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​ = 12 ΩR3​ = 5 Ω+−V1​18 VR2​10 ΩIR2​AR4​3 ΩB

Find the current IR2 through R2.

V1 = 18 VR1 = 12 ΩR2 = 10 ΩR3 = 5 ΩR4 = 3 Ω
IR2 = A
2.Series & parallel resistorsHard
R1​ = 20 ΩR3​ = 5 ΩR5​ = 6 Ω+−V1​5 VR2​2 ΩIR2​AR4​10 ΩBR6​15 Ω

Find the current IR2 through R2.

V1 = 5 VR1 = 20 ΩR2 = 2 ΩR3 = 5 ΩR4 = 10 ΩR5 = 6 ΩR6 = 15 Ω
IR2 = A
3.Nodal analysisMedium
−+Vs​ = 6 VIs​3 AAR1​15 ΩR2​20 ΩB

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 VA (measured from ground).

Is = 3 AVs = 6 VR1 = 15 ΩR2 = 20 Ω
VA = V
4.Nodal analysisHard
R1​ = 4 Ω−+Vs​ = 4 V+−V1​36 VR2​4 ΩAR3​3 ΩBIs​2.5 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 VA (measured from ground).

V1 = 36 VIs = 2.5 AVs = 4 VR1 = 4 ΩR2 = 4 ΩR3 = 3 Ω
VA = V
5.Mesh analysisMedium
R1​ = 20 ΩR2​ = 5 Ω+−V1​30 VR3​15 ΩA+−V2​12 VI1​I2​

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

V1 = 30 VV2 = 12 VR1 = 20 ΩR2 = 5 ΩR3 = 15 Ω
I2 = A
6.Mesh analysisHard
R1​ = 10 ΩR2​ = 5 ΩIR2​R3​ = 4 Ω+−V1​30 VR4​12 ΩAR5​5 ΩB+−V2​24 VI1​I2​I3​

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

V1 = 30 VV2 = 24 VR1 = 10 ΩR2 = 5 ΩR3 = 4 ΩR4 = 12 ΩR5 = 5 Ω
IR2 = 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 = 15 V and Is = 2 mA, I = 4.35 mA; with Vs = 4 V and Is = 2 mA, I = 1.6 mA. Find the load current I when Vs = 4 V and Is = 8 mA.

15 V, 2 mA → 4.35 mA4 V, 2 mA → 1.6 mAVs = 4 V, Is = 8 mA
I = mA
8.SuperpositionHard
R1​ = 3 ΩR2​ = 4 ΩIR2​R4​ = 20 Ω+−V1​18 VR3​3 ΩAIs​4 AB+−V2​36 V

Use superposition to find the power PR2 absorbed by R2.

V1 = 18 VIs = 4 AV2 = 36 VR1 = 3 ΩR2 = 4 ΩR3 = 3 ΩR4 = 20 Ω
PR2 = W
9.Thévenin & NortonMedium
R1​ = 10 ΩR3​ = 20 Ω+−V1​30 VR2​12 ΩAab

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

V1 = 30 VR1 = 10 ΩR2 = 12 ΩR3 = 20 Ω
IN = A
10.Thévenin & NortonHard
R1​ = 12 ΩR3​ = 8 Ω+−V1​24 VR2​4 ΩAIs​5 ABab

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

V1 = 24 VIs = 5 AR1 = 12 ΩR2 = 4 ΩR3 = 8 Ω
Vth = V
11.RC & RL transientsMedium
+−Vs​5 Vt = 0R = 50 ΩL47 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the voltage vL across the inductor at t = 1 ms.

Vs = 5 VR = 50 ΩL = 47 mHt = 1 ms
vL = V
12.RC & RL transientsHard
+−Vs​12 Vt = 0R = 2.2 kΩC22 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 7.2 V? Find t.

