|
Angle of departure
θd = 180° − Σ∠(from other poles) + Σ∠(from zeros)

Method

  1. Write the closed-loop characteristic polynomial: the denominator of 1 + K·G(s) = 0, multiplied out.
  2. Put the coefficients alternately into the first two rows of the Routh array and build the remaining rows.
  3. Count the sign changes in the first column: each one is a pole in the right half-plane. With K in the array, demand that every first-column entry stays positive.
  4. For a root locus, mark the poles and zeros, the real-axis segments, the asymptotes from the centroid and the break points where dK/ds = 0. Use K = 1/|G(s)| for the gain at a point.

Steady-state error & PID control

System type
number of poles of G(s) at s = 0
Error constants
Kp = lim G(s), Kv = lim s·G(s), Ka = lim s²·G(s) (s → 0)
Step A
ess = A/(1 + Kp)
Ramp A·t
ess = A/Kv
Parabola A·t²/2
ess = A/Ka
PID controller
C(s) = Kp + Ki/s + Kd·s
Second-order target
s² + 2ζωns + ωn²
Overshoot and timing
%OS = 100·e−ζπ/√(1−ζ²), tp = π/ωd, ts ≈ 4/(ζωn)

Method

  1. Count the poles of the open loop C(s)·G(s) at s = 0: that is the system type.
  2. Work out the error constant that matches the input: Kp for a step, Kv for a ramp, Ka for a parabola.
  3. Apply ess = A/(1 + Kp), A/Kv or A/Ka, or solve it for the gain you need.
  4. For a damping or speed specification, write the closed-loop characteristic polynomial and match it with s² + 2ζω_n·s + ω_n².

2Practice problems

1.Series & parallel resistorsMedium
R1​ = 20 ΩR3​ = 12 Ω+−V1​24 VR2​5 ΩAR4​20 ΩB

Find the equivalent resistance Req seen by the source V1.

V1 = 24 VR1 = 20 ΩR2 = 5 ΩR3 = 12 ΩR4 = 20 Ω
Req = Ω
2.Series & parallel resistorsHard
R1​ = 25 ΩIR1​R3​ = 3 ΩR5​ = 15 Ω+−V1​5 VR2​4 ΩAR4​30 ΩBR6​4 Ω

Find the current IR1 delivered by the source, which flows through R1.

V1 = 5 VR1 = 25 ΩR2 = 4 ΩR3 = 3 ΩR4 = 30 ΩR5 = 15 ΩR6 = 4 Ω
IR1 = A
3.Nodal analysisMedium
R1​10 Ω+−V1​6 VI1​R2​20 Ω+−V2​12 VR3​20 Ω+−V3​30 VA

Three branches are connected between node A and ground, as shown. Use nodal analysis to find the current I1 in branch 1, flowing up into node A in the direction of the arrow.

V1 = 6 VR1 = 10 ΩV2 = 12 VR2 = 20 ΩV3 = 30 VR3 = 20 Ω
I1 = A
4.Nodal analysisHard
R1​ = 10 ΩR3​ = 5 Ω+−V1​15 VR2​3 ΩAR4​25 ΩB0.25·VA​

The diamond is a voltage-controlled current source that pushes a current of 0.25·VA (in A, with VA in V) up into node B. Use nodal analysis to find the node voltage VA (measured from ground).

V1 = 15 VR1 = 10 ΩR2 = 3 ΩR3 = 5 ΩR4 = 25 Ωg = 0.25 A/V
VA = V
5.Mesh analysisMedium
R1​ = 2 ΩR2​ = 6 Ω+−VR2​+−V1​12 VR3​15 ΩA+−V2​30 VI1​I2​

Use mesh analysis to find the voltage VR2 across R2 with the polarity shown.

V1 = 12 VV2 = 30 VR1 = 2 ΩR2 = 6 ΩR3 = 15 Ω
VR2 = V
6.Mesh analysisHard
R1​ = 8 ΩR2​ = 5 Ω+−VR2​R3​ = 2 Ω+−V1​24 VR4​15 ΩAR5​10 ΩB+−V2​15 VI1​I2​I3​

Use mesh analysis to find the voltage VR2 across R2 with the polarity shown.

V1 = 24 VV2 = 15 VR1 = 8 ΩR2 = 5 ΩR3 = 2 ΩR4 = 15 ΩR5 = 10 Ω
VR2 = V
7.SuperpositionMedium
R1​ = 20 ΩR3​ = 15 Ω+−V1​30 VR2​6 ΩIR2​AR4​10 ΩBIs​1.5 A

Use superposition to find the current IR2 through R2 in the direction of the arrow.

