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​ = 3 ΩR3​ = 15 Ω+−V1​6 VR2​5 ΩIR2​AR4​8 ΩB

Find the current IR2 through R2.

V1 = 6 VR1 = 3 ΩR2 = 5 ΩR3 = 15 ΩR4 = 8 Ω
IR2 = A
2.Series & parallel resistorsHard
R1​ = 20 ΩR3​ = 5 ΩR5​ = 25 Ω+−V1​15 VR2​20 ΩIR2​AR4​2 ΩBR6​4 Ω

Find the current IR2 through R2.

V1 = 15 VR1 = 20 ΩR2 = 20 ΩR3 = 5 ΩR4 = 2 ΩR5 = 25 ΩR6 = 4 Ω
IR2 = A
3.Nodal analysisMedium
R1​ = 2 ΩIR1​R3​ = 4 Ω+−V1​9 VR2​20 ΩAR4​15 ΩBIs​2 A

Use nodal analysis to find the current IR1 through R1 in the direction of the arrow.

V1 = 9 VIs = 2 AR1 = 2 ΩR2 = 20 ΩR3 = 4 ΩR4 = 15 Ω
IR1 = A
4.Nodal analysisHard
R1​ = 12 Ω−+Vs​ = 6 V+−V1​6 VR2​6 ΩAR3​20 ΩBIs​0.5 A

The voltage source Vₛ between A and B has its + terminal at B, so it makes VB − VA = Vₛ. Use nodal analysis with a supernode to find the node voltage VB (measured from ground).

V1 = 6 VIs = 0.5 AVs = 6 VR1 = 12 ΩR2 = 6 ΩR3 = 20 Ω
VB = V
5.Mesh analysisMedium
R1​ = 12 ΩR2​ = 5 Ω+−V1​6 VR3​15 ΩA+−V2​10 VI1​I2​

Use mesh analysis to find the mesh current I2 (clockwise).

V1 = 6 VV2 = 10 VR1 = 12 ΩR2 = 5 ΩR3 = 15 Ω
I2 = A
6.Mesh analysisHard
R1​ = 25 ΩR2​ = 8 ΩR3​ = 2 Ω+−V1​30 VR4​5 ΩAR5​8 ΩIR5​B+−V2​5 VI1​I2​I3​

Use mesh analysis to find the current IR5 through R5 in the direction of the arrow.

V1 = 30 VV2 = 5 VR1 = 25 ΩR2 = 8 ΩR3 = 2 ΩR4 = 5 ΩR5 = 8 Ω
IR5 = A
7.SuperpositionMedium
linear circuitVs​Is​IRL​

A linear resistive circuit contains one voltage source Vs and one current source Is. The current I through a load resistor is measured twice: with Vs = 8 V and Is = 1 mA, I = 8.2 mA; with Vs = 2 V and Is = 4 mA, I = 2.8 mA. Find the load current I when Vs = 12 V and Is = 5 mA.

8 V, 1 mA → 8.2 mA2 V, 4 mA → 2.8 mAVs = 12 V, Is = 5 mA
I = mA
8.SuperpositionHard
R1​ = 15 ΩR2​ = 30 ΩR4​ = 2 Ω+−V1​10 VR3​5 ΩIR3​AIs​6 AB+−V2​6 V

Use superposition to find the current IR3 through R3 in the direction of the arrow.

V1 = 10 VIs = 6 AV2 = 6 VR1 = 15 ΩR2 = 30 ΩR3 = 5 ΩR4 = 2 Ω
IR3 = A
9.Thévenin & NortonMedium
R1​ = 8 ΩR3​ = 15 Ω+−V1​10 VR2​4 ΩARL​12 ΩIL​B

A load RL = 12 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 10 VR1 = 8 ΩR2 = 4 ΩR3 = 15 ΩRL = 12 Ω
IL = A
10.Thévenin & NortonHard
R1​ = 20 ΩR3​ = 4 Ω+−V1​24 VR2​4 ΩAIs​5 ABab

Find the Norton current IN: the current that flows from a to b if the terminals are shorted.

