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.

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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​ = 5 ΩIR1​R3​ = 6 Ω+−V1​12 VR2​4 ΩAR4​5 ΩB

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

V1 = 12 VR1 = 5 ΩR2 = 4 ΩR3 = 6 ΩR4 = 5 Ω
IR1 = A
2.Series & parallel resistorsHard
R1​ = 6 ΩR3​ = 8 ΩR5​ = 15 Ω+−V1​15 VR2​2 ΩAR4​2 ΩBR6​10 Ω

Find the equivalent resistance Req seen by the source V1.

V1 = 15 VR1 = 6 ΩR2 = 2 ΩR3 = 8 ΩR4 = 2 ΩR5 = 15 ΩR6 = 10 Ω
Req = Ω
3.Nodal analysisMedium
−+Vs​ = 4 VIs​3 AAR1​12 ΩR2​12 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 VB (measured from ground).

Is = 3 AVs = 4 VR1 = 12 ΩR2 = 12 Ω
VB = V
4.Nodal analysisHard
R1​ = 20 ΩR2​ = 6 ΩR4​ = 6 Ω+−V1​15 VR3​4 ΩAR5​10 ΩBIs​6 A+−V2​18 V

Use nodal analysis to find the node voltage VA (measured from ground).

V1 = 15 VIs = 6 AV2 = 18 VR1 = 20 ΩR2 = 6 ΩR3 = 4 ΩR4 = 6 ΩR5 = 10 Ω
VA = V
5.Mesh analysisMedium
R1​ = 10 ΩR2​ = 15 Ω+−V1​15 VR3​12 ΩA+−V2​15 VI1​I2​

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

V1 = 15 VV2 = 15 VR1 = 10 ΩR2 = 15 ΩR3 = 12 Ω
I2 = A
6.Mesh analysisHard
R1​ = 20 ΩIR1​R2​ = 10 ΩR3​ = 3 Ω+−V1​10 VR4​3 ΩAR5​8 ΩB+−V2​9 VI1​I2​I3​

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

V1 = 10 VV2 = 9 VR1 = 20 ΩR2 = 10 ΩR3 = 3 ΩR4 = 3 ΩR5 = 8 Ω
IR1 = A
7.SuperpositionMedium
R1​ = 15 ΩR3​ = 8 Ω+−V1​10 VR2​12 ΩAR4​2 ΩIR4​BIs​1 A

Use superposition to find the power PR4 absorbed by R4.

V1 = 10 VIs = 1 AR1 = 15 ΩR2 = 12 ΩR3 = 8 ΩR4 = 2 Ω
PR4 = W
8.SuperpositionHard
R1​ = 6 ΩR2​ = 15 ΩR4​ = 30 Ω+−V1​12 VR3​8 ΩIR3​AIs​3 AB+−V2​15 V

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

V1 = 12 VIs = 3 AV2 = 15 VR1 = 6 ΩR2 = 15 ΩR3 = 8 ΩR4 = 30 Ω
IR3 = A
9.Thévenin & NortonMedium
R1​ = 20 ΩR3​ = 20 Ω+−V1​10 VR2​3 ΩARL​15 ΩIL​B

A load RL = 15 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 = 20 ΩR2 = 3 ΩR3 = 20 ΩRL = 15 Ω
IL = A
10.Thévenin & NortonHard
R1​ = 3 ΩR3​ = 10 Ω+−V1​12 VR2​10 ΩAIs​2.5 ABRL​3 ΩIL​

A load RL = 3 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.

V1 = 12 VIs = 2.5 AR1 = 3 ΩR2 = 10 ΩR3 = 10 ΩRL = 3 Ω
PL = W
11.RC & RL transientsMedium
+−Vs​24 Vt = 0R1​ = 6.8 kΩR2​ = 2.2 kΩC68 µF+−vC​

The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the time constant τ of the circuit after the switch closes.

Vs = 24 VR1 = 6.8 kΩR2 = 2.2 kΩC = 68 µF
τ = ms
12.RC & RL transientsHard
+−Vs​6 VR1​ = 33 ΩR2​ = 10 Ωcloses at t = 0L22 mHi(t)

The switch across R2 has been open for a long time, so the inductor current is steady. At t = 0 the switch closes and shorts out R2. Find the inductor current i at t = 1 ms.