Vs = 12 VR = 2.2 kΩC = 22 µFvC(0) = 4 V
t = ms
13.AC impedance & phasorsMedium
50 V1000 HzR = 60 ΩL = 50 mHC = 10 µF

A series RLC circuit is connected to a 50 V (rms), 1000 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 50 V (rms)f = 1000 HzR = 60 ΩL = 50 mHC = 10 µF
φ = °
14.AC impedance & phasorsHard
10 Vf0​R = 5 ΩL = 200 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 = 5 ΩL = 200 mHC = 10 µF
VL = V
15.AC power & power factorMedium
230 V118 kW, 0.75217 kW, 0.95I

Two loads are connected in parallel to a 230 V single-phase supply: an 18 kW load at a power factor of 0.75 lagging and a 17 kW load at a power factor of 0.95 leading. Find the overall power factor cos φ seen by the supply.

V = 230 VP1 = 18 kW, pf1 = 0.75 laggingP2 = 17 kW, pf2 = 0.95 leading
cos φ =
16.AC power & power factorHard
P = 15 kWQ1​S1​φ1​QC​

A single-phase load takes 15 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 0.98. Find the capacitance C needed.

P = 15 kWpf1 = 0.7pf2 = 0.98V = 400 Vf = 50 Hz
C = µF
17.Three-phase powerMedium
ABCVL​ = 480 VZ = 12 + j16 Ωper phase

A balanced Δ-connected load with an impedance of 12 + j16 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.

VL = 480 VZ = 12 + j16 Ω per phase
P = kW
18.Three-phase powerHard
P = 100 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 100 kW at a power factor of 0.65 lagging from a 690 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 100 kWpf1 = 0.65pf2 = 0.98VL = 690 Vf = 60 Hz
Cph = µF
19.TransformersMedium
4160 V480 VIHV​ = ?ILV​10 kVA

A single-phase 10 kVA, 4160/480 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.

S = 10 kVAV1 = 4160 VV2 = 480 V
IHV = A
20.TransformersHard
pf 0.9525 kVA, 75% load

A single-phase 25 kVA transformer has a core loss of 150 W and a copper loss of 450 W at rated load. It supplies 75% of its rated kVA at a power factor of 0.95 lagging. Find the total losses Ploss in watts at this load.

S = 25 kVAPcore = 150 WPcu,fl = 450 Wx = 0.75pf = 0.95
Ploss = W
21.Op-amp circuitsMedium
−+Vout​Rf​ = ?R1​ = 100 kΩVin​

Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 25.

R1 = 100 kΩAv = 25
Rf = kΩ
22.Op-amp circuitsHard
Vin​V1​Vout​stage 1: invertingR1​ = 47 kΩRf1​ = 100 kΩstage 2: non-invertingR2​ = 100 kΩRf2​ = 22 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. The input is Vin = −0.85 V. Find the output voltage Vout.

R1 = 47 kΩRf1 = 100 kΩR2 = 100 kΩRf2 = 22 kΩVin = −0.85 V
Vout = V
23.Diodes & transistorsMedium
RC​ = 1.5 kΩβ = 75VCC​ = 12 VRB​ = 150 kΩ

An NPN transistor with β = 75 is biased by a base resistor RB = 150 kΩ from the supply VCC = 12 V. The collector resistor is RC = 1.5 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 = 12 VRB = 150 kΩRC = 1.5 kΩβ = 75VBE = 0.7 V
VCE = V
24.Diodes & transistorsHard
+−≥ 12 VR = ?3.3 V≤ 40 mA

A 3.3 V Zener regulator must supply up to 40 mA to a load. The input can drop to 12 V, and the Zener needs at least 5 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 12 VVZ = 3.3 VIL,max = 40 mAIZ,min = 5 mA
Rmax = Ω
25.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −112 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1128 bit
U =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1110 0101 << 2. What is the result's value R in decimal?

1110 0101<< 2
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = A′BC′ + AB′C′ + A′B′C + A′BC to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = A′BC′ + AB′C′ + A′B′C + A′BC
L =
28.Digital logic: gates, K-maps & flip-flopsHard
CLKJK1234567

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0JK
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
29.Laplace transformsMedium
ReIm−1−2−3−4−5−61

F(s) = 8/(s² + 9s + 18). Its inverse transform has the form f(t) = A·e−3t + B·e−6t. Find the coefficient A by partial fractions.