V1 = 24 VIs = 5 AR1 = 20 ΩR2 = 4 ΩR3 = 4 Ω
IN = A
11.RC & RL transientsMedium
R = 20 Ωt = 0L47 mHi(t)

The inductor carries a steady current of 3 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. Find the current i at t = 5 ms.

i(0) = 3 AR = 20 ΩL = 47 mHt = 5 ms
i = A
12.RC & RL transientsHard
+−Vs​24 Vt = 0R = 4.7 kΩC220 µF+−vC​vC​(0) = 4 V

Just before the switch closes at t = 0, the capacitor holds 4 V with the polarity shown. How long does it take for vC to reach 16 V? Find t.

Vs = 24 VR = 4.7 kΩC = 220 µFvC(0) = 4 V
t = ms
13.AC impedance & phasorsMedium
Vf0​ = 2000 HzRL = 50 mHC = ?

A series RLC circuit with L = 50 mH must resonate at 2000 Hz. Find the capacitance C needed.

f0 = 2000 HzL = 50 mH
C = µF
14.AC impedance & phasorsHard
10 Vf0​R = 10 ΩL = 50 mHC = 10 µF

A series RLC circuit is driven by a 10 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 10 V (rms)R = 10 ΩL = 50 mHC = 10 µF
VL = V
15.AC power & power factorMedium
230 VZ7 + j24 ΩI

A load with impedance Z = 7 + j24 Ω is connected to a 230 V (rms) single-phase supply. Find the real power P it absorbs.

V = 230 VZ = 7 + j24 Ω
P = W
16.AC power & power factorHard
P = 45 kWQ1​S1​φ1​QC​

A single-phase load takes 45 kW at a power factor of 0.7 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the capacitance C needed.

P = 45 kWpf1 = 0.7pf2 = 0.95V = 400 Vf = 50 Hz
C = µF
17.Three-phase powerMedium
ABCVL​ = 690 VIL​ = ?M11 kW, η = 95%cos φ = 0.84

A three-phase induction motor delivers 11 kW of mechanical output power at full load. At this load its efficiency is 95% and its power factor is 0.84 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.

Pout = 11 kWη = 95%cos φ = 0.84VL = 690 V
IL = A
18.Three-phase powerHard
P = 150 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 150 kW at a power factor of 0.65 lagging from a 480 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 150 kWpf1 = 0.65pf2 = 0.95VL = 480 Vf = 50 Hz
QC = kvar
19.TransformersMedium
RL​ = 30 Ω2014800 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 20:1 has 4800 V applied to its primary winding. A 30 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 20:1V1 = 4800 VRL = 30 Ω
I1 = A
20.TransformersHard
250 kVA

A single-phase 250 kVA transformer has a core loss of 1000 W and a copper loss of 3000 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 250 kVAPcore = 1000 WPcu,fl = 3000 W
Sη,max = kVA
21.Op-amp circuitsMedium
−+Vout​R2​ = 2 kΩV1​ = 0.5 VR1​ = 4.7 kΩV2​ = 2.2 VR3​ = 4.7 kΩR4​ = 2 kΩ

The difference amplifier uses an ideal op-amp with ±15 V supplies, with R3 = R1 and R4 = R2. Find the output voltage Vout.

V1 = 0.5 VV2 = 2.2 VR1 = R3 = 4.7 kΩR2 = R4 = 2 kΩ
Vout = V
22.Op-amp circuitsHard
−+Vout​Rf​ = 2 kΩV1​ = 1.3 VR1​ = 47 kΩV2​ = 1.8 VR2​ = 2 kΩIf​

The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the current If through the feedback resistor, flowing towards the output.

V1 = 1.3 VV2 = 1.8 VR1 = 47 kΩR2 = 2 kΩRf = 2 kΩ
If = mA
23.Diodes & transistorsMedium
RC​ = 1.2 kΩβ = 75VCC​ = 15 VRB​ = 330 kΩ

An NPN transistor with β = 75 is biased by a base resistor RB = 330 kΩ from the supply VCC = 15 V. The collector resistor is RC = 1.2 kΩ and the emitter is grounded. Take VBE = 0.7 V and assume the transistor is in the active region. Find the collector–emitter voltage VCE.