Vs = 6 VR1 = 33 ΩR2 = 10 ΩL = 22 mHt = 1 ms
i = A
13.AC impedance & phasorsMedium
100 V200 HzR = 100 ΩL = 250 mHC = 100 µFI

A series RLC circuit is connected to a 100 V (rms), 200 Hz supply. Find the real power P taken from the supply.

V = 100 V (rms)f = 200 HzR = 100 ΩL = 250 mHC = 100 µF
P = W
14.AC impedance & phasorsHard
VfR = 100 ΩL = 50 mHC = 0.01 µF

A series RLC circuit has R = 100 Ω, L = 50 mH and C = 0.01 µF. Find its quality factor Q at resonance.

R = 100 ΩL = 50 mHC = 0.01 µF
Q =
15.AC power & power factorMedium
120 VR + jXP = 315 W3.5 A

An inductive load takes 3.5 A and 315 W from a 120 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.

V = 120 VI = 3.5 AP = 315 W
R = Ω
16.AC power & power factorHard
P = 15 kWQ1​S1​φ1​QC​

A single-phase load takes 15 kW at a power factor of 0.75 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 = 15 kWpf1 = 0.75pf2 = 0.95V = 400 Vf = 50 Hz
C = µF
17.Three-phase powerMedium
ABCVL​ = 480 VIL​ = ?M18.5 kW, η = 94%cos φ = 0.86

A three-phase induction motor delivers 18.5 kW of mechanical output power at full load. At this load its efficiency is 94% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 18.5 kWη = 94%cos φ = 0.86VL = 480 V
IL = A
18.Three-phase powerHard
P = 120 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 120 kW at a power factor of 0.8 lagging from a 400 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 120 kWpf1 = 0.8pf2 = 0.9VL = 400 Vf = 50 Hz
Cph = µF
19.TransformersMedium
RL​ = 10 Ω41960 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 4:1 has 960 V applied to its primary winding. A 10 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 4:1V1 = 960 VRL = 10 Ω
I1 = A
20.TransformersHard
pf 0.85250 kVA, 25% load

A single-phase 250 kVA transformer has a core loss of 2000 W and a copper loss of 3000 W at rated load. It supplies 25% of its rated kVA at a power factor of 0.85 lagging. Find the total losses Ploss in watts at this load.

S = 250 kVAPcore = 2000 WPcu,fl = 3000 Wx = 0.25pf = 0.85
Ploss = W
21.Op-amp circuitsMedium
−+Vout​Rf​ = 10 kΩR1​ = 4.7 kΩVin​ = −1.4 V

The op-amp is ideal and runs from ±15 V supplies. Find the output voltage Vout.

Vin = −1.4 VR1 = 4.7 kΩRf = 10 kΩ
Vout = V
22.Op-amp circuitsHard
Vin​V1​Vout​stage 1: invertingR1​ = 47 kΩRf1​ = 100 kΩstage 2: non-invertingR2​ = 1 kΩRf2​ = 4.7 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 = 1 kΩRf2 = 4.7 kΩVin = 0.85 V
Vout = V
23.Diodes & transistorsMedium
RC​ = 3.3 kΩβ = 120VCC​ = 15 VRB​ = 470 kΩ

An NPN transistor with β = 120 is biased by a base resistor RB = 470 kΩ from the supply VCC = 15 V. The collector resistor is RC = 3.3 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 = 470 kΩRC = 3.3 kΩβ = 120VBE = 0.7 V
VCE = V
24.Diodes & transistorsHard
+−≥ 20 VR = ?3.3 V≤ 30 mA

A 3.3 V Zener regulator must supply up to 30 mA to a load. The input can drop to 20 V, and the Zener needs at least 5 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 20 VVZ = 3.3 VIL,max = 30 mAIZ,min = 5 mA
Rmax = Ω
25.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −43 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −438 bit
U =
26.Number systemsHard
4406 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 4406 items its own binary code of the same length. What is the smallest number of bits b needed?

4406
b = bit
27.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = ABC + BC + AB′ + B′C′
L =
28.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​FFQ3​Q3​FFQ4​Q4​CLKMOD-25

A mod-25 counter is built from the minimum number of flip-flops. How many of the possible states U are never used?

