FE Electrical and Computer exam practice: cram pack

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

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1Formulas and methods

Series & parallel resistors

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

Method

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

Nodal analysis

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

Method

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

Mesh analysis

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

Method

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

Superposition

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

Method

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

Thévenin & Norton

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

Method

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

RC & RL transients

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

Method

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

AC impedance & phasors

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

Method

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

AC power & power factor

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

Three-phase power

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

Transformers

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

Op-amp circuits

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

Diodes & transistors

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

Number systems

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

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

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

Method

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

Stability: Routh–Hurwitz & root locus

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

Method

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

Steady-state error & PID control

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

Method

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

2Practice problems

1.Series & parallel resistorsMedium
R1​ = 10 ΩIR1​R3​ = 12 Ω+−V1​30 VR2​5 ΩAR4​4 ΩB

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

V1 = 30 VR1 = 10 ΩR2 = 5 ΩR3 = 12 ΩR4 = 4 Ω
IR1 = A
2.Series & parallel resistorsHard
R1​ = 2 ΩR3​ = 3 ΩR5​ = 10 Ω+−V1​18 VR2​30 ΩAR4​6 ΩBR6​3 Ω

Find the equivalent resistance Req seen by the source V1.

V1 = 18 VR1 = 2 ΩR2 = 30 ΩR3 = 3 ΩR4 = 6 ΩR5 = 10 ΩR6 = 3 Ω
Req = Ω
3.Nodal analysisMedium
R1​ = 5 ΩR3​ = 20 Ω+−V1​15 VR2​10 ΩAR4​2 ΩBIs​1 A

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

V1 = 15 VIs = 1 AR1 = 5 ΩR2 = 10 ΩR3 = 20 ΩR4 = 2 Ω
PIs = W
4.Nodal analysisHard
R1​ = 30 ΩR2​ = 10 ΩIR2​R4​ = 6 Ω+−V1​9 VR3​10 ΩAR5​25 ΩBIs​4 A+−V2​24 V

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

V1 = 9 VIs = 4 AV2 = 24 VR1 = 30 ΩR2 = 10 ΩR3 = 10 ΩR4 = 6 ΩR5 = 25 Ω
IR2 = A
5.Mesh analysisMedium
R1​ = 3 Ω+−VR1​R2​ = 2 Ω+−V1​24 VR3​4 ΩA+−V2​24 VI1​I2​

Use mesh analysis to find the voltage VR1 across R1 with the polarity shown.

V1 = 24 VV2 = 24 VR1 = 3 ΩR2 = 2 ΩR3 = 4 Ω
VR1 = V
6.Mesh analysisHard
R1​ = 5 ΩR2​ = 30 ΩR3​ = 20 Ω+−V1​18 VR4​4 ΩAR5​4 ΩB+−V2​18 VI1​I2​I3​

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

V1 = 18 VV2 = 18 VR1 = 5 ΩR2 = 30 ΩR3 = 20 ΩR4 = 4 ΩR5 = 4 Ω
I2 = A
7.SuperpositionMedium
R1​ = 5 ΩR3​ = 2 ΩIR3​+−V1​18 VR2​15 ΩAR4​5 ΩBIs​?

Find the value of the current source Is (direction as shown) that makes the current IR3 through R3 equal to zero.

V1 = 18 VR1 = 5 ΩR2 = 15 ΩR3 = 2 ΩR4 = 5 Ω
Is = A
8.SuperpositionHard
R1​ = 20 ΩR2​ = 4 ΩR4​ = 6 ΩIR4​+−V1​5 VR3​4 ΩAIs​4 AB−+V2​12 V

Use superposition to find the power PR4 absorbed by R4.

V1 = 5 VIs = 4 AV2 = 12 VR1 = 20 ΩR2 = 4 ΩR3 = 4 ΩR4 = 6 Ω
PR4 = W
9.Thévenin & NortonMedium
R1​ = 3 ΩR3​ = 15 Ω+−V1​12 VR2​3 ΩAab

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

V1 = 12 VR1 = 3 ΩR2 = 3 ΩR3 = 15 Ω
IN = A
10.Thévenin & NortonHard
R1​ = 5 ΩR3​ = 8 Ω+−V1​15 VR2​30 ΩAIs​1.5 ABRL​12 ΩIL​

A load RL = 12 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.

V1 = 15 VIs = 1.5 AR1 = 5 ΩR2 = 30 ΩR3 = 8 ΩRL = 12 Ω
PL = W
11.RC & RL transientsMedium
+−Vs​15 Vt = 0R1​ = 2.2 kΩR2​ = 1 kΩC33 µF+−vC​

The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the capacitor voltage vC at t = 10 ms.

Vs = 15 VR1 = 2.2 kΩR2 = 1 kΩC = 33 µFt = 10 ms
vC = V
12.RC & RL transientsHard
+−Vs​10 Vt = 0R = 1 kΩC68 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 = 80 ms.