VCC = 15 VRB = 330 kΩRC = 1.2 kΩβ = 75VBE = 0.7 V
VCE = V
24.Diodes & transistorsHard
+−≤ 22 V100 Ω5.1 V

A 5.1 V Zener regulator has a 100 Ω series resistor. The input can rise to 22 V, and the load may be disconnected. Find the largest power PZ,max the Zener must dissipate.

Vin,max = 22 VVZ = 5.1 VR = 100 Ω
PZ,max = mW
25.Number systemsMedium
place values in two's complement1−12816413201608141201

The 8-bit pattern 1110 0110 stores a signed integer in two's complement. What is its value N in decimal?

1110 01108 bit
N =
26.Number systemsHard
32 bit1 px640 px480 pxM = ? KiB

An uncompressed image is 640 × 480 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

640 × 480 px32 bit/px1 KiB = 1024 B
M = KiB
27.Digital logic: gates, K-maps & flip-flopsMedium
MUXC′I0​0I1​0I2​C′I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = C′, I₁ = 0, I₂ = 0, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = C′I₁ = 0I₂ = 0I₃ = C′
N =
28.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​FFQ3​Q3​CLKMOD-9

A mod-9 counter is built from the minimum number of flip-flops. How many of the possible states U are never used?

N = 9
U =
29.Laplace transformsMedium
t (s)f(t)t = 0.5f(0.5) = ?

Find f(t) for F(s) = (3s + 32)/(s² + 6s + 9) and evaluate f(0.5).

F(s) = (3s + 32)/(s² + 6s + 9)t = 0.5 s
f(0.5) =
30.Laplace transformsHard
t (s)y(t)t = 0.5y(0.5) = ?

A system obeys y″ + 7y′ + 10y = 25·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 1 for t ≥ 0. Use the Laplace transform to find the output y(0.5).

y″ + 7y′ + 10y = 25·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.5 s
y(0.5) =
31.Block diagramsMedium
R(s)+3s + 5G(s)C(s)0.25H−

A plant G(s) = 3/(s + 5) is controlled in a negative feedback loop whose sensor has the gain H = 0.25. Find the time constant τ of the closed loop.

G(s) = 3/(s + 5)H = 0.25
τ = s
32.Block diagramsHard
R(s)+10s + 3G₁+5s + 6G₂C(s)2H₁−0.2H₂−

Reduce the block diagram with G₁ = 10/(s + 3), G₂ = 5/(s + 6), an inner negative feedback H₁ = 2 around G₂ and an outer negative feedback H₂ = 0.2. Find the DC gain T(0) of C(s)/R(s).

G₁ = 10/(s + 3)G₂ = 5/(s + 6)H₁ = 2H₂ = 0.2
T(0) =
33.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 18/(s² + 3s + 18) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 18/(s² + 3s + 18)
ts = s
34.Step responseHard
k = 800 N/mc = 17 N·s/mm = 4 kgF = 25 Nx(t)

A mass m = 4 kg hangs on a spring k = 800 N/m with a viscous damper c = 17 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 25 N is suddenly applied. Find the time tp at which the displacement is largest.

m = 4 kgk = 800 N/mc = 17 N·s/mF = 25 N
tp = s
35.Bode plots & marginsMedium
vin​R = 3.3 kΩC = 2.2 nFvout​

The RC low-pass filter shown has R = 3.3 kΩ, C = 2.2 nF. A sinusoidal input of 66 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 3.3 kΩC = 2.2 nFf = 66 kHz
φ = °
36.Bode plots & marginsHard
|H| (dB)0.10.111101010010010001000−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 160/(s(s + 20)). Find the phase margin PM in degrees.