N = 25
U =
29.Laplace transformsMedium
t (s)f(t)t = 0.2f(0.2) = ?

Find f(t) for F(s) = (6s + 13)/(s² + 8s + 16) and evaluate f(0.2).

F(s) = (6s + 13)/(s² + 8s + 16)t = 0.2 s
f(0.2) =
30.Laplace transformsHard
t (s)y(t)t = 0.2y(0.2) = ?

Solve y″ + 9y′ + 18y = 0 with y(0) = 4 and y′(0) = 13 using the Laplace transform, and find y(0.2).

y″ + 9y′ + 18y = 0y(0) = 4y′(0) = 13t = 0.2 s
y(0.2) =
31.Block diagramsMedium
R(s)+8(s + 9)(s + 3)(s + 8)G(s)C(s)−

A unity negative feedback loop has G(s) = 8(s + 9)/((s + 3)(s + 8)). A step of size 10 is applied at the reference. Find the steady-state error ess.

G(s) = 8(s + 9)/((s + 3)(s + 8))R = 10
ess =
32.Block diagramsHard
R(s)+80s(s + 2)G(s)C(s)2H−

The plant G(s) = 80/(s(s + 2)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their real part σ.

G(s) = 80/(s(s + 2))H = 2
σ =
33.Step responseMedium
t (s)y(t)2?

A system has the transfer function G(s) = 162/(s² + 6s + 81) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 162/(s² + 6s + 81)
%OS = %
34.Step responseHard
R(s)+Ks(s + 16)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 16)). Find the gain K that gives a step response with 2 % overshoot.

G(s) = K/(s(s + 16))%OS = 2 %
K =
35.Bode plots & marginsMedium
vin​R = 1000 ΩL = 47 mHvout​

The RL high-pass filter shown has R = 1000 Ω, L = 47 mH. A sinusoidal input of 850 Hz is applied. Find the gain |H| = |vout/vin| in dB.

R = 1000 ΩL = 47 mHf = 850 Hz
|H| = 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) = 70/(s(s + 3)(s + 10)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 70/(s(s + 3)(s + 10))
ω180 = rad/s
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 3)(s + 4)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 3)(s + 4)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.

L(s) = K/(s(s + 3)(s + 4))
ω = rad/s
38.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6j1

The root locus of L(s) = K(s + 5)/(s(s + 3)) leaves the real axis between s = 0 and s = −3, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 5)/(s(s + 3))
si =
39.Steady-state error & PID controlMedium
r(t) = 10 + 1t+5(s + 2)s(s + 4)(s + 10)C(s)−

The input r(t) = 10 + 1t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 5(s + 2)/(s(s + 4)(s + 10)). Find the steady-state error ess.

G(s) = 5(s + 2)/(s(s + 4)(s + 10))r(t) = 10 + 1t
ess =
40.Steady-state error & PID controlHard
r(t) = 5t+5 + 5/s20(s + 1)(s + 10)C(s)−

A PI controller C(s) = 5 + 5/s drives the plant G(s) = 20/((s + 1)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 5t.

C(s) = 5 + 5/sG(s) = 20/((s + 1)(s + 10))r(t) = 5t
ess =

3Solutions

1.Series & parallel resistors

Answer: IR1 = 1.513 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​ = 5 ΩR3​ = 6 Ω+−V1​12 VR2​4 ΩAR4​5 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 = 6 + 5 = 11 Ω
R2 ∥ 11 = 4·11/(4+11) = 2.933 Ω
Req = R1 + 2.933 = 5 + 2.933 = 7.933 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 = 12/7.933 = 1.513 A

Answer

IR1 = 1.513 A
2.Series & parallel resistors

Answer: Req = 7.662 Ω

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

R1​ = 6 ΩR3​ = 8 ΩR5​ = 15 Ω+−V1​15 VR2​2 ΩAR4​2 ΩBR6​10 Ω

Combine resistors from the far end

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

R5 + R6 = 15 + 10 = 25 Ω
R4 ∥ 25 = 1.852 Ω
R3 + 1.852 = 9.852 Ω
R2 ∥ 9.852 = 1.663 Ω
Req = R1 + 1.663 = 6 + 1.663 = 7.662 Ω

Answer

Req = 7.662 Ω
3.Nodal analysis

Answer: VB = 20 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 + 4
VA/R1 + VB/R2 = Is

Solve

VA/12 + (VA + 4)/12 = 3
VA = (3 − 4/12)/(1/12 + 1/12) = 16 V
VB = 16 + 4 = 20 V

Answer

VB = 20 V
4.Nodal analysis

Answer: VA = 10.46 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.

