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​ = 6 ΩIR1​R3​ = 10 Ω+−V1​18 VR2​5 ΩAR4​2 ΩB

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

V1 = 18 VR1 = 6 ΩR2 = 5 ΩR3 = 10 ΩR4 = 2 Ω
IR1 = A
2.Series & parallel resistorsHard
R1​ = 20 ΩR3​ = 12 ΩR5​ = 20 Ω+−V1​5 VR2​8 ΩIR2​AR4​5 ΩBR6​10 Ω

Find the current IR2 through R2.

V1 = 5 VR1 = 20 ΩR2 = 8 ΩR3 = 12 ΩR4 = 5 ΩR5 = 20 ΩR6 = 10 Ω
IR2 = A
3.Nodal analysisMedium
−+Vs​ = 12 VIs​5 AAR1​4 ΩR2​8 ΩB

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

Is = 5 AVs = 12 VR1 = 4 ΩR2 = 8 Ω
VA = V
4.Nodal analysisHard
R1​ = 5 Ω−+Vs​ = 12 V+−V1​12 VR2​6 ΩAR3​3 ΩBIs​4 A

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

V1 = 12 VIs = 4 AVs = 12 VR1 = 5 ΩR2 = 6 ΩR3 = 3 Ω
VA = V
5.Mesh analysisMedium
R1​ = 2 ΩR2​ = 20 Ω+−V1​15 VR3​20 ΩIR3​A+−V2​15 VI1​I2​

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

V1 = 15 VV2 = 15 VR1 = 2 ΩR2 = 20 ΩR3 = 20 Ω
IR3 = A
6.Mesh analysisHard
R1​ = 8 ΩR2​ = 12 ΩR3​ = 10 Ω+−V1​5 VR4​30 ΩAR5​15 ΩB+−V2​18 VI1​I2​I3​

Use mesh analysis to find the mesh current I3 (clockwise).

V1 = 5 VV2 = 18 VR1 = 8 ΩR2 = 12 ΩR3 = 10 ΩR4 = 30 ΩR5 = 15 Ω
I3 = A
7.SuperpositionMedium
R1​ = 3 ΩIR1​R3​ = 20 Ω+−V1​9 VR2​2 ΩAR4​10 ΩBIs​5 A

Use superposition to find the power PR1 absorbed by R1.

V1 = 9 VIs = 5 AR1 = 3 ΩR2 = 2 ΩR3 = 20 ΩR4 = 10 Ω
PR1 = W
8.SuperpositionHard
R1​ = 15 ΩR2​ = 10 ΩR4​ = 12 Ω+−V1​5 VR3​25 ΩIR3​AIs​4 AB+−V2​?

Find the value of the voltage source V2 (polarity as shown) that makes the current IR3 through R3 equal to zero.

V1 = 5 VIs = 4 AR1 = 15 ΩR2 = 10 ΩR3 = 25 ΩR4 = 12 Ω
V2 = V
9.Thévenin & NortonMedium
R1​ = 5 ΩR3​ = 8 Ω+−V1​9 VR2​6 ΩAab

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

V1 = 9 VR1 = 5 ΩR2 = 6 ΩR3 = 8 Ω
Rth = Ω
10.Thévenin & NortonHard
R1​ = 10 ΩR3​ = 20 Ω+−V1​36 VR2​20 ΩAIs​5 ABab

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

V1 = 36 VIs = 5 AR1 = 10 ΩR2 = 20 ΩR3 = 20 Ω
Vth = V
11.RC & RL transientsMedium
R = 68 Ωt = 0L100 mHi(t)

The inductor carries a steady current of 3 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. Find the voltage vR across R at t = 1 ms.

i(0) = 3 AR = 68 ΩL = 100 mHt = 1 ms
vR = V
12.RC & RL transientsHard
+−Vs​20 Vt = 0R = 3.3 kΩC10 µF+−vC​vC​(0) = 4 V

Just before the switch closes at t = 0, the capacitor holds 4 V with the polarity shown. How long does it take for vC to reach 15.2 V? Find t.