L(s) = 160/(s(s + 20))
PM = °
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 7201s(s + 8)(s + 10)C(s)−

The loop in the diagram runs with K = 720 and L(s) = K/(s(s + 8)(s + 10)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 8)(s + 10))K = 720
m =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−81j1j2

The root locus of L(s) = K(s + 7)/(s(s + 2)) leaves the real axis between s = 0 and s = −2, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 7)/(s(s + 2))
si =
39.Steady-state error & PID controlMedium
r(t) = 10·t²/2+4(s + 3)s²(s + 1)C(s)−

The parabolic input r(t) = 10·t²/2 is applied to a unity-feedback loop with G(s) = 4(s + 3)/(s²(s + 1)). Find the steady-state error ess.

G(s) = 4(s + 3)/(s²(s + 1))r(t) = 10·t²/2
ess =
40.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s2s + 1C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 2/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 3 rad/s and ζ = 0.8. Find the integral gain Ki.

G(s) = 2/(s + 1)ωₙ = 3 rad/sζ = 0.8
Ki =

3Solutions

1.Series & parallel resistors

Answer: IR2 = 0.6935 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​ = 3 ΩR3​ = 15 Ω+−V1​6 VR2​5 ΩAR4​8 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 = 15 + 8 = 23 Ω
R2 ∥ 23 = 5·23/(5+23) = 4.107 Ω
Req = R1 + 4.107 = 3 + 4.107 = 7.107 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 = 6/7.107 = 0.8442 A

Current through R₂

Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.

VA = V1 − IR1·R1 = 6 − 0.8442·3 = 3.467 V
IR2 = VA/R2 = 3.467/5 = 0.6935 A

Answer

IR2 = 0.6935 A
2.Series & parallel resistors

Answer: IR2 = 0.1527 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​ = 20 ΩR3​ = 5 ΩR5​ = 25 Ω+−V1​15 VR2​20 ΩAR4​2 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 = 25 + 4 = 29 Ω
R4 ∥ 29 = 1.871 Ω
R3 + 1.871 = 6.871 Ω
R2 ∥ 6.871 = 5.114 Ω
Req = R1 + 5.114 = 20 + 5.114 = 25.11 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 = 15/25.11 = 0.5973 A

Current through R₂

Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.

VA = V1 − IR1·R1 = 15 − 0.5973·20 = 3.054 V
IR2 = VA/R2 = 3.054/20 = 0.1527 A

Answer

IR2 = 0.1527 A
3.Nodal analysis

Answer: IR1 = −0.5437 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.

R1​ = 2 ΩR3​ = 4 Ω+−V1​9 VR2​20 ΩAR4​15 ΩBIs​2 A

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 9 V). Unknowns: VA, VB.

KCL at node A

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

VA/R2 + (VA − V1)/R1 + (VA − VB)/R3 = 0
VA/20 + (VA − 9)/2 + (VA − VB)/4 = 0
0.8·VA − 0.25·VB = 4.5

KCL at node B

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

VB/R4 + (VB − VA)/R3 = Is
VB/15 + (VB − VA)/4 = 2
−0.25·VA + 0.3167·VB = 2

Solve

Δ = (0.8)(0.3167) − (−0.25)(−0.25) = 0.1908
VA = Δ1/Δ = 1.925/0.1908 = 10.09 V
VB = Δ2/Δ = 2.725/0.1908 = 14.28 V

Answer

IR1 = (V1 − VA)/R1 = (9 − 10.09)/2 = −0.5437 A
4.Nodal analysis

Answer: VB = 8.333 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.

Form a supernode

The voltage source between A and B carries an unknown current, so KCL cannot be written at A or B separately. Enclose A, B and the source in one supernode: the current entering it equals the current leaving it.

VB = VA + 6
(VA − V1)/R1 + VA/R2 + VB/R3 = Is

Solve

(VA − 6)/12 + VA/6 + (VA + 6)/20 = 0.5
VA·(1/12 + 1/6 + 1/20) = 0.5 + 6/12 − 6/20 = 0.7
VA = 0.7/0.3 = 2.333 V
VB = 8.333 V

Answer

VB = 8.333 V
5.Mesh analysis

Answer: I2 = −0.5714 A

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​ = 12 ΩR2​ = 5 Ω+−V1​6 VR3​15 ΩA+−V2​10 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
12·I1 + 15(I1 − I2) = 6
27·I1 − 15·I2 = 6