V1 = 30 VIs = 1.5 AR1 = 20 ΩR2 = 6 ΩR3 = 15 ΩR4 = 10 Ω
IR2 = A
8.SuperpositionHard
R1​ = 12 ΩR2​ = 8 ΩR4​ = 5 ΩIR4​+−V1​6 VR3​5 ΩAIs​5 AB+−V2​36 V

Use superposition to find the current IR4 through R4 in the direction of the arrow.

V1 = 6 VIs = 5 AV2 = 36 VR1 = 12 ΩR2 = 8 ΩR3 = 5 ΩR4 = 5 Ω
IR4 = A
9.Thévenin & NortonMedium
R1​ = 20 ΩR3​ = 8 Ω+−V1​24 VR2​2 ΩAab

Find the Thévenin resistance Rth seen from terminals a–b.

V1 = 24 VR1 = 20 ΩR2 = 2 ΩR3 = 8 Ω
Rth = Ω
10.Thévenin & NortonHard
R1​ = 4 ΩR3​ = 30 Ω+−V1​9 VR2​10 ΩAIs​2.5 ABab

Find the Thévenin voltage Vth seen from terminals a–b.

V1 = 9 VIs = 2.5 AR1 = 4 ΩR2 = 10 ΩR3 = 30 Ω
Vth = V
11.RC & RL transientsMedium
R = 50 Ωt = 0L68 mHi(t)

The inductor carries a steady current of 2.5 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. How long does it take for the current to fall to 0.625 A? Find t.

i(0) = 2.5 AR = 50 ΩL = 68 mH
t = ms
12.RC & RL transientsHard
+−Vs​10 Vopens at t = 0R1​ = 3.3 kΩR2​ = 4.7 kΩC10 µF+−vC​

The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the capacitor voltage vC at t = 15 ms.

Vs = 10 VR1 = 3.3 kΩR2 = 4.7 kΩC = 10 µFt = 15 ms
vC = V
13.AC impedance & phasorsMedium
24 V50 HzR = 10 ΩL = 50 mHI

A resistor and an inductor are connected in parallel across a 24 V (rms), 50 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 24 V (rms)f = 50 HzR = 10 ΩL = 50 mH
φ = °
14.AC impedance & phasorsHard
100 V50 HzR = 100 ΩL = 200 mHC = 100 µF

A series RLC circuit is connected to a 100 V (rms), 50 Hz supply. Find the resonant frequency f0 of the circuit.

V = 100 V (rms)f = 50 HzR = 100 ΩL = 200 mHC = 100 µF
f0 = Hz
15.AC power & power factorMedium
P = 18 kWQSφ

A single-phase motor takes 18 kW at a power factor of 0.7 lagging from a 230 V supply. Find the line current I.

P = 18 kWpf = 0.7V = 230 V
I = A
16.AC power & power factorHard
P = 5 kWQ1​S1​φ1​QC​

A single-phase load takes 5 kW at a power factor of 0.65 lagging from a 240 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the reactive power QC the capacitor must supply.

P = 5 kWpf1 = 0.65pf2 = 0.98V = 240 Vf = 50 Hz
QC = kvar
17.Three-phase powerMedium
ABCVL​ = 415 VZ = 15 + j20 Ωper phase

A balanced Δ-connected load with an impedance of 15 + j20 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 415 V. Find the total real power P absorbed by the load in kW.

VL = 415 VZ = 15 + j20 Ω per phase
P = kW
18.Three-phase powerHard
P = 300 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 300 kW at a power factor of 0.75 lagging from a 415 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 300 kWpf1 = 0.75pf2 = 0.92VL = 415 Vf = 60 Hz
Cph = µF
19.TransformersMedium
7200 V240 VIHV​ILV​ = ?10 kVA

A single-phase 10 kVA, 7200/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 10 kVAV1 = 7200 VV2 = 240 V
ILV = A
20.TransformersHard
100 kVA

A single-phase 100 kVA transformer has a core loss of 400 W and a copper loss of 1200 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 100 kVAPcore = 400 WPcu,fl = 1200 W
Sη,max = kVA
21.Op-amp circuitsMedium
−+Vout​Rf​ = ?R1​ = 22 kΩVin​

Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 31.

R1 = 22 kΩAv = 31
Rf = kΩ
22.Op-amp circuitsHard
Vin​V1​Vout​stage 1: invertingR1​ = 10 kΩRf1​ = 10 kΩstage 2: non-invertingR2​ = 47 kΩRf2​ = 47 kΩ

Two ideal op-amp stages are connected in cascade: an inverting amplifier (R1, Rf1) followed by a non-inverting amplifier (R2 to ground, Rf2 in the feedback). Both run from ±15 V supplies. Find the overall voltage gain Av = Vout/Vin.

R1 = 10 kΩRf1 = 10 kΩR2 = 47 kΩRf2 = 47 kΩ
Av =
23.Diodes & transistorsMedium
+−12 VR = ?2 × 3 V5 mA

2 white LEDs (forward voltage 3 V each) are connected in series with a resistor to a 12 V supply. Find the resistance R that sets the current to 5 mA.