Vs = 20 VR = 3.3 kΩC = 10 µFvC(0) = 4 V
t = ms
13.AC impedance & phasorsMedium
12 V60 HzR = 50 ΩL = 100 mHI

A resistor and an inductor are connected in parallel across a 12 V (rms), 60 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 12 V (rms)f = 60 HzR = 50 ΩL = 100 mH
φ = °
14.AC impedance & phasorsHard
120 V200 HzR = 30 ΩL = 200 mHC = 2.2 µFI

A series RLC circuit is connected to a 120 V (rms), 200 Hz supply. Find the rms current I.

V = 120 V (rms)f = 200 HzR = 30 ΩL = 200 mHC = 2.2 µF
I = A
15.AC power & power factorMedium
120 VZ8 + j6 ΩI

A load with impedance Z = 8 + j6 Ω is connected to a 120 V (rms) single-phase supply. Find the real power P it absorbs.

V = 120 VZ = 8 + j6 Ω
P = W
16.AC power & power factorHard
P = 15 kWQ1​S1​φ1​QC​

A single-phase load takes 15 kW at a power factor of 0.6 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the capacitance C needed.

P = 15 kWpf1 = 0.6pf2 = 0.92V = 230 Vf = 60 Hz
C = µF
17.Three-phase powerMedium
ABCVL​ = 415 VIL​ = ?M45 kW, η = 92%cos φ = 0.86

A three-phase induction motor delivers 45 kW of mechanical output power at full load. At this load its efficiency is 92% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 415 V. Find the line current IL.

Pout = 45 kWη = 92%cos φ = 0.86VL = 415 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 0.72 lagging from a 690 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 120 kWpf1 = 0.72pf2 = 0.9VL = 690 Vf = 60 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
250 kVA

A single-phase 250 kVA transformer has a core loss of 1500 W and a copper loss of 3750 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 250 kVAPcore = 1500 WPcu,fl = 3750 W
Sη,max = kVA
21.Op-amp circuitsMedium
−+Vout​R2​ = 100 kΩV1​ = 1 VR1​ = 22 kΩV2​ = 2.1 VR3​ = 22 kΩR4​ = 100 kΩ

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

V1 = 1 VV2 = 2.1 VR1 = R3 = 22 kΩR2 = R4 = 100 kΩ
V− = V
22.Op-amp circuitsHard
−+Vout​Rf​ = 10 kΩV1​ = 0.7 VR1​ = 100 kΩV2​ = 0.2 VR2​ = 1 kΩ

The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the output voltage Vout.

V1 = 0.7 VV2 = 0.2 VR1 = 100 kΩR2 = 1 kΩRf = 10 kΩ
Vout = V
23.Diodes & transistorsMedium
+−5 VR = ?2 × 2 V25 mA

2 red LEDs (forward voltage 2 V each) are connected in series with a resistor to a 5 V supply. Find the resistance R that sets the current to 25 mA.

Vs = 5 VVF = 2 Vn = 2I = 25 mA
R = Ω
24.Diodes & transistorsHard
RC​ = 680 Ωβ = 150IC​ = ?VCC​ = 12 V5 VRB​ = 22 kΩ

An NPN transistor with β = 150 switches a 680 Ω load resistor connected between VCC = 12 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB = 22 kΩ. Take VBE = 0.7 V and VCE(sat) = 0.2 V. Find the collector current IC in mA when the input is high. Decide first whether the transistor is saturated.

VCC = 12 VRC = 680 Ωβ = 150Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 22 kΩ
IC = mA
25.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −119 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1198 bit
U =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0100 0110 AND 1101 1110. What is the result's value R in decimal?

0100 0110AND1101 1110
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010001031214151716

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(2, 4, 5, 6, 7). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(2, 4, 5, 6, 7)
L =
28.Digital logic: gates, K-maps & flip-flopsHard
CLKD1234567

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

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

F(s) = (2s + 8)/(s² + 7s + 6). Its inverse transform has the form f(t) = A·e−t + B·e−6t. Find the coefficient A by partial fractions.