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
5·I2 + 15(I2 − I1) = (−10)
−15·I1 + 20·I2 = −10

Solve

Δ = (27)(20) − (−15)(−15) = 315
I1 = Δ1/Δ = −30/315 = −0.09524 A
I2 = Δ2/Δ = −180/315 = −0.5714 A

Answer

I2 = −0.5714 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: IR5 = 0.5145 A

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​ = 25 ΩR2​ = 8 ΩR3​ = 2 Ω+−V1​30 VR4​5 ΩAR5​8 ΩB+−V2​5 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
25·I1 + 5(I1 − I2) = 30
30·I1 − 5·I2 = 30

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
8·I2 + 5(I2 − I1) + 8(I2 − I3) = 0
−5·I1 + 21·I2 − 8·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 + 8(I3 − I2) = (−5)
−8·I2 + 10·I3 = −5

Solve

Δ = det of the coefficient matrix = 4130
I1 = Δ1/Δ = 4180/4130 = 1.012 A
I2 = Δ2/Δ = 300/4130 = 0.07264 A
I3 = Δ3/Δ = −1825/4130 = −0.4419 A

Answer

IR5 = I2 − I3 = 0.5145 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: I = 13 mA

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.

Write the response as a sum of contributions

Each contribution is proportional to its own source, with a constant that depends only on the resistors.

I = k₁·Vs + k₂·Is
8·k₁ + 1·k₂ = 8.2
2·k₁ + 4·k₂ = 2.8

Solve for the constants

Δ = (8)(4) − (1)(2) = 30
k₁ = Δ1/Δ = 30/30 = 1
k₂ = Δ2/Δ = 6/30 = 0.2
k₁ = 1 mA/V, k₂ = 0.2

Predict the new current

I = 1·12 + 0.2·5 = 13 mA

Answer

I = 13 mA
8.Superposition

Answer: IR3 = −0.8252 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​ = 15 ΩR2​ = 30 ΩR4​ = 2 Ω+−V1​10 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 ∥ 32 = 4.324 Ω
VA = 10·4.324/(15+4.324) = 2.238 V
VB = VA·R4/(R2+R4) = 2.238·2/32 = 0.1399 V
IR3′ = (0 − VA)/R3 = (0 − 2.238)/5 = −0.4476 A
R1​ = 15 ΩR2​ = 30 ΩR4​ = 2 ΩSHORTR3​5 ΩAIs​6 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 = 3.75 Ω, so the left path is R2+3.75 = 33.75 Ω
Req = R4 ∥ 33.75 = 1.888 Ω
VB = Is·Req = 6·1.888 = 11.33 V
VA = VB·3.75/33.75 = 1.259 V
IR3″ = (0 − VA)/R3 = (0 − 1.259)/5 = −0.2517 A
R1​ = 15 ΩR2​ = 30 ΩR4​ = 2 ΩSHORTR3​5 ΩAOPENB+−V2​6 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 = 15·5/(15+5) = 3.75 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 6·33.75/35.75 = 5.664 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 5.664·3.75/33.75 = 0.6294 V
IR3‴ = (0 − VA)/R3 = (0 − 0.6294)/5 = −0.1259 A

Add the contributions

IR3 = IR3′ + IR3″ + IR3‴
IR3 = (−0.4476) + (−0.2517) + (−0.1259) = −0.8252 A

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

9.Thévenin & Norton

Answer: IL = 0.1124 A

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​ = 8 ΩR3​ = 15 ΩSHORTR2​4 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 = 8·4/(8+4) = 2.667 Ω
Rth = (R1 ∥ R2) + R3 = 2.667 + 15 = 17.67 Ω
R1​ = 8 ΩR3​ = 15 Ω+−V1​10 VR2​4 ΩAab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, no current flows through R₃, so there is no drop across it and V_b = V_A. R₁ and R₂ form a voltage divider.

Vth = V1·R2/(R1+R2) = 10·4/(8+4) = 3.333 V
Rth​ = 17.67 Ω+−Vth​3.333 VRL​12 ΩIL​

Connect the load to the Thévenin equivalent

The Thévenin source and R_th in series behave exactly like the original circuit at a–b, so the load sees a simple series loop.