R1​ = 20 ΩR2​ = 6 ΩR4​ = 6 Ω+−V1​15 VR3​4 ΩAR5​10 ΩBIs​6 A+−V2​18 V

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 15 V, V2 = 18 V). Unknowns: VA, VB.

KCL at node A

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

VA/R3 + (VA − V1)/R1 + (VA − VB)/R2 = 0
VA/4 + (VA − 15)/20 + (VA − VB)/6 = 0
0.4667·VA − 0.1667·VB = 0.75

KCL at node B

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

VB/R5 + (VB − VA)/R2 + (VB − V2)/R4 = Is
VB/10 + (VB − VA)/6 + (VB − 18)/6 = 6
−0.1667·VA + 0.4333·VB = 9

Solve

Δ = (0.4667)(0.4333) − (−0.1667)(−0.1667) = 0.1744
VA = Δ1/Δ = 1.825/0.1744 = 10.46 V
VB = Δ2/Δ = 4.325/0.1744 = 24.79 V

Answer

VA = 10.46 V
5.Mesh analysis

Answer: I2 = −0.3333 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​ = 15 Ω+−V1​15 VR3​12 ΩA+−V2​15 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
10·I1 + 12(I1 − I2) = 15
22·I1 − 12·I2 = 15

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
15·I2 + 12(I2 − I1) = (−15)
−12·I1 + 27·I2 = −15

Solve

Δ = (22)(27) − (−12)(−12) = 450
I1 = Δ1/Δ = 225/450 = 0.5 A
I2 = Δ2/Δ = −150/450 = −0.3333 A

Answer

I2 = −0.3333 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: IR1 = 0.3886 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​ = 10 ΩR3​ = 3 Ω+−V1​10 VR4​3 ΩAR5​8 ΩB+−V2​9 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
20·I1 + 3(I1 − I2) = 10
23·I1 − 3·I2 = 10

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
10·I2 + 3(I2 − I1) + 8(I2 − I3) = 0
−3·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
3·I3 + 8(I3 − I2) = (−9)
−8·I2 + 11·I3 = −9

Solve

Δ = det of the coefficient matrix = 3742
I1 = Δ1/Δ = 1454/3742 = 0.3886 A
I2 = Δ2/Δ = −1326/3742 = −0.3544 A
I3 = Δ3/Δ = −4026/3742 = −1.076 A

Answer

IR1 = I1 = 0.3886 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: PR4 = 0.7524 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​ = 15 ΩR3​ = 8 Ω+−V1​10 VR2​12 ΩAR4​2 ΩBOPEN

Case 1 · V1 acting alone

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

R3+R4 = 8+2 = 10 Ω
R2 ∥ (R3+R4) = 12·10/(12+10) = 5.455 Ω
VA = 10·5.455/(15+5.455) = 2.667 V
VB = VA·R4/(R3+R4) = 2.667·2/10 = 0.5333 V
IR4′ = (VB − 0)/R4 = (0.5333 − 0)/2 = 0.2667 A
R1​ = 15 ΩR3​ = 8 ΩSHORTR2​12 ΩAR4​2 ΩBIs​1 A

Case 2 · Is acting alone

V₁ becomes a short, so R₁ is in parallel with R₂. Iₛ splits at B between R₄ and the path through R₃. 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 ∥ R2 = 15·12/(15+12) = 6.667 Ω
Req = R4 ∥ (R3+R1∥R2) = 2 ∥ 14.67 = 1.76 Ω
VB = −Is·Req = −1·1.76 = −1.76 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = −1.76·6.667/14.67 = −0.8 V
IR4″ = (VB − 0)/R4 = (−1.76 − 0)/2 = −0.88 A

Add the current contributions

IR4 = IR4′ + IR4″ = 0.2667 + (−0.88) = −0.6133 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.

PR4 = IR4²·R4 = (−0.6133)²·2 = 0.7524 W

Answer

PR4 = 0.7524 W

Adding the powers of the separate cases would give 1.691 W, which is wrong.

