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​ = 12 ΩR3​ = 12 Ω+−V1​24 VR2​3 ΩAR4​12 ΩB

Find the equivalent resistance Req seen by the source V1.

V1 = 24 VR1 = 12 ΩR2 = 3 ΩR3 = 12 ΩR4 = 12 Ω
Req = Ω
2.Series & parallel resistorsHard
R1​ = 5 ΩR3​ = 20 ΩR5​ = 8 Ω+−V1​24 VR2​5 ΩAR4​5 ΩBR6​2 Ω

Find the equivalent resistance Req seen by the source V1.

V1 = 24 VR1 = 5 ΩR2 = 5 ΩR3 = 20 ΩR4 = 5 ΩR5 = 8 ΩR6 = 2 Ω
Req = Ω
3.Nodal analysisMedium
R1​15 Ω+−V1​15 VI1​R2​2 Ω+−V2​9 VR3​4 Ω+−V3​6 VA

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

V1 = 15 VR1 = 15 ΩV2 = 9 VR2 = 2 ΩV3 = 6 VR3 = 4 Ω
I1 = A
4.Nodal analysisHard
R1​ = 5 ΩR2​ = 6 ΩR4​ = 10 ΩIR4​+−V1​18 VR3​5 ΩAR5​25 ΩBIs​6 A+−V2​5 V

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

V1 = 18 VIs = 6 AV2 = 5 VR1 = 5 ΩR2 = 6 ΩR3 = 5 ΩR4 = 10 ΩR5 = 25 Ω
IR4 = A
5.Mesh analysisMedium
R1​ = 15 ΩR2​ = 20 Ω+−V1​6 VR3​4 ΩA+−V2​12 VI1​I2​

Use mesh analysis to find the mesh current I2 (clockwise).

V1 = 6 VV2 = 12 VR1 = 15 ΩR2 = 20 ΩR3 = 4 Ω
I2 = A
6.Mesh analysisHard
R1​ = 4 ΩR2​ = 8 ΩR3​ = 10 Ω+−V1​10 VR4​25 ΩAR5​3 ΩB+−V2​15 VI1​I2​I3​

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

V1 = 10 VV2 = 15 VR1 = 4 ΩR2 = 8 ΩR3 = 10 ΩR4 = 25 ΩR5 = 3 Ω
I1 = A
7.SuperpositionMedium
R1​ = 3 ΩR3​ = 3 Ω+−V1​15 VR2​4 Ω+−VR2​AR4​15 ΩBIs​2.5 A

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

V1 = 15 VIs = 2.5 AR1 = 3 ΩR2 = 4 ΩR3 = 3 ΩR4 = 15 Ω
VR2 = V
8.SuperpositionHard
R1​ = 6 ΩR2​ = 6 ΩR4​ = 5 ΩIR4​+−V1​6 VR3​5 ΩAIs​2.5 AB+−V2​10 V

Use superposition to find the power PR4 absorbed by R4.

V1 = 6 VIs = 2.5 AV2 = 10 VR1 = 6 ΩR2 = 6 ΩR3 = 5 ΩR4 = 5 Ω
PR4 = W
9.Thévenin & NortonMedium
R1​ = 6 ΩR3​ = 8 Ω+−V1​6 VR2​6 ΩAab

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

V1 = 6 VR1 = 6 ΩR2 = 6 ΩR3 = 8 Ω
IN = A
10.Thévenin & NortonHard
R1​ = 8 ΩR3​ = 30 Ω+−V1​24 VR2​20 ΩAIs​1.5 ABab

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

V1 = 24 VIs = 1.5 AR1 = 8 ΩR2 = 20 ΩR3 = 30 Ω
IN = A
11.RC & RL transientsMedium
+−Vs​15 Vt = 0R = 10 ΩL10 mHi(t)

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

Vs = 15 VR = 10 ΩL = 10 mH
t = ms
12.RC & RL transientsHard
+−Vs​20 Vt = 0R = 2.2 kΩC10 µF+−vC​vC​(0) = 30 V

Just before the switch closes at t = 0, the capacitor holds 30 V with the polarity shown. Find the capacitor voltage vC at t = 25 ms.