F(s) = 8/(s² + 9s + 18)
A =
30.Laplace transformsHard
t (s)y(t)t = 0.75y(0.75) = ?

A system obeys y″ + 5y′ + 4y = 4·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(0.75).

y″ + 5y′ + 4y = 4·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.75 s
y(0.75) =
31.Block diagramsMedium
R(s)+11(s + 5)(s + 3)(s + 4)G(s)C(s)−

A unity negative feedback loop has G(s) = 11(s + 5)/((s + 3)(s + 4)). A step of size 5 is applied at the reference. Find the steady-state error ess.

G(s) = 11(s + 5)/((s + 3)(s + 4))R = 5
ess =
32.Block diagramsHard
R(s)+Kcontroller+D(s)+3s + 6plantC(s)−

A proportional controller K drives the plant G(s) = 3/(s + 6) in a unity negative feedback loop. A disturbance step of size 0.5 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.005.

G(s) = 3/(s + 6)D = 0.5css ≤ 0.005
K =
33.Step responseMedium
t (s)y(t)3ts​ = ?

A system has the transfer function G(s) = 195/(s² + 7s + 65) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 195/(s² + 7s + 65)
ts = s
34.Step responseHard
R(s)+55s(s + 2)G(s)C(s)−

A unity negative feedback loop has G(s) = 55/(s(s + 2)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 55/(s(s + 2))
ts = s
35.Bode plots & marginsMedium
|H| (dB)0.11101001000−80−60−40−2002040ω (rad/s)ω = 5

Find the exact magnitude |G(jω)| of G(s) = 462/((s + 3)(s + 14)) at ω = 5 rad/s, in dB.

G(s) = 462/((s + 3)(s + 14))ω = 5 rad/s
|G(jω)| = dB
36.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​

A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 5)(s + 6)). Find the gain K that gives a gain margin of 12 dB.

L(s) = K/(s(s + 5)(s + 6))GM = 12 dB
K =
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 93.751s(s + 5)(s + 10)C(s)−

The loop in the diagram runs with K = 93.75 and L(s) = K/(s(s + 5)(s + 10)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 5)(s + 10))K = 93.75
m =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−8−9−101j1j2−1.5

For the loop L(s) = K/(s(s + 3)(s + 8)), find the gain K that puts a closed-loop pole at s = −1.5.

L(s) = K/(s(s + 3)(s + 8))s = −1.5
K =
39.Steady-state error & PID controlMedium
r(t) = 2·t²/2+4(s + 5)s²(s + 10)C(s)−

The parabolic input r(t) = 2·t²/2 is applied to a unity-feedback loop with G(s) = 4(s + 5)/(s²(s + 10)). Find the steady-state error ess.

G(s) = 4(s + 5)/(s²(s + 10))r(t) = 2·t²/2
ess =
40.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s1s + 2C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 1/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 10 rad/s and ζ = 0.5. Find the integral gain Ki.

G(s) = 1/(s + 2)ωₙ = 10 rad/sζ = 0.5
Ki =

3Solutions

1.Series & parallel resistors

Answer: IR2 = 0.4865 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​ = 12 ΩR3​ = 5 Ω+−V1​18 VR2​10 ΩAR4​3 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 = 5 + 3 = 8 Ω
R2 ∥ 8 = 10·8/(10+8) = 4.444 Ω
Req = R1 + 4.444 = 12 + 4.444 = 16.44 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 = 18/16.44 = 1.095 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 = 18 − 1.095·12 = 4.865 V
IR2 = VA/R2 = 4.865/10 = 0.4865 A

Answer

IR2 = 0.4865 A
2.Series & parallel resistors

Answer: IR2 = 0.1969 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​ = 6 Ω+−V1​5 VR2​2 ΩAR4​10 ΩBR6​15 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 = 6 + 15 = 21 Ω
R4 ∥ 21 = 6.774 Ω
R3 + 6.774 = 11.77 Ω
R2 ∥ 11.77 = 1.71 Ω
Req = R1 + 1.71 = 20 + 1.71 = 21.71 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 = 5/21.71 = 0.2303 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 = 5 − 0.2303·20 = 0.3937 V
IR2 = VA/R2 = 0.3937/2 = 0.1969 A