IL = Vth/(Rth + RL) = 3.333/(17.67 + 12) = 0.1124 A
10.Thévenin & Norton

Answer: IN = 5.545 A

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​ = 4 ΩSHORTR2​4 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 = 20·4/(20+4) = 3.333 Ω
Rth = (R1 ∥ R2) + R3 = 3.333 + 4 = 7.333 Ω
R1​ = 20 ΩR3​ = 4 Ω+−V1​24 VR2​4 ΩAIs​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) = 4 + 5·3.333 = 20.67 V
Vth = VB = VA + Is·R3 = 20.67 + 5·4 = 40.67 V
IN​5.545 ARth​7.333 Ωab

Norton equivalent

A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.

IN = Vth/Rth = 40.67/7.333 = 5.545 A
11.RC & RL transients

Answer: i = 0.3573 A

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 = 47 mH / 20 Ω = 2.35 ms

Decay of the current

i(t) = 3·e−5/2.35 = 0.3573 A

Answer

i = 0.3573 A
12.RC & RL transients

Answer: t = 947.4 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.

Initial value, final value, time constant

The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.

vC(0) = 4 V, vC(∞) = Vs = 24 V
τ = R·C = 4.7 kΩ · 220 µF = 1034 ms
vC(t) = 24 + (4 − 24)·e−t/τ

Solve for the time

16 = 24 + (−20)·e−t/τ → e−t/τ = (16 − 24)/(4 − 24) = 0.4
t = τ·ln((4 − 24)/(16 − 24)) = 1034·ln(2.5) = 947.4 ms

Answer

t = 947.4 ms
13.AC impedance & phasors

Answer: C = 0.1267 µF

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 condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·2000 = 12566 rad/s
C = 1/(ω₀²·L) = 1/(12566²·0.05) = 0.1267 µF

Answer

C = 0.1267 µF
14.AC impedance & phasors

Answer: VL = 70.71 V

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 reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.05·10·10⁻⁶) = 1414 rad/s
I = V/R = 10/10 = 1 A

Voltage across the inductor

XL = ω₀L = 1414·0.05 = 70.71 Ω
VL = I·XL = 1·70.71 = 70.71 V

Answer

VL = 70.71 V

This is Q = 7.071 times the supply voltage.

15.AC power & power factor

Answer: P = 592.5 W

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.

Current

|Z| = √(7² + 24²) = 25 Ω
I = V/|Z| = 230/25 = 9.2 A

Real power

P = I²·R = 9.2²·7 = 592.5 W

Answer

P = 592.5 W
16.AC power & power factor

Answer: C = 619.1 µF

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 = 45·tan(45.57°) = 45.91 kvar
Q2 = P·tan φ2 = 45·tan(18.19°) = 14.79 kvar
QC = Q1 − Q2 = 31.12 kvar

Capacitance

A capacitor across the supply delivers Q = V²·ωC.

C = QC/(2πf·V²) = 31118/(2π·50·400²) = 619.1 µF

Answer

C = 619.1 µF
17.Three-phase power

Answer: IL = 11.53 A

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.

Electrical input power

Pin = Pout/η = 11/0.95 = 11.58 kW

Line current

IL = Pin/(√3·VL·cos φ) = 11579/(√3·690·0.84) = 11.53 A

Answer

IL = 11.53 A
18.Three-phase power

Answer: QC = 126.1 kvar

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 = 150·tan(acos 0.65) = 150·1.169 = 175.4 kvar
Q2 = 150·tan(acos 0.95) = 150·0.3287 = 49.3 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 126.1 kvar

Answer

QC = 126.1 kvar
19.Transformers

Answer: I1 = 0.4 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.

Secondary voltage and current

V2 = 4800/20 = 240 V
I2 = 240/30 = 8 A

Primary current

An ideal transformer has no losses: V₁·I₁ = V₂·I₂.

I1 = I2/20 = 0.4 A

Answer

I1 = 0.4 A

Check: the load seen from the primary is a²·RL = 12000 Ω, and 4800/12000 gives the same current.