Vs = 20 VR = 2.2 kΩC = 10 µFvC(0) = 30 Vt = 25 ms
vC = V
13.AC impedance & phasorsMedium
Vf0​ = 2000 HzRL = ?C = 10 µF

A series RLC circuit with C = 10 µF must resonate at 2000 Hz. Find the inductance L needed.

f0 = 2000 HzC = 10 µF
L = mH
14.AC impedance & phasorsHard
12 V500 HzR = 25 ΩL = 100 mHC = 100 µF

A series RLC circuit is connected to a 12 V (rms), 500 Hz supply. Find the resonant frequency f0 of the circuit.

V = 12 V (rms)f = 500 HzR = 25 ΩL = 100 mHC = 100 µF
f0 = Hz
15.AC power & power factorMedium
230 VR + jXP = 2013 W17.5 A

An inductive load takes 17.5 A and 2013 W from a 230 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find X.

V = 230 VI = 17.5 AP = 2013 W
X = Ω
16.AC power & power factorHard
P = 15 kWQ1​S1​φ1​QC​

A single-phase load takes 15 kW at a power factor of 0.8 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the new line current Inew.

P = 15 kWpf1 = 0.8pf2 = 0.92V = 230 Vf = 60 Hz
Inew = A
17.Three-phase powerMedium
ABCVL​ = 415 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 415 V. Find the total real power P absorbed by the load in kW.

VL = 415 VZ = 8 + j6 Ω per phase
P = kW
18.Three-phase powerHard
P = 100 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 100 kW at a power factor of 0.8 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 = 100 kWpf1 = 0.8pf2 = 0.92VL = 400 Vf = 60 Hz
Cph = µF
19.TransformersMedium
75 Ω21

A 75 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 2:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 2:1RL = 75 Ω
R1′ = Ω
20.TransformersHard
pf 0.9525 kVA, 25% load

A single-phase 25 kVA transformer has a core loss of 200 W and a copper loss of 400 W at rated load. It supplies 25% of its rated kVA at a power factor of 0.95 lagging. Find the total losses Ploss in watts at this load.

S = 25 kVAPcore = 200 WPcu,fl = 400 Wx = 0.25pf = 0.95
Ploss = W
21.Op-amp circuitsMedium
−+Vout​Rf​ = 100 kΩR1​ = 10 kΩVin​

The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±12 V. The non-inverting amplifier is driven by a sine wave. Find the largest input amplitude Vin,max (peak value) for which the output does not clip.

R1 = 10 kΩRf = 100 kΩVsat = ±12 V
Vin,max = V
22.Op-amp circuitsHard
−+Vout​Rf​ = 2 kΩV1​ = 0.7 VR1​ = 2 kΩV2​ = 1.9 VR2​ = 22 kΩ

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

V1 = 0.7 VV2 = 1.9 VR1 = 2 kΩR2 = 22 kΩRf = 2 kΩ
Vout = V
23.Diodes & transistorsMedium
RC​ = 3.3 kΩβ = 100RE​ = 2.2 kΩVCC​ = 18 VR1​ = 68 kΩR2​ = 15 kΩ

An NPN transistor with β = 100 has voltage-divider bias from VCC = 18 V: R₁ = 68 kΩ from VCC to the base and R₂ = 15 kΩ from the base to ground. The collector resistor is RC = 3.3 kΩ and the emitter resistor is RE = 2.2 kΩ. 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 = 68 kΩR2 = 15 kΩRC = 3.3 kΩRE = 2.2 kΩβ = 100
VCE = V
24.Diodes & transistorsHard
RC​ = 330 Ωβ = 80VCC​ = 24 V3.3 VRB​ = ?

An NPN transistor with β = 80 switches a 330 Ω load resistor connected between VCC = 24 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 2 times the minimum needed for saturation. Find RB in kΩ.

VCC = 24 VRC = 330 Ωβ = 80Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 Voverdrive = 2
RB = kΩ
25.Number systemsMedium
place values in two's complement1−12816403201618040211

The 8-bit pattern 1100 1001 stores a signed integer in two's complement. What is its value N in decimal?

1100 10018 bit
N =
26.Number systemsHard
3580 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 3580 items its own binary code of the same length. What is the smallest number of bits b needed?

3580
b = bit
27.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010011302141507160120131151141809111010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 3, 4, 5, 6, 8, 11, 14, 15). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(0, 3, 4, 5, 6, 8, 11, 14, 15)
L =
28.Digital logic: gates, K-maps & flip-flopsHard
CLKD123456

A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown at 6 rising clock edges. How many times N does Q change state?