Vs = 12 VVF = 3 Vn = 2I = 5 mA
R = Ω
24.Diodes & transistorsHard
RC​ = 220 Ωβ = 150IC​ = ?VCC​ = 12 V5 VRB​ = 100 kΩ

An NPN transistor with β = 150 switches a 220 Ω load resistor connected between VCC = 12 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB = 100 kΩ. Take VBE = 0.7 V and VCE(sat) = 0.2 V. Find the collector current IC in mA when the input is high. Decide first whether the transistor is saturated.

VCC = 12 VRC = 220 Ωβ = 150Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 100 kΩ
IC = mA
25.Number systemsMedium
8-bit two's complement?−128?64?32?16?8?4?2?1

Write −44 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −448 bit
U =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0111 1011 << 1. What is the result's value R in decimal?

0111 1011<< 1
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010110312040507061120130150141819111110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 1, 2, 8, 9, 10, 11, 12). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(0, 1, 2, 8, 9, 10, 11, 12)
L =
28.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKDOWN

A 3-bit binary down counter starts at 2 (decimal, Q = 010). Which count Q (in decimal) does it show after 28 clock pulses?

Q0 = 228n = 3
Q =
29.Laplace transformsMedium
t (s)f(t)t = 0.3f(0.3) = ?

Find f(t) for F(s) = 45/(s(s + 1)(s + 5)) by partial fractions, and evaluate f(0.3).

F(s) = 45/(s(s + 1)(s + 5))t = 0.3 s
f(0.3) =
30.Laplace transformsHard
ReIm−224

A system is described by the differential equation y″ + 6y′ + 25y = 2u′ + 6u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 0.25 s. Find g(0.25).

y″ + 6y′ + 25y = 2u′ + 6ut = 0.25 s
g(0.25) =
31.Block diagramsMedium
R(s)+8s(s + 3)G(s)C(s)−

A unity negative feedback loop has G(s) = 8/(s(s + 3)). The reference is a ramp r(t) = 2·t. Find the steady-state error ess.

G(s) = 8/(s(s + 3))r(t) = 2·t
ess =
32.Block diagramsHard
R(s)+Ks(s + 7)(s + 8)G(s)C(s)−

The plant G(s) = K/(s(s + 7)(s + 8)) is in a unity negative feedback loop. At the largest stable gain the loop oscillates. Find the frequency of oscillation ω in rad/s.

G(s) = K/(s(s + 7)(s + 8))unity negative feedback
ω = rad/s
33.Step responseMedium
t (s)y(t)0.5

A system has the transfer function G(s) = 37.5/(s² + 12s + 75) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 37.5/(s² + 12s + 75)
σ =
34.Step responseHard
R(s)+45s(s + 2)G(s)C(s)−

A unity negative feedback loop has G(s) = 45/(s(s + 2)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 45/(s(s + 2))
ts = s
35.Bode plots & marginsMedium
vin​C = 47 nFR = 3.3 kΩvout​

The RC high-pass filter shown has R = 3.3 kΩ, C = 47 nF. A sinusoidal input of 260 Hz is applied. Find the gain |H| = |vout/vin| in dB.

R = 3.3 kΩC = 47 nFf = 260 Hz
|H| = dB
36.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 80/(s(s + 4)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 80/(s(s + 4))
ωc = rad/s
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 10(s − 3)(s + 2)C(s)−

The plant G(s) = (s + 10)/((s − 3)(s + 2)) has an unstable pole at s = 3. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.

G(s) = (s + 10)/((s − 3)(s + 2))
Kmin =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10j2sb​

The root locus of L(s) = K/((s + 3)(s + 9)) is shown. Find the gain K at the break-away point sb, where the two closed-loop poles meet on the real axis.

L(s) = K/((s + 3)(s + 9))
K =
39.Steady-state error & PID controlMedium
r(t) = 0.5·t²/2+10(s + 4)s²(s + 1)C(s)−

The parabolic input r(t) = 0.5·t²/2 is applied to a unity-feedback loop with G(s) = 10(s + 4)/(s²(s + 1)). Find the steady-state error ess.

G(s) = 10(s + 4)/(s²(s + 1))r(t) = 0.5·t²/2
ess =
40.Steady-state error & PID controlHard
R(s)+25 + Kd​s4s(s + 2)C(s)−

A PD controller C(s) = 25 + Kd·s drives the plant G(s) = 4/(s(s + 2)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.707.

