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​ = 8 Ω+−V1​6 VR2​8 ΩAR4​2 ΩB

Find the equivalent resistance Req seen by the source V1.

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

Find the equivalent resistance Req seen by the source V1.

V1 = 10 VR1 = 5 ΩR2 = 4 ΩR3 = 3 ΩR4 = 15 ΩR5 = 6 ΩR6 = 4 Ω
Req = Ω
3.Nodal analysisMedium
R1​ = 6 ΩR3​ = 4 Ω+−V1​18 VR2​2 ΩAR4​15 ΩBIs​2.5 A

Use nodal analysis to find the power PIs delivered by the current source Is.

V1 = 18 VIs = 2.5 AR1 = 6 ΩR2 = 2 ΩR3 = 4 ΩR4 = 15 Ω
PIs = W
4.Nodal analysisHard
R1​ = 25 ΩR2​ = 3 ΩR4​ = 20 Ω+−V1​12 VR3​3 ΩAR5​10 ΩBIs​2 A+−V2​5 V

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

V1 = 12 VIs = 2 AV2 = 5 VR1 = 25 ΩR2 = 3 ΩR3 = 3 ΩR4 = 20 ΩR5 = 10 Ω
VB = V
5.Mesh analysisMedium
R1​ = 2 ΩR2​ = 15 Ω+−V1​12 VR3​5 ΩIR3​A+−V2​6 VI1​I2​

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

V1 = 12 VV2 = 6 VR1 = 2 ΩR2 = 15 ΩR3 = 5 Ω
IR3 = A
6.Mesh analysisHard
R1​ = 2 ΩR2​ = 10 ΩR3​ = 20 Ω+−VR3​+−V1​10 VR4​20 ΩAR5​25 ΩB+−V2​9 VI1​I2​I3​

Use mesh analysis to find the voltage VR3 across R3 with the polarity shown.

V1 = 10 VV2 = 9 VR1 = 2 ΩR2 = 10 ΩR3 = 20 ΩR4 = 20 ΩR5 = 25 Ω
VR3 = V
7.SuperpositionMedium
R1​ = 20 ΩR3​ = 12 Ω+−V1​9 VR2​5 Ω+−VR2​AR4​3 ΩBIs​3 A

Use superposition to find the voltage VR2 across R2 with the polarity shown.

V1 = 9 VIs = 3 AR1 = 20 ΩR2 = 5 ΩR3 = 12 ΩR4 = 3 Ω
VR2 = V
8.SuperpositionHard
R1​ = 15 ΩR2​ = 12 ΩR4​ = 8 Ω+−V1​15 VR3​8 ΩIR3​AIs​4 AB+−V2​6 V

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

V1 = 15 VIs = 4 AV2 = 6 VR1 = 15 ΩR2 = 12 ΩR3 = 8 ΩR4 = 8 Ω
IR3 = A
9.Thévenin & NortonMedium
R1​ = 4 ΩR3​ = 4 Ω+−V1​24 VR2​15 ΩAab

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

V1 = 24 VR1 = 4 ΩR2 = 15 ΩR3 = 4 Ω
Vth = V
10.Thévenin & NortonHard
R1​ = 25 ΩR3​ = 15 Ω+−V1​5 VR2​2 ΩAIs​3 ABab

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

V1 = 5 VIs = 3 AR1 = 25 ΩR2 = 2 ΩR3 = 15 Ω
Vth = V
11.RC & RL transientsMedium
+−Vs​15 Vt = 0R = 20 ΩL22 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the current i at t = 1 ms.

Vs = 15 VR = 20 ΩL = 22 mHt = 1 ms
i = A
12.RC & RL transientsHard
+−Vs​10 Vt = 0R = 3.3 kΩC47 µF+−vC​vC​(0) = 6 V

Just before the switch closes at t = 0, the capacitor holds 6 V with the polarity shown. Find the capacitor voltage vC at t = 120 ms.

