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​ = 3 ΩR3​ = 4 Ω+−V1​18 VR2​5 ΩIR2​AR4​20 ΩB

Find the current IR2 through R2.

V1 = 18 VR1 = 3 ΩR2 = 5 ΩR3 = 4 ΩR4 = 20 Ω
IR2 = A
2.Series & parallel resistorsHard
R1​ = 6 ΩIR1​R3​ = 5 ΩR5​ = 30 Ω+−V1​18 VR2​25 ΩAR4​8 ΩBR6​30 Ω

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

V1 = 18 VR1 = 6 ΩR2 = 25 ΩR3 = 5 ΩR4 = 8 ΩR5 = 30 ΩR6 = 30 Ω
IR1 = A
3.Nodal analysisMedium
R1​2 Ω+−V1​30 VI1​R2​8 Ω+−V2​6 VR3​5 Ω+−V3​9 VA

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

V1 = 30 VR1 = 2 ΩV2 = 6 VR2 = 8 ΩV3 = 9 VR3 = 5 Ω
I1 = A
4.Nodal analysisHard
R1​ = 15 ΩR3​ = 10 Ω+−V1​15 VR2​2 ΩAR4​20 ΩBIs​?

The current source is adjusted until the node voltage at A is VA = 2 V. Use nodal analysis to find the source current Is (direction as shown).

V1 = 15 VR1 = 15 ΩR2 = 2 ΩR3 = 10 ΩR4 = 20 ΩVA = 2 V
Is = A
5.Mesh analysisMedium
R1​ = 3 ΩR2​ = 15 Ω+−V1​12 VR3​15 ΩIR3​A+−V2​6 VI1​I2​

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

V1 = 12 VV2 = 6 VR1 = 3 ΩR2 = 15 ΩR3 = 15 Ω
IR3 = A
6.Mesh analysisHard
R1​ = 15 ΩR2​ = 8 ΩR3​ = 25 Ω+−V1​24 VR4​30 ΩAR5​15 Ω+−VR5​B+−V2​10 VI1​I2​I3​

Use mesh analysis to find the voltage VR5 across R5 with the polarity shown.

V1 = 24 VV2 = 10 VR1 = 15 ΩR2 = 8 ΩR3 = 25 ΩR4 = 30 ΩR5 = 15 Ω
VR5 = V
7.SuperpositionMedium
R1​ = 8 ΩR3​ = 6 Ω+−V1​10 VR2​3 ΩAR4​15 Ω−+VR4​BIs​2.5 A

Use superposition to find the voltage VR4 across R4 with the polarity shown.

V1 = 10 VIs = 2.5 AR1 = 8 ΩR2 = 3 ΩR3 = 6 ΩR4 = 15 Ω
VR4 = V
8.SuperpositionHard
linear circuitV1​V2​Is​IRL​

A linear resistive circuit contains three independent sources: V1, V2 and Is. The current I through a load is measured three times. With all sources at their normal values, I = 21.5 mA. With V1 doubled and the other two unchanged, I = 31 mA. With the direction of Is reversed and the other two unchanged, I = 11.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 21.5 mA2·V1: I = 31 mA−Is: I = 11.5 mA
I = mA
9.Thévenin & NortonMedium
R1​ = 15 ΩR3​ = 4 Ω+−V1​6 VR2​12 ΩARL​5 ΩIL​B

A load RL = 5 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 6 VR1 = 15 ΩR2 = 12 ΩR3 = 4 ΩRL = 5 Ω
IL = A
10.Thévenin & NortonHard
R1​ = 8 ΩR3​ = 25 Ω+−V1​18 VR2​15 ΩAIs​2 ABab

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

V1 = 18 VIs = 2 AR1 = 8 ΩR2 = 15 ΩR3 = 25 Ω
IN = A
11.RC & RL transientsMedium
+−Vs​24 Vt = 0R = 22 ΩL100 mHi(t)

The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 95% of its final value? Find t.