Answer

IR2 = 0.1969 A
3.Nodal analysis

Answer: VA = 23.14 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 + Vs = VA + 6
VA/R1 + VB/R2 = Is

Solve

VA/15 + (VA + 6)/20 = 3
VA = (3 − 6/20)/(1/15 + 1/20) = 23.14 V
VB = 23.14 + 6 = 29.14 V

Answer

VA = 23.14 V
4.Nodal analysis

Answer: VA = 12.2 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 + 4
(VA − V1)/R1 + VA/R2 + VB/R3 = Is

Solve

(VA − 36)/4 + VA/4 + (VA + 4)/3 = 2.5
VA·(1/4 + 1/4 + 1/3) = 2.5 + 36/4 − 4/3 = 10.17
VA = 10.17/0.8333 = 12.2 V
VB = 16.2 V

Answer

VA = 12.2 V
5.Mesh analysis

Answer: I2 = 0.06316 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​ = 20 ΩR2​ = 5 Ω+−V1​30 VR3​15 ΩA+−V2​12 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
20·I1 + 15(I1 − I2) = 30
35·I1 − 15·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 + R3(I2 − I1) = −V2
5·I2 + 15(I2 − I1) = (−12)
−15·I1 + 20·I2 = −12

Solve

Δ = (35)(20) − (−15)(−15) = 475
I1 = Δ1/Δ = 420/475 = 0.8842 A
I2 = Δ2/Δ = 30/475 = 0.06316 A

Answer

I2 = 0.06316 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: IR2 = 0.239 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​ = 10 ΩR2​ = 5 ΩR3​ = 4 Ω+−V1​30 VR4​12 ΩAR5​5 ΩB+−V2​24 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
10·I1 + 12(I1 − I2) = 30
22·I1 − 12·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
5·I2 + 12(I2 − I1) + 5(I2 − I3) = 0
−12·I1 + 22·I2 − 5·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
4·I3 + 5(I3 − I2) = (−24)
−5·I2 + 9·I3 = −24

Solve

Δ = det of the coefficient matrix = 2510
I1 = Δ1/Δ = 3750/2510 = 1.494 A
I2 = Δ2/Δ = 600/2510 = 0.239 A
I3 = Δ3/Δ = −6360/2510 = −2.534 A

Answer

IR2 = I2 = 0.239 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: I = 3.4 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
15·k₁ + 2·k₂ = 4.35
4·k₁ + 2·k₂ = 1.6

Solve for the constants

Δ = (15)(2) − (2)(4) = 22
k₁ = Δ1/Δ = 5.5/22 = 0.25
k₂ = Δ2/Δ = 6.6/22 = 0.3
k₁ = 0.25 mA/V, k₂ = 0.3

Predict the new current

I = 0.25·4 + 0.3·8 = 3.4 mA

Answer

I = 3.4 mA
8.Superposition

Answer: PR2 = 17.28 W

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​ = 3 ΩR2​ = 4 ΩR4​ = 20 Ω+−V1​18 VR3​3 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) = 3 ∥ 24 = 2.667 Ω
VA = 18·2.667/(3+2.667) = 8.471 V
VB = VA·R4/(R2+R4) = 8.471·20/24 = 7.059 V
IR2′ = (VA − VB)/R2 = (8.471 − 7.059)/4 = 0.3529 A
R1​ = 3 ΩR2​ = 4 ΩR4​ = 20 ΩSHORTR3​3 ΩAIs​4 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. The arrow of Iₛ points down, away from its node, so Iₛ draws current out of that node: its term enters with a minus sign.