Vs = 10 VR = 3.3 kΩC = 47 µFvC(0) = 6 Vt = 120 ms
vC = V
13.AC impedance & phasorsMedium
230 V200 HzR = 15 ΩL = 50 mHC = 47 µF

A series RLC circuit is connected to a 230 V (rms), 200 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 230 V (rms)f = 200 HzR = 15 ΩL = 50 mHC = 47 µF
φ = °
14.AC impedance & phasorsHard
50 V500 HzR1​ = 40 ΩR2​ = 80 ΩC = 4.7 µF

A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 50 V (rms), 500 Hz supply. Find the magnitude of the impedance |Z|.

V = 50 V (rms)f = 500 HzR1 = 40 ΩR2 = 80 ΩC = 4.7 µF
|Z| = Ω
15.AC power & power factorMedium
P = 9 kWQSφ

A single-phase motor takes 9 kW at a power factor of 0.9 lagging from a 240 V supply. Find the line current I.

P = 9 kWpf = 0.9V = 240 V
I = A
16.AC power & power factorHard
240 V60 Hzload21 kW, 0.75C150 µFI = ?

A single-phase load takes 21 kW at a power factor of 0.75 lagging from a 240 V, 60 Hz supply. A 150 µF capacitor is connected in parallel with the load. Find the new line current I.

P = 21 kWpf1 = 0.75V = 240 Vf = 60 HzC = 150 µF
I = A
17.Three-phase powerMedium
ABCVL​ = 208 VNZ = 6 + j8 Ωper phase

A balanced Y-connected load with an impedance of 6 + j8 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 = 6 + j8 Ω per phase
P = kW
18.Three-phase powerHard
P = 60 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 60 kW at a power factor of 0.78 lagging from a 415 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 60 kWpf1 = 0.78pf2 = 0.92VL = 415 Vf = 50 Hz
Cph = µF
19.TransformersMedium
4 Ω51

A 4 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 5:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 5:1RL = 4 Ω
R1′ = Ω
20.TransformersHard
pf 0.925 kVA, η = 98%

A single-phase 25 kVA transformer has a core loss of 150 W. At rated load and a power factor of 0.9 lagging its efficiency is 98%. Find the full-load copper loss Pcu,fl in watts.

S = 25 kVAPcore = 150 Wpf = 0.9η = 98%
Pcu,fl = W
21.Op-amp circuitsMedium
−+Vout​Rf​ = 100 kΩR1​ = 100 kΩVin​ = 2.5 VRa​ = 22 kΩRb​ = 47 kΩ

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

Vin = 2.5 VRa = 22 kΩRb = 47 kΩR1 = 100 kΩRf = 100 kΩ
Vout = V
22.Op-amp circuitsHard
−+Vout​R2​ = 47 kΩV1​ = 0.7 VR1​ = 4.7 kΩV2​ = 0.8 VR3​ = 2 kΩR4​ = 4.7 kΩ

The op-amp is ideal and runs from ±15 V supplies. The resistor ratios are not matched. Find the output voltage Vout.

V1 = 0.7 VV2 = 0.8 VR1 = 4.7 kΩR2 = 47 kΩR3 = 2 kΩR4 = 4.7 kΩ
Vout = V
23.Diodes & transistorsMedium
RC​ = 2.7 kΩβ = 100IC​ = ?RE​ = 1.2 kΩVCC​ = 20 VR1​ = 33 kΩR2​ = 8.2 kΩ

An NPN transistor with β = 100 has voltage-divider bias from VCC = 20 V: R₁ = 33 kΩ from VCC to the base and R₂ = 8.2 kΩ from the base to ground. The collector resistor is RC = 2.7 kΩ and the emitter resistor is RE = 1.2 kΩ. Take VBE = 0.7 V and use the Thévenin equivalent of the divider (do not neglect the base current). Find the collector current IC in mA.