G(s) = 4/(s(s + 2))Kp = 25ζ = 0.707
Kd = s

3Solutions

1.Series & parallel resistors

Answer: Req = 24.32 Ω

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 20 ΩR3​ = 12 Ω+−V1​24 VR2​5 ΩAR4​20 ΩB

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R3 + R4 = 12 + 20 = 32 Ω
R2 ∥ 32 = 5·32/(5+32) = 4.324 Ω
Req = R1 + 4.324 = 20 + 4.324 = 24.32 Ω

Answer

Req = 24.32 Ω
2.Series & parallel resistors

Answer: IR1 = 0.1777 A

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 25 ΩR3​ = 3 ΩR5​ = 15 Ω+−V1​5 VR2​4 ΩAR4​30 ΩBR6​4 Ω

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R5 + R6 = 15 + 4 = 19 Ω
R4 ∥ 19 = 11.63 Ω
R3 + 11.63 = 14.63 Ω
R2 ∥ 14.63 = 3.141 Ω
Req = R1 + 3.141 = 25 + 3.141 = 28.14 Ω

Current from the source

The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.

IR1 = V1/Req = 5/28.14 = 0.1777 A

Answer

IR1 = 0.1777 A
3.Nodal analysis

Answer: I1 = −0.75 A

Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.

KCL at node A

Take the bottom wire as ground. Each branch carries (VA − Vk)/Rk away from A, where Vk is the voltage of its source (zero for a plain resistor). The currents leaving A add up to zero.

(VA − 6)/10 + (VA − 12)/20 + (VA − 30)/20 = 0

Solve for VA

VA = (6/10 + 12/20 + 30/20) / (1/10 + 1/20 + 1/20) = 2.7/0.2 = 13.5 V

Branch current

The arrow points up through the branch, into node A.

I1 = (6 − VA)/R1 = (6 − 13.5)/10 = −0.75 A

Answer

I1 = −0.75 A

A negative result means the real current flows against the arrow.

4.Nodal analysis

Answer: VA = 5.806 V

Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.

KCL at node A

Currents leaving A through the resistors equal the current pushed into A by sources.

(VA − V1)/R1 + VA/R2 + (VA − VB)/R3 = 0
0.6333·VA − 0.2·VB = 1.5

KCL at node B

The dependent source pushes g·VA into node B. Treat it like an independent source, then move the VA term to the left-hand side.

(VB − VA)/R3 + VB/R4 = 0.25·VA
−0.45·VA + 0.24·VB = 0

Solve

Δ = (0.6333)(0.24) − (−0.2)(−0.45) = 0.062
VA = Δ1/Δ = 0.36/0.062 = 5.806 V
VB = Δ2/Δ = 0.675/0.062 = 10.89 V

Answer

VA = 5.806 V
5.Mesh analysis

Answer: VR2 = −15 V

Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.

R1​ = 2 ΩR2​ = 6 Ω+−V1​12 VR3​15 ΩA+−V2​30 VI1​I2​

Assign mesh currents

Draw a clockwise current in each window: I1, I2. A resistor shared by two meshes carries the difference of their currents.

KVL around mesh 1

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R1·I1 + R3(I1 − I2) = V1
2·I1 + 15(I1 − I2) = 12
17·I1 − 15·I2 = 12

KVL around mesh 2

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R2·I2 + R3(I2 − I1) = −V2
6·I2 + 15(I2 − I1) = (−30)
−15·I1 + 21·I2 = −30

Solve

Δ = (17)(21) − (−15)(−15) = 132
I1 = Δ1/Δ = −198/132 = −1.5 A
I2 = Δ2/Δ = −330/132 = −2.5 A

Answer

IR2 = I2 = −2.5 A
VR2 = IR2·R2 = −2.5·6 = −15 V

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: VR2 = 1.326 V

Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.

R1​ = 8 ΩR2​ = 5 ΩR3​ = 2 Ω+−V1​24 VR4​15 ΩAR5​10 ΩB+−V2​15 VI1​I2​I3​

Assign mesh currents

Draw a clockwise current in each window: I1, I2, I3. A resistor shared by two meshes carries the difference of their currents.

KVL around mesh 1

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R1·I1 + R4(I1 − I2) = V1
8·I1 + 15(I1 − I2) = 24
23·I1 − 15·I2 = 24

KVL around mesh 2

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R2·I2 + R4(I2 − I1) + R5(I2 − I3) = 0
5·I2 + 15(I2 − I1) + 10(I2 − I3) = 0
−15·I1 + 30·I2 − 10·I3 = 0

KVL around mesh 3

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R3·I3 + R5(I3 − I2) = −V2
2·I3 + 10(I3 − I2) = (−15)
−10·I2 + 12·I3 = −15

Solve

Δ = det of the coefficient matrix = 3280
I1 = Δ1/Δ = 3990/3280 = 1.216 A
I2 = Δ2/Δ = 870/3280 = 0.2652 A
I3 = Δ3/Δ = −3375/3280 = −1.029 A

Answer

IR2 = I2 = 0.2652 A
VR2 = IR2·R2 = 0.2652·5 = 1.326 V

A negative mesh current flows anticlockwise.