Q₀ = 0D
N =
29.Laplace transformsMedium
ReIm5

Find f(t) for F(s) = 16/(s² + 2s + 37) by completing the square, and evaluate f(0.75).

F(s) = 16/(s² + 2s + 37)t = 0.75 s
f(0.75) =
30.Laplace transformsHard
t (s)y(t)t = 0.3y(0.3) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 2 and y′(0) = 9 using the Laplace transform, and find y(0.3).

y″ + 4y′ + 3y = 0y(0) = 2y′(0) = 9t = 0.3 s
y(0.3) =
31.Block diagramsMedium
R(s)+10(s + 8)(s + 1)(s + 7)G(s)C(s)−

A unity negative feedback loop has G(s) = 10(s + 8)/((s + 1)(s + 7)). A step of size 1 is applied at the reference. Find the steady-state error ess.

G(s) = 10(s + 8)/((s + 1)(s + 7))R = 1
ess =
32.Block diagramsHard
R(s)+Ks(s + 3)(s + 5)G(s)C(s)−

The plant G(s) = K/(s(s + 3)(s + 5)) 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(s + 3)(s + 5))unity negative feedback
ω = rad/s
33.Step responseMedium
t (s)y(t)0.5

A system has the transfer function G(s) = 11.5/(s² + 6s + 23) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 11.5/(s² + 6s + 23)
σ =
34.Step responseHard
R(s)+135s(s + 6)G(s)C(s)−

A unity negative feedback loop has G(s) = 135/(s(s + 6)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 135/(s(s + 6))
ts = s
35.Bode plots & marginsMedium
|H| (dB)0.11101001000−100−80−60−40−200ω (rad/s)ω = 30

Find the exact magnitude |G(jω)| of G(s) = 18/((s + 2)(s + 9)) at ω = 30 rad/s, in dB.

G(s) = 18/((s + 2)(s + 9))ω = 30 rad/s
|G(jω)| = dB
36.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 2/(s(s + 1)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 2/(s(s + 1))
ωc = rad/s
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K = 8.41s(s + 3)(s + 4)C(s)−

The loop in the diagram runs with K = 8.4 and L(s) = K/(s(s + 3)(s + 4)). By what factor m can the gain be raised before the closed loop becomes unstable?

L(s) = K/(s(s + 3)(s + 4))K = 8.4
m =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2

A unity-feedback loop has L(s) = K(s + 5)/((s + 4)(s + 8)(s + 10)(s + 6)). Find the point σa where the asymptotes of the root locus cross the real axis.

L(s) = K(s + 5)/((s + 4)(s + 8)(s + 10)(s + 6))
σa =
39.Steady-state error & PID controlMedium
r(t) = 0.5·t²/2+20(s + 1)s²(s + 4)C(s)−

The parabolic input r(t) = 0.5·t²/2 is applied to a unity-feedback loop with G(s) = 20(s + 1)/(s²(s + 4)). Find the steady-state error ess.

G(s) = 20(s + 1)/(s²(s + 4))r(t) = 0.5·t²/2
ess =
40.Steady-state error & PID controlHard
R(s)+10 + Kd​s4s(s + 1)C(s)−

A PD controller C(s) = 10 + Kd·s drives the plant G(s) = 4/(s(s + 1)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.6.

G(s) = 4/(s(s + 1))Kp = 10ζ = 0.6
Kd = s

3Solutions

1.Series & parallel resistors

Answer: Req = 14.67 Ω

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

R1​ = 12 ΩR3​ = 12 Ω+−V1​24 VR2​3 ΩAR4​12 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 = 12 + 12 = 24 Ω
R2 ∥ 24 = 3·24/(3+24) = 2.667 Ω
Req = R1 + 2.667 = 12 + 2.667 = 14.67 Ω

Answer

Req = 14.67 Ω
2.Series & parallel resistors

Answer: Req = 9.118 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​ = 20 ΩR5​ = 8 Ω+−V1​24 VR2​5 ΩAR4​5 ΩBR6​2 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 = 8 + 2 = 10 Ω
R4 ∥ 10 = 3.333 Ω
R3 + 3.333 = 23.33 Ω
R2 ∥ 23.33 = 4.118 Ω
Req = R1 + 4.118 = 5 + 4.118 = 9.118 Ω