VCC = 20 VR1 = 33 kΩR2 = 8.2 kΩRC = 2.7 kΩRE = 1.2 kΩβ = 100
IC = mA
24.Diodes & transistorsHard
RC​ = 680 Ωβ = 200VCC​ = 12 V3.3 VRB​ = ?

An NPN transistor with β = 200 switches a 680 Ω load resistor connected between VCC = 12 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. For reliable switching, design for a base current 4 times the minimum needed for saturation. Find RB in kΩ.

VCC = 12 VRC = 680 Ωβ = 200Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 Voverdrive = 4
RB = kΩ
25.Number systemsMedium
8-bit two's complement?−128?64?32?16?8?4?2?1

Write −108 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −1088 bit
U =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0011 0001 XOR 0011 1101. What is the result's value R in decimal?

0011 0001XOR0011 1101
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
ABCD00011110000111101111111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(0, 4, 7, 8, 10, 12, 13)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
28.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​TQ4​Q4​TQ5​Q5​TQ6​Q6​TQ7​Q7​CLK

An 8-bit ripple (asynchronous) counter is built from flip-flops with a propagation delay of 8 ns each. All outputs must have settled before the next clock edge. Find the maximum clock frequency fmax.

n = 8tpd = 8 ns
fmax = MHz
29.Laplace transformsMedium
ReIm−1−2−3−4−5−6−7−81

F(s) = (4s + 30)/(s² + 5s + 4). Its inverse transform has the form f(t) = A·e−t + B·e−4t. Find the coefficient A by partial fractions.

F(s) = (4s + 30)/(s² + 5s + 4)
A =
30.Laplace transformsHard
ReIm−0.5−1−1.5−2−2.5−3−3.5−4−4.50.51

A system is described by the differential equation y″ + 6y′ + 8y = 8u, 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 = 0.4 s. Find g(0.4).

y″ + 6y′ + 8y = 8ut = 0.4 s
g(0.4) =
31.Block diagramsMedium
R(s)+7K+4s + 4G(s)C(s)5H₁−−

The plant G(s) = 4/(s + 4) has an inner (minor) negative feedback loop with gain H₁ = 5. A controller gain K = 7 and an outer unity negative feedback loop close the system. Find the closed-loop pole p.

G(s) = 4/(s + 4)H₁ = 5K = 7
p =
32.Block diagramsHard
R(s)+50s(s + 4)G(s)C(s)0.5H−

The plant G(s) = 50/(s(s + 4)) is in a negative feedback loop with sensor gain H = 0.5. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.

G(s) = 50/(s(s + 4))H = 0.5
ωd = rad/s
33.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 16/(s² + 2s + 16) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 16/(s² + 2s + 16)
ωn = rad/s
34.Step responseHard
+−10 VR = 36 ΩL = 20 mHC = 10 µFvC​

A series RLC circuit with R = 36 Ω, L = 20 mH and C = 10 µF is connected to a 10 V step at t = 0. The capacitor is uncharged and no current flows before the step. Find the percent overshoot %OS of the capacitor voltage.

R = 36 ΩL = 20 mHC = 10 µFV = 10 V
%OS = %
35.Bode plots & marginsMedium
vin​L = 47 mHR = 22 Ωvout​

The RL low-pass filter shown has R = 22 Ω, L = 47 mH. A sinusoidal input of 37 Hz is applied. Find the gain |H| = |vout/vin| in dB.

R = 22 ΩL = 47 mHf = 37 Hz
|H| = dB
36.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 20/(s(s + 8)). Find the phase margin PM in degrees.

L(s) = 20/(s(s + 8))
PM = °
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 1)(s + 3)(s + 5)C(s)−

A unity-feedback loop has L(s) = K/((s + 1)(s + 3)(s + 5)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 1)(s + 3)(s + 5))
Kmax =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2−1.5

For the loop L(s) = K/(s(s + 3)(s + 10)), find the gain K that puts a closed-loop pole at s = −1.5.