7.Superposition

Answer: IR2 = −1.364 A

Superposition: in a linear circuit, any voltage or current is the sum of the responses to each independent source acting alone. To switch a source off, replace an ideal voltage source with a short circuit and an ideal current source with an open circuit. Resistors stay where they are.

R1​ = 20 ΩR3​ = 15 Ω+−V1​30 VR2​6 ΩAR4​10 ΩBOPEN

Case 1 · V1 acting alone

Iₛ becomes an open circuit, so R₃ and R₄ form one series branch from A to ground, in parallel with R₂.

R3+R4 = 15+10 = 25 Ω
R2 ∥ (R3+R4) = 6·25/(6+25) = 4.839 Ω
VA = 30·4.839/(20+4.839) = 5.844 V
VB = VA·R4/(R3+R4) = 5.844·10/25 = 2.338 V
IR2′ = (0 − VA)/R2 = (0 − 5.844)/6 = −0.974 A
R1​ = 20 ΩR3​ = 15 ΩSHORTR2​6 ΩAR4​10 ΩBIs​1.5 A

Case 2 · Is acting alone

V₁ becomes a short, so R₁ is in parallel with R₂. Iₛ splits at B between R₄ and the path through R₃.

R1 ∥ R2 = 20·6/(20+6) = 4.615 Ω
Req = R4 ∥ (R3+R1∥R2) = 10 ∥ 19.62 = 6.623 Ω
VB = Is·Req = 1.5·6.623 = 9.935 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 9.935·4.615/19.62 = 2.338 V
IR2″ = (0 − VA)/R2 = (0 − 2.338)/6 = −0.3896 A

Add the contributions

IR2 = IR2′ + IR2″
IR2 = (−0.974) + (−0.3896) = −1.364 A

A negative result means the real current flows against the arrow.

8.Superposition

Answer: IR4 = 1.416 A

Superposition: in a linear circuit, any voltage or current is the sum of the responses to each independent source acting alone. To switch a source off, replace an ideal voltage source with a short circuit and an ideal current source with an open circuit. Resistors stay where they are.

R1​ = 12 ΩR2​ = 8 ΩR4​ = 5 Ω+−V1​6 VR3​5 ΩAOPENBSHORT

Case 1 · V1 acting alone

V₂ becomes a short and Iₛ an open circuit. Seen from A, R₃ is in parallel with the series pair R₂+R₄.

R3 ∥ (R2+R4) = 5 ∥ 13 = 3.611 Ω
VA = 6·3.611/(12+3.611) = 1.388 V
VB = VA·R4/(R2+R4) = 1.388·5/13 = 0.5338 V
IR4′ = (VB − V2)/R4 = (0.5338 − 0)/5 = 0.1068 A
R1​ = 12 ΩR2​ = 8 ΩR4​ = 5 ΩSHORTR3​5 ΩAIs​5 ABSHORT

Case 2 · Is acting alone

Both voltage sources become shorts. Iₛ splits at B between R₄ (now to ground) and R₂ in series with R₁ ∥ R₃.

R1 ∥ R3 = 3.529 Ω, so the left path is R2+3.529 = 11.53 Ω
Req = R4 ∥ 11.53 = 3.488 Ω
VB = Is·Req = 5·3.488 = 17.44 V
VA = VB·3.529/11.53 = 5.338 V
IR4″ = (VB − V2)/R4 = (17.44 − 0)/5 = 3.488 A
R1​ = 12 ΩR2​ = 8 ΩR4​ = 5 ΩSHORTR3​5 ΩAOPENB+−V2​36 V

Case 3 · V2 acting alone

V₁ becomes a short and Iₛ an open circuit. R₁ ∥ R₃ hangs from A, in series with R₂ back to B.

R1 ∥ R3 = 12·5/(12+5) = 3.529 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 36·11.53/16.53 = 25.11 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 25.11·3.529/11.53 = 7.687 V
IR4‴ = (VB − V2)/R4 = (25.11 − 36)/5 = −2.178 A

Add the contributions

IR4 = IR4′ + IR4″ + IR4‴
IR4 = 0.1068 + 3.488 + (−2.178) = 1.416 A

A negative result means the real current flows against the arrow.