R1 ∥ R3 = 1.5 Ω, so the left path is R2+1.5 = 5.5 Ω
Req = R4 ∥ 5.5 = 4.314 Ω
VB = −Is·Req = −4·4.314 = −17.25 V
VA = VB·1.5/5.5 = −4.706 V
IR2″ = (VA − VB)/R2 = (−4.706 − −17.25)/4 = 3.137 A
R1​ = 3 ΩR2​ = 4 ΩR4​ = 20 ΩSHORTR3​3 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 = 3·3/(3+3) = 1.5 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 36·5.5/25.5 = 7.765 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 7.765·1.5/5.5 = 2.118 V
IR2‴ = (VA − VB)/R2 = (2.118 − 7.765)/4 = −1.412 A

Add the current contributions

IR2 = IR2′ + IR2″ + IR2‴ = 0.3529 + 3.137 + (−1.412) = 2.078 A

Power in the resistor

Power depends on the square of the current, so it does not obey superposition: add the currents first, then square.

PR2 = IR2²·R2 = 2.078²·4 = 17.28 W

Answer

PR2 = 17.28 W

Adding the powers of the separate cases would give 47.84 W, which is wrong.

9.Thévenin & Norton

Answer: IN = 0.6429 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​ = 10 ΩR3​ = 20 ΩSHORTR2​12 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 = 10·12/(10+12) = 5.455 Ω
Rth = (R1 ∥ R2) + R3 = 5.455 + 20 = 25.45 Ω
R1​ = 10 ΩR3​ = 20 Ω+−V1​30 VR2​12 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) = 30·12/(10+12) = 16.36 V
IN​0.6429 ARth​25.45 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 = 16.36/25.45 = 0.6429 A
10.Thévenin & Norton

Answer: Vth = 61 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​ = 12 ΩR3​ = 8 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 = 12·4/(12+4) = 3 Ω
Rth = (R1 ∥ R2) + R3 = 3 + 8 = 11 Ω
R1​ = 12 ΩR3​ = 8 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) = 6 + 5·3 = 21 V
Vth = VB = VA + Is·R3 = 21 + 5·8 = 61 V
11.RC & RL transients

Answer: vL = 1.726 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.

Time constant and final current

Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.

τ = L/R = 47 mH / 50 Ω = 0.94 ms
i(∞) = Vs/R = 5/50 = 0.1 A

Inductor voltage

At t = 0 the whole supply voltage appears across the inductor; it then decays to zero.

vL(t) = Vs·e−t/τ = 5·e−1/0.94 = 1.726 V

Answer

vL = 1.726 V
12.RC & RL transients

Answer: t = 24.72 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 = 12 V
τ = R·C = 2.2 kΩ · 22 µF = 48.4 ms
vC(t) = 12 + (4 − 12)·e−t/τ

Solve for the time

7.2 = 12 + (−8)·e−t/τ → e−t/τ = (7.2 − 12)/(4 − 12) = 0.6
t = τ·ln((4 − 12)/(7.2 − 12)) = 48.4·ln(1.667) = 24.72 ms

Answer

t = 24.72 ms
13.AC impedance & phasors

Answer: φ = 78.63 °

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|.

Reactances

ω = 2πf = 2π·1000 = 6283 rad/s
XL = ωL = 6283·0.05 = 314.2 Ω
XC = 1/(ωC) = 1/(6283·10·10⁻⁶) = 15.92 Ω

Impedance

In series the impedances add as complex numbers: Z = R + j(XL − XC).

|Z| = √(R² + (XL − XC)²) = √(60² + (298.2)²) = 304.2 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(298.2/60) = 78.63°

Answer

φ = 78.63°
14.AC impedance & phasors

Answer: VL = 282.8 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/√(0.2·10·10⁻⁶) = 707.1 rad/s
I = V/R = 10/5 = 2 A

Voltage across the inductor

XL = ω₀L = 707.1·0.2 = 141.4 Ω
VL = I·XL = 2·141.4 = 282.8 V

Answer

VL = 282.8 V

This is Q = 28.28 times the supply voltage.