L(s) = K/(s(s + 3)(s + 10))s = −1.5
K =
39.Steady-state error & PID controlMedium
r(t) = 2t+Ks(s + 4)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 4)) must follow the ramp r(t) = 2t with a steady-state error of 0.5. Find the gain K.

G(s) = K/(s(s + 4))r(t) = 2tess = 0.5
K =
40.Steady-state error & PID controlHard
R(s)+K10s(s + 10)C(s)−

A proportional controller K drives the plant G(s) = 10/(s(s + 10)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.4.

G(s) = 10/(s(s + 10))ζ = 0.4
K =

3Solutions

1.Series & parallel resistors

Answer: Req = 9.444 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​ = 8 Ω+−V1​6 VR2​8 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 = 8 + 2 = 10 Ω
R2 ∥ 10 = 8·10/(8+10) = 4.444 Ω
Req = R1 + 4.444 = 5 + 4.444 = 9.444 Ω

Answer

Req = 9.444 Ω
2.Series & parallel resistors

Answer: Req = 7.769 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​ = 3 ΩR5​ = 6 Ω+−V1​10 VR2​4 ΩAR4​15 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 = 6 + 4 = 10 Ω
R4 ∥ 10 = 6 Ω
R3 + 6 = 9 Ω
R2 ∥ 9 = 2.769 Ω
Req = R1 + 2.769 = 5 + 2.769 = 7.769 Ω

Answer

Req = 7.769 Ω
3.Nodal analysis

Answer: PIs = 33.38 W

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

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)/6 + (VA − VB)/4 = 0
0.9167·VA − 0.25·VB = 3

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/15 + (VB − VA)/4 = 2.5
−0.25·VA + 0.3167·VB = 2.5

Solve

Δ = (0.9167)(0.3167) − (−0.25)(−0.25) = 0.2278
VA = Δ1/Δ = 1.575/0.2278 = 6.915 V
VB = Δ2/Δ = 3.042/0.2278 = 13.35 V

Power delivered by the current source

The source pushes Iₛ up into node B, from ground (0 V) to VB. A source whose current leaves its + end delivers P = V·I.

PIs = VB·Is = 13.35·2.5 = 33.38 W

Answer

PIs = 33.38 W
4.Nodal analysis

Answer: VB = 7.594 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​ = 25 ΩR2​ = 3 ΩR4​ = 20 Ω+−V1​12 VR3​3 ΩAR5​10 ΩBIs​2 A+−V2​5 V

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 12 V, V2 = 5 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/3 + (VA − 12)/25 + (VA − VB)/3 = 0
0.7067·VA − 0.3333·VB = 0.48

KCL at node B

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

VB/R5 + (VB − VA)/R2 + (VB − V2)/R4 = Is
VB/10 + (VB − VA)/3 + (VB − 5)/20 = 2
−0.3333·VA + 0.4833·VB = 2.25

Solve

Δ = (0.7067)(0.4833) − (−0.3333)(−0.3333) = 0.2304
VA = Δ1/Δ = 0.982/0.2304 = 4.261 V
VB = Δ2/Δ = 1.75/0.2304 = 7.594 V

Answer

VB = 7.594 V
5.Mesh analysis

Answer: IR3 = 1.67 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​ = 15 Ω+−V1​12 VR3​5 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
2·I1 + 5(I1 − I2) = 12
7·I1 − 5·I2 = 12

KVL around mesh 2

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

R2·I2 + R3(I2 − I1) = −V2
15·I2 + 5(I2 − I1) = (−6)
−5·I1 + 20·I2 = −6

Solve

Δ = (7)(20) − (−5)(−5) = 115
I1 = Δ1/Δ = 210/115 = 1.826 A
I2 = Δ2/Δ = 18/115 = 0.1565 A

Answer

IR3 = I1 − I2 = 1.67 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: VR3 = −2.018 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​ = 2 ΩR2​ = 10 ΩR3​ = 20 Ω+−V1​10 VR4​20 ΩAR5​25 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
2·I1 + 20(I1 − I2) = 10
22·I1 − 20·I2 = 10