9.Thévenin & Norton

Answer: Rth = 9.818 Ω

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 20 ΩR3​ = 8 ΩSHORTR2​2 ΩAabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 20·2/(20+2) = 1.818 Ω
Rth = (R1 ∥ R2) + R3 = 1.818 + 8 = 9.818 Ω
10.Thévenin & Norton

Answer: Vth = 88.57 V

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 4 ΩR3​ = 30 ΩSHORTR2​10 ΩAOPENBabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 4·10/(4+10) = 2.857 Ω
Rth = (R1 ∥ R2) + R3 = 2.857 + 30 = 32.86 Ω
R1​ = 4 ΩR3​ = 30 Ω+−V1​9 VR2​10 ΩAIs​2.5 ABab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 6.429 + 2.5·2.857 = 13.57 V
Vth = VB = VA + Is·R3 = 13.57 + 2.5·30 = 88.57 V
11.RC & RL transients

Answer: t = 1.885 ms

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Time constant

τ = L/R = 68 mH / 50 Ω = 1.36 ms

Solve for the time

0.625 = 2.5·e−t/τ → t = τ·ln(2.5/0.625)
t = 1.36·ln(4) = 1.885 ms

Answer

t = 1.885 ms
12.RC & RL transients

Answer: vC = 4.27 V

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Initial capacitor voltage

In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.

vC(0) = 10·4.7/(3.3 + 4.7) = 5.875 V

Time constant after the switch opens

τ = R2·C = 4.7·10 = 47 ms

Discharge

vC(t) = 5.875·e−15/47 = 4.27 V

Answer

vC = 4.27 V
13.AC impedance & phasors

Answer: φ = 32.48 °

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Reactance

XL = ωL = 2π·50·0.05 = 15.71 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

IR = V/R = 24/10 = 2.4 A
IX = V/XL = 24/15.71 = 1.528 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(IR² + IX²) = √(2.4² + 1.528²) = 2.845 A

Phase angle

The inductor current lags V, so the total current lags and the angle is positive.

φ = arctan(IX/IR) = arctan(1.528/2.4) = 32.48°

Answer

φ = 32.48°
14.AC impedance & phasors

Answer: f0 = 35.59 Hz

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Resonance

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.

f0 = 1/(2π√(LC)) = 1/(2π·√(0.2·100·10⁻⁶)) = 35.59 Hz

Answer

f0 = 35.59 Hz
15.AC power & power factor

Answer: I = 111.8 A

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Line current

Only the real part of the power does work, so divide by the power factor.

I = P/(V·cos φ) = 18000/(230·0.7) = 111.8 A

Answer

I = 111.8 A
16.AC power & power factor

Answer: QC = 4.83 kvar

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Reactive power before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = P·tan φ1 = 5·tan(49.46°) = 5.846 kvar
Q2 = P·tan φ2 = 5·tan(11.48°) = 1.015 kvar
QC = Q1 − Q2 = 4.83 kvar

Answer

QC = 4.83 kvar
17.Three-phase power

Answer: P = 12.4 kW

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Phase voltage

Vph = VL = 415 V

Phase current

|Z| = √(15² + 20²) = 25 Ω
Iph = 415/25 = 16.6 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·16.6²·15 = 12.4 kW

Answer

P = 12.4 kW

Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.

18.Three-phase power

Answer: Cph = 702.2 µF

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.

Reactive power before and after

The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.

Q1 = 300·tan(acos 0.75) = 300·0.8819 = 264.6 kvar
Q2 = 300·tan(acos 0.92) = 300·0.426 = 127.8 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 136.8 kvar

Capacitance per phase

In a Δ bank each capacitor is connected between two lines.

V = 415 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 45592/(377·415²) = 702.2 µF

Answer

Cph = 702.2 µF
19.Transformers

Answer: ILV = 41.67 A

Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.

Rated current

Both windings carry the same apparent power.

ILV = S/V = 10000/240 = 41.67 A

Answer

ILV = 41.67 A
20.Transformers

Answer: Sη,max = 57.74 kVA

Transformer losses: the core loss (hysteresis and eddy currents) depends on the voltage and is the same at any load. The copper loss in the windings goes with the current squared, so at a fraction x of rated load it is x²·Pcu,fl. Efficiency η = Pout/(Pout + Pcore + Pcu), with Pout = x·S·pf. The efficiency is highest at the load where the copper loss equals the core loss.

Copper loss equals core loss

x²·1200 = 400 → x = √(400/1200) = 0.5774

Load at maximum efficiency

Sη,max = x·S = 0.5774·100 = 57.74 kVA

Answer

Sη,max = 57.74 kVA
21.Op-amp circuits

Answer: Rf = 660 kΩ

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Gain of a non-inverting amplifier

V₋ = V₊ = Vin, and the current through R₁ also flows through Rf, so Vout = Vin·(1 + Rf/R₁).

Av = 1 + Rf/R1
Rf = (31 − 1)·22 = 660 kΩ

Answer

Rf = 660 kΩ
22.Op-amp circuits

Answer: Av = −2

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Gain of each stage

A1 = −Rf1/R1 = −10/10 = −1
A2 = 1 + Rf2/R2 = 1 + 47/47 = 2

Overall gain

Av = A1·A2 = −1·2 = −2

Answer

Av = −2
23.Diodes & transistors

Answer: R = 1200 Ω

LED and diode circuits: a forward-biased LED drops roughly its forward voltage VF (about 2 V for red, 3 V for blue or white), whatever the current. A series resistor sets the current: R = (Vs − n·VF)/I for n LEDs in series. With a more exact diode model V = V₀ + I·rd, the operating point is where the diode line meets the load line I = (Vs − V)/R, so I = (Vs − V₀)/(R + rd).