F(s) = (2s + 8)/(s² + 7s + 6)
A =
30.Laplace transformsHard
ReIm−1−2−3−4−51

A system is described by the differential equation y″ + 6y′ + 10y = u′ + 5u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 0.25 s. Find g(0.25).

y″ + 6y′ + 10y = u′ + 5ut = 0.25 s
g(0.25) =
31.Block diagramsMedium
R(s)+26s(s + 2)G(s)C(s)−

A unity negative feedback loop has G(s) = 26/(s(s + 2)). The reference is a ramp r(t) = 10·t. Find the steady-state error ess.

G(s) = 26/(s(s + 2))r(t) = 10·t
ess =
32.Block diagramsHard
R(s)+3s + 4G₁+6s + 8G₂C(s)0.5H₁−1H₂−

Reduce the block diagram with G₁ = 3/(s + 4), G₂ = 6/(s + 8), an inner negative feedback H₁ = 0.5 around G₂ and an outer negative feedback H₂ = 1. Find the DC gain T(0) of C(s)/R(s).

G₁ = 3/(s + 4)G₂ = 6/(s + 8)H₁ = 0.5H₂ = 1
T(0) =
33.Step responseMedium
t (s)y(t)2tp​ = ?

A system has the transfer function G(s) = 28/(s² + 6s + 14) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 28/(s² + 6s + 14)
tp = s
34.Step responseHard
t (s)y(t)120 %

A second-order system without zeros must have a step response with 20 % overshoot. Find the damping ratio ζ.

%OS = 20 %
ζ =
35.Bode plots & marginsMedium
vin​L = 2 mHC = 1 µFR = 2 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 2 Ω, with L = 2 mH and C = 1 µF. Find the centre (resonant) frequency f0 in Hz.

R = 2 ΩL = 2 mHC = 1 µF
f0 = Hz
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) = 54/(s(s + 6)(s + 7)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.

L(s) = 54/(s(s + 6)(s + 7))
ω180 = rad/s
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+51s(s + 3)(s + 6)C(s)+

The loop in the diagram uses positive feedback, with L(s) = 5/(s(s + 3)(s + 6)). How many closed-loop poles N lie in the right half-plane?

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

A unity-feedback loop has L(s) = K(s + 1)/(s(s + 2)(s + 5)). Find the point σa where the asymptotes of the root locus cross the real axis.

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

A unity-feedback loop with G(s) = K/(s(s + 1)) must follow the ramp r(t) = 2t with a steady-state error of 0.5. Find the gain K.

G(s) = K/(s(s + 1))r(t) = 2tess = 0.5
K =
40.Steady-state error & PID controlHard
R(s)+20 + Kd​s5s(s + 1)C(s)−

A PD controller C(s) = 20 + Kd·s drives the plant G(s) = 5/(s(s + 1)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.5.

G(s) = 5/(s(s + 1))Kp = 20ζ = 0.5
Kd = s

3Solutions

1.Series & parallel resistors

Answer: IR1 = 1.889 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​ = 10 Ω+−V1​18 VR2​5 ΩAR4​2 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 = 10 + 2 = 12 Ω
R2 ∥ 12 = 5·12/(5+12) = 3.529 Ω
Req = R1 + 3.529 = 6 + 3.529 = 9.529 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 = 18/9.529 = 1.889 A

Answer

IR1 = 1.889 A
2.Series & parallel resistors

Answer: IR2 = 0.1322 A

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

R1​ = 20 ΩR3​ = 12 ΩR5​ = 20 Ω+−V1​5 VR2​8 ΩAR4​5 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 = 20 + 10 = 30 Ω
R4 ∥ 30 = 4.286 Ω
R3 + 4.286 = 16.29 Ω
R2 ∥ 16.29 = 5.365 Ω
Req = R1 + 5.365 = 20 + 5.365 = 25.36 Ω