20.Transformers

Answer: Sη,max = 144.3 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

x²·3000 = 1000 → x = √(1000/3000) = 0.5774

Load at maximum efficiency

Sη,max = x·S = 0.5774·250 = 144.3 kVA

Answer

Sη,max = 144.3 kVA
21.Op-amp circuits

Answer: Vout = 0.7234 V

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.

Voltage at the inputs

R₃ and R₄ carry the same current, so they form a voltage divider; feedback makes V₋ = V₊.

V₊ = V2·R4/(R3 + R4) = 2.2·2/(4.7 + 2) = 0.6567 V
V₋ = V₊

KCL at the inverting input

(0.5 − 0.6567)/4.7 = (0.6567 − Vout)/2
Vout = (R2/R1)·(V2 − V1) = (2/4.7)·(2.2 − 0.5) = 0.7234 V

Answer

Vout = 0.7234 V
22.Op-amp circuits

Answer: If = 0.9277 mA

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.

Input currents

V₋ = 0, so each input resistor has its full input voltage across it.

I1 = V1/R1 = 1.3/47 = 0.02766 mA
I2 = V2/R2 = 1.8/2 = 0.9 mA
If = I1 + I2 = 0.9277 mA

Answer

If = 0.9277 mA
23.Diodes & transistors

Answer: VCE = 11.1 V

BJT bias point: in the active region VBE = 0.7 V, IC = β·IB and IE = (β + 1)·IB. Find IB from the base loop with KVL, then IC, then VCE from the collector–emitter loop. A voltage divider at the base is replaced by its Thévenin equivalent VTH = VCC·R₂/(R₁ + R₂), RTH = R₁∥R₂; an emitter resistor appears (β + 1) times larger in the base loop.

Base current

IB = (VCC − VBE)/RB = (15 − 0.7)/330 kΩ = 43.33 µA

Collector current

IC = β·IB = 75·43.33 µA = 3.25 mA

Collector–emitter voltage

VCE = 15 − 3.25 mA·1.2 kΩ = 11.1 V

Answer

VCE = 11.1 V
24.Diodes & transistors

Answer: PZ,max = 861.9 mW

Zener regulator: a reverse-biased Zener diode holds its voltage VZ while its current stays above a minimum IZ,min. The series resistor R takes the difference between input and output: IR = (Vin − VZ)/R. That current splits between the load, IL = VZ/RL, and the Zener: IZ = IR − IL. Design for the worst cases: the lowest input with the largest load must still leave IZ,min, and the highest input with no load gives the largest Zener power PZ = VZ·IZ.

Worst-case Zener current

IZ = 16.9/100 = 169 mA

Zener power

PZ = 5.1·169 = 861.9 mW

Answer

PZ,max = 861.9 mW
25.Number systems

Answer: N = −26

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.

Add the place values

The other bits keep their normal positive values.

−128 + 64 + 32 + 4 + 2 = −26

Answer

N = −26
26.Number systems

Answer: M = 1200 KiB

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.

Bits and bytes

640·480·32 = 9830400 bit = 1228800 B

Convert to KiB

1228800/1024 = 1200 KiB

Answer

M = 1200 KiB
27.Digital logic: gates, K-maps & flip-flops

Answer: N = 2

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′C′ + ABC′

Rows with F = 1

ABC = 000, 110
N = 2

Answer

N = 2
28.Digital logic: gates, K-maps & flip-flops

Answer: U = 7

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.

Number of flip-flops

2ⁿ ≥ 9 → n = 4 (2⁴ = 16)

Unused states

U = 16 − 9 = 7

Answer

U = 7
29.Laplace transforms

Answer: f(0.5) = 3.235

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).

Factor the denominator

s² + 6s + 9 = (s + 3)²

Partial fractions

3s + 32 = 3(s + 3) + 23
F(s) = 3/(s + 3) + 23/(s + 3)²

Inverse transform

f(t) = 3·e−3t + 23·t·e−3t
f(0.5) = (3 + 23·0.5)·e−3·0.5 = 3.235

Answer

f(0.5) = 3.235
30.Laplace transforms

Answer: y(0.5) = 1.104

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).