Voltage across the resistor

VR = 12 − 2·3 = 6 V

Resistance

R = 6/0.005 = 1200 Ω

Answer

R = 1200 Ω

The resistor dissipates 30 mW.

24.Diodes & transistors

Answer: IC = 6.45 mA

BJT as a switch: to switch a load fully on, the transistor is driven into saturation, where VCE = VCE(sat) and the collector current is set by the load: IC(sat) = (VCC − VCE(sat))/RC. Saturation needs at least IB = IC(sat)/β. If the base current is smaller, the transistor stays active and IC = β·IB.

Base current

IB = (5 − 0.7)/100 kΩ = 43 µA

Saturated or active?

β·IB is smaller than IC(sat), so the transistor stays active.

β·IB = 6.45 mA
IC(sat) = (12 − 0.2)/220 = 53.64 mA

Collector current

IC = β·IB = 6.45 mA

Answer

IC = 6.45 mA

The transistor is active, so the load is not fully switched on.

25.Number systems

Answer: U = 212

Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.

Invert and add one

44 = 0010 1100
inverted: 1101 0011
+1 → 1101 0100

Read it as unsigned

1101 0100₂ = 256 − 44 = 212

Answer

U = 212
26.Number systems

Answer: R = 246

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Bit by bit

Shifting left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.

0111 1011 << 1
= 1111 0110

Value of the result

1111 0110₂ = 246

Answer

R = 246
27.Digital logic: gates, K-maps & flip-flops

Answer: L = 9

Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.

ABCD000111100001111010110312040507061120130150141819111110

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

B′D′
B′C′
AC′D′
AB′

Minimal SOP form

F = B′D′ + B′C′ + AC′D′ + AB′
L = 9

Answer

L = 9
28.Digital logic: gates, K-maps & flip-flops

Answer: Q = 6

Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

Full cycles

Every 8 pulses bring the counter back to the same state.

28 = 3·8 + 4

Count after the remaining pulses

Q = (2 − 4) mod 8 = 6
Q = 110₂

Answer

Q = 6
29.Laplace transforms

Answer: f(0.3) = 1.168

Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).

Partial fractions

F(s) = A/s + B/(s + 1) + C/(s + 5)
A = 45/(1·5) = 9
B = 45/((−1)·4) = −11.25
C = 45/((−5)·(−4)) = 2.25

Inverse transform

f(t) = 9 − 11.25·e−t + 2.25·e−5t

Evaluate

f(0.3) = 9 − 11.25·e−0.3 + 2.25·e−5·0.3 = 1.168

Answer

f(0.3) = 1.168
30.Laplace transforms

Answer: g(0.25) = 0.5104

Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).

Transfer function

G(s) = Y(s)/U(s) = (2s + 6)/(s² + 6s + 25)

Partial fractions

s² + 6s + 25 = (s + 3)² + 4²
G(s) = 2·(s + 3)/((s + 3)² + 4²) + (0/4)·4/((s + 3)² + 4²)

Impulse response

g(t) = e−3t·(2·cos 4t)
g(0.25) = 0.5104

Answer

g(0.25) = 0.5104
31.Block diagrams

Answer: ess = 0.75

Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).

Velocity error constant

Kᵥ = lims→0 s·G(s) = 8/3 = 2.667

Steady-state error

ess = A/Kᵥ = 2/2.667 = 0.75

Answer

ess = 0.75
32.Block diagrams

Answer: ω = 7.483 rad/s

Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).

Characteristic equation

s³ + 15s² + 56s + K = 0

Routh array

s³ | 1 56
s² | 15 K
s¹ | (15·56 − K)/15
s⁰ | K

Stability limit

All entries of the first column must be positive.

Kmax = 15·56 = 840

Frequency of oscillation

15·(jω)² + 840 = 0 → ω² = 840/15 = 56
ω = 7.483 rad/s

Answer

ω = 7.483 rad/s
33.Step response

Answer: σ = −6

Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 75 → ωₙ = 8.66 rad/s
2ζωₙ = 12 → ζ = 12/(2·8.66) = 0.6928

Result

σ = −ζωₙ = −12/2 = −6

Answer

σ = −6
34.Step response

Answer: ts = 4 s

Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).

Closed-loop transfer function

T(s) = 45/(s² + 2s + 45): ωₙ = 6.708 rad/s, ζ = 0.1491

Settling time

ζωₙ = 2/2 = 1
ts ≈ 4/(ζωₙ) = 4 s

Answer

ts = 4 s

For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.