KVL around mesh 2

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

R2·I2 + R4(I2 − I1) + R5(I2 − I3) = 0
10·I2 + 20(I2 − I1) + 25(I2 − I3) = 0
−20·I1 + 55·I2 − 25·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
20·I3 + 25(I3 − I2) = (−9)
−25·I2 + 45·I3 = −9

Solve

Δ = det of the coefficient matrix = 22700
I1 = Δ1/Δ = 14000/22700 = 0.6167 A
I2 = Δ2/Δ = 4050/22700 = 0.1784 A
I3 = Δ3/Δ = −2290/22700 = −0.1009 A

Answer

IR3 = I3 = −0.1009 A
VR3 = IR3·R3 = −0.1009·20 = −2.018 V

A negative mesh current flows anticlockwise.

7.Superposition

Answer: VR2 = −0.4737 V

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

R1​ = 20 ΩR3​ = 12 Ω+−V1​9 VR2​5 ΩAR4​3 ΩBOPEN

Case 1 · V1 acting alone

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

R3+R4 = 12+3 = 15 Ω
R2 ∥ (R3+R4) = 5·15/(5+15) = 3.75 Ω
VA = 9·3.75/(20+3.75) = 1.421 V
VB = VA·R4/(R3+R4) = 1.421·3/15 = 0.2842 V
VR2′ = VA − 0 = 1.421 − 0 = 1.421 V
R1​ = 20 ΩR3​ = 12 ΩSHORTR2​5 ΩAR4​3 ΩBIs​3 A

Case 2 · Is acting alone

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

R1 ∥ R2 = 20·5/(20+5) = 4 Ω
Req = R4 ∥ (R3+R1∥R2) = 3 ∥ 16 = 2.526 Ω
VB = Is·Req = (−3)·2.526 = −7.579 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = −7.579·4/16 = −1.895 V
VR2″ = VA − 0 = −1.895 − 0 = −1.895 V

Add the contributions

VR2 = VR2′ + VR2″
VR2 = 1.421 + (−1.895) = −0.4737 V

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

8.Superposition

Answer: IR3 = −1.5 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​ = 15 ΩR2​ = 12 ΩR4​ = 8 Ω+−V1​15 VR3​8 ΩAOPENBSHORT

Case 1 · V1 acting alone

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

R3 ∥ (R2+R4) = 8 ∥ 20 = 5.714 Ω
VA = 15·5.714/(15+5.714) = 4.138 V
VB = VA·R4/(R2+R4) = 4.138·8/20 = 1.655 V
IR3′ = (0 − VA)/R3 = (0 − 4.138)/8 = −0.5172 A
R1​ = 15 ΩR2​ = 12 ΩR4​ = 8 ΩSHORTR3​8 ΩAIs​4 ABSHORT

Case 2 · Is acting alone

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

R1 ∥ R3 = 5.217 Ω, so the left path is R2+5.217 = 17.22 Ω
Req = R4 ∥ 17.22 = 5.462 Ω
VB = Is·Req = 4·5.462 = 21.85 V
VA = VB·5.217/17.22 = 6.621 V
IR3″ = (0 − VA)/R3 = (0 − 6.621)/8 = −0.8276 A
R1​ = 15 ΩR2​ = 12 ΩR4​ = 8 ΩSHORTR3​8 ΩAOPENB+−V2​6 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 = 15·8/(15+8) = 5.217 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 6·17.22/25.22 = 4.097 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 4.097·5.217/17.22 = 1.241 V
IR3‴ = (0 − VA)/R3 = (0 − 1.241)/8 = −0.1552 A