Answer

Req = 9.118 Ω
3.Nodal analysis

Answer: I1 = 0.4286 A

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

KCL at node A

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

(VA − 15)/15 + (VA − 9)/2 + (VA − 6)/4 = 0

Solve for VA

VA = (15/15 + 9/2 + 6/4) / (1/15 + 1/2 + 1/4) = 7/0.8167 = 8.571 V

Branch current

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

I1 = (15 − VA)/R1 = (15 − 8.571)/15 = 0.4286 A

Answer

I1 = 0.4286 A

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

4.Nodal analysis

Answer: IR4 = 2.434 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​ = 5 ΩR2​ = 6 ΩR4​ = 10 Ω+−V1​18 VR3​5 ΩAR5​25 ΩBIs​6 A+−V2​5 V

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 18 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/5 + (VA − 18)/5 + (VA − VB)/6 = 0
0.5667·VA − 0.1667·VB = 3.6

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/25 + (VB − VA)/6 + (VB − 5)/10 = 6
−0.1667·VA + 0.3067·VB = 6.5

Solve

Δ = (0.5667)(0.3067) − (−0.1667)(−0.1667) = 0.146
VA = Δ1/Δ = 2.187/0.146 = 14.98 V
VB = Δ2/Δ = 4.283/0.146 = 29.34 V

Answer

IR4 = (VB − V2)/R4 = (29.34 − 5)/10 = 2.434 A
5.Mesh analysis

Answer: I2 = −0.4636 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​ = 15 ΩR2​ = 20 Ω+−V1​6 VR3​4 ΩA+−V2​12 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
15·I1 + 4(I1 − I2) = 6
19·I1 − 4·I2 = 6

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 + 4(I2 − I1) = (−12)
−4·I1 + 24·I2 = −12

Solve

Δ = (19)(24) − (−4)(−4) = 440
I1 = Δ1/Δ = 96/440 = 0.2182 A
I2 = Δ2/Δ = −204/440 = −0.4636 A

Answer

I2 = −0.4636 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: I1 = 0.6681 A

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

R1​ = 4 ΩR2​ = 8 ΩR3​ = 10 Ω+−V1​10 VR4​25 ΩAR5​3 ΩB+−V2​15 VI1​I2​I3​

Assign mesh currents

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

KVL around mesh 1

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

R1·I1 + R4(I1 − I2) = V1
4·I1 + 25(I1 − I2) = 10
29·I1 − 25·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
8·I2 + 25(I2 − I1) + 3(I2 − I3) = 0
−25·I1 + 36·I2 − 3·I3 = 0

KVL around mesh 3

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

R3·I3 + R5(I3 − I2) = −V2
10·I3 + 3(I3 − I2) = (−15)
−3·I2 + 13·I3 = −15

Solve

Δ = det of the coefficient matrix = 5186
I1 = Δ1/Δ = 3465/5186 = 0.6681 A
I2 = Δ2/Δ = 1945/5186 = 0.375 A
I3 = Δ3/Δ = −5535/5186 = −1.067 A

Answer

I1 = 0.6681 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: VR2 = 11.09 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​ = 3 ΩR3​ = 3 Ω+−V1​15 VR2​4 ΩAR4​15 ΩBOPEN

Case 1 · V1 acting alone

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

R3+R4 = 3+15 = 18 Ω
R2 ∥ (R3+R4) = 4·18/(4+18) = 3.273 Ω
VA = 15·3.273/(3+3.273) = 7.826 V
VB = VA·R4/(R3+R4) = 7.826·15/18 = 6.522 V
VR2′ = VA − 0 = 7.826 − 0 = 7.826 V
R1​ = 3 ΩR3​ = 3 ΩSHORTR2​4 ΩAR4​15 ΩBIs​2.5 A

Case 2 · Is acting alone

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

R1 ∥ R2 = 3·4/(3+4) = 1.714 Ω
Req = R4 ∥ (R3+R1∥R2) = 15 ∥ 4.714 = 3.587 Ω
VB = Is·Req = 2.5·3.587 = 8.967 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 8.967·1.714/4.714 = 3.261 V
VR2″ = VA − 0 = 3.261 − 0 = 3.261 V