8.Superposition

Answer: IR3 = 1.726 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​ = 6 ΩR2​ = 15 ΩR4​ = 30 Ω+−V1​12 VR3​8 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) = 8 ∥ 45 = 6.792 Ω
VA = 12·6.792/(6+6.792) = 6.372 V
VB = VA·R4/(R2+R4) = 6.372·30/45 = 4.248 V
IR3′ = (VA − 0)/R3 = (6.372 − 0)/8 = 0.7965 A
R1​ = 6 ΩR2​ = 15 ΩR4​ = 30 ΩSHORTR3​8 ΩAIs​3 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 = 3.429 Ω, so the left path is R2+3.429 = 18.43 Ω
Req = R4 ∥ 18.43 = 11.42 Ω
VB = Is·Req = 3·11.42 = 34.25 V
VA = VB·3.429/18.43 = 6.372 V
IR3″ = (VA − 0)/R3 = (6.372 − 0)/8 = 0.7965 A
R1​ = 6 ΩR2​ = 15 ΩR4​ = 30 ΩSHORTR3​8 ΩAOPENB+−V2​15 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 = 6·8/(6+8) = 3.429 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 15·18.43/48.43 = 5.708 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 5.708·3.429/18.43 = 1.062 V
IR3‴ = (VA − 0)/R3 = (1.062 − 0)/8 = 0.1327 A

Add the contributions

IR3 = IR3′ + IR3″ + IR3‴
IR3 = 0.7965 + 0.7965 + 0.1327 = 1.726 A

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

9.Thévenin & Norton

Answer: IL = 0.03468 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​ = 20 ΩSHORTR2​3 ΩAabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 20·3/(20+3) = 2.609 Ω
Rth = (R1 ∥ R2) + R3 = 2.609 + 20 = 22.61 Ω
R1​ = 20 ΩR3​ = 20 Ω+−V1​10 VR2​3 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) = 10·3/(20+3) = 1.304 V
Rth​ = 22.61 Ω+−Vth​1.304 VRL​15 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) = 1.304/(22.61 + 15) = 0.03468 A
10.Thévenin & Norton

Answer: PL = 20.48 W

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​ = 3 ΩR3​ = 10 ΩSHORTR2​10 ΩAOPENBabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 3·10/(3+10) = 2.308 Ω
Rth = (R1 ∥ R2) + R3 = 2.308 + 10 = 12.31 Ω
R1​ = 3 ΩR3​ = 10 Ω+−V1​12 VR2​10 ΩAIs​2.5 ABab+−Vth​

Find Vth, the open-circuit voltage

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

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 9.231 + 2.5·2.308 = 15 V
Vth = VB = VA + Is·R3 = 15 + 2.5·10 = 40 V
Rth​ = 12.31 Ω+−Vth​40 VRL​3 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) = 40/(12.31 + 3) = 2.613 A
PL = IL²·RL = 2.613²·3 = 20.48 W
11.RC & RL transients

Answer: τ = 113 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.

Thévenin equivalent seen by the capacitor

Short the source to find Rth; open the capacitor to find the voltage it will settle at.

Rth = R1 ∥ R2 = 6.8·2.2/(6.8 + 2.2) = 1.662 kΩ
vC(∞) = 24·2.2/(6.8 + 2.2) = 5.867 V

Time constant

τ = Rth·C = 1.662·68 = 113 ms

Answer

τ = 113 ms
12.RC & RL transients

Answer: i = 0.1724 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.

Initial value, final value, time constant

Before t = 0 both resistors are in the loop; afterwards the closed switch bypasses R₂.

i(0) = 6/(33 + 10) = 0.1395 A
i(∞) = 6/33 = 0.1818 A
τ = L/R = 22/33 = 0.6667 ms
i(t) = 0.1818 + (0.1395 − 0.1818)·e−t/τ

Current at the given time

i(1) = 0.1818 + (−0.04228)·e−1/0.6667 = 0.1724 A

Answer

i = 0.1724 A
13.AC impedance & phasors

Answer: P = 9.638 W

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

XL = 314.2 Ω, XC = 7.958 Ω

Impedance

|Z| = √(100² + (306.2)²) = 322.1 Ω

Real power

Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.

