FE Electrical and Computer exam practice: cram pack

Formula sheets, the key method steps for every topic and 40 practice problems, with full solutions at the back.

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1Formulas and methods

Series & parallel resistors

Resistors in series
Req = R1 + R2 + …
Resistors in parallel
1/Req = 1/R1 + 1/R2 + …
Two in parallel
Req = R1R2 / (R1 + R2)
Ohm's law
V = I·R
Power
P = V·I = I²R = V²/R

Method

  1. Start at the end of the network furthest from the terminals.
  2. Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
  3. Redraw the circuit after each step.
  4. Repeat until one resistor is left between the terminals.

Nodal analysis

Kirchhoff’s current law
ΣIin = ΣIout
Branch current
I = (Va − Vb)/R
Reference node
Vground = 0

Method

  1. Choose a reference node (ground) and label the other node voltages.
  2. Write Kirchhoff’s current law at every unknown node: currents leaving through each branch equal the currents entering.
  3. Express every branch current as (Va − Vb)/R.
  4. Solve the system of equations, then find any current you need.

Mesh analysis

Kirchhoff’s voltage law
ΣV = 0 around every loop
Shared resistor
current = I1 − I2
Current source in a mesh
fixes that mesh current

Method

  1. Draw a clockwise current in every mesh (window) of the circuit.
  2. Apply Kirchhoff’s voltage law around each mesh: the sum of voltage drops is zero.
  3. In a resistor shared by two meshes the current is the difference of the two mesh currents.
  4. Solve the equations; a negative result means the current flows the other way.

Superposition

Superposition
x = x1 + x2 + … (one source at a time)
Switched-off voltage source
short circuit (0 V)
Switched-off current source
open circuit (0 A)
Voltage divider
V2 = V·R2/(R1 + R2)
Current divider
I1 = I·R2/(R1 + R2)

Method

  1. Count the independent sources. You will solve one simpler circuit per source.
  2. Keep one source and switch off the others: a voltage source becomes a short circuit (a wire), a current source becomes an open circuit (a gap).
  3. Solve the simpler circuit for the voltage or current you need, keeping track of its direction.
  4. Repeat for every source, then add the contributions with their signs.

Thévenin & Norton

Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)

Method

  1. Remove the load and mark the two terminals a and b.
  2. Find the open-circuit voltage between a and b: this is Vth.
  3. Switch off all independent sources and find the resistance seen from a–b: this is Rth.
  4. Put the load back on the equivalent circuit: Vth in series with Rth.
  5. For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.

RC & RL transients

Time constant
τ = RC | τ = L/R
Response
x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ
Capacitor at DC
open circuit; vC cannot jump
Inductor at DC
short circuit; iL cannot jump
Settled
t ≈ 5τ

Method

  1. Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
  2. Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
  3. Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
  4. Use x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.

AC impedance & phasors

Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z

Method

  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.

AC power & power factor

Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)

Three-phase power

Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η

Transformers

Turns ratio
a = N1/N2 = V1/V2
Current ratio
I1/I2 = N2/N1 = 1/a
Rated current
I = S/V
Referred impedance
Z1′ = a²·ZL
Copper loss at part load
Pcu = x²·Pcu,fl
Efficiency
η = x·S·pf/(x·S·pf + Pcore + x²·Pcu,fl)
Maximum efficiency
x = √(Pcore/Pcu,fl)

Op-amp circuits

Inverting amplifier
Vout = −(Rf/Rin)·Vin
Non-inverting amplifier
Vout = (1 + Rf/R1)·Vin
Summing amplifier
Vout = −Rf(V1/R1 + V2/R2)
Ideal op-amp
V+ = V−, Iin = 0

Diodes & transistors

Diode (constant drop)
VD = 0.7 V when conducting
Active region
VBE = 0.7 V, IC = β·IB, IE = (β + 1)·IB
Fixed bias
IB = (VCC − 0.7)/RB
Divider bias
IB = (VTH − 0.7)/(RTH + (β + 1)RE)
Thévenin of the divider
VTH = VCC·R2/(R1 + R2), RTH = R1∥R2
Collector loop
VCE = VCC − ICRC − IERE
Saturation
IC(sat) = (VCC − VCE(sat))/RC, IB ≥ IC(sat)/β

Number systems

Place values
binary 2k, octal 8k, hex 16k
To decimal
Σ dk·bk
From decimal
divide by the base, read the remainders bottom-up
Hex ↔ binary
one hex digit = 4 bits
Two’s complement
invert the bits and add 1

Digital logic: gates, K-maps & flip-flops

Basic gates
A·B, A + B, A′, (A·B)′, (A + B)′, A ⊕ B
Boolean rules
X + X′ = 1, X·X′ = 0, X + XY = X, XY + XY′ = X
De Morgan
(X·Y)′ = X′ + Y′, (X + Y)′ = X′·Y′
Karnaugh map order
00, 01, 11, 10 (Gray code)
Group of 2ᵏ cells
loses k variables
4-to-1 multiplexer
F = A′B′I0 + A′BI1 + AB′I2 + ABI3
Flip-flops
D: Q⁺ = D, T: Q⁺ = T ⊕ Q, JK: Q⁺ = JQ′ + K′Q
Counters
n-bit: 2ⁿ states; mod-N: smallest n with 2ⁿ ≥ N; fout = fclk/N
Ripple counter
fmax = 1/(n·tpd)
Ring and Johnson counters
n and 2n states