Add the contributions

VR2 = VR2′ + VR2″
VR2 = 7.826 + 3.261 = 11.09 V

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

8.Superposition

Answer: PR4 = 5.614 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​ = 6 ΩR2​ = 6 ΩR4​ = 5 Ω+−V1​6 VR3​5 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) = 5 ∥ 11 = 3.438 Ω
VA = 6·3.438/(6+3.438) = 2.185 V
VB = VA·R4/(R2+R4) = 2.185·5/11 = 0.9934 V
IR4′ = (VB − V2)/R4 = (0.9934 − 0)/5 = 0.1987 A
R1​ = 6 ΩR2​ = 6 ΩR4​ = 5 ΩSHORTR3​5 ΩAIs​2.5 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 = 2.727 Ω, so the left path is R2+2.727 = 8.727 Ω
Req = R4 ∥ 8.727 = 3.179 Ω
VB = Is·Req = 2.5·3.179 = 7.947 V
VA = VB·2.727/8.727 = 2.483 V
IR4″ = (VB − V2)/R4 = (7.947 − 0)/5 = 1.589 A
R1​ = 6 ΩR2​ = 6 ΩR4​ = 5 ΩSHORTR3​5 ΩAOPENB+−V2​10 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 = 6·5/(6+5) = 2.727 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 10·8.727/13.73 = 6.358 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 6.358·2.727/8.727 = 1.987 V
IR4‴ = (VB − V2)/R4 = (6.358 − 10)/5 = −0.7285 A

Add the current contributions

IR4 = IR4′ + IR4″ + IR4‴ = 0.1987 + 1.589 + (−0.7285) = 1.06 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 = 1.06²·5 = 5.614 W

Answer

PR4 = 5.614 W

Adding the powers of the separate cases would give 15.48 W, which is wrong.

9.Thévenin & Norton

Answer: IN = 0.2727 A

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

R1​ = 6 ΩR3​ = 8 ΩSHORTR2​6 ΩAabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 6·6/(6+6) = 3 Ω
Rth = (R1 ∥ R2) + R3 = 3 + 8 = 11 Ω
R1​ = 6 ΩR3​ = 8 Ω+−V1​6 VR2​6 ΩAab+−Vth​

Find Vth, the open-circuit voltage

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

Vth = V1·R2/(R1+R2) = 6·6/(6+6) = 3 V
IN​0.2727 ARth​11 Ωab

Norton equivalent

A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.

IN = Vth/Rth = 3/11 = 0.2727 A
10.Thévenin & Norton

Answer: IN = 1.98 A

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

R1​ = 8 ΩR3​ = 30 ΩSHORTR2​20 ΩAOPENBabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 8·20/(8+20) = 5.714 Ω
Rth = (R1 ∥ R2) + R3 = 5.714 + 30 = 35.71 Ω
R1​ = 8 ΩR3​ = 30 Ω+−V1​24 VR2​20 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) = 17.14 + 1.5·5.714 = 25.71 V
Vth = VB = VA + Is·R3 = 25.71 + 1.5·30 = 70.71 V
IN​1.98 ARth​35.71 Ωab

Norton equivalent

A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.

IN = Vth/Rth = 70.71/35.71 = 1.98 A
11.RC & RL transients

Answer: t = 2.996 ms

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

Time constant and final current

Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.

τ = L/R = 10 mH / 10 Ω = 1 ms
i(∞) = Vs/R = 15/10 = 1.5 A

Solve for the time

0.95 = 1 − e−t/τ → t = −τ·ln(1 − 0.95)
t = −1·ln(0.05) = 2.996 ms

Answer

t = 2.996 ms
12.RC & RL transients

Answer: vC = 23.21 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) = 30 V, vC(∞) = Vs = 20 V
τ = R·C = 2.2 kΩ · 10 µF = 22 ms
vC(t) = 20 + (30 − 20)·e−t/τ

Voltage at the given time

vC(25) = 20 + (10)·e−25/22 = 23.21 V

Answer

vC = 23.21 V
13.AC impedance & phasors

Answer: L = 0.6333 mH

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 condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·2000 = 12566 rad/s
L = 1/(ω₀²·C) = 1/(12566²·10·10⁻⁶) = 0.6333 mH

Answer

L = 0.6333 mH
14.AC impedance & phasors

Answer: f0 = 50.33 Hz

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

Resonance

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

f0 = 1/(2π√(LC)) = 1/(2π·√(0.1·100·10⁻⁶)) = 50.33 Hz

Answer

f0 = 50.33 Hz
15.AC power & power factor

Answer: X = 11.38 Ω

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.