Current from the source

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

IR1 = V1/Req = 5/25.36 = 0.1971 A

Current through R₂

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

VA = V1 − IR1·R1 = 5 − 0.1971·20 = 1.058 V
IR2 = VA/R2 = 1.058/8 = 0.1322 A

Answer

IR2 = 0.1322 A
3.Nodal analysis

Answer: VA = 9.333 V

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

Form a supernode

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

VB = VA + Vs = VA + 12
VA/R1 + VB/R2 = Is

Solve

VA/4 + (VA + 12)/8 = 5
VA = (5 − 12/8)/(1/4 + 1/8) = 9.333 V
VB = 9.333 + 12 = 21.33 V

Answer

VA = 9.333 V
4.Nodal analysis

Answer: VA = 3.429 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 + 12
(VA − V1)/R1 + VA/R2 + VB/R3 = Is

Solve

(VA − 12)/5 + VA/6 + (VA + 12)/3 = 4
VA·(1/5 + 1/6 + 1/3) = 4 + 12/5 − 12/3 = 2.4
VA = 2.4/0.7 = 3.429 V
VB = 15.43 V

Answer

VA = 3.429 V
5.Mesh analysis

Answer: IR3 = 0.6875 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​ = 2 ΩR2​ = 20 Ω+−V1​15 VR3​20 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
2·I1 + 20(I1 − I2) = 15
22·I1 − 20·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
20·I2 + 20(I2 − I1) = (−15)
−20·I1 + 40·I2 = −15

Solve

Δ = (22)(40) − (−20)(−20) = 480
I1 = Δ1/Δ = 300/480 = 0.625 A
I2 = Δ2/Δ = −30/480 = −0.0625 A

Answer

IR3 = I1 − I2 = 0.6875 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: I3 = −0.8891 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​ = 8 ΩR2​ = 12 ΩR3​ = 10 Ω+−V1​5 VR4​30 ΩAR5​15 ΩB+−V2​18 VI1​I2​I3​

Assign mesh currents

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

KVL around mesh 1

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

R1·I1 + R4(I1 − I2) = V1
8·I1 + 30(I1 − I2) = 5
38·I1 − 30·I2 = 5

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
12·I2 + 30(I2 − I1) + 15(I2 − I3) = 0
−30·I1 + 57·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
10·I3 + 15(I3 − I2) = (−18)
−15·I2 + 25·I3 = −18

Solve

Δ = det of the coefficient matrix = 23100
I1 = Δ1/Δ = −2100/23100 = −0.09091 A
I2 = Δ2/Δ = −6510/23100 = −0.2818 A
I3 = Δ3/Δ = −20538/23100 = −0.8891 A

Answer

I3 = −0.8891 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: PR1 = 18.56 W

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

R1​ = 3 ΩR3​ = 20 Ω+−V1​9 VR2​2 ΩAR4​10 ΩBOPEN

Case 1 · V1 acting alone

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

R3+R4 = 20+10 = 30 Ω
R2 ∥ (R3+R4) = 2·30/(2+30) = 1.875 Ω
VA = 9·1.875/(3+1.875) = 3.462 V
VB = VA·R4/(R3+R4) = 3.462·10/30 = 1.154 V
IR1′ = (V1 − VA)/R1 = (9 − 3.462)/3 = 1.846 A
R1​ = 3 ΩR3​ = 20 ΩSHORTR2​2 ΩAR4​10 ΩBIs​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. 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 = 3·2/(3+2) = 1.2 Ω
Req = R4 ∥ (R3+R1∥R2) = 10 ∥ 21.2 = 6.795 Ω
VB = −Is·Req = −5·6.795 = −33.97 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = −33.97·1.2/21.2 = −1.923 V
IR1″ = (V1 − VA)/R1 = (0 − −1.923)/3 = 0.641 A

Add the current contributions

IR1 = IR1′ + IR1″ = 1.846 + 0.641 = 2.487 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.

PR1 = IR1²·R1 = 2.487²·3 = 18.56 W

Answer

PR1 = 18.56 W

Adding the powers of the separate cases would give 11.46 W, which is wrong.