Transform the equation

Y(s) = 25/(s(s + 2)(s + 5))

Partial fractions

A = 25/(2·5) = 2.5
B = 25/((−2)·3) = −4.167
C = 25/((−5)·(−3)) = 1.667

Output

y(t) = 2.5 − 4.167·e−2t + 1.667·e−5t
y(0.5) = 1.104

Answer

y(0.5) = 1.104
31.Block diagrams

Answer: τ = 0.1739 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).

Closed-loop transfer function

T(s) = G/(1 + G·H) = 3/(s + 5 + 0.75) = 3/(s + 5.75)

Result

τ = 1/5.75 = 0.1739 s

Answer

τ = 0.1739 s
32.Block diagrams

Answer: T(0) = 0.8621

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).

Inner loop

Gᵢ = G₂/(1 + G₂H₁) = 5/(s + 16)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 50/((s + 3)(s + 16) + 10)

DC gain

T(0) = 50/(3·16 + 10) = 50/58 = 0.8621

Answer

T(0) = 0.8621
33.Step response

Answer: ts = 2.667 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).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 18 → ωₙ = 4.243 rad/s
2ζωₙ = 3 → ζ = 3/(2·4.243) = 0.3536

Result

ζωₙ = 1.5 → ts ≈ 4/1.5 = 2.667 s

Answer

ts = 2.667 s
34.Step response

Answer: tp = 0.2247 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).

Natural frequency and damping ratio

ωₙ = √(k/m) = √(800/4) = 14.14 rad/s
ζ = c/(2√(km)) = 17/(2·√(800·4)) = 0.1503

Result

ωd = 13.98 rad/s
tp = π/ωd = 0.2247 s

Answer

tp = 0.2247 s
35.Bode plots & margins

Answer: φ = −71.63 °

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/(3300·2.2 × 10⁻⁹) = 137741 rad/s
fc = 21922 Hz → f/fc = 66000/21922 = 3.011

Phase

φ = − atan(3.011) = −71.63°

Answer

φ = −71.63°
36.Bode plots & margins

Answer: PM = 69.46 °

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 + 400ω² − 25600 = 0
ω² = (−400 + √(160000 + 4·25600))/2 = 56.12
ωc = 7.492 rad/s

Phase margin

∠L(jωc) = −90° − atan(7.492/20) = −110.5°
PM = 180° + ∠L = 69.46°

Answer

PM = 69.46°
37.Stability: Routh–Hurwitz & root locus

Answer: m = 2

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.

Routh array

s³ + 18s² + 80s + K = 0
s³: 1, 80
s²: 18, K
s¹: (18·80 − K)/18
s⁰: K

Largest stable gain

Kmax = 18·80 = 1440

Factor

m = 1440/720 = 2

Answer

m = 2

This factor is the gain margin as a ratio: 6.021 dB.

38.Stability: Routh–Hurwitz & root locus

Answer: si = −12.92

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.

Gain along the real axis

K(s) = −(s² + 2s)/(s + 7)

Stationary points

Differentiate with the quotient rule and set the numerator to zero.

(2s + 2)(s + 7) − (s² + 2s) = s² + 14s + 14 = 0
s = −7 ± √35 = −1.084, −12.92

Break-in point

si = −12.92

Answer

si = −12.92

Between the two break points the locus is a circle centred on the zero, with radius √35 = 5.916.

39.Steady-state error & PID control

Answer: ess = 0.8333

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 = 4·3/1 = 12

Parabola error

ess = 10/12 = 0.8333

Answer

ess = 0.8333
40.Steady-state error & PID control

Answer: Ki = 4.5

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

1 + C(s)G(s) = 0, multiplied by s(s + 1).

s² + (1 + 2Kp)s + 2Ki

Target polynomial

s² + 2·0.8·3s + 3² = s² + 4.8s + 9

Match the coefficients

2Ki = 9 → Ki = 4.5

Answer

Ki = 4.5

The integral term makes the loop type 1, so a step is followed without steady-state error.

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