Vs = 24 VR = 22 ΩL = 100 mH
t = ms
12.RC & RL transientsHard
+−Vs​24 Vt = 0R = 2.2 kΩC68 µF+−vC​vC​(0) = 15 V

Just before the switch closes at t = 0, the capacitor holds 15 V with the polarity shown. Find the capacitor voltage vC at t = 100 ms.

Vs = 24 VR = 2.2 kΩC = 68 µFvC(0) = 15 Vt = 100 ms
vC = V
13.AC impedance & phasorsMedium
100 V100 HzR = 50 ΩL = 200 mHC = 10 µFI

A series RLC circuit is connected to a 100 V (rms), 100 Hz supply. Find the power factor cos φ of the circuit.

V = 100 V (rms)f = 100 HzR = 50 ΩL = 200 mHC = 10 µF
cos φ =
14.AC impedance & phasorsHard
100 V500 HzR = 80 ΩL = 150 mHC = 100 µF

A series RLC circuit is connected to a 100 V (rms), 500 Hz supply. Find the resonant frequency f0 of the circuit.

V = 100 V (rms)f = 500 HzR = 80 ΩL = 150 mHC = 100 µF
f0 = Hz
15.AC power & power factorMedium
P = 40 kWQSφ

A single-phase motor takes 40 kW at a power factor of 0.6 lagging from a 240 V supply. Find the line current I.

P = 40 kWpf = 0.6V = 240 V
I = A
16.AC power & power factorHard
P = 5 kWQ1​S1​φ1​QC​

A single-phase load takes 5 kW at a power factor of 0.8 lagging from a 240 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the capacitance C needed.

P = 5 kWpf1 = 0.8pf2 = 0.92V = 240 Vf = 50 Hz
C = µF
17.Three-phase powerMedium
ABCVL​ = 400 VIL​ = 10 Acos φ = 0.95lagging

A balanced three-phase load draws a line current of 10 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.95 lagging. Find the total real power P drawn by the load in kW.

VL = 400 VIL = 10 Acos φ = 0.95
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 0.7 lagging from a 415 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 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 = 0.7pf2 = 0.9VL = 415 Vf = 60 Hz
Cph = µF
19.TransformersMedium
5 Ω81

A 5 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 8:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 8:1RL = 5 Ω
R1′ = Ω
20.TransformersHard
pf 0.9525 kVA, η = 99%

A single-phase 25 kVA transformer has a core loss of 150 W. At rated load and a power factor of 0.95 lagging its efficiency is 99%. Find the full-load copper loss Pcu,fl in watts.

S = 25 kVAPcore = 150 Wpf = 0.95η = 99%
Pcu,fl = W
21.Op-amp circuitsMedium
−+Vout​Rf​ = 10 kΩR1​ = 1 kΩVin​

The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±14 V. The non-inverting amplifier is driven by a sine wave. Find the largest input amplitude Vin,max (peak value) for which the output does not clip.

R1 = 1 kΩRf = 10 kΩVsat = ±14 V
Vin,max = V
22.Op-amp circuitsHard
Vin​V1​Vout​stage 1: invertingR1​ = 1 kΩRf1​ = 4.7 kΩstage 2: non-invertingR2​ = 47 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 = 0.4 V. Find the output voltage Vout.

R1 = 1 kΩRf1 = 4.7 kΩR2 = 47 kΩRf2 = 22 kΩVin = 0.4 V
Vout = V
23.Diodes & transistorsMedium
RC​ = 5.6 kΩβ = 120IC​ = ?RE​ = 680 ΩVCC​ = 18 VR1​ = 56 kΩR2​ = 8.2 kΩ

An NPN transistor with β = 120 has voltage-divider bias from VCC = 18 V: R₁ = 56 kΩ from VCC to the base and R₂ = 8.2 kΩ from the base to ground. The collector resistor is RC = 5.6 kΩ and the emitter resistor is RE = 680 Ω. Take VBE = 0.7 V and use the Thévenin equivalent of the divider (do not neglect the base current). Find the collector current IC in mA.