8.Superposition

Answer: V2 = 40.67 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​ = 15 ΩR2​ = 10 ΩR4​ = 12 Ω+−V1​5 VR3​25 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) = 25 ∥ 22 = 11.7 Ω
VA = 5·11.7/(15+11.7) = 2.191 V
VB = VA·R4/(R2+R4) = 2.191·12/22 = 1.195 V
R1​ = 15 ΩR2​ = 10 ΩR4​ = 12 ΩSHORTR3​25 ΩAIs​4 ABSHORT

Case 2 · Is acting alone

Both voltage sources become shorts. Iₛ splits at B between R₄ (now to ground) and R₂ in series with R₁ ∥ R₃. The arrow of Iₛ points down, away from its node, so Iₛ draws current out of that node: its term enters with a minus sign.

R1 ∥ R3 = 9.375 Ω, so the left path is R2+9.375 = 19.38 Ω
Req = R4 ∥ 19.38 = 7.41 Ω
VB = −Is·Req = −4·7.41 = −29.64 V
VA = VB·9.375/19.38 = −14.34 V

Case 3 · V2 acting alone, per volt

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

R1 ∥ R3 = 9.375 Ω
VA‴ = V2·9.375/(12 + 10 + 9.375) = 0.2988·V2

Set VA to zero

With VA = 0 there is no voltage across R₃, so its current is zero.

VA′ + VA″ + VA‴ = 2.191 + (−14.34) + 0.2988·V2 = 0
V2 = 12.15/0.2988 = 40.67 V

Answer

V2 = 40.67 V
9.Thévenin & Norton

Answer: Rth = 10.73 Ω

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

R1​ = 5 ΩR3​ = 8 ΩSHORTR2​6 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 = 5·6/(5+6) = 2.727 Ω
Rth = (R1 ∥ R2) + R3 = 2.727 + 8 = 10.73 Ω
10.Thévenin & Norton

Answer: Vth = 157.3 V

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

R1​ = 10 ΩR3​ = 20 ΩSHORTR2​20 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 = 10·20/(10+20) = 6.667 Ω
Rth = (R1 ∥ R2) + R3 = 6.667 + 20 = 26.67 Ω
R1​ = 10 ΩR3​ = 20 Ω+−V1​36 VR2​20 ΩAIs​5 ABab+−Vth​

Find Vth, the open-circuit voltage

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

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 24 + 5·6.667 = 57.33 V
Vth = VB = VA + Is·R3 = 57.33 + 5·20 = 157.3 V
11.RC & RL transients

Answer: vR = 103.3 V

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

Time constant

τ = L/R = 100 mH / 68 Ω = 1.471 ms

Decay of the current

i(t) = 3·e−1/1.471 = 1.52 A
vR = R·i = 68·1.52 = 103.3 V

Answer

vR = 103.3 V
12.RC & RL transients

Answer: t = 39.73 ms

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

Initial value, final value, time constant

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

vC(0) = 4 V, vC(∞) = Vs = 20 V
τ = R·C = 3.3 kΩ · 10 µF = 33 ms
vC(t) = 20 + (4 − 20)·e−t/τ

Solve for the time

15.2 = 20 + (−16)·e−t/τ → e−t/τ = (15.2 − 20)/(4 − 20) = 0.3
t = τ·ln((4 − 20)/(15.2 − 20)) = 33·ln(3.333) = 39.73 ms

Answer

t = 39.73 ms
13.AC impedance & phasors

Answer: φ = 52.98 °

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

Reactance

XL = ωL = 2π·60·0.1 = 37.7 Ω

Branch currents

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

IR = V/R = 12/50 = 0.24 A
IX = V/XL = 12/37.7 = 0.3183 A

Total current

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

I = √(IR² + IX²) = √(0.24² + 0.3183²) = 0.3986 A

Phase angle

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

φ = arctan(IX/IR) = arctan(0.3183/0.24) = 52.98°

Answer

φ = 52.98°
14.AC impedance & phasors

Answer: I = 1.049 A

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

Reactances

ω = 2πf = 2π·200 = 1257 rad/s
XL = ωL = 1257·0.2 = 251.3 Ω
XC = 1/(ωC) = 1/(1257·2.2·10⁻⁶) = 361.7 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(30² + (−110.4)²) = 114.4 Ω

Current

The same current flows through every element of a series circuit.