35.Bode plots & margins

Answer: |H| = −12.19 dB

Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.

Corner frequency

ωc = 1/(RC) = 1/(3300·4.7 × 10⁻⁸) = 6447 rad/s
fc = 1026 Hz → f/fc = 260/1026 = 0.2534

Gain

|H| = 0.2534/√(1 + 0.2534²) = 0.2456
20·log₁₀(0.2456) = −12.19 dB

Answer

|H| = −12.19 dB
36.Bode plots & margins

Answer: ωc = 8.509 rad/s

Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.

Gain crossover

ω⁴ + 16ω² − 6400 = 0
ω² = (−16 + √(256 + 4·6400))/2 = 72.4
ωc = 8.509 rad/s

Answer

ωc = 8.509 rad/s
37.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 1

Routh–Hurwitz criterion: a linear system is stable when every root of its characteristic polynomial lies in the left half of the s-plane. Write the coefficients alternately into the first two rows of the Routh array. Each entry of a new row comes from the two rows above it; for a cubic a₃s³ + a₂s² + a₁s + a₀ the s¹ entry is b₁ = (a₂·a₁ − a₃·a₀)/a₂. The number of sign changes in the first column equals the number of roots in the right half-plane. With a gain K in the coefficients, every first-column entry must stay positive: the gain that makes the s¹ entry zero is the stability limit, and the s² row then gives the frequency of the sustained oscillation.

Characteristic polynomial

(s − 3)(s + 2) + K(s + 10) = s² + (-1 + K)s + (10K − 6)

Conditions

For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.

-1 + K > 0 → K > 1
10K − 6 > 0 → K > 0.6
Kmin = 1

Answer

Kmin = 1
38.Stability: Routh–Hurwitz & root locus

Answer: K = 9

Root locus: as the gain K goes from 0 to ∞, the closed-loop poles, the roots of 1 + K·G(s) = 0, move from the open-loop poles to the open-loop zeros or to infinity. A point on the real axis belongs to the locus when the number of real poles and zeros to its right is odd. The n − m branches that go to infinity follow asymptotes at the angles (2q + 1)·180°/(n − m), which start from the centroid σₐ = (Σ poles − Σ zeros)/(n − m). Branches leave or join the real axis where dK/ds = 0. The angle condition ∠G(s) = 180° fixes the shape, and the magnitude condition K = 1/|G(s)| gives the gain at any point of the locus.

Break-away point

dK/ds = 0 for K = −(s + 3)(s + 9) gives 2s + 12 = 0.

sb = −6

Magnitude condition

K = |−6 + 3|·|−6 + 9| = 3·3 = 9

Answer

K = 9

For this gain the closed loop is critically damped (ζ = 1).

39.Steady-state error & PID control

Answer: ess = 0.0125

Steady-state error: in a unity-feedback loop with open-loop transfer function G(s) the error is E(s) = R(s)/(1 + G(s)), and by the final value theorem ess = lims→0 s·E(s). The system type is the number of poles of G(s) at s = 0. The error constants are Kp = lim G(s), Kv = lim s·G(s) and Ka = lim s²·G(s), all for s → 0. A step of size A gives ess = A/(1 + Kp), a ramp A·t gives A/Kv and a parabola A·t²/2 gives A/Ka. A type-0 loop follows a step with a finite error, a type-1 loop follows it exactly and a ramp with a finite error.

Acceleration constant

Ka = 10·4/1 = 40

Parabola error

ess = 0.5/40 = 0.0125

Answer

ess = 0.0125
40.Steady-state error & PID control

Answer: Kd = 3.035 s

P, PI and PID control: the controller C(s) = Kp + Ki/s + Kd·s acts on the error. A proportional gain makes the loop faster and reduces the error of a type-0 plant, but never removes it. The integral term adds a pole at s = 0: the system type rises by one, and a step is followed without error. The derivative term adds damping. To choose gains, write the closed-loop characteristic polynomial and compare it with s² + 2ζωns + ωn². Then %OS = 100·e−ζπ/√(1−ζ²), tp = π/ωd with ωd = ωn√(1 − ζ²), and ts ≈ 4/(ζωn).

Characteristic polynomial

s(s + 2) + 4(25 + Kds) = s² + (2 + 4Kd)s + 100

Natural frequency

ωₙ = √100 = 10 rad/s

Derivative gain

2 + 4Kd = 2·0.707·10 = 14.14
Kd = 3.035 s

Answer

Kd = 3.035 s

The derivative term adds damping without changing ωₙ.

This is independent practice material. ExamProblems is not affiliated with or endorsed by the organisation that runs this exam, and the exam name is used only to describe the content. The FE exam is run by NCEES, which is not affiliated with ExamProblems and does not endorse it.

ExamProblems · examproblems.com/courses/fe-exam-electrical