Method

  1. For a circuit, work from the inputs to the output one gate at a time and write the value on every wire, or write its expression and fill in the truth table.
  2. For simplification, put the 1s (and the don’t-cares as X) into a Karnaugh map with Gray-code rows and columns: 00, 01, 11, 10.
  3. Cover the 1s with as few and as large groups of 1, 2, 4, 8 or 16 cells as possible, wrapping round the edges; each group is one product term made of the variables that stay constant. Group the 0s for the POS form.
  4. For flip-flops and counters, read the inputs at each active clock edge, write the state after every edge, and remember that an n-bit counter wraps round after 2ⁿ states.

Stability: Routh–Hurwitz & root locus

Stability
all roots of the characteristic polynomial in the left half-plane
Routh array, s¹ entry of a cubic
b1 = (a2a1 − a3a0)/a2
Right half-plane roots
number of sign changes in the first column
Cubic s³ + a₂s² + a₁s + a₀
stable when all a > 0 and a₂·a₁ > a₀
K/(s(s + a)(s + b))
Kmax = ab(a + b), ω = √(ab)
Characteristic equation
1 + K·G(s) = 0
Asymptotes
σa = (Σp − Σz)/(n − m), θ = (2q + 1)·180°/(n − m)
Break-away and break-in points
dK/ds = 0
Magnitude condition
K = 1/|G(s)|
Angle of departure
θd = 180° − Σ∠(from other poles) + Σ∠(from zeros)

Method

  1. Write the closed-loop characteristic polynomial: the denominator of 1 + K·G(s) = 0, multiplied out.
  2. Put the coefficients alternately into the first two rows of the Routh array and build the remaining rows.
  3. Count the sign changes in the first column: each one is a pole in the right half-plane. With K in the array, demand that every first-column entry stays positive.
  4. For a root locus, mark the poles and zeros, the real-axis segments, the asymptotes from the centroid and the break points where dK/ds = 0. Use K = 1/|G(s)| for the gain at a point.

Steady-state error & PID control

System type
number of poles of G(s) at s = 0
Error constants
Kp = lim G(s), Kv = lim s·G(s), Ka = lim s²·G(s) (s → 0)
Step A
ess = A/(1 + Kp)
Ramp A·t
ess = A/Kv
Parabola A·t²/2
ess = A/Ka
PID controller
C(s) = Kp + Ki/s + Kd·s
Second-order target
s² + 2ζωns + ωn²
Overshoot and timing
%OS = 100·e−ζπ/√(1−ζ²), tp = π/ωd, ts ≈ 4/(ζωn)

Method

  1. Count the poles of the open loop C(s)·G(s) at s = 0: that is the system type.
  2. Work out the error constant that matches the input: Kp for a step, Kv for a ramp, Ka for a parabola.
  3. Apply ess = A/(1 + Kp), A/Kv or A/Ka, or solve it for the gain you need.
  4. For a damping or speed specification, write the closed-loop characteristic polynomial and match it with s² + 2ζω_n·s + ω_n².

2Practice problems

1.Series & parallel resistorsMedium
R1​ = 5 ΩR3​ = 4 Ω+−V1​12 VR2​20 ΩAR4​6 ΩB

Find the equivalent resistance Req seen by the source V1.

V1 = 12 VR1 = 5 ΩR2 = 20 ΩR3 = 4 ΩR4 = 6 Ω
Req = Ω
2.Series & parallel resistorsHard
R1​ = 8 ΩIR1​R3​ = 10 ΩR5​ = 3 Ω+−V1​6 VR2​15 ΩAR4​3 ΩBR6​20 Ω

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

V1 = 6 VR1 = 8 ΩR2 = 15 ΩR3 = 10 ΩR4 = 3 ΩR5 = 3 ΩR6 = 20 Ω
IR1 = A
3.Nodal analysisMedium
R1​ = 3 ΩR3​ = 20 Ω+−V1​18 VR2​2 ΩAR4​10 ΩBIs​4 A

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

V1 = 18 VIs = 4 AR1 = 3 ΩR2 = 2 ΩR3 = 20 ΩR4 = 10 Ω
VB = V
4.Nodal analysisHard
R1​ = 30 ΩR2​ = 4 ΩR4​ = 20 ΩIR4​+−V1​36 VR3​12 ΩAR5​2 ΩBIs​3 A+−V2​30 V

Use nodal analysis to find the current IR4 through R4 in the direction of the arrow.

