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​ = 10 ΩR3​ = 5 Ω+−V1​15 VR2​3 ΩIR2​AR4​2 ΩB

Find the current IR2 through R2.

V1 = 15 VR1 = 10 ΩR2 = 3 ΩR3 = 5 ΩR4 = 2 Ω
IR2 = A
2.Series & parallel resistorsHard
R1​ = 4 ΩR3​ = 12 ΩR5​ = 5 Ω+−V1​30 VR2​15 ΩIR2​AR4​6 ΩBR6​4 Ω

Find the current IR2 through R2.

V1 = 30 VR1 = 4 ΩR2 = 15 ΩR3 = 12 ΩR4 = 6 ΩR5 = 5 ΩR6 = 4 Ω
IR2 = A
3.Nodal analysisMedium
R1​ = 8 ΩIR1​R3​ = 2 Ω+−V1​12 VR2​12 ΩAR4​4 ΩBIs​2 A

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

V1 = 12 VIs = 2 AR1 = 8 ΩR2 = 12 ΩR3 = 2 ΩR4 = 4 Ω
IR1 = A
4.Nodal analysisHard
R1​ = 5 ΩR3​ = 12 Ω+−V1​10 VR2​12 ΩAR4​15 ΩBIs​?

The current source is adjusted until the node voltage at A is VA = 7.5 V. Use nodal analysis to find the source current Is (direction as shown).

V1 = 10 VR1 = 5 ΩR2 = 12 ΩR3 = 12 ΩR4 = 15 ΩVA = 7.5 V
Is = A
5.Mesh analysisMedium
R1​ = 2 ΩR2​ = 20 Ω+−V1​18 VR3​15 ΩA+−V2​9 VI1​I2​

Use mesh analysis to find the mesh current I1 (clockwise).

V1 = 18 VV2 = 9 VR1 = 2 ΩR2 = 20 ΩR3 = 15 Ω
I1 = A
6.Mesh analysisHard
R1​ = 12 ΩR2​ = 4 ΩR3​ = 15 Ω+−V1​15 VR4​5 ΩIR4​AR5​30 ΩB+−V2​18 VI1​I2​I3​

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

V1 = 15 VV2 = 18 VR1 = 12 ΩR2 = 4 ΩR3 = 15 ΩR4 = 5 ΩR5 = 30 Ω
IR4 = A
7.SuperpositionMedium
linear circuitVs​Is​IRL​

A linear resistive circuit contains one voltage source Vs and one current source Is. The current I through a load resistor is measured twice: with Vs = 4 V and Is = 4 mA, I = 3 mA; with Vs = 10 V and Is = 6 mA, I = 10.5 mA. Find the load current I when Vs = 2 V and Is = 6 mA.

4 V, 4 mA → 3 mA10 V, 6 mA → 10.5 mAVs = 2 V, Is = 6 mA
I = mA
8.SuperpositionHard
R1​ = 25 ΩR2​ = 6 ΩR4​ = 3 ΩIR4​+−V1​36 VR3​3 ΩAIs​3 AB+−V2​18 V

Use superposition to find the power PR4 absorbed by R4.

V1 = 36 VIs = 3 AV2 = 18 VR1 = 25 ΩR2 = 6 ΩR3 = 3 ΩR4 = 3 Ω
PR4 = W
9.Thévenin & NortonMedium
R1​ = 10 ΩR3​ = 4 Ω+−V1​18 VR2​8 ΩAab

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

V1 = 18 VR1 = 10 ΩR2 = 8 ΩR3 = 4 Ω
Rth = Ω
10.Thévenin & NortonHard
R1​ = 4 ΩR3​ = 4 Ω+−V1​5 VR2​4 ΩAIs​1.5 ABab

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

V1 = 5 VIs = 1.5 AR1 = 4 ΩR2 = 4 ΩR3 = 4 Ω
Vth = V
11.RC & RL transientsMedium
+−Vs​6 Vt = 0R = 10 ΩL150 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 40 ms, in mJ.