Resistance from the real power

R = P/I² = 2013/17.5² = 6.573 Ω

Reactance from the impedance

|Z| = V/I = 230/17.5 = 13.14 Ω
X = √(|Z|² − R²) = √(13.14² − 6.573²) = 11.38 Ω

Answer

X = 11.38 Ω
16.AC power & power factor

Answer: Inew = 70.89 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 = P·tan φ1 = 15·tan(36.87°) = 11.25 kvar
Q2 = P·tan φ2 = 15·tan(23.07°) = 6.39 kvar
QC = Q1 − Q2 = 4.86 kvar

New line current

Inew = P/(V·cos φ2) = 15000/(230·0.92) = 70.89 A

Answer

Inew = 70.89 A
17.Three-phase power

Answer: P = 41.33 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 = 415 V

Phase current

|Z| = √(8² + 6²) = 10 Ω
Iph = 415/10 = 41.5 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·41.5²·8 = 41.33 kW

Answer

P = 41.33 kW

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

18.Three-phase power

Answer: Cph = 179.1 µF

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

Reactive power before and after

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

Q1 = 100·tan(acos 0.8) = 100·0.75 = 75 kvar
Q2 = 100·tan(acos 0.92) = 100·0.426 = 42.6 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 32.4 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) = 10800/(377·400²) = 179.1 µF

Answer

Cph = 179.1 µF
19.Transformers

Answer: R1′ = 300 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 = 2²·75 = 300 Ω

Answer

R1′ = 300 Ω
20.Transformers

Answer: Ploss = 225 W

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

Copper loss at this load

Pcu = x²·Pcu,fl = 0.0625·400 = 25 W

Total losses

The core loss is the same at any load.

Ploss = 200 + 25 = 225 W

Answer

Ploss = 225 W

The efficiency at this load is η = 96.35%.

21.Op-amp circuits

Answer: Vin,max = 1.091 V

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

Gain

Av = 1 + Rf/R1 = 1 + 100/10 = 11

Largest input before clipping

Vin,max = Vsat/|Av| = 12/11 = 1.091 V

Answer

Vin,max = 1.091 V
22.Op-amp circuits

Answer: Vout = −0.8727 V

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

Input currents

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

I1 = V1/R1 = 0.7/2 = 0.35 mA
I2 = V2/R2 = 1.9/22 = 0.08636 mA
If = I1 + I2 = 0.4364 mA

Output voltage

Vout = −Rf·If = −2·0.4364 = −0.8727 V

Answer

Vout = −0.8727 V
23.Diodes & transistors

Answer: VCE = 11.99 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 = 18·15/(68 + 15) = 3.253 V
RTH = 68·15/(68 + 15) = 12.29 kΩ

Base current

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

IB = (3.253 − 0.7)/(12.29 + 101·2.2) = 10.89 µA

Collector current

IC = 100·10.89 µA = 1.089 mA
IE = 101·10.89 µA = 1.1 mA

Collector–emitter voltage

VCE = 18 − 1.089·3.3 − 1.1·2.2 = 11.99 V

Answer

VCE = 11.99 V
24.Diodes & transistors

Answer: RB = 1.442 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) = (24 − 0.2)/330 = 72.12 mA

Minimum base current

IB(min) = IC(sat)/β = 72.12/80 = 901.5 µA

Design base current

IB = 2·901.5 µA = 1803 µA

Base resistor

RB = (3.3 − 0.7)/IB = 2.6 V/1.803 mA = 1.442 kΩ

Answer

RB = 1.442 kΩ
25.Number systems

Answer: N = −55

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.

Add the place values

The other bits keep their normal positive values.

−128 + 64 + 8 + 1 = −55

Answer

N = −55
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 < 3580
212 = 4096 ≥ 3580

Answer

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

Answer: L = 15

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.

ABCD000111100001111010011302141507160120131151141809111010

Group the 1s

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

B′C′D′
B′CD
BCD′
ACD
A′BC′

Minimal SOP form

F = B′C′D′ + B′CD + BCD′ + ACD + A′BC′
L = 15

Answer

L = 15

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

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

Answer: N = 2

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

Q after every edge

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

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

Count the changes

0 → 0 → 0 → 1 → 1 → 1 → 0
N = 2

Answer

N = 2
29.Laplace transforms

Answer: f(0.75) = −1.231

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).