V1 = 36 VIs = 3 AV2 = 30 VR1 = 30 ΩR2 = 4 ΩR3 = 12 ΩR4 = 20 ΩR5 = 2 Ω
IR4 = A
5.Mesh analysisMedium
R1​ = 4 ΩR2​ = 15 Ω+−V1​18 VR3​20 ΩIR3​A+−V2​6 VI1​I2​

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

V1 = 18 VV2 = 6 VR1 = 4 ΩR2 = 15 ΩR3 = 20 Ω
IR3 = A
6.Mesh analysisHard
R1​ = 4 ΩR2​ = 6 ΩR3​ = 12 Ω+−V1​12 VR4​25 Ω+−VR4​AR5​5 ΩB+−V2​9 VI1​I2​I3​

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

V1 = 12 VV2 = 9 VR1 = 4 ΩR2 = 6 ΩR3 = 12 ΩR4 = 25 ΩR5 = 5 Ω
VR4 = V
7.SuperpositionMedium
linear circuitVs​Is​IRL​

A linear resistive circuit contains one voltage source Vs and one current source Is. The current I through a load resistor is measured twice: with Vs = 10 V and Is = 10 mA, I = 6 mA; with Vs = 2 V and Is = 5 mA, I = 1.8 mA. Find the load current I when Vs = 20 V and Is = 3 mA.

10 V, 10 mA → 6 mA2 V, 5 mA → 1.8 mAVs = 20 V, Is = 3 mA
I = mA
8.SuperpositionHard
R1​ = 5 ΩR2​ = 4 ΩIR2​R4​ = 6 Ω+−V1​18 VR3​2 ΩAIs​3 AB+−V2​24 V

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

V1 = 18 VIs = 3 AV2 = 24 VR1 = 5 ΩR2 = 4 ΩR3 = 2 ΩR4 = 6 Ω
IR2 = A
9.Thévenin & NortonMedium
R1​ = 8 ΩR3​ = 8 Ω+−V1​12 VR2​3 ΩARL​12 ΩIL​B

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

V1 = 12 VR1 = 8 ΩR2 = 3 ΩR3 = 8 ΩRL = 12 Ω
IL = A
10.Thévenin & NortonHard
R1​ = 30 ΩR3​ = 2 Ω+−V1​15 VR2​20 ΩAIs​2 ABab

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

V1 = 15 VIs = 2 AR1 = 30 ΩR2 = 20 ΩR3 = 2 Ω
Vth = V
11.RC & RL transientsMedium
+−Vs​5 Vt = 0R = 10 ΩL33 mHi(t)

The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 75% of its final value? Find t.

Vs = 5 VR = 10 ΩL = 33 mH
t = ms
12.RC & RL transientsHard
+−Vs​10 Vopens at t = 0R1​ = 10 kΩR2​ = 2.2 kΩi2​C47 µF+−vC​

The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the current i2 through R2 at t = 75 ms.

Vs = 10 VR1 = 10 kΩR2 = 2.2 kΩC = 47 µFt = 75 ms
i2 = mA
13.AC impedance & phasorsMedium
24 V200 HzR = 30 ΩL = 20 mHC = 47 µFI

A series RLC circuit is connected to a 24 V (rms), 200 Hz supply. Find the real power P taken from the supply.

V = 24 V (rms)f = 200 HzR = 30 ΩL = 20 mHC = 47 µF
P = W
14.AC impedance & phasorsHard
12 Vf0​R = 15 ΩL = 20 mHC = 0.47 µF

A series RLC circuit is driven by a 12 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 12 V (rms)R = 15 ΩL = 20 mHC = 0.47 µF
VL = V
15.AC power & power factorMedium
P = 25 kWQ = ?Sφ

A single-phase motor takes 25 kW at a power factor of 0.75 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 25 kWpf = 0.75V = 230 V
Q = kvar
16.AC power & power factorHard
230 V60 Hzload22 kW, 0.75C100 µFI

A single-phase load takes 22 kW at a power factor of 0.75 lagging from a 230 V, 60 Hz supply. A 100 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 22 kWpf1 = 0.75V = 230 Vf = 60 HzC = 100 µF
cos φ =
17.Three-phase powerMedium
ABCVL​ = 690 VIL​ = ?M90 kW, η = 91%cos φ = 0.88

A three-phase induction motor delivers 90 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.88 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.

Pout = 90 kWη = 91%cos φ = 0.88VL = 690 V
IL = A
18.Three-phase powerHard
P = 100 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 100 kW at a power factor of 0.78 lagging from a 400 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 100 kWpf1 = 0.78pf2 = 0.9VL = 400 Vf = 50 Hz
QC = kvar
19.TransformersMedium
4160 V480 VIHV​ILV​ = ?150 kVA

A single-phase 150 kVA, 4160/480 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 150 kVAV1 = 4160 VV2 = 480 V
ILV = A
20.TransformersHard
pf 0.8575 kVA, 25% load

A single-phase 75 kVA transformer has a core loss of 450 W and a copper loss of 1800 W at rated load. It supplies 25% of its rated kVA at a power factor of 0.85 lagging. Find the total losses Ploss in watts at this load.