Vs = 6 VR = 10 ΩL = 150 mHt = 40 ms
W = mJ
12.RC & RL transientsHard
+−Vs​15 Vt = 0R = 1 kΩC220 µF+−vC​vC​(0) = 30 V

Just before the switch closes at t = 0, the capacitor holds 30 V with the polarity shown. How long does it take for vC to reach 22.5 V? Find t.

Vs = 15 VR = 1 kΩC = 220 µFvC(0) = 30 V
t = ms
13.AC impedance & phasorsMedium
50 V100 HzR = 15 ΩL = 200 mHC = 4.7 µFI

A series RLC circuit is connected to a 50 V (rms), 100 Hz supply. Find the rms current I.

V = 50 V (rms)f = 100 HzR = 15 ΩL = 200 mHC = 4.7 µF
I = A
14.AC impedance & phasorsHard
12 Vf0​R = 20 ΩL = 5 mHC = 2.2 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 = 20 ΩL = 5 mHC = 2.2 µF
VL = V
15.AC power & power factorMedium
240 V12 kW, 0.6230 kW, 0.95I

Two loads are connected in parallel to a 240 V single-phase supply: a 2 kW load at a power factor of 0.6 lagging and a 30 kW load at a power factor of 0.95 leading. Find the overall power factor cos φ seen by the supply.

V = 240 VP1 = 2 kW, pf1 = 0.6 laggingP2 = 30 kW, pf2 = 0.95 leading
cos φ =
16.AC power & power factorHard
P = 35 kWQ1​S1​φ1​QC​

A single-phase load takes 35 kW at a power factor of 0.65 lagging from a 230 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the capacitance C needed.

P = 35 kWpf1 = 0.65pf2 = 0.9V = 230 Vf = 50 Hz
C = µF
17.Three-phase powerMedium
ABCVL​ = 208 VZ = 8 + j6 Ωper phase

A balanced Δ-connected load with an impedance of 8 + j6 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.

VL = 208 VZ = 8 + j6 Ω per phase
P = kW
18.Three-phase powerHard
P = 80 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 80 kW at a power factor of 0.78 lagging from a 400 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 80 kWpf1 = 0.78pf2 = 0.92VL = 400 Vf = 60 Hz
Cph = µF
19.TransformersMedium
4 Ω41

A 4 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 4:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 4:1RL = 4 Ω
R1′ = Ω
20.TransformersHard
pf 0.9525 kVA, η = 98.2%

A single-phase 25 kVA transformer has a core loss of 125 W. At rated load and a power factor of 0.95 lagging its efficiency is 98.2%. Find the full-load copper loss Pcu,fl in watts.

S = 25 kVAPcore = 125 Wpf = 0.95η = 98.2%
Pcu,fl = W
21.Op-amp circuitsMedium
−+Vout​Rf​ = 22 kΩR1​ = 10 kΩVin​ = 1.2 V

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

Vin = 1.2 VR1 = 10 kΩRf = 22 kΩ
Vout = V
22.Op-amp circuitsHard
−+Vout​C = 4.7 µFVin​ = −2.1 VR = 47 kΩ

The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±14 V. The capacitor is uncharged when a constant input Vin = −2.1 V is applied at t = 0. Find the output voltage Vout at t = 500 ms.

Vin = −2.1 VR = 47 kΩC = 4.7 µFVsat = ±14 Vt = 500 ms
Vout = V
23.Diodes & transistorsMedium
RC​ = 1 kΩβ = 120RE​ = 470 ΩVCC​ = 18 VR1​ = 56 kΩR2​ = 8.2 kΩ

An NPN transistor with β = 120 has voltage-divider bias from VCC = 18 V: R₁ = 56 kΩ from VCC to the base and R₂ = 8.2 kΩ from the base to ground. The collector resistor is RC = 1 kΩ and the emitter resistor is RE = 470 Ω. Take VBE = 0.7 V and use the Thévenin equivalent of the divider (do not neglect the base current). Find the collector–emitter voltage VCE.