VCC = 18 VR1 = 56 kΩR2 = 8.2 kΩRC = 5.6 kΩRE = 680 Ωβ = 120
IC = mA
24.Diodes & transistorsHard
+−≥ 20 VR = ?12 V≤ 25 mA

A 12 V Zener regulator must supply up to 25 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 = 12 VIL,max = 25 mAIZ,min = 5 mA
Rmax = Ω
25.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −18 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −188 bit
U =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0011 0000 >> 1. What is the result's value R in decimal?

0011 0000>> 1
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = A′B′ + A′B′C + AB′C
L =
28.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKDOWN

A 4-bit binary down counter starts at 12 (decimal, Q = 1100). Which count Q (in decimal) does it show after 40 clock pulses?

Q0 = 1240n = 4
Q =
29.Laplace transformsMedium
t (s)f(t)t = 0.5f(0.5) = ?

Find f(t) for F(s) = 45/(s(s + 2)(s + 5)) by partial fractions, and evaluate f(0.5).

F(s) = 45/(s(s + 2)(s + 5))t = 0.5 s
f(0.5) =
30.Laplace transformsHard
t (s)y(t)t = 0.3y(0.3) = ?

A system obeys y″ + 5y′ + 4y = 2·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 5 for t ≥ 0. Use the Laplace transform to find the output y(0.3).

y″ + 5y′ + 4y = 2·u(t)u(t) = 5y(0) = 0, y′(0) = 0t = 0.3 s
y(0.3) =
31.Block diagramsMedium
R(s)+4s + 2G(s)C(s)0.1H−

A plant G(s) = 4/(s + 2) is controlled in a negative feedback loop whose sensor has the gain H = 0.1. Find the time constant τ of the closed loop.

G(s) = 4/(s + 2)H = 0.1
τ = s
32.Block diagramsHard
R(s)+Kcontroller+D(s)+2s + 2plantC(s)−

A proportional controller K drives the plant G(s) = 2/(s + 2) in a unity negative feedback loop. A disturbance step of size 2 enters at the plant input while R = 0. Find the smallest gain K that keeps the steady-state output at or below 0.4.

G(s) = 2/(s + 2)D = 2css ≤ 0.4
K =
33.Step responseMedium
t (s)y(t)3?

A system has the transfer function G(s) = 135/(s² + 6s + 45) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 135/(s² + 6s + 45)
%OS = %
34.Step responseHard
R(s)+160s(s + 8)G(s)C(s)−

A unity negative feedback loop has G(s) = 160/(s(s + 8)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 160/(s(s + 8))
ts = s
35.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−40−200204060ω (rad/s)φ (°)ω (rad/s)−90−450ω = 5

Find the phase ∠G(jω) of G(s) = 17(s + 12)/(s(s + 14)) at ω = 5 rad/s, in degrees.

G(s) = 17(s + 12)/(s(s + 14))ω = 5 rad/s
∠G(jω) = °
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) = 50/(s(s + 2)(s + 5)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 50/(s(s + 2)(s + 5))
ω180 = rad/s
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 1.21s(s + 1)(s + 2)C(s)−

The loop in the diagram runs with K = 1.2 and L(s) = K/(s(s + 1)(s + 2)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 1)(s + 2))K = 1.2
m =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2j1j2

For L(s) = K/((s + 1)(s² + 4s + 8)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.

L(s) = K/((s + 1)(s² + 4s + 8))
θd = °
39.Steady-state error & PID controlMedium
r(t) = 10t+Ks(s + 10)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 10)) must follow the ramp r(t) = 10t with a steady-state error of 0.2. Find the gain K.

G(s) = K/(s(s + 10))r(t) = 10tess = 0.2
K =
40.Steady-state error & PID controlHard
R(s)+K4s(s + 5)C(s)−

A proportional controller K drives the plant G(s) = 4/(s(s + 5)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.5.