I = 100/322.1 = 0.3104 A
P = I²·R = 0.3104²·100 = 9.638 W

Answer

P = 9.638 W
14.AC impedance & phasors

Answer: Q = 22.36

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|.

Resonant frequency

f0 = 1/(2π·√(0.05·0.01·10⁻⁶)) = 7118 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/100)·√(0.05/(0.01·10⁻⁶)) = 22.36

Answer

Q = 22.36
15.AC power & power factor

Answer: R = 25.71 Ω

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.

Resistance from the real power

R = P/I² = 315/3.5² = 25.71 Ω

Answer

R = 25.71 Ω
16.AC power & power factor

Answer: C = 165.1 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 = 15·tan(41.41°) = 13.23 kvar
Q2 = P·tan φ2 = 15·tan(18.19°) = 4.93 kvar
QC = Q1 − Q2 = 8.298 kvar

Capacitance

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

C = QC/(2πf·V²) = 8298/(2π·50·400²) = 165.1 µF

Answer

C = 165.1 µF
17.Three-phase power

Answer: IL = 27.53 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 = 18.5/0.94 = 19.68 kW

Line current

IL = Pin/(√3·VL·cos φ) = 19681/(√3·480·0.86) = 27.53 A

Answer

IL = 27.53 A
18.Three-phase power

Answer: Cph = 634.3 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 = 120·tan(acos 0.8) = 120·0.75 = 90 kvar
Q2 = 120·tan(acos 0.9) = 120·0.4843 = 58.12 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 31.88 kvar

Capacitance per phase

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

V = 400/√3 = 230.9 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 10627/(314.2·230.9²) = 634.3 µF

Answer

Cph = 634.3 µF
19.Transformers

Answer: I1 = 6 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 = 960/4 = 240 V
I2 = 240/10 = 24 A

Primary current

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

I1 = I2/4 = 6 A

Answer

I1 = 6 A

Check: the load seen from the primary is a²·RL = 160 Ω, and 960/160 gives the same current.

20.Transformers

Answer: Ploss = 2188 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 = 0.0625·3000 = 187.5 W

Total losses

The core loss is the same at any load.

Ploss = 2000 + 187.5 = 2188 W

Answer

Ploss = 2188 W

The efficiency at this load is η = 96.05%.

21.Op-amp circuits

Answer: Vout = −4.379 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.

Inverting input follows the input

V₋ = V₊ = −1.4 V
I = V₋/R1 = −1.4/4.7 = −0.2979 mA

Output voltage

The current through R₁ also flows through Rf.

Vout = V₋ + I·Rf = −1.4 + (−0.2979)·10 = −4.379 V

Answer

Vout = −4.379 V
22.Op-amp circuits

Answer: Vout = −10.31 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 + 4.7/1 = 5.7

Stage by stage

V1 = −2.128·0.85 = −1.809 V
Vout = 5.7·(−1.809) = −10.31 V

Answer

Vout = −10.31 V
23.Diodes & transistors

Answer: VCE = 2.951 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)/470 kΩ = 30.43 µA

Collector current

IC = β·IB = 120·30.43 µA = 3.651 mA

Collector–emitter voltage

VCE = 15 − 3.651 mA·3.3 kΩ = 2.951 V

Answer

VCE = 2.951 V
24.Diodes & transistors

Answer: Rmax = 477.1 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 = 30 + 5 = 35 mA

Largest resistance

Rmax = 16.7/0.035 = 477.1 Ω

Answer

Rmax = 477.1 Ω
25.Number systems

Answer: U = 213

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

43 = 0010 1011
inverted: 1101 0100
+1 → 1101 0101

Read it as unsigned

1101 0101₂ = 256 − 43 = 213

Answer

U = 213
26.Number systems

Answer: b = 13 bit

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.

Powers of two

212 = 4096 < 4406
213 = 8192 ≥ 4406

Answer

b = 13
27.Digital logic: gates, K-maps & flip-flops

Answer: L = 6

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

ABC → m(7)
BC → m(3, 7)
AB′ → m(4, 5)
B′C′ → m(0, 4)
ABC00011110011001130214151706

Group and read off

F = B′C′ + BC + AC

Count the literals

L = 6

Answer

L = 6

Another grouping of the same size exists; it has the same number of literals.