I = V/|Z| = 120/114.4 = 1.049 A

Answer

I = 1.049 A
15.AC power & power factor

Answer: P = 1152 W

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

Current

|Z| = √(8² + 6²) = 10 Ω
I = V/|Z| = 120/10 = 12 A

Real power

P = I²·R = 12²·8 = 1152 W

Answer

P = 1152 W
16.AC power & power factor

Answer: C = 682.5 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(53.13°) = 20 kvar
Q2 = P·tan φ2 = 15·tan(23.07°) = 6.39 kvar
QC = Q1 − Q2 = 13.61 kvar

Capacitance

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

C = QC/(2πf·V²) = 13610/(2π·60·230²) = 682.5 µF

Answer

C = 682.5 µF
17.Three-phase power

Answer: IL = 79.13 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 = 45/0.92 = 48.91 kW

Line current

IL = Pin/(√3·VL·cos φ) = 48913/(√3·415·0.86) = 79.13 A

Answer

IL = 79.13 A
18.Three-phase power

Answer: Cph = 106.9 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 0.72) = 120·0.9639 = 115.7 kvar
Q2 = 120·tan(acos 0.9) = 120·0.4843 = 58.12 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 57.54 kvar

Capacitance per phase

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

V = 690 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 19181/(377·690²) = 106.9 µF

Answer

Cph = 106.9 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: Sη,max = 158.1 kVA

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

Copper loss equals core loss

x²·3750 = 1500 → x = √(1500/3750) = 0.6325

Load at maximum efficiency

Sη,max = x·S = 0.6325·250 = 158.1 kVA

Answer

Sη,max = 158.1 kVA
21.Op-amp circuits

Answer: V− = 1.721 V

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

Voltage at the inputs

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

V₊ = V2·R4/(R3 + R4) = 2.1·100/(22 + 100) = 1.721 V
V₋ = V₊

Answer

V− = 1.721 V
22.Op-amp circuits

Answer: Vout = −2.07 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.

Input currents

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

I1 = V1/R1 = 0.7/100 = 0.007 mA
I2 = V2/R2 = 0.2/1 = 0.2 mA
If = I1 + I2 = 0.207 mA

Output voltage

Vout = −Rf·If = −10·0.207 = −2.07 V

Answer

Vout = −2.07 V
23.Diodes & transistors

Answer: R = 40 Ω

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

Voltage across the resistor

VR = 5 − 2·2 = 1 V

Resistance

R = 1/0.025 = 40 Ω

Answer

R = 40 Ω

The resistor dissipates 25 mW.

24.Diodes & transistors

Answer: IC = 17.35 mA

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

Base current

IB = (5 − 0.7)/22 kΩ = 195.5 µA

Saturated or active?

β·IB is larger than IC(sat), so the transistor saturates.

β·IB = 29.32 mA
IC(sat) = (12 − 0.2)/680 = 17.35 mA

Collector current

IC = IC(sat) = 17.35 mA

Answer

IC = 17.35 mA

The transistor is saturated.

25.Number systems

Answer: U = 137

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

119 = 0111 0111
inverted: 1000 1000
+1 → 1000 1001

Read it as unsigned

1000 1001₂ = 256 − 119 = 137

Answer

U = 137
26.Number systems

Answer: R = 70

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

AND gives 1 only where both bits are 1.

0100 0110
AND 1101 1110
= 0100 0110

Value of the result

0100 0110₂ = 70

Answer

R = 70
27.Digital logic: gates, K-maps & flip-flops

Answer: L = 3

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

ABC00011110010001031214151716

Group the 1s

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

BC′
A

Minimal SOP form

F = BC′ + A
L = 3

Answer

L = 3
28.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 6

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

Q after every edge

A D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 0 → Q = 0
2: D = 0 → Q = 0
3: D = 0 → Q = 0
4: D = 0 → Q = 0
5: D = 1 → Q = 1
6: D = 0 → Q = 0
7: D = 1 → Q = 1

First change from 1 to 0

clock edge 6

Answer

clock edge 6
29.Laplace transforms

Answer: A = 1.2

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 + 7s + 6 = (s + 1)(s + 6)
F(s) = A/(s + 1) + B/(s + 6)

Cover-up rule

Multiply by (s + 1) and set s = −1: the B-term vanishes.