S = 75 kVAPcore = 450 WPcu,fl = 1800 Wx = 0.25pf = 0.85
Ploss = W
21.Op-amp circuitsMedium
−+Vout​R2​ = 2 kΩV1​ = 0.1 VR1​ = 4.7 kΩV2​ = −2.4 VR3​ = 4.7 kΩR4​ = 2 kΩ

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

V1 = 0.1 VV2 = −2.4 VR1 = R3 = 4.7 kΩR2 = R4 = 2 kΩ
Vout = V
22.Op-amp circuitsHard
Vin​V1​Vout​stage 1: invertingR1​ = 2 kΩRf1​ = 10 kΩstage 2: non-invertingR2​ = 22 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 = 0.3 V. Find the output voltage Vout.

R1 = 2 kΩRf1 = 10 kΩR2 = 22 kΩRf2 = 22 kΩVin = 0.3 V
Vout = V
23.Diodes & transistorsMedium
+−18 V270 Ω5.1 V2200 ΩIZ​ = ?

A 5.1 V Zener regulator is fed from 18 V through 270 Ω and supplies a 2200 Ω load. Find the Zener current IZ.

Vin = 18 VVZ = 5.1 VR = 270 ΩRL = 2200 Ω
IZ = mA
24.Diodes & transistorsHard
RC​ = 330 Ωβ = 200VCC​ = 9 V5 VRB​ = ?

An NPN transistor with β = 200 switches a 330 Ω load resistor connected between VCC = 9 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB. Take VBE = 0.7 V and VCE(sat) = 0.2 V. Find the largest base resistor RB, in kΩ, that still drives the transistor just into saturation.

VCC = 9 VRC = 330 Ωβ = 200Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 V
RB = kΩ
25.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −10 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −108 bit
U =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1001 1111 << 2. What is the result's value R in decimal?

1001 1111<< 2
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010010302141517061121130150141819011110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 4, 5, 7, 8, 9, 10, 12, 13). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 4, 5, 7, 8, 9, 10, 12, 13)
N =
28.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKJOHNSON

A 4-bit Johnson counter Q₃Q₂Q₁Q₀ starts at 0000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives the inverse of Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 7 clock edges.

Q₃Q₂Q₁Q₀ = 00007
Q =
29.Laplace transformsMedium
t (s)f(t)t = 0.25f(0.25) = ?

Find f(t) for F(s) = 15/(s(s + 2)(s + 3)) by partial fractions, and evaluate f(0.25).

F(s) = 15/(s(s + 2)(s + 3))t = 0.25 s
f(0.25) =
30.Laplace transformsHard
ReIm−0.5−1−1.5−20.51

A system is described by the differential equation y″ + 3y′ + 2y = 11u, 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 = 0.3 s. Find g(0.3).

y″ + 3y′ + 2y = 11ut = 0.3 s
g(0.3) =
31.Block diagramsMedium
R(s)+13K+4s + 7G(s)C(s)0.5H₁−−

The plant G(s) = 4/(s + 7) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 13 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 4/(s + 7)H₁ = 0.5K = 13
T(0) =
32.Block diagramsHard
R(s)+K(s + 4)(s + 6)(s + 8)G(s)C(s)−

The plant G(s) = K/((s + 4)(s + 6)(s + 8)) is in a unity negative feedback loop. At the largest stable gain the loop oscillates. Find the frequency of oscillation ω in rad/s.

G(s) = K/((s + 4)(s + 6)(s + 8))unity negative feedback
ω = rad/s
33.Step responseMedium
t (s)y(t)0.5

A system has the transfer function G(s) = 8.5/(s² + 5s + 17) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 8.5/(s² + 5s + 17)
ωn = rad/s
34.Step responseHard
R(s)+25s(s + 2)G(s)C(s)−

A unity negative feedback loop has G(s) = 25/(s(s + 2)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 25/(s(s + 2))
ts = s
35.Bode plots & marginsMedium
|H| (dB)11101010010010001000−40−2002040ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 30

Find the phase ∠G(jω) of G(s) = 36(s + 18)/(s(s + 12)) at ω = 30 rad/s, in degrees.

G(s) = 36(s + 18)/(s(s + 12))ω = 30 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) = 118/(s(s + 3)(s + 7)). Find the gain margin GM in dB.

L(s) = 118/(s(s + 3)(s + 7))
GM = dB
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 1)(s + 2)(s + 5)C(s)−

A unity-feedback loop has L(s) = K/((s + 1)(s + 2)(s + 5)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 1)(s + 2)(s + 5))
Kmax =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12−14−162j2j4

The root locus of L(s) = K(s + 14)/(s(s + 8)) leaves the real axis between s = 0 and s = −8, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 14)/(s(s + 8))
si =
39.Steady-state error & PID controlMedium
r(t) = 10 + 4t+50(s + 1)s(s + 5)(s + 10)C(s)−

The input r(t) = 10 + 4t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 50(s + 1)/(s(s + 5)(s + 10)). Find the steady-state error ess.