15.AC power & power factor

Answer: cos φ = 0.9594

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 of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 18·tan(arccos 0.75) = 15.87 kvar
Q2 = −17·tan(arccos 0.95) = −5.588 kvar

Totals

P = 18 + 17 = 35 kW
Q = 15.87 + (−5.588) = 10.29 kvar
S = √(35² + 10.29²) = 36.48 kVA

Overall power factor

cos φ = P/S = 35/36.48 = 0.9594 lagging

Answer

cos φ = 0.9594
16.AC power & power factor

Answer: C = 243.8 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 = 15·tan(45.57°) = 15.3 kvar
Q2 = P·tan φ2 = 15·tan(11.48°) = 3.046 kvar
QC = Q1 − Q2 = 12.26 kvar

Capacitance

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

C = QC/(2πf·V²) = 12257/(2π·50·400²) = 243.8 µF

Answer

C = 243.8 µF
17.Three-phase power

Answer: P = 20.74 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 = 480 V

Phase current

|Z| = √(12² + 16²) = 20 Ω
Iph = 480/20 = 24 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·24²·12 = 20.74 kW

Answer

P = 20.74 kW

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

18.Three-phase power

Answer: Cph = 538.2 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 = 100·tan(acos 0.65) = 100·1.169 = 116.9 kvar
Q2 = 100·tan(acos 0.98) = 100·0.2031 = 20.31 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 96.61 kvar

Capacitance per phase

In a Y bank each capacitor is connected between a line and the star point.

V = 690/√3 = 398.4 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 32202/(377·398.4²) = 538.2 µF

Answer

Cph = 538.2 µF
19.Transformers

Answer: IHV = 2.404 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.

IHV = S/V = 10000/4160 = 2.404 A

Answer

IHV = 2.404 A
20.Transformers

Answer: Ploss = 403.1 W

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 at this load

Pcu = x²·Pcu,fl = 0.5625·450 = 253.1 W

Total losses

The core loss is the same at any load.

Ploss = 150 + 253.1 = 403.1 W

Answer

Ploss = 403.1 W

The efficiency at this load is η = 97.79%.

21.Op-amp circuits

Answer: Rf = 2400 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 = (25 − 1)·100 = 2400 kΩ

Answer

Rf = 2400 kΩ
22.Op-amp circuits

Answer: Vout = 2.206 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.

Gain of each stage

A1 = −Rf1/R1 = −100/47 = −2.128
A2 = 1 + Rf2/R2 = 1 + 22/100 = 1.22

Stage by stage

V1 = −2.128·(−0.85) = 1.809 V
Vout = 1.22·1.809 = 2.206 V

Answer

Vout = 2.206 V
23.Diodes & transistors

Answer: VCE = 3.525 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 = (12 − 0.7)/150 kΩ = 75.33 µA

Collector current

IC = β·IB = 75·75.33 µA = 5.65 mA

Collector–emitter voltage

VCE = 12 − 5.65 mA·1.5 kΩ = 3.525 V

Answer

VCE = 3.525 V
24.Diodes & transistors

Answer: Rmax = 193.3 Ω

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 current

IR = 40 + 5 = 45 mA

Largest resistance

Rmax = 8.7/0.045 = 193.3 Ω

Answer

Rmax = 193.3 Ω
25.Number systems

Answer: U = 144

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.

Invert and add one

112 = 0111 0000
inverted: 1000 1111
+1 → 1001 0000

Read it as unsigned

1001 0000₂ = 256 − 112 = 144

Answer

U = 144
26.Number systems

Answer: R = 148

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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.

1110 0101 << 2
= 1001 0100

Value of the result

1001 0100₂ = 148

Answer

R = 148
27.Digital logic: gates, K-maps & flip-flops

Answer: L = 7

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

A′BC′ → m(2)
AB′C′ → m(4)
A′B′C → m(1)
A′BC → m(3)
ABC00011110010011131214050706

Group and read off

F = A′C + A′B + AB′C′

Count the literals

L = 7

Answer

L = 7
28.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 5

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.