VCC = 18 VR1 = 56 kΩR2 = 8.2 kΩRC = 1 kΩRE = 470 Ωβ = 120
VCE = V
24.Diodes & transistorsHard
RC​ = 220 Ωβ = 150VCC​ = 9 V3.3 VRB​ = ?

An NPN transistor with β = 150 switches a 220 Ω load resistor connected between VCC = 9 V and its collector. The emitter is grounded. The base is driven from a 3.3 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 = 220 Ωβ = 150Vin = 3.3 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 −59 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −598 bit
U =
26.Number systemsHard
2709 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 2709 items its own binary code of the same length. What is the smallest number of bits b needed?

2709
b = bit
27.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011001130214051716

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(0, 3, 4, 6, 7). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(0, 3, 4, 6, 7)
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 5 clock edges.

Q₃Q₂Q₁Q₀ = 00005
Q =
29.Laplace transformsMedium
t (s)f(t)t = 0.1f(0.1) = ?

Find f(t) for F(s) = (s + 3)/(s² + 8s + 16) and evaluate f(0.1).

F(s) = (s + 3)/(s² + 8s + 16)t = 0.1 s
f(0.1) =
30.Laplace transformsHard
t (s)y(t)t = 0.1y(0.1) = ?

Solve y″ + 6y′ + 8y = 0 with y(0) = −9 and y′(0) = −3 using the Laplace transform, and find y(0.1).

y″ + 6y′ + 8y = 0y(0) = −9y′(0) = −3t = 0.1 s
y(0.1) =
31.Block diagramsMedium
R(s)+10K+1s + 5G(s)C(s)0.5H₁−−

The plant G(s) = 1/(s + 5) has an inner (minor) negative feedback loop with gain H₁ = 0.5. A controller gain K = 10 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 1/(s + 5)H₁ = 0.5K = 10
T(0) =
32.Block diagramsHard
R(s)+K(s + 2)(s + 5)(s + 7)G(s)C(s)−

The plant G(s) = K/((s + 2)(s + 5)(s + 7)) 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 + 2)(s + 5)(s + 7))unity negative feedback
ω = rad/s
33.Step responseMedium
t (s)y(t)1tp​ = ?

A system has the transfer function G(s) = 65/(s² + 6s + 65) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 65/(s² + 6s + 65)
tp = s
34.Step responseHard
+−12 VR = 58 ΩL = 100 mHC = 4.7 µFvC​

A series RLC circuit with R = 58 Ω, L = 100 mH and C = 4.7 µF is connected to a 12 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the peak capacitor voltage vC,max.

R = 58 ΩL = 100 mHC = 4.7 µFV = 12 V
vC,max = V
35.Bode plots & marginsMedium
vin​L = 10 mHC = 4.7 µFR = 1 Ωvout​

In the series RLC band-pass filter shown the output is taken across R = 1 Ω, with L = 10 mH and C = 4.7 µF. Find the centre (resonant) frequency f0 in Hz.

R = 1 ΩL = 10 mHC = 4.7 µF
f0 = Hz
36.Bode plots & marginsHard
|H| (dB)11101010010010001000−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 320/(s(s + 10)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 320/(s(s + 10))
ωc = rad/s
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 10(s − 4)(s + 2)C(s)−

The plant G(s) = (s + 10)/((s − 4)(s + 2)) has an unstable pole at s = 4. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.

G(s) = (s + 10)/((s − 4)(s + 2))
Kmin =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2

The root locus of L(s) = K/(s(s + 5)(s + 10)) leaves the real axis between s = 0 and s = −5. Find the break-away point sb.