G(s) = 4/(s(s + 5))ζ = 0.5
K =

3Solutions

1.Series & parallel resistors

Answer: IR2 = 2.087 A

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

R1​ = 3 ΩR3​ = 4 Ω+−V1​18 VR2​5 ΩAR4​20 ΩB

Combine resistors from the far end

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

R3 + R4 = 4 + 20 = 24 Ω
R2 ∥ 24 = 5·24/(5+24) = 4.138 Ω
Req = R1 + 4.138 = 3 + 4.138 = 7.138 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 = 18/7.138 = 2.522 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 − 2.522·3 = 10.43 V
IR2 = VA/R2 = 10.43/5 = 2.087 A

Answer

IR2 = 2.087 A
2.Series & parallel resistors

Answer: IR1 = 1.273 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​ = 6 ΩR3​ = 5 ΩR5​ = 30 Ω+−V1​18 VR2​25 ΩAR4​8 ΩBR6​30 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 = 30 + 30 = 60 Ω
R4 ∥ 60 = 7.059 Ω
R3 + 7.059 = 12.06 Ω
R2 ∥ 12.06 = 8.135 Ω
Req = R1 + 8.135 = 6 + 8.135 = 14.13 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 = 18/14.13 = 1.273 A

Answer

IR1 = 1.273 A
3.Nodal analysis

Answer: I1 = 4.364 A

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

KCL at node A

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

(VA − 30)/2 + (VA − 6)/8 + (VA − 9)/5 = 0

Solve for VA

VA = (30/2 + 6/8 + 9/5) / (1/2 + 1/8 + 1/5) = 17.55/0.825 = 21.27 V

Branch current

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

I1 = (30 − VA)/R1 = (30 − 21.27)/2 = 4.364 A

Answer

I1 = 4.364 A

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

4.Nodal analysis

Answer: Is = 0.3 A

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

KCL at node A

VA is known, so this equation gives VB.

(VA − V1)/R1 + VA/R2 + (VA − VB)/R3 = 0
(2 − 15)/15 + 2/2 = (VB − 2)/10
0.1333 = (VB − 2)/10 → VB = 3.333 V

KCL at node B

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

Is = (VB − VA)/R3 + VB/R4 = (3.333 − 2)/10 + 3.333/20 = 0.3 A

Answer

Is = 0.3 A
5.Mesh analysis

Answer: IR3 = 0.6286 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​ = 3 ΩR2​ = 15 Ω+−V1​12 VR3​15 ΩA+−V2​6 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
3·I1 + 15(I1 − I2) = 12
18·I1 − 15·I2 = 12

KVL around mesh 2

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

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

Solve

Δ = (18)(30) − (−15)(−15) = 315
I1 = Δ1/Δ = 270/315 = 0.8571 A
I2 = Δ2/Δ = 72/315 = 0.2286 A

Answer

IR3 = I1 − I2 = 0.6286 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: VR5 = 7.945 V

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

R1​ = 15 ΩR2​ = 8 ΩR3​ = 25 Ω+−V1​24 VR4​30 ΩAR5​15 ΩB+−V2​10 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
15·I1 + 30(I1 − I2) = 24
45·I1 − 30·I2 = 24

KVL around mesh 2

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

R2·I2 + R4(I2 − I1) + R5(I2 − I3) = 0
8·I2 + 30(I2 − I1) + 15(I2 − I3) = 0
−30·I1 + 53·I2 − 15·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
25·I3 + 15(I3 − I2) = (−10)
−15·I2 + 40·I3 = −10

Solve

Δ = det of the coefficient matrix = 49275
I1 = Δ1/Δ = 40980/49275 = 0.8317 A
I2 = Δ2/Δ = 22050/49275 = 0.4475 A
I3 = Δ3/Δ = −4050/49275 = −0.08219 A

Answer

IR5 = I2 − I3 = 0.5297 A
VR5 = IR5·R5 = 0.5297·15 = 7.945 V

A negative mesh current flows anticlockwise.