Add the contributions

IR3 = IR3′ + IR3″ + IR3‴
IR3 = (−0.5172) + (−0.8276) + (−0.1552) = −1.5 A

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

9.Thévenin & Norton

Answer: Vth = 18.95 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​15 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 = 4·15/(4+15) = 3.158 Ω
Rth = (R1 ∥ R2) + R3 = 3.158 + 4 = 7.158 Ω
R1​ = 4 ΩR3​ = 4 Ω+−V1​24 VR2​15 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) = 24·15/(4+15) = 18.95 V
10.Thévenin & Norton

Answer: Vth = 50.93 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​ = 25 ΩR3​ = 15 ΩSHORTR2​2 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 = 25·2/(25+2) = 1.852 Ω
Rth = (R1 ∥ R2) + R3 = 1.852 + 15 = 16.85 Ω
R1​ = 25 ΩR3​ = 15 Ω+−V1​5 VR2​2 ΩAIs​3 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) = 0.3704 + 3·1.852 = 5.926 V
Vth = VB = VA + Is·R3 = 5.926 + 3·15 = 50.93 V
11.RC & RL transients

Answer: i = 0.4478 A

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

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 = 22 mH / 20 Ω = 1.1 ms
i(∞) = Vs/R = 15/20 = 0.75 A

Current at the given time

i(t) = i(∞)·(1 − e−t/τ) = 0.75·(1 − e−1/1.1) = 0.4478 A

Answer

i = 0.4478 A
12.RC & RL transients

Answer: vC = 8.155 V

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

Initial value, final value, time constant

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

vC(0) = 6 V, vC(∞) = Vs = 10 V
τ = R·C = 3.3 kΩ · 47 µF = 155.1 ms
vC(t) = 10 + (6 − 10)·e−t/τ

Voltage at the given time

vC(120) = 10 + (−4)·e−120/155.1 = 8.155 V

Answer

vC = 8.155 V
13.AC impedance & phasors

Answer: φ = 71.9 °

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

Reactances

ω = 2πf = 2π·200 = 1257 rad/s
XL = ωL = 1257·0.05 = 62.83 Ω
XC = 1/(ωC) = 1/(1257·47·10⁻⁶) = 16.93 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(15² + (45.9)²) = 48.29 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(45.9/15) = 71.9°

Answer

φ = 71.9°
14.AC impedance & phasors

Answer: |Z| = 83.33 Ω

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

Reactance

XC = 1/(2π·500·4.7·10⁻⁶) = 67.73 Ω

Parallel combination

Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.

Zp = R₂·(−jXC)/(R₂ − jXC) = (R₂·XC² − jR₂²·XC)/(R₂² + XC²)
Zp = 33.4 − j39.45 Ω

Total impedance

Z = 40 + 33.4 − j39.45 = 73.4 − j39.45 Ω
|Z| = √(73.4² + 39.45²) = 83.33 Ω

Answer

|Z| = 83.33 Ω
15.AC power & power factor

Answer: I = 41.67 A

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

Line current

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

I = P/(V·cos φ) = 9000/(240·0.9) = 41.67 A

Answer

I = 41.67 A
16.AC power & power factor

Answer: I = 108.2 A

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

Reactive power before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = 21·tan(arccos 0.75) = 18.52 kvar
QC = V²·2πf·C = 240²·2π·60·150·10⁻⁶ = 3257 var = 3.257 kvar
Q2 = 18.52 − 3.257 = 15.26 kvar

New apparent power

S2 = √(21² + 15.26²) = 25.96 kVA

New line current

I = S2/V = 25961/240 = 108.2 A

Answer

I = 108.2 A
17.Three-phase power

Answer: P = 2.596 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 = 208/√3 = 120.1 V

Phase current

|Z| = √(6² + 8²) = 10 Ω
Iph = 120.1/10 = 12.01 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·12.01²·6 = 2.596 kW

Answer

P = 2.596 kW

Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.