L(s) = K/(s(s + 5)(s + 10))
sb =
39.Steady-state error & PID controlMedium
R(s)+K = 241s + 1C(s)−

A process G(s) = 4/(1s + 1) is controlled by a proportional gain K = 2 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 4/(1s + 1)K = 2
ess = %
40.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s1s + 2C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 1/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 3 rad/s and ζ = 0.7. Find the proportional gain Kp.

G(s) = 1/(s + 2)ωₙ = 3 rad/sζ = 0.7
Kp =

3Solutions

1.Series & parallel resistors

Answer: IR2 = 0.8678 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​ = 10 ΩR3​ = 5 Ω+−V1​15 VR2​3 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 = 5 + 2 = 7 Ω
R2 ∥ 7 = 3·7/(3+7) = 2.1 Ω
Req = R1 + 2.1 = 10 + 2.1 = 12.1 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 = 15/12.1 = 1.24 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 = 15 − 1.24·10 = 2.603 V
IR2 = VA/R2 = 2.603/3 = 0.8678 A

Answer

IR2 = 0.8678 A
2.Series & parallel resistors

Answer: IR2 = 1.313 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​ = 4 ΩR3​ = 12 ΩR5​ = 5 Ω+−V1​30 VR2​15 ΩAR4​6 ΩBR6​4 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 = 5 + 4 = 9 Ω
R4 ∥ 9 = 3.6 Ω
R3 + 3.6 = 15.6 Ω
R2 ∥ 15.6 = 7.647 Ω
Req = R1 + 7.647 = 4 + 7.647 = 11.65 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 = 30/11.65 = 2.576 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 = 30 − 2.576·4 = 19.7 V
IR2 = VA/R2 = 19.7/15 = 1.313 A

Answer

IR2 = 1.313 A
3.Nodal analysis

Answer: IR1 = 0.5556 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​ = 8 ΩR3​ = 2 Ω+−V1​12 VR2​12 ΩAR4​4 ΩBIs​2 A

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 12 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/12 + (VA − 12)/8 + (VA − VB)/2 = 0
0.7083·VA − 0.5·VB = 1.5

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/4 + (VB − VA)/2 = 2
−0.5·VA + 0.75·VB = 2

Solve

Δ = (0.7083)(0.75) − (−0.5)(−0.5) = 0.2813
VA = Δ1/Δ = 2.125/0.2813 = 7.556 V
VB = Δ2/Δ = 2.167/0.2813 = 7.704 V

Answer

IR1 = (V1 − VA)/R1 = (12 − 7.556)/8 = 0.5556 A
4.Nodal analysis

Answer: Is = 0.725 A

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

KCL at node A

VA is known, so this equation gives VB.

(VA − V1)/R1 + VA/R2 + (VA − VB)/R3 = 0
(7.5 − 10)/5 + 7.5/12 = (VB − 7.5)/12
0.125 = (VB − 7.5)/12 → VB = 9 V

KCL at node B

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

Is = (VB − VA)/R3 + VB/R4 = (9 − 7.5)/12 + 9/15 = 0.725 A

Answer

Is = 0.725 A
5.Mesh analysis

Answer: I1 = 1.338 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​18 VR3​15 ΩA+−V2​9 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 + 15(I1 − I2) = 18
17·I1 − 15·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
20·I2 + 15(I2 − I1) = (−9)
−15·I1 + 35·I2 = −9

Solve

Δ = (17)(35) − (−15)(−15) = 370
I1 = Δ1/Δ = 495/370 = 1.338 A
I2 = Δ2/Δ = 117/370 = 0.3162 A

Answer

I1 = 1.338 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: IR4 = 1.188 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​ = 12 ΩR2​ = 4 ΩR3​ = 15 Ω+−V1​15 VR4​5 ΩAR5​30 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
12·I1 + 5(I1 − I2) = 15
17·I1 − 5·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 + R4(I2 − I1) + R5(I2 − I3) = 0
4·I2 + 5(I2 − I1) + 30(I2 − I3) = 0
−5·I1 + 39·I2 − 30·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
15·I3 + 30(I3 − I2) = (−18)
−30·I2 + 45·I3 = −18