Q after every edge

A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 1, K = 1 → Q = 1
2: J = 0, K = 0 → Q = 1
3: J = 1, K = 0 → Q = 1
4: J = 0, K = 0 → Q = 1
5: J = 1, K = 1 → Q = 0
6: J = 0, K = 1 → Q = 0
7: J = 0, K = 0 → Q = 0

First change from 1 to 0

clock edge 5

Answer

clock edge 5
29.Laplace transforms

Answer: A = 2.667

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 + 9s + 18 = (s + 3)(s + 6)
F(s) = A/(s + 3) + B/(s + 6)

Cover-up rule

Multiply by (s + 3) and set s = −3: the B-term vanishes.

A = 8/(s + 6) |s = −3 = 8/3 = 2.667

Answer

A = 2.667
(B = −2.667, so f(t) = 2.667·e−3t − 2.667·e−6t)
30.Laplace transforms

Answer: y(0.75) = 0.3868

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) = 4/(s(s + 1)(s + 4))

Partial fractions

A = 4/(1·4) = 1
B = 4/((−1)·3) = −1.333
C = 4/((−4)·(−3)) = 0.3333

Output

y(t) = 1 − 1.333·e−t + 0.3333·e−4t
y(0.75) = 0.3868

Answer

y(0.75) = 0.3868
31.Block diagrams

Answer: ess = 0.8955

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).

Position error constant

Kₚ = G(0) = 11·5/(3·4) = 4.583

Steady-state error

ess = A/(1 + Kₚ) = 5/(1 + 4.583) = 0.8955

Answer

ess = 0.8955
32.Block diagrams

Answer: K = 98

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).

Disturbance response

css = D·G(0)/(1 + K·G(0)) = 0.5·3/(6 + 3K)

Solve for K

6 + 3K = 0.5·3/0.005 = 300
K = (300 − 6)/3 = 98

Answer

K = 98
33.Step response

Answer: ts = 1.143 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 = 65 → ωₙ = 8.062 rad/s
2ζωₙ = 7 → ζ = 7/(2·8.062) = 0.4341

Result

ζωₙ = 3.5 → ts ≈ 4/3.5 = 1.143 s

Answer

ts = 1.143 s
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) = 55/(s² + 2s + 55): ωₙ = 7.416 rad/s, ζ = 0.1348

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: |G(jω)| = 14.53 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.

Multiply the magnitudes

|G(j5)| = 462/(√(5² + 3²)·√(5² + 14²)) = 462/(5.831·14.87) = 5.33

Result

20·log₁₀(5.33) = 14.53 dB

Answer

|G(jω)| = 14.53 dB
36.Bode plots & margins

Answer: K = 82.89

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.

Phase crossover

(ω/5)·(ω/6) = 1 → ω180 = √(5·6) = 5.477 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 25)·√(ω² + 36)) = K/(5·6·(5 + 6)) = K/330

Required gain

330/K = 1012/20 = 3.981
K = 330/3.981 = 82.89

Answer

K = 82.89
37.Stability: Routh–Hurwitz & root locus

Answer: m = 8

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 + 15s² + 50s + K = 0
s³: 1, 50
s²: 15, K
s¹: (15·50 − K)/15
s⁰: K

Largest stable gain

Kmax = 15·50 = 750

Factor

m = 750/93.75 = 8

Answer

m = 8

This factor is the gain margin as a ratio: 18.06 dB.

38.Stability: Routh–Hurwitz & root locus

Answer: K = 14.63

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.

Magnitude condition

K = |s|·|s + 3|·|s + 8|

Evaluate at the point

K = 1.5·1.5·6.5 = 14.63

Answer

K = 14.63

Check: s(s + 3)(s + 8) + K is zero at s = −1.5.

39.Steady-state error & PID control

Answer: ess = 1

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 = 4·5/10 = 2

Parabola error

ess = 2/2 = 1

Answer

ess = 1
40.Steady-state error & PID control

Answer: Ki = 100

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 + 2).

s² + (2 + 1Kp)s + 1Ki

Target polynomial

s² + 2·0.5·10s + 10² = s² + 10s + 100

Match the coefficients

1Ki = 100 → Ki = 100

Answer

Ki = 100

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