7.Superposition

Answer: VR4 = 11.47 V

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​ = 8 ΩR3​ = 6 Ω+−V1​10 VR2​3 ΩAR4​15 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 = 6+15 = 21 Ω
R2 ∥ (R3+R4) = 3·21/(3+21) = 2.625 Ω
VA = 10·2.625/(8+2.625) = 2.471 V
VB = VA·R4/(R3+R4) = 2.471·15/21 = 1.765 V
VR4′ = 0 − VB = 0 − 1.765 = −1.765 V
R1​ = 8 ΩR3​ = 6 ΩSHORTR2​3 ΩAR4​15 ΩBIs​2.5 A

Case 2 · Is acting alone

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

R1 ∥ R2 = 8·3/(8+3) = 2.182 Ω
Req = R4 ∥ (R3+R1∥R2) = 15 ∥ 8.182 = 5.294 Ω
VB = Is·Req = (−2.5)·5.294 = −13.24 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = −13.24·2.182/8.182 = −3.529 V
VR4″ = 0 − VB = 0 − −13.24 = 13.24 V

Add the contributions

VR4 = VR4′ + VR4″
VR4 = (−1.765) + 13.24 = 11.47 V

A negative result means the real polarity is opposite to the one marked.

8.Superposition

Answer: I = 14.5 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 measurements with contributions

Each contribution is proportional to its own source.

I₁ + I₂ + I₃ = 21.5
2·I₁ + I₂ + I₃ = 31
I₁ + I₂ − I₃ = 11.5

Solve for the contributions

I₁ = 31 − 21.5 = 9.5 mA
I₃ = (21.5 − 11.5)/2 = 5 mA
I₂ = 21.5 − 9.5 − 5 = 7 mA

Switch V₂ off

I = I₁ + I₃ = 9.5 + 5 = 14.5 mA

Answer

I = 14.5 mA
9.Thévenin & Norton

Answer: IL = 0.1702 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​ = 15 ΩR3​ = 4 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 = 15·12/(15+12) = 6.667 Ω
Rth = (R1 ∥ R2) + R3 = 6.667 + 4 = 10.67 Ω
R1​ = 15 ΩR3​ = 4 Ω+−V1​6 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) = 6·12/(15+12) = 2.667 V
Rth​ = 10.67 Ω+−Vth​2.667 VRL​5 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) = 2.667/(10.67 + 5) = 0.1702 A
10.Thévenin & Norton

Answer: IN = 2.388 A

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

R1​ = 8 ΩR3​ = 25 ΩSHORTR2​15 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 = 8·15/(8+15) = 5.217 Ω
Rth = (R1 ∥ R2) + R3 = 5.217 + 25 = 30.22 Ω
R1​ = 8 ΩR3​ = 25 Ω+−V1​18 VR2​15 ΩAIs​2 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) = 11.74 + 2·5.217 = 22.17 V
Vth = VB = VA + Is·R3 = 22.17 + 2·25 = 72.17 V
IN​2.388 ARth​30.22 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 = 72.17/30.22 = 2.388 A
11.RC & RL transients

Answer: t = 13.62 ms

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

Time constant 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 = 100 mH / 22 Ω = 4.545 ms
i(∞) = Vs/R = 24/22 = 1.091 A

Solve for the time

0.95 = 1 − e−t/τ → t = −τ·ln(1 − 0.95)
t = −4.545·ln(0.05) = 13.62 ms

Answer

t = 13.62 ms
12.RC & RL transients

Answer: vC = 19.39 V

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

Initial 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) = 15 V, vC(∞) = Vs = 24 V
τ = R·C = 2.2 kΩ · 68 µF = 149.6 ms
vC(t) = 24 + (15 − 24)·e−t/τ

Voltage at the given time

vC(100) = 24 + (−9)·e−100/149.6 = 19.39 V

Answer

vC = 19.39 V
13.AC impedance & phasors

Answer: cos φ = 0.8308

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 = 125.7 Ω, XC = 159.2 Ω

Impedance

|Z| = √(50² + (−33.49)²) = 60.18 Ω

Power factor

cos φ = R/|Z| = 50/60.18 = 0.8308
The circuit is capacitive, so the power factor is leading.