Solve

Δ = det of the coefficient matrix = 13410
I1 = Δ1/Δ = 10125/13410 = 0.755 A
I2 = Δ2/Δ = −5805/13410 = −0.4329 A
I3 = Δ3/Δ = −9234/13410 = −0.6886 A

Answer

IR4 = I1 − I2 = 1.188 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: I = −1.5 mA

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

Write the 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
4·k₁ + 4·k₂ = 3
10·k₁ + 6·k₂ = 10.5

Solve for the constants

Δ = (4)(6) − (4)(10) = −16
k₁ = Δ1/Δ = −24/−16 = 1.5
k₂ = Δ2/Δ = 12/−16 = −0.75
k₁ = 1.5 mA/V, k₂ = −0.75

Predict the new current

I = 1.5·2 + (−0.75)·6 = −1.5 mA

Answer

I = −1.5 mA
8.Superposition

Answer: PR4 = 3.111 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​ = 25 ΩR2​ = 6 ΩR4​ = 3 Ω+−V1​36 VR3​3 ΩAOPENBSHORT

Case 1 · V1 acting alone

V₂ becomes a short and Iₛ an open circuit. Seen from A, R₃ is in parallel with the series pair R₂+R₄.

R3 ∥ (R2+R4) = 3 ∥ 9 = 2.25 Ω
VA = 36·2.25/(25+2.25) = 2.972 V
VB = VA·R4/(R2+R4) = 2.972·3/9 = 0.9908 V
IR4′ = (VB − V2)/R4 = (0.9908 − 0)/3 = 0.3303 A
R1​ = 25 ΩR2​ = 6 ΩR4​ = 3 ΩSHORTR3​3 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 = 2.679 Ω, so the left path is R2+2.679 = 8.679 Ω
Req = R4 ∥ 8.679 = 2.229 Ω
VB = Is·Req = 3·2.229 = 6.688 V
VA = VB·2.679/8.679 = 2.064 V
IR4″ = (VB − V2)/R4 = (6.688 − 0)/3 = 2.229 A
R1​ = 25 ΩR2​ = 6 ΩR4​ = 3 ΩSHORTR3​3 ΩAOPENB+−V2​18 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 = 25·3/(25+3) = 2.679 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 18·8.679/11.68 = 13.38 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 13.38·2.679/8.679 = 4.128 V
IR4‴ = (VB − V2)/R4 = (13.38 − 18)/3 = −1.541 A

Add the current contributions

IR4 = IR4′ + IR4″ + IR4‴ = 0.3303 + 2.229 + (−1.541) = 1.018 A

Power in the resistor

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

PR4 = IR4²·R4 = 1.018²·3 = 3.111 W

Answer

PR4 = 3.111 W

Adding the powers of the separate cases would give 22.36 W, which is wrong.

9.Thévenin & Norton

Answer: Rth = 8.444 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​ = 10 ΩR3​ = 4 ΩSHORTR2​8 ΩAabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 10·8/(10+8) = 4.444 Ω
Rth = (R1 ∥ R2) + R3 = 4.444 + 4 = 8.444 Ω
10.Thévenin & Norton

Answer: Vth = 11.5 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​ = 4 ΩR3​ = 4 ΩSHORTR2​4 ΩAOPENBabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 4·4/(4+4) = 2 Ω
Rth = (R1 ∥ R2) + R3 = 2 + 4 = 6 Ω
R1​ = 4 ΩR3​ = 4 Ω+−V1​5 VR2​4 ΩAIs​1.5 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) = 2.5 + 1.5·2 = 5.5 V
Vth = VB = VA + Is·R3 = 5.5 + 1.5·4 = 11.5 V
11.RC & RL transients