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 ≥ 25 → n = 5 (2⁵ = 32)

Unused states

U = 32 − 25 = 7

Answer

U = 7
29.Laplace transforms

Answer: f(0.2) = 1.707

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 + 8s + 16 = (s + 4)²

Partial fractions

6s + 13 = 6(s + 4) + −11
F(s) = 6/(s + 4) − 11/(s + 4)²

Inverse transform

f(t) = 6·e−4t − 11·t·e−4t
f(0.2) = (6 − 11·0.2)·e−4·0.2 = 1.707

Answer

f(0.2) = 1.707
30.Laplace transforms

Answer: y(0.2) = 4.259

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

s²Y − 4s − 13 + 9(sY − 4) + 18Y = 0
Y(s) = (4s + 49)/((s + 3)(s + 6))

Partial fractions

A = 37/3 = 12.33
B = 25/(−3) = −8.333

Result

y(t) = 12.33·e−3t − 8.333·e−6t
y(0.2) = 4.259

Answer

y(0.2) = 4.259

Check: y(0) = A + B = 4.

31.Block diagrams

Answer: ess = 2.5

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) = 8·9/(3·8) = 3

Steady-state error

ess = A/(1 + Kₚ) = 10/(1 + 3) = 2.5

Answer

ess = 2.5
32.Block diagrams

Answer: σ = −1

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

Characteristic equation

s(s + 2) + 80·2 = 0 → s² + 2s + 160 = 0

Roots

s = −1 ± j·√(160 − 1) = −1 ± j12.61
σ = −1

Answer

σ = −1
33.Step response

Answer: %OS = 32.93 %

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 = 81 → ωₙ = 9 rad/s
2ζωₙ = 6 → ζ = 6/(2·9) = 0.3333

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−1.111 = 32.93 %

Answer

%OS = 32.93 %
34.Step response

Answer: K = 105.3

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) = K/(s² + 16s + K)

Damping ratio from the overshoot

ζ = −ln(0.02)/√(π² + ln²(0.02)) = 0.7797

Gain

ωₙ = 16/(2ζ) = 10.26 rad/s
K = ωₙ² = 105.3

Answer

K = 105.3
35.Bode plots & margins

Answer: |H| = −12.27 dB

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

Corner frequency

ωc = R/L = 1000/0.047 = 21277 rad/s
fc = 3386 Hz → f/fc = 850/3386 = 0.251

Gain

|H| = 0.251/√(1 + 0.251²) = 0.2435
20·log₁₀(0.2435) = −12.27 dB

Answer

|H| = −12.27 dB
36.Bode plots & margins

Answer: ω180 = 5.477 rad/s

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

Phase crossover

(ω/3)·(ω/10) = 1 → ω180 = √(3·10) = 5.477 rad/s

Answer

ω180 = 5.477 rad/s
37.Stability: Routh–Hurwitz & root locus

Answer: ω = 3.464 rad/s

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 + 7s² + 12s + K = 0
s³: 1, 12
s²: 7, K
s¹: (7·12 − K)/7
s⁰: K

Critical gain

K = 7·12 = 84

Auxiliary equation

At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.

7s² + 84 = 0 → s = ±j√12
ω = 3.464 rad/s

Answer

ω = 3.464 rad/s
38.Stability: Routh–Hurwitz & root locus

Answer: si = −8.162

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 + 3s)/(s + 5)

Stationary points

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

(2s + 3)(s + 5) − (s² + 3s) = s² + 10s + 15 = 0
s = −5 ± √10 = −1.838, −8.162

Break-in point

si = −8.162

Answer

si = −8.162

Between the two break points the locus is a circle centred on the zero, with radius √10 = 3.162.

39.Steady-state error & PID control

Answer: ess = 4

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.

Velocity constant

Kv = lim s·G(s) = 0.25

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 1/0.25 = 4

Answer

ess = 4
40.Steady-state error & PID control

Answer: ess = 0.5

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).

Velocity constant

s·C(s) = 5s + 5 → 5 as s → 0.

Kv = 5·20/10 = 10

Ramp error

ess = 5/10 = 0.5

Answer

ess = 0.5

The PI controller removes the step error completely; a ramp is followed with a constant lag.

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