G(s) = 50(s + 1)/(s(s + 5)(s + 10))r(t) = 10 + 4t
ess =
40.Steady-state error & PID controlHard
R(s)+K = 110s(s + 2)C(s)−

A proportional controller K = 1 drives the plant G(s) = 10/(s(s + 2)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 10/(s(s + 2))K = 1
ts = s

3Solutions

1.Series & parallel resistors

Answer: Req = 11.67 Ω

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

R1​ = 5 ΩR3​ = 4 Ω+−V1​12 VR2​20 ΩAR4​6 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 + 6 = 10 Ω
R2 ∥ 10 = 20·10/(20+10) = 6.667 Ω
Req = R1 + 6.667 = 5 + 6.667 = 11.67 Ω

Answer

Req = 11.67 Ω
2.Series & parallel resistors

Answer: IR1 = 0.4037 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​ = 8 ΩR3​ = 10 ΩR5​ = 3 Ω+−V1​6 VR2​15 ΩAR4​3 ΩBR6​20 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 = 3 + 20 = 23 Ω
R4 ∥ 23 = 2.654 Ω
R3 + 2.654 = 12.65 Ω
R2 ∥ 12.65 = 6.864 Ω
Req = R1 + 6.864 = 8 + 6.864 = 14.86 Ω

Current from the source

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

IR1 = V1/Req = 6/14.86 = 0.4037 A

Answer

IR1 = 0.4037 A
3.Nodal analysis

Answer: VB = 29.49 V

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

R1​ = 3 ΩR3​ = 20 Ω+−V1​18 VR2​2 ΩAR4​10 ΩBIs​4 A

Set up the nodes

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

KCL at node A

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

VA/R2 + (VA − V1)/R1 + (VA − VB)/R3 = 0
VA/2 + (VA − 18)/3 + (VA − VB)/20 = 0
0.8833·VA − 0.05·VB = 6

KCL at node B

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

VB/R4 + (VB − VA)/R3 = Is
VB/10 + (VB − VA)/20 = 4
−0.05·VA + 0.15·VB = 4

Solve

Δ = (0.8833)(0.15) − (−0.05)(−0.05) = 0.13
VA = Δ1/Δ = 1.1/0.13 = 8.462 V
VB = Δ2/Δ = 3.833/0.13 = 29.49 V

Answer

VB = 29.49 V
4.Nodal analysis

Answer: IR4 = −1.554 A

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

R1​ = 30 ΩR2​ = 4 ΩR4​ = 20 Ω+−V1​36 VR3​12 ΩAR5​2 ΩBIs​3 A+−V2​30 V

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 36 V, V2 = 30 V). Unknowns: VA, VB.

KCL at node A

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

VA/R3 + (VA − V1)/R1 + (VA − VB)/R2 = 0
VA/12 + (VA − 36)/30 + (VA − VB)/4 = 0
0.3667·VA − 0.25·VB = 1.2

KCL at node B

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

VB/R5 + (VB − VA)/R2 + (VB − V2)/R4 = Is
VB/2 + (VB − VA)/4 + (VB − 30)/20 = (−3)
−0.25·VA + 0.8·VB = −1.5

Solve

Δ = (0.3667)(0.8) − (−0.25)(−0.25) = 0.2308
VA = Δ1/Δ = 0.585/0.2308 = 2.534 V
VB = Δ2/Δ = −0.25/0.2308 = −1.083 V

Answer

IR4 = (VB − V2)/R4 = (−1.083 − 30)/20 = −1.554 A
5.Mesh analysis

Answer: IR3 = 0.6682 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​ = 4 ΩR2​ = 15 Ω+−V1​18 VR3​20 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
4·I1 + 20(I1 − I2) = 18
24·I1 − 20·I2 = 18

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 + 20(I2 − I1) = (−6)
−20·I1 + 35·I2 = −6

Solve

Δ = (24)(35) − (−20)(−20) = 440
I1 = Δ1/Δ = 510/440 = 1.159 A
I2 = Δ2/Δ = 216/440 = 0.4909 A

Answer

IR3 = I1 − I2 = 0.6682 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: VR4 = 8.299 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​ = 4 ΩR2​ = 6 ΩR3​ = 12 Ω+−V1​12 VR4​25 ΩAR5​5 ΩB+−V2​9 VI1​I2​I3​

Assign mesh currents

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

KVL around mesh 1

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

R1·I1 + R4(I1 − I2) = V1
4·I1 + 25(I1 − I2) = 12
29·I1 − 25·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 + R4(I2 − I1) + R5(I2 − I3) = 0
6·I2 + 25(I2 − I1) + 5(I2 − I3) = 0
−25·I1 + 36·I2 − 5·I3 = 0

KVL around mesh 3

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

R3·I3 + R5(I3 − I2) = −V2
12·I3 + 5(I3 − I2) = (−9)
−5·I2 + 17·I3 = −9

Solve

Δ = det of the coefficient matrix = 6398
I1 = Δ1/Δ = 5919/6398 = 0.9251 A
I2 = Δ2/Δ = 3795/6398 = 0.5932 A
I3 = Δ3/Δ = −2271/6398 = −0.355 A

Answer

IR4 = I1 − I2 = 0.332 A
VR4 = IR4·R4 = 0.332·25 = 8.299 V

A negative mesh current flows anticlockwise.