Answer: W = 23.38 mJ

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 = 150 mH / 10 Ω = 15 ms
i(∞) = Vs/R = 6/10 = 0.6 A

Current at the given time

i(t) = 0.6·(1 − e−40/15) = 0.5583 A

Stored energy

W = ½·L·i² = ½·150·0.5583² = 23.38 mJ

Answer

W = 23.38 mJ
12.RC & RL transients

Answer: t = 152.5 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) = 30 V, vC(∞) = Vs = 15 V
τ = R·C = 1 kΩ · 220 µF = 220 ms
vC(t) = 15 + (30 − 15)·e−t/τ

Solve for the time

22.5 = 15 + (15)·e−t/τ → e−t/τ = (22.5 − 15)/(30 − 15) = 0.5
t = τ·ln((30 − 15)/(22.5 − 15)) = 220·ln(2) = 152.5 ms

Answer

t = 152.5 ms
13.AC impedance & phasors

Answer: I = 0.2342 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·100 = 628.3 rad/s
XL = ωL = 628.3·0.2 = 125.7 Ω
XC = 1/(ωC) = 1/(628.3·4.7·10⁻⁶) = 338.6 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(15² + (−213)²) = 213.5 Ω

Current

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

I = V/|Z| = 50/213.5 = 0.2342 A

Answer

I = 0.2342 A
14.AC impedance & phasors

Answer: VL = 28.6 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/√(0.005·2.2·10⁻⁶) = 9535 rad/s
I = V/R = 12/20 = 0.6 A

Voltage across the inductor

XL = ω₀L = 9535·0.005 = 47.67 Ω
VL = I·XL = 0.6·47.67 = 28.6 V

Answer

VL = 28.6 V

This is Q = 2.384 times the supply voltage.

15.AC power & power factor

Answer: cos φ = 0.9756

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

Reactive power of each load

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

Q1 = 2·tan(arccos 0.6) = 2.667 kvar
Q2 = −30·tan(arccos 0.95) = −9.861 kvar

Totals

P = 2 + 30 = 32 kW
Q = 2.667 + (−9.861) = −7.194 kvar
S = √(32² + (−7.194)²) = 32.8 kVA

Overall power factor

cos φ = P/S = 32/32.8 = 0.9756 leading

Answer

cos φ = 0.9756
16.AC power & power factor

Answer: C = 1442 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 = 35·tan(49.46°) = 40.92 kvar
Q2 = P·tan φ2 = 35·tan(25.84°) = 16.95 kvar
QC = Q1 − Q2 = 23.97 kvar

Capacitance

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

C = QC/(2πf·V²) = 23968/(2π·50·230²) = 1442 µF

Answer

C = 1442 µF
17.Three-phase power

Answer: P = 10.38 kW

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Phase voltage

Vph = VL = 208 V

Phase current

|Z| = √(8² + 6²) = 10 Ω
Iph = 208/10 = 20.8 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·20.8²·8 = 10.38 kW

Answer

P = 10.38 kW

Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.

18.Three-phase power

Answer: Cph = 166.4 µF

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

Reactive power before and after

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

Q1 = 80·tan(acos 0.78) = 80·0.8023 = 64.18 kvar
Q2 = 80·tan(acos 0.92) = 80·0.426 = 34.08 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 30.1 kvar

Capacitance per phase

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

V = 400 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 10034/(377·400²) = 166.4 µF

Answer

Cph = 166.4 µF
19.Transformers

Answer: R1′ = 64 Ω

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

Referred resistance

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

R1′ = a²·RL = 4²·4 = 64 Ω

Answer

R1′ = 64 Ω
20.Transformers

Answer: Pcu,fl = 310.3 W

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

Output and input power

Pout = 25·0.95 = 23.75 kW
Pin = 23750/0.982 = 24185 W

Total and copper losses

Ploss = 24185 − 23750 = 435.3 W
Pcu,fl = 435.3 − 125 = 310.3 W

Answer

Pcu,fl = 310.3 W
21.Op-amp circuits

Answer: Vout = 3.84 V

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

Inverting input follows the input

V₋ = V₊ = 1.2 V
I = V₋/R1 = 1.2/10 = 0.12 mA

Output voltage

The current through R₁ also flows through Rf.