18.Three-phase power

Answer: Cph = 417.3 µF

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

Reactive power before and after

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

Q1 = 60·tan(acos 0.78) = 60·0.8023 = 48.14 kvar
Q2 = 60·tan(acos 0.92) = 60·0.426 = 25.56 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 22.58 kvar

Capacitance per phase

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

V = 415/√3 = 239.6 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 7526/(314.2·239.6²) = 417.3 µF

Answer

Cph = 417.3 µF
19.Transformers

Answer: R1′ = 100 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 = 5²·4 = 100 Ω

Answer

R1′ = 100 Ω
20.Transformers

Answer: Pcu,fl = 309.2 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 = 25·0.9 = 22.5 kW
Pin = 22500/0.98 = 22959 W

Total and copper losses

Ploss = 22959 − 22500 = 459.2 W
Pcu,fl = 459.2 − 150 = 309.2 W

Answer

Pcu,fl = 309.2 W
21.Op-amp circuits

Answer: Vout = 3.406 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 + input

V₊ = Vin·Rb/(Ra + Rb) = 2.5·47/(22 + 47) = 1.703 V

Non-inverting gain

V₋ = V₊, and the current through R₁ also flows through Rf.

Vout = V₊·(1 + Rf/R1) = 1.703·(1 + 100/100) = 3.406 V

Answer

Vout = 3.406 V
22.Op-amp circuits

Answer: Vout = −0.8269 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 + input

V₊ = V2·R4/(R3 + R4) = 0.8·4.7/(2 + 4.7) = 0.5612 V

Superpose the two inputs

With V₂ = 0, V₁ sees an inverting amplifier. With V₁ = 0, V₊ is amplified by the non-inverting gain 1 + R₂/R₁.

Vout = −(47/4.7)·0.7 + (1 + 47/4.7)·0.5612 = −7 + 6.173 = −0.8269 V

Answer

Vout = −0.8269 V
23.Diodes & transistors

Answer: IC = 2.568 mA

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

Thévenin equivalent of the divider

VTH = 20·8.2/(33 + 8.2) = 3.981 V
RTH = 33·8.2/(33 + 8.2) = 6.568 kΩ

Base current

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

IB = (3.981 − 0.7)/(6.568 + 101·1.2) = 25.68 µA

Collector current

IC = 100·25.68 µA = 2.568 mA
IE = 101·25.68 µA = 2.593 mA

Answer

IC = 2.568 mA
24.Diodes & transistors

Answer: RB = 7.492 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) = (12 − 0.2)/680 = 17.35 mA

Minimum base current

IB(min) = IC(sat)/β = 17.35/200 = 86.76 µA

Design base current

IB = 4·86.76 µA = 347.1 µA

Base resistor

RB = (3.3 − 0.7)/IB = 2.6 V/0.3471 mA = 7.492 kΩ

Answer

RB = 7.492 kΩ
25.Number systems

Answer: U = 148

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.

Invert and add one

108 = 0110 1100
inverted: 1001 0011
+1 → 1001 0100

Read it as unsigned

1001 0100₂ = 256 − 108 = 148

Answer

U = 148
26.Number systems

Answer: R = 12

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

XOR gives 1 where the two bits differ.

0011 0001
XOR 0011 1101
= 0000 1100

Value of the result

0000 1100₂ = 12

Answer

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

Answer: AB = 01, CD = 11

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m7: m6, m5, m3, m15 → 0
ABCD00011110000111101111111

The isolated 1

m7 = A′BCD → AB = 01, CD = 11

Answer

AB = 01, CD = 11
28.Digital logic: gates, K-maps & flip-flops

Answer: fmax = 15.63 MHz

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.