7.Superposition

Answer: I = 8.6 mA

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

Write the response as a sum of contributions

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

I = k₁·Vs + k₂·Is
10·k₁ + 10·k₂ = 6
2·k₁ + 5·k₂ = 1.8

Solve for the constants

Δ = (10)(5) − (10)(2) = 30
k₁ = Δ1/Δ = 12/30 = 0.4
k₂ = Δ2/Δ = 6/30 = 0.2
k₁ = 0.4 mA/V, k₂ = 0.2

Predict the new current

I = 0.4·20 + 0.2·3 = 8.6 mA

Answer

I = 8.6 mA
8.Superposition

Answer: IR2 = −3.225 A

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

R1​ = 5 ΩR2​ = 4 ΩR4​ = 6 Ω+−V1​18 VR3​2 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) = 2 ∥ 10 = 1.667 Ω
VA = 18·1.667/(5+1.667) = 4.5 V
VB = VA·R4/(R2+R4) = 4.5·6/10 = 2.7 V
IR2′ = (VA − VB)/R2 = (4.5 − 2.7)/4 = 0.45 A
R1​ = 5 ΩR2​ = 4 ΩR4​ = 6 ΩSHORTR3​2 ΩAIs​3 ABSHORT

Case 2 · Is acting alone

Both voltage sources become shorts. Iₛ splits at B between R₄ (now to ground) and R₂ in series with R₁ ∥ R₃.

R1 ∥ R3 = 1.429 Ω, so the left path is R2+1.429 = 5.429 Ω
Req = R4 ∥ 5.429 = 2.85 Ω
VB = Is·Req = 3·2.85 = 8.55 V
VA = VB·1.429/5.429 = 2.25 V
IR2″ = (VA − VB)/R2 = (2.25 − 8.55)/4 = −1.575 A
R1​ = 5 ΩR2​ = 4 ΩR4​ = 6 ΩSHORTR3​2 ΩAOPENB+−V2​24 V

Case 3 · V2 acting alone

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

R1 ∥ R3 = 5·2/(5+2) = 1.429 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 24·5.429/11.43 = 11.4 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 11.4·1.429/5.429 = 3 V
IR2‴ = (VA − VB)/R2 = (3 − 11.4)/4 = −2.1 A

Add the contributions

IR2 = IR2′ + IR2″ + IR2‴
IR2 = 0.45 + (−1.575) + (−2.1) = −3.225 A

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

9.Thévenin & Norton

Answer: IL = 0.1475 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​ = 8 ΩSHORTR2​3 ΩAabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 8·3/(8+3) = 2.182 Ω
Rth = (R1 ∥ R2) + R3 = 2.182 + 8 = 10.18 Ω
R1​ = 8 ΩR3​ = 8 Ω+−V1​12 VR2​3 ΩAab+−Vth​

Find Vth, the open-circuit voltage

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

Vth = V1·R2/(R1+R2) = 12·3/(8+3) = 3.273 V
Rth​ = 10.18 Ω+−Vth​3.273 VRL​12 ΩIL​

Connect the load to the Thévenin equivalent

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

IL = Vth/(Rth + RL) = 3.273/(10.18 + 12) = 0.1475 A
10.Thévenin & Norton

Answer: Vth = 34 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​ = 30 ΩR3​ = 2 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 = 30·20/(30+20) = 12 Ω
Rth = (R1 ∥ R2) + R3 = 12 + 2 = 14 Ω
R1​ = 30 ΩR3​ = 2 Ω+−V1​15 VR2​20 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) = 6 + 2·12 = 30 V
Vth = VB = VA + Is·R3 = 30 + 2·2 = 34 V
11.RC & RL transients

Answer: t = 4.575 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 = 33 mH / 10 Ω = 3.3 ms
i(∞) = Vs/R = 5/10 = 0.5 A

Solve for the time

0.75 = 1 − e−t/τ → t = −τ·ln(1 − 0.75)
t = −3.3·ln(0.25) = 4.575 ms

Answer

t = 4.575 ms
12.RC & RL transients

Answer: i2 = 0.3969 mA

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 capacitor voltage

In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.

vC(0) = 10·2.2/(10 + 2.2) = 1.803 V

Time constant after the switch opens

τ = R2·C = 2.2·47 = 103.4 ms

Discharge

vC(t) = 1.803·e−75/103.4 = 0.8731 V
i2 = vC/R2 = 0.8731/2.2 = 0.3969 mA

Answer

i2 = 0.3969 mA
13.AC impedance & phasors

Answer: P = 17.86 W

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

Reactances

XL = 25.13 Ω, XC = 16.93 Ω

Impedance

|Z| = √(30² + (8.201)²) = 31.1 Ω

Real power

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

I = 24/31.1 = 0.7717 A
P = I²·R = 0.7717²·30 = 17.86 W

Answer

P = 17.86 W
14.AC impedance & phasors

Answer: VL = 165 V

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

Resonance

At resonance the reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.02·0.47·10⁻⁶) = 10314 rad/s
I = V/R = 12/15 = 0.8 A

Voltage across the inductor

XL = ω₀L = 10314·0.02 = 206.3 Ω
VL = I·XL = 0.8·206.3 = 165 V

Answer

VL = 165 V

This is Q = 13.75 times the supply voltage.