Vout = V₋ + I·Rf = 1.2 + 0.12·22 = 3.84 V

Answer

Vout = 3.84 V
22.Op-amp circuits

Answer: Vout = 4.753 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 current and ramp rate

kΩ times µF gives milliseconds.

RC = 47 kΩ · 4.7 µF = 220.9 ms
ramp rate = −Vin/(RC) = −(−2.1)/220.9 = 0.009507 V/ms

Output at the given time

The ideal value lies inside ±Vsat, so the integrator has not clipped yet.

Vout = −Vin·t/(RC) = −(−2.1)·500/220.9 = 4.753 V

Answer

Vout = 4.753 V
23.Diodes & transistors

Answer: VCE = 13.58 V

BJT bias point: in the active region VBE = 0.7 V, IC = β·IB and IE = (β + 1)·IB. Find IB from the base loop with KVL, then IC, then VCE from the collector–emitter loop. A voltage divider at the base is replaced by its Thévenin equivalent VTH = VCC·R₂/(R₁ + R₂), RTH = R₁∥R₂; an emitter resistor appears (β + 1) times larger in the base loop.

Thévenin equivalent of the divider

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

Base current

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

IB = (2.299 − 0.7)/(7.153 + 121·0.47) = 24.98 µA

Collector current

IC = 120·24.98 µA = 2.997 mA
IE = 121·24.98 µA = 3.022 mA

Collector–emitter voltage

VCE = 18 − 2.997·1 − 3.022·0.47 = 13.58 V

Answer

VCE = 13.58 V
24.Diodes & transistors

Answer: RB = 9.75 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)/220 = 40 mA

Minimum base current

IB(min) = IC(sat)/β = 40/150 = 266.7 µA

Base resistor

RB = (3.3 − 0.7)/IB = 2.6 V/0.2667 mA = 9.75 kΩ

Answer

RB = 9.75 kΩ

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

25.Number systems

Answer: U = 197

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

59 = 0011 1011
inverted: 1100 0100
+1 → 1100 0101

Read it as unsigned

1100 0101₂ = 256 − 59 = 197

Answer

U = 197
26.Number systems

Answer: b = 12 bit

Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.

Powers of two

211 = 2048 < 2709
212 = 4096 ≥ 2709

Answer

b = 12
27.Digital logic: gates, K-maps & flip-flops

Answer: N = 2

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.

ABC00011110011001130214051716

Group the 0s

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

(B + C′)
(A + B′ + C)

Minimal POS form

F = (B + C′)(A + B′ + C)
N = 2

Answer

N = 2
28.Digital logic: gates, K-maps & flip-flops

Answer: Q = 7

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

State after every edge

0000 → 1000 → 1100 → 1110 → 1111 → 0111

Decimal value

0111₂ = 7

Answer

Q = 7
29.Laplace transforms

Answer: f(0.1) = 0.6033

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

Factor the denominator

s² + 8s + 16 = (s + 4)²

Partial fractions

s + 3 = 1(s + 4) + −1
F(s) = 1/(s + 4) − 1/(s + 4)²

Inverse transform

f(t) = e−4t − t·e−4t
f(0.1) = (1 − 1·0.1)·e−4·0.1 = 0.6033

Answer

f(0.1) = 0.6033
30.Laplace transforms

Answer: y(0.1) = −8.927

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

Transform the equation

s²Y − (−9)s − (−3) + 6(sY − (−9)) + 8Y = 0
Y(s) = (−9s − 57)/((s + 2)(s + 4))

Partial fractions

A = −39/2 = −19.5
B = −21/(−2) = 10.5

Result

y(t) = −19.5·e−2t + 10.5·e−4t
y(0.1) = −8.927

Answer

y(0.1) = −8.927

Check: y(0) = A + B = −9.