A = (2s + 8)/(s + 6) |s = −1 = 6/5 = 1.2

Answer

A = 1.2
(B = 0.8, so f(t) = 1.2·e−t + 0.8·e−6t)
30.Laplace transforms

Answer: g(0.25) = 0.6914

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

Transfer function

G(s) = Y(s)/U(s) = (s + 5)/(s² + 6s + 10)

Partial fractions

s² + 6s + 10 = (s + 3)² + 1²
G(s) = 1·(s + 3)/((s + 3)² + 1²) + (2/1)·1/((s + 3)² + 1²)

Impulse response

g(t) = e−3t·(cos 1t + 2·sin 1t)
g(0.25) = 0.6914

Answer

g(0.25) = 0.6914
31.Block diagrams

Answer: ess = 0.7692

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

Velocity error constant

Kᵥ = lims→0 s·G(s) = 26/2 = 13

Steady-state error

ess = A/Kᵥ = 10/13 = 0.7692

Answer

ess = 0.7692
32.Block diagrams

Answer: T(0) = 0.2903

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

Inner loop

Gᵢ = G₂/(1 + G₂H₁) = 6/(s + 11)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 18/((s + 4)(s + 11) + 18)

DC gain

T(0) = 18/(4·11 + 18) = 18/62 = 0.2903

Answer

T(0) = 0.2903
33.Step response

Answer: tp = 1.405 s

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

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 14 → ωₙ = 3.742 rad/s
2ζωₙ = 6 → ζ = 6/(2·3.742) = 0.8018

Result

ωd = 3.742·√(1 − 0.8018²) = 2.236 rad/s → tp = π/ωd = 1.405 s

Answer

tp = 1.405 s
34.Step response

Answer: ζ = 0.4559

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).

Damping ratio

ζ = −ln(0.2)/√(π² + ln²(0.2)) = 1.609/√(9.87 + 2.59) = 0.4559

Answer

ζ = 0.4559
35.Bode plots & margins

Answer: f0 = 3559 Hz

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.

Resonant frequency

ω₀ = 1/√(LC) = 1/√(0.002·1 × 10⁻⁶) = 22361 rad/s
f₀ = ω₀/(2π) = 3559 Hz

Answer

f0 = 3559 Hz
36.Bode plots & margins

Answer: ω180 = 6.481 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/6)·(ω/7) = 1 → ω180 = √(6·7) = 6.481 rad/s

Answer

ω180 = 6.481 rad/s
37.Stability: Routh–Hurwitz & root locus

Answer: N = 1

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

Characteristic polynomial

s³ + 9s² + 18s − 5 = 0

Routh array

s³: 1, 18
s²: 9, −5
s¹: 18.56
s⁰: −5

Sign changes in the first column

1 → 9 → 18.56 → −5
N = 1

Answer

N = 1

Positive feedback makes the constant coefficient negative: one real pole always moves into the right half-plane.

38.Stability: Routh–Hurwitz & root locus

Answer: σa = −3

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.

Poles and zeros

poles: 0, −2, −5
zeros: −1
n − m = 3 − 1 = 2

Centroid

σa = (−7 − (−1))/2 = −3

Answer

σa = −3

The 2 asymptotes leave the centroid at angles of 90°, 270°.

39.Steady-state error & PID control

Answer: K = 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.

Required velocity constant

Kv = A/ess = 2/0.5 = 4

Gain

K = 1·Kv = 4

Answer

K = 4

A large gain makes this loop very lightly damped: ζ = 0.25. That conflict is why compensators are used.

40.Steady-state error & PID control

Answer: Kd = 1.8 s

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

Characteristic polynomial

s(s + 1) + 5(20 + Kds) = s² + (1 + 5Kd)s + 100

Natural frequency

ωₙ = √100 = 10 rad/s

Derivative gain

1 + 5Kd = 2·0.5·10 = 10
Kd = 1.8 s

Answer

Kd = 1.8 s

The derivative term adds damping without changing ωₙ.

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