Answer

cos φ = 0.8308
14.AC impedance & phasors

Answer: f0 = 41.09 Hz

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

Resonance

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

f0 = 1/(2π√(LC)) = 1/(2π·√(0.15·100·10⁻⁶)) = 41.09 Hz

Answer

f0 = 41.09 Hz
15.AC power & power factor

Answer: I = 277.8 A

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

Line current

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

I = P/(V·cos φ) = 40000/(240·0.6) = 277.8 A

Answer

I = 277.8 A
16.AC power & power factor

Answer: C = 89.53 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 = 5·tan(36.87°) = 3.75 kvar
Q2 = P·tan φ2 = 5·tan(23.07°) = 2.13 kvar
QC = Q1 − Q2 = 1.62 kvar

Capacitance

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

C = QC/(2πf·V²) = 1620/(2π·50·240²) = 89.53 µF

Answer

C = 89.53 µF
17.Three-phase power

Answer: P = 6.582 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.

Apparent power

S = √3·VL·IL = √3·400·10 = 6.928 kVA

Real power

P = S·cos φ = 6.928·0.95 = 6.582 kW

Answer

P = 6.582 kW
18.Three-phase power

Answer: Cph = 825.4 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 0.7) = 100·1.02 = 102 kvar
Q2 = 100·tan(acos 0.9) = 100·0.4843 = 48.43 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 53.59 kvar

Capacitance per phase

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

V = 415/√3 = 239.6 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 17863/(377·239.6²) = 825.4 µF

Answer

Cph = 825.4 µF
19.Transformers

Answer: R1′ = 320 Ω

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.

Referred resistance

The voltage goes up by a and the current down by a, so the ratio V/I goes up by a².

R1′ = a²·RL = 8²·5 = 320 Ω

Answer

R1′ = 320 Ω
20.Transformers

Answer: Pcu,fl = 89.9 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.

Output and input power

Pout = 25·0.95 = 23.75 kW
Pin = 23750/0.99 = 23990 W

Total and copper losses

Ploss = 23990 − 23750 = 239.9 W
Pcu,fl = 239.9 − 150 = 89.9 W

Answer

Pcu,fl = 89.9 W
21.Op-amp circuits

Answer: Vin,max = 1.273 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

Av = 1 + Rf/R1 = 1 + 10/1 = 11

Largest input before clipping

Vin,max = Vsat/|Av| = 14/11 = 1.273 V

Answer

Vin,max = 1.273 V
22.Op-amp circuits

Answer: Vout = −2.76 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 = −4.7/1 = −4.7
A2 = 1 + Rf2/R2 = 1 + 22/47 = 1.468

Stage by stage

V1 = −4.7·0.4 = −1.88 V
Vout = 1.468·(−1.88) = −2.76 V

Answer

Vout = −2.76 V
23.Diodes & transistors

Answer: IC = 2.146 mA

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.

Thévenin equivalent of the divider

VTH = 18·8.2/(56 + 8.2) = 2.299 V
RTH = 56·8.2/(56 + 8.2) = 7.153 kΩ

Base current

KVL around the base loop; the emitter current is (β + 1)·IB.

IB = (2.299 − 0.7)/(7.153 + 121·0.68) = 17.88 µA

Collector current

IC = 120·17.88 µA = 2.146 mA
IE = 121·17.88 µA = 2.163 mA

Answer

IC = 2.146 mA
24.Diodes & transistors

Answer: Rmax = 266.7 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 = 25 + 5 = 30 mA

Largest resistance

Rmax = 8/0.03 = 266.7 Ω

Answer

Rmax = 266.7 Ω
25.Number systems

Answer: U = 238

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

18 = 0001 0010
inverted: 1110 1101
+1 → 1110 1110

Read it as unsigned

1110 1110₂ = 256 − 18 = 238

Answer

U = 238
26.Number systems

Answer: R = 24

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

0011 0000 >> 1
= 0001 1000

Value of the result

0001 1000₂ = 24

Answer

R = 24
27.Digital logic: gates, K-maps & flip-flops

Answer: L = 4

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′B′ → m(0, 1)
A′B′C → m(1)
AB′C → m(5)
ABC00011110011011030204150706

Group and read off

F = B′C + A′B′

Count the literals

L = 4

Answer

L = 4
28.Digital logic: gates, K-maps & flip-flops

Answer: Q = 4

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

Full cycles

Every 16 pulses bring the counter back to the same state.