Complete the square

s² + 2s + 37 = (s + 1)² + 36 = (s + 1)² + 6²

Match the numerator

F(s) = (16/6)·6/((s + 1)² + 6²)

Inverse transform

f(t) = e−t·(2.667·sin 6t)
f(0.75) = e−0.75·(2.667·sin(6·0.75)) = −1.231

Answer

f(0.75) = −1.231
30.Laplace transforms

Answer: y(0.3) = 3.32

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 − 2s − 9 + 4(sY − 2) + 3Y = 0
Y(s) = (2s + 17)/((s + 1)(s + 3))

Partial fractions

A = 15/2 = 7.5
B = 11/(−2) = −5.5

Result

y(t) = 7.5·e−t − 5.5·e−3t
y(0.3) = 3.32

Answer

y(0.3) = 3.32

Check: y(0) = A + B = 2.

31.Block diagrams

Answer: ess = 0.08046

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).

Position error constant

Kₚ = G(0) = 10·8/(1·7) = 11.43

Steady-state error

ess = A/(1 + Kₚ) = 1/(1 + 11.43) = 0.08046

Answer

ess = 0.08046
32.Block diagrams

Answer: ω = 3.873 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 + 8s² + 15s + K = 0

Routh array

s³ | 1 15
s² | 8 K
s¹ | (8·15 − K)/8
s⁰ | K

Stability limit

All entries of the first column must be positive.

Kmax = 8·15 = 120

Frequency of oscillation

8·(jω)² + 120 = 0 → ω² = 120/8 = 15
ω = 3.873 rad/s

Answer

ω = 3.873 rad/s
33.Step response

Answer: σ = −3

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 = 23 → ωₙ = 4.796 rad/s
2ζωₙ = 6 → ζ = 6/(2·4.796) = 0.6255

Result

σ = −ζωₙ = −6/2 = −3

Answer

σ = −3
34.Step response

Answer: ts = 1.333 s

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

Closed-loop transfer function

T(s) = 135/(s² + 6s + 135): ωₙ = 11.62 rad/s, ζ = 0.2582

Settling time

ζωₙ = 6/2 = 3
ts ≈ 4/(ζωₙ) = 1.333 s

Answer

ts = 1.333 s

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

35.Bode plots & margins

Answer: |G(jω)| = −34.37 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.

Multiply the magnitudes

|G(j30)| = 18/(√(30² + 2²)·√(30² + 9²)) = 18/(30.07·31.32) = 0.01911

Result

20·log₁₀(0.01911) = −34.37 dB

Answer

|G(jω)| = −34.37 dB
36.Bode plots & margins

Answer: ωc = 1.25 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 + 1ω² − 4 = 0
ω² = (−1 + √(1 + 4·4))/2 = 1.562
ωc = 1.25 rad/s

Answer

ωc = 1.25 rad/s
37.Stability: Routh–Hurwitz & root locus

Answer: m = 10

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

Routh array

s³ + 7s² + 12s + K = 0
s³: 1, 12
s²: 7, K
s¹: (7·12 − K)/7
s⁰: K

Largest stable gain

Kmax = 7·12 = 84

Factor

m = 84/8.4 = 10

Answer

m = 10

This factor is the gain margin as a ratio: 20 dB.

38.Stability: Routh–Hurwitz & root locus

Answer: σa = −7.667

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

Poles and zeros

poles: −4, −8, −10, −6
zeros: −5
n − m = 4 − 1 = 3

Centroid

σa = (−28 − (−5))/3 = −7.667

Answer

σa = −7.667

The 3 asymptotes leave the centroid at angles of 60°, 180°, 300°.

39.Steady-state error & PID control

Answer: ess = 0.1

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.

Acceleration constant

Ka = 20·1/4 = 5

Parabola error

ess = 0.5/5 = 0.1

Answer

ess = 0.1
40.Steady-state error & PID control

Answer: Kd = 1.647 s

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

Characteristic polynomial

s(s + 1) + 4(10 + Kds) = s² + (1 + 4Kd)s + 40

Natural frequency

ωₙ = √40 = 6.325 rad/s

Derivative gain

1 + 4Kd = 2·0.6·6.325 = 7.589
Kd = 1.647 s

Answer

Kd = 1.647 s

The derivative term adds damping without changing ωₙ.

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