Worst-case settling time

T = 8·8 ns = 64 ns

Maximum frequency

fmax = 1/64 ns = 15.63 MHz

Answer

fmax = 15.63 MHz
29.Laplace transforms

Answer: A = 8.667

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 + 5s + 4 = (s + 1)(s + 4)
F(s) = A/(s + 1) + B/(s + 4)

Cover-up rule

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

A = (4s + 30)/(s + 4) |s = −1 = 26/3 = 8.667

Answer

A = 8.667
(B = −4.667, so f(t) = 8.667·e−t − 4.667·e−4t)
30.Laplace transforms

Answer: g(0.4) = 0.9897

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) = 8/(s² + 6s + 8)

Partial fractions

s² + 6s + 8 = (s + 2)(s + 4)
A = 4, B = −4

Impulse response

g(t) = 4·e−2t − 4·e−4t
g(0.4) = 0.9897

Answer

g(0.4) = 0.9897
31.Block diagrams

Answer: p = −52

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 + 4 + 4·5) = 4/(s + 24)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 28/(s + 24 + 28) = 28/(s + 52)

Result

p = −52

Answer

p = −52
32.Block diagrams

Answer: ωd = 4.583 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

s(s + 4) + 50·0.5 = 0 → s² + 4s + 25 = 0

Roots

s = −2 ± j·√(25 − 4) = −2 ± j4.583
ωd = 4.583 rad/s

Answer

ωd = 4.583 rad/s
33.Step response

Answer: ωn = 4 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 = 16 → ωₙ = 4 rad/s
2ζωₙ = 2 → ζ = 2/(2·4) = 0.25

Result

ωₙ = 4 rad/s

Answer

ωn = 4 rad/s
34.Step response

Answer: %OS = 25.13 %

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/√(0.02·1 × 10⁻⁵) = 2236 rad/s
ζ = (R/2)·√(C/L) = (36/2)·√(1 × 10⁻⁵/0.02) = 0.4025

Result

%OS = 100·e−ζπ/√(1−ζ²) = 25.13 %

Answer

%OS = 25.13 %
35.Bode plots & margins

Answer: |H| = −0.9575 dB

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

Corner frequency

ωc = R/L = 22/0.047 = 468.1 rad/s
fc = 74.5 Hz → f/fc = 37/74.5 = 0.4967

Gain

|H| = 1/√(1 + 0.4967²) = 0.8956
20·log₁₀(0.8956) = −0.9575 dB

Answer

|H| = −0.9575 dB
36.Bode plots & margins

Answer: PM = 73.33 °

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 + 64ω² − 400 = 0
ω² = (−64 + √(4096 + 4·400))/2 = 5.736
ωc = 2.395 rad/s

Phase margin

∠L(jωc) = −90° − atan(2.395/8) = −106.7°
PM = 180° + ∠L = 73.33°

Answer

PM = 73.33°
37.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 192

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

Characteristic polynomial

s³ + 9s² + 23s + 15 + K = 0

Routh array

s³: 1, 23
s²: 9, 15 + K
s¹: (9·23 − 15 − K)/9
s⁰: 15 + K

Condition

The s¹ entry must stay positive.

K < 9·23 − 15 = 192

Answer

Kmax = 192

At the limit the loop oscillates at ω = √23 = 4.796 rad/s.

38.Stability: Routh–Hurwitz & root locus

Answer: K = 19.13

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.

Magnitude condition

K = |s|·|s + 3|·|s + 10|

Evaluate at the point

K = 1.5·1.5·8.5 = 19.13

Answer

K = 19.13

Check: s(s + 3)(s + 10) + K is zero at s = −1.5.

39.Steady-state error & PID control

Answer: K = 16

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

Required velocity constant

Kv = A/ess = 2/0.5 = 4

Gain

K = 4·Kv = 16

Answer

K = 16

A large gain makes this loop very lightly damped: ζ = 0.5. That conflict is why compensators are used.

40.Steady-state error & PID control

Answer: K = 15.62

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 + 10) + 10K = s² + 10s + 10K

Natural frequency

2·0.4·ωₙ = 10 → ωₙ = 12.5 rad/s

Gain

10K = ωₙ² = 156.2 → K = 15.62

Answer

K = 15.62

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