40 = 2·16 + 8

Count after the remaining pulses

Q = (12 − 8) mod 16 = 4
Q = 0100₂

Answer

Q = 4
29.Laplace transforms

Answer: f(0.5) = 1.987

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

Partial fractions

F(s) = A/s + B/(s + 2) + C/(s + 5)
A = 45/(2·5) = 4.5
B = 45/((−2)·3) = −7.5
C = 45/((−5)·(−3)) = 3

Inverse transform

f(t) = 4.5 − 7.5·e−2t + 3·e−5t

Evaluate

f(0.5) = 4.5 − 7.5·e−2·0.5 + 3·e−5·0.5 = 1.987

Answer

f(0.5) = 1.987
30.Laplace transforms

Answer: y(0.3) = 0.2816

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) = 10/(s(s + 1)(s + 4))

Partial fractions

A = 10/(1·4) = 2.5
B = 10/((−1)·3) = −3.333
C = 10/((−4)·(−3)) = 0.8333

Output

y(t) = 2.5 − 3.333·e−t + 0.8333·e−4t
y(0.3) = 0.2816

Answer

y(0.3) = 0.2816
31.Block diagrams

Answer: τ = 0.4167 s

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

Closed-loop transfer function

T(s) = G/(1 + G·H) = 4/(s + 2 + 0.4) = 4/(s + 2.4)

Result

τ = 1/2.4 = 0.4167 s

Answer

τ = 0.4167 s
32.Block diagrams

Answer: K = 4

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)) = 2·2/(2 + 2K)

Solve for K

2 + 2K = 2·2/0.4 = 10
K = (10 − 2)/2 = 4

Answer

K = 4
33.Step response

Answer: %OS = 20.79 %

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 = 45 → ωₙ = 6.708 rad/s
2ζωₙ = 6 → ζ = 6/(2·6.708) = 0.4472

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−1.571 = 20.79 %

Answer

%OS = 20.79 %
34.Step response

Answer: ts = 1 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) = 160/(s² + 8s + 160): ωₙ = 12.65 rad/s, ζ = 0.3162

Settling time

ζωₙ = 8/2 = 4
ts ≈ 4/(ζωₙ) = 1 s

Answer

ts = 1 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) = −87.03 °

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.

Add the angles

∠G = atan(5/12) − 90° − atan(5/14) = 22.62° − 90° − 19.65°

Result

∠G(j5) = −87.03°

Answer

∠G(jω) = −87.03°
36.Bode plots & margins

Answer: ω180 = 3.162 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/2)·(ω/5) = 1 → ω180 = √(2·5) = 3.162 rad/s

Answer

ω180 = 3.162 rad/s
37.Stability: Routh–Hurwitz & root locus

Answer: m = 5

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 + 3s² + 2s + K = 0
s³: 1, 2
s²: 3, K
s¹: (3·2 − K)/3
s⁰: K

Largest stable gain

Kmax = 3·2 = 6

Factor

m = 6/1.2 = 5

Answer

m = 5

This factor is the gain margin as a ratio: 13.98 dB.

38.Stability: Routh–Hurwitz & root locus

Answer: θd = −26.57 °

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.

Pole locations

s² + 4s + 8 = 0 → s = −2 ± j2
s = −1

Angles from the other poles

∠(−2 + j2 + 1) = atan2(2, -1) = 116.6°
∠(−2 + j2 − (−2 − j2)) = ∠(j4) = 90°

Angle of departure

θd = 180° − 116.6° − 90° = −26.57°

Answer

θd = −26.57°
39.Steady-state error & PID control

Answer: K = 500

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.

Required velocity constant

Kv = A/ess = 10/0.2 = 50

Gain

K = 10·Kv = 500

Answer

K = 500

A large gain makes this loop very lightly damped: ζ = 0.2236. That conflict is why compensators are used.

40.Steady-state error & PID control

Answer: K = 6.25

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

Characteristic polynomial

s(s + 5) + 4K = s² + 5s + 4K

Natural frequency

2·0.5·ωₙ = 5 → ωₙ = 5 rad/s

Gain

4K = ωₙ² = 25 → K = 6.25

Answer

K = 6.25

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