31.Block diagrams

Answer: T(0) = 0.6452

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) = 1/(s + 5 + 1·0.5) = 1/(s + 5.5)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 10/(s + 5.5 + 10) = 10/(s + 15.5)

Result

T(0) = 10/15.5 = 0.6452

Answer

T(0) = 0.6452
32.Block diagrams

Answer: ω = 7.681 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 + 14s² + 59s + 70 + K = 0

Routh array

s³ | 1 59
s² | 14 70 + K
s¹ | (14·59 − 70 − K)/14
s⁰ | 70 + K

Stability limit

All entries of the first column must be positive.

Kmax = 14·59 − 70 = 756

Frequency of oscillation

14·(jω)² + 70 + 756 = 0 → ω² = 826/14 = 59
ω = 7.681 rad/s

Answer

ω = 7.681 rad/s
33.Step response

Answer: tp = 0.4198 s

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

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 65 → ωₙ = 8.062 rad/s
2ζωₙ = 6 → ζ = 6/(2·8.062) = 0.3721

Result

ωd = 8.062·√(1 − 0.3721²) = 7.483 rad/s → tp = π/ωd = 0.4198 s

Answer

tp = 0.4198 s
34.Step response

Answer: vC,max = 18.34 V

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).

Natural frequency and damping ratio

ωₙ = 1/√(LC) = 1/√(0.1·4.7 × 10⁻⁶) = 1459 rad/s
ζ = (R/2)·√(C/L) = (58/2)·√(4.7 × 10⁻⁶/0.1) = 0.1988

Result

%OS = 52.87 %
vC,max = 12·(1 + 0.5287) = 18.34 V

Answer

vC,max = 18.34 V
35.Bode plots & margins

Answer: f0 = 734.1 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/√(0.01·4.7 × 10⁻⁶) = 4613 rad/s
f₀ = ω₀/(2π) = 734.1 Hz

Answer

f0 = 734.1 Hz
36.Bode plots & margins

Answer: ωc = 16.55 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.

Gain crossover

ω⁴ + 100ω² − 102400 = 0
ω² = (−100 + √(10000 + 4·102400))/2 = 273.9
ωc = 16.55 rad/s

Answer

ωc = 16.55 rad/s
37.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 2

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

(s − 4)(s + 2) + K(s + 10) = s² + (-2 + K)s + (10K − 8)

Conditions

For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.

-2 + K > 0 → K > 2
10K − 8 > 0 → K > 0.8
Kmin = 2

Answer

Kmin = 2
38.Stability: Routh–Hurwitz & root locus

Answer: sb = −2.113

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) = −s(s + 5)(s + 10) = −(s³ + 15s² + 50s)

Stationary points

dK/ds = −(3s² + 30s + 50) = 0
s = (−30 ± √(900 − 600))/6 = −2.113, −7.887

Pick the point on the locus

The locus lies on the real axis between 0 and −5; the other root is not on the locus.

sb = −2.113

Answer

sb = −2.113
39.Steady-state error & PID control

Answer: ess = 11.11 %

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.

Position constant

Kp = 2·4 = 8

Step error

ess = 100/(1 + 8) = 11.11 %

Answer

ess = 11.11 %
40.Steady-state error & PID control

Answer: Kp = 2.2

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

Characteristic polynomial

1 + C(s)G(s) = 0, multiplied by s(s + 2).

s² + (2 + 1Kp)s + 1Ki

Target polynomial

s² + 2·0.7·3s + 3² = s² + 4.2s + 9

Match the coefficients

2 + 1Kp = 4.2 → Kp = 2.2

Answer

Kp = 2.2

The integral term makes the loop type 1, so a step is followed without steady-state error.

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