15.AC power & power factor

Answer: Q = 22.05 kvar

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.

Angle

φ = arccos(0.75) = 41.41°, tan φ = 0.8819

Reactive power

Q = P·tan φ = 25·0.8819 = 22.05 kvar

Answer

Q = 22.05 kvar
16.AC power & power factor

Answer: cos φ = 0.7842

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 = 22·tan(arccos 0.75) = 19.4 kvar
QC = V²·2πf·C = 230²·2π·60·100·10⁻⁶ = 1994 var = 1.994 kvar
Q2 = 19.4 − 1.994 = 17.41 kvar

New apparent power

S2 = √(22² + 17.41²) = 28.05 kVA

New power factor

cos φ = P/S2 = 22/28.05 = 0.7842

Answer

cos φ = 0.7842
17.Three-phase power

Answer: IL = 94.04 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 = 90/0.91 = 98.9 kW

Line current

IL = Pin/(√3·VL·cos φ) = 98901/(√3·690·0.88) = 94.04 A

Answer

IL = 94.04 A
18.Three-phase power

Answer: QC = 31.8 kvar

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.

Reactive power before and after

The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.

Q1 = 100·tan(acos 0.78) = 100·0.8023 = 80.23 kvar
Q2 = 100·tan(acos 0.9) = 100·0.4843 = 48.43 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 31.8 kvar

Answer

QC = 31.8 kvar
19.Transformers

Answer: ILV = 312.5 A

Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.

Rated current

Both windings carry the same apparent power.

ILV = S/V = 150000/480 = 312.5 A

Answer

ILV = 312.5 A
20.Transformers

Answer: Ploss = 562.5 W

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

Copper loss at this load

Pcu = x²·Pcu,fl = 0.0625·1800 = 112.5 W

Total losses

The core loss is the same at any load.

Ploss = 450 + 112.5 = 562.5 W

Answer

Ploss = 562.5 W

The efficiency at this load is η = 96.59%.

21.Op-amp circuits

Answer: Vout = −1.064 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) = −2.4·2/(4.7 + 2) = −0.7164 V
V₋ = V₊

KCL at the inverting input

(0.1 − −0.7164)/4.7 = (−0.7164 − Vout)/2
Vout = (R2/R1)·(V2 − V1) = (2/4.7)·(−2.4 − 0.1) = −1.064 V

Answer

Vout = −1.064 V
22.Op-amp circuits

Answer: Vout = −3 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 = −10/2 = −5
A2 = 1 + Rf2/R2 = 1 + 22/22 = 2

Stage by stage

V1 = −5·0.3 = −1.5 V
Vout = 2·(−1.5) = −3 V

Answer

Vout = −3 V
23.Diodes & transistors

Answer: IZ = 45.46 mA

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.

Resistor current

IR = 12.9/270 = 47.78 mA

Load current

IL = 5.1/2200 = 2.318 mA

Zener current

IZ = 47.78 − 2.318 = 45.46 mA

Answer

IZ = 45.46 mA
24.Diodes & transistors

Answer: RB = 32.25 kΩ

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.

Collector current in saturation

IC(sat) = (9 − 0.2)/330 = 26.67 mA

Minimum base current

IB(min) = IC(sat)/β = 26.67/200 = 133.3 µA

Base resistor

RB = (5 − 0.7)/IB = 4.3 V/0.1333 mA = 32.25 kΩ

Answer

RB = 32.25 kΩ

Any smaller base resistor also saturates the transistor; in practice you pick one a few times smaller.

25.Number systems

Answer: U = 246

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

10 = 0000 1010
inverted: 1111 0101
+1 → 1111 0110

Read it as unsigned

1111 0110₂ = 256 − 10 = 246

Answer

U = 246
26.Number systems

Answer: R = 124

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Bit by bit

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

1001 1111 << 2
= 0111 1100

Value of the result

0111 1100₂ = 124

Answer

R = 124
27.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

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.

ABCD000111100001111010010302141517061121130150141819011110

Group the 0s

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

(B′ + C′ + D)
(A + C′ + D)
(A′ + C′ + D′)
(A + B + D′)

Minimal POS form

F = (B′ + C′ + D)(A + C′ + D)(A′ + C′ + D′)(A + B + D′)
N = 4

Answer

N = 4

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

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

Answer: Q = 1

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.

State after every edge

0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001

Decimal value

0001₂ = 1

Answer

Q = 1
29.Laplace transforms

Answer: f(0.25) = 0.3129

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 + 3)
A = 15/(2·3) = 2.5
B = 15/((−2)·1) = −7.5
C = 15/((−3)·(−1)) = 5

Inverse transform

f(t) = 2.5 − 7.5·e−2t + 5·e−3t

Evaluate

f(0.25) = 2.5 − 7.5·e−2·0.25 + 5·e−3·0.25 = 0.3129

Answer

f(0.25) = 0.3129
30.Laplace transforms

Answer: g(0.3) = 2.112

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) = 11/(s² + 3s + 2)

Partial fractions

s² + 3s + 2 = (s + 1)(s + 2)
A = 11, B = −11

Impulse response

g(t) = 11·e−t − 11·e−2t
g(0.3) = 2.112

Answer

g(0.3) = 2.112
31.Block diagrams

Answer: T(0) = 0.8525

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 = G/(1 + G·H₁) = 4/(s + 7 + 4·0.5) = 4/(s + 9)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 52/(s + 9 + 52) = 52/(s + 61)

Result

T(0) = 52/61 = 0.8525

Answer

T(0) = 0.8525
32.Block diagrams

Answer: ω = 10.2 rad/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).

Characteristic equation

s³ + 18s² + 104s + 192 + K = 0

Routh array

s³ | 1 104
s² | 18 192 + K
s¹ | (18·104 − 192 − K)/18
s⁰ | 192 + K

Stability limit

All entries of the first column must be positive.

Kmax = 18·104 − 192 = 1680

Frequency of oscillation

18·(jω)² + 192 + 1680 = 0 → ω² = 1872/18 = 104
ω = 10.2 rad/s

Answer

ω = 10.2 rad/s
33.Step response

Answer: ωn = 4.123 rad/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 = 17 → ωₙ = 4.123 rad/s
2ζωₙ = 5 → ζ = 5/(2·4.123) = 0.6063

Result

ωₙ = 4.123 rad/s

Answer

ωn = 4.123 rad/s
34.Step response

Answer: ts = 4 s

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

Closed-loop transfer function

T(s) = 25/(s² + 2s + 25): ωₙ = 5 rad/s, ζ = 0.2

Settling time

ζωₙ = 2/2 = 1
ts ≈ 4/(ζωₙ) = 4 s

Answer

ts = 4 s

For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.

35.Bode plots & margins

Answer: ∠G(jω) = −99.16 °

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(30/18) − 90° − atan(30/12) = 59.04° − 90° − 68.2°

Result

∠G(j30) = −99.16°

Answer

∠G(jω) = −99.16°
36.Bode plots & margins

Answer: GM = 5.007 dB

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

Phase crossover

(ω/3)·(ω/7) = 1 → ω180 = √(3·7) = 4.583 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 9)·√(ω² + 49)) = K/(3·7·(3 + 7)) = K/210

Gain margin

GM = 20·log₁₀(210/118) = 5.007 dB

Answer

GM = 5.007 dB
37.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 126

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 + 8s² + 17s + 10 + K = 0

Routh array

s³: 1, 17
s²: 8, 10 + K
s¹: (8·17 − 10 − K)/8
s⁰: 10 + K

Condition

The s¹ entry must stay positive.

K < 8·17 − 10 = 126

Answer

Kmax = 126

At the limit the loop oscillates at ω = √17 = 4.123 rad/s.

38.Stability: Routh–Hurwitz & root locus

Answer: si = −23.17

Root locus: as the gain K goes from 0 to ∞, the closed-loop poles, the roots of 1 + K·G(s) = 0, move from the open-loop poles to the open-loop zeros or to infinity. A point on the real axis belongs to the locus when the number of real poles and zeros to its right is odd. The n − m branches that go to infinity follow asymptotes at the angles (2q + 1)·180°/(n − m), which start from the centroid σₐ = (Σ poles − Σ zeros)/(n − m). Branches leave or join the real axis where dK/ds = 0. The angle condition ∠G(s) = 180° fixes the shape, and the magnitude condition K = 1/|G(s)| gives the gain at any point of the locus.

Gain along the real axis

K(s) = −(s² + 8s)/(s + 14)

Stationary points

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

(2s + 8)(s + 14) − (s² + 8s) = s² + 28s + 112 = 0
s = −14 ± √84 = −4.835, −23.17

Break-in point

si = −23.17

Answer

si = −23.17

Between the two break points the locus is a circle centred on the zero, with radius √84 = 9.165.

39.Steady-state error & PID control

Answer: ess = 4

Steady-state error: in a unity-feedback loop with open-loop transfer function G(s) the error is E(s) = R(s)/(1 + G(s)), and by the final value theorem ess = lims→0 s·E(s). The system type is the number of poles of G(s) at s = 0. The error constants are Kp = lim G(s), Kv = lim s·G(s) and Ka = lim s²·G(s), all for s → 0. A step of size A gives ess = A/(1 + Kp), a ramp A·t gives A/Kv and a parabola A·t²/2 gives A/Ka. A type-0 loop follows a step with a finite error, a type-1 loop follows it exactly and a ramp with a finite error.

Velocity constant

Kv = lim s·G(s) = 1

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 4/1 = 4

Answer

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

Answer: ts = 4 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 + 2) + 1·10 = s² + 2s + 10

ωₙ and ζ

ωₙ = √10 = 3.162 rad/s
ζ = 2/(2·3.162) = 0.3162

Settling time

ζωₙ = 2/2 = 1
ts ≈ 4/1 = 4 s

Answer

ts = 4 s

The settling time does not depend on K here: raising K only makes the response oscillate faster.

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