Vs = 10 VR = 1 kΩC = 68 µFvC(0) = 6 Vt = 80 ms
vC = V
13.AC impedance & phasorsMedium
24 V200 HzR = 25 ΩL = 250 mHC = 2.2 µF

A series RLC circuit is connected to a 24 V (rms), 200 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 24 V (rms)f = 200 HzR = 25 ΩL = 250 mHC = 2.2 µF
φ = °
14.AC impedance & phasorsHard
VfR = 5 ΩL = 2 mHC = 10 µF

A series RLC circuit has R = 5 Ω, L = 2 mH and C = 10 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.

R = 5 ΩL = 2 mHC = 10 µF
B = Hz
15.AC power & power factorMedium
120 VZ6 + j8 ΩI

A load with impedance Z = 6 + j8 Ω is connected to a 120 V (rms) single-phase supply. Find the apparent power S.

V = 120 VZ = 6 + j8 Ω
S = VA
16.AC power & power factorHard
120 V50 HzRs​ = 10 ΩXs​ = j6 ΩZL​ = ?

A 120 V (rms), 50 Hz source has an internal impedance Zs = 10 + j6 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.

Vs = 120 V (rms)f = 50 HzZs = 10 + j6 Ω
C = µF
17.Three-phase powerMedium
ABCVL​ = 415 VZ = 9 + j12 Ωper phase

A balanced Δ-connected load with an impedance of 9 + j12 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 = 9 + j12 Ω per phase
P = kW
18.Three-phase powerHard
P = 250 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 250 kW at a power factor of 0.7 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 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 250 kWpf1 = 0.7pf2 = 0.95VL = 415 Vf = 50 Hz
Cph = µF
19.TransformersMedium
75 ΩN1​N2​1200 Ω

An amplifier with an output resistance of 1200 Ω drives a 75 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 1200 Ω from the amplifier. Find the turns ratio a = N₁/N₂.

Rs = 1200 ΩRL = 75 Ω
a =
20.TransformersHard
pf 0.9525 kVA, 50% load

A single-phase 25 kVA transformer has a core loss of 125 W and a copper loss of 312.5 W at rated load. It supplies 50% 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 = 125 WPcu,fl = 312.5 Wx = 0.5pf = 0.95
Ploss = W
21.Op-amp circuitsMedium
−+Vout​R2​ = 10 kΩV1​ = −2.9 VR1​ = 22 kΩV2​ = 0.2 VR3​ = 22 kΩR4​ = 10 kΩ

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

V1 = −2.9 VV2 = 0.2 VR1 = R3 = 22 kΩR2 = R4 = 10 kΩ
V− = V
22.Op-amp circuitsHard
−+Vout​Rf​ = 47 kΩV1​ = 0.3 VR1​ = 22 kΩV2​ = −1.8 VR2​ = 10 kΩ

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

V1 = 0.3 VV2 = −1.8 VR1 = 22 kΩR2 = 10 kΩRf = 47 kΩ
Vout = V
23.Diodes & transistorsMedium
+−24 V330 Ω12 V2200 ΩIZ​ = ?

A 12 V Zener regulator is fed from 24 V through 330 Ω and supplies a 2200 Ω load. Find the Zener current IZ.

Vin = 24 VVZ = 12 VR = 330 ΩRL = 2200 Ω
IZ = mA
24.Diodes & transistorsHard
RC​ = 470 Ωβ = 50VCC​ = 5 V5 VRB​ = ?

An NPN transistor with β = 50 switches a 470 Ω load resistor connected between VCC = 5 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB. Take VBE = 0.7 V and VCE(sat) = 0.2 V. For reliable switching, design for a base current 10 times the minimum needed for saturation. Find RB in kΩ.

VCC = 5 VRC = 470 Ωβ = 50Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 Voverdrive = 10
RB = kΩ
25.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −99 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −998 bit
U =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0010 0110 AND 0100 0101. What is the result's value R in decimal?

0010 0110AND0100 0101
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
MUXC′I0​0I1​0I2​C′I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = C′, I₁ = 0, I₂ = 0, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = C′I₁ = 0I₂ = 0I₃ = C′
N =
28.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKJOHNSON

A 4-bit Johnson counter Q₃Q₂Q₁Q₀ starts at 0000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives the inverse of Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 10 clock edges.

Q₃Q₂Q₁Q₀ = 000010
Q =
29.Laplace transformsMedium
t (s)f(t)t = 0.3f(0.3) = ?

Find f(t) for F(s) = 10/(s(s + 1)(s + 4)) by partial fractions, and evaluate f(0.3).

F(s) = 10/(s(s + 1)(s + 4))t = 0.3 s
f(0.3) =
30.Laplace transformsHard
t (s)y(t)t = 1y(1) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 2 and y′(0) = −11 using the Laplace transform, and find y(1).

y″ + 4y′ + 3y = 0y(0) = 2y′(0) = −11t = 1 s
y(1) =
31.Block diagramsMedium
R(s)+Ks(s + 10)G(s)C(s)−

The plant G(s) = K/(s(s + 10)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/(s(s + 10))unity negative feedback
K =
32.Block diagramsHard
R(s)+13s + 4G₁+5s + 1G₂C(s)4H₁−0.2H₂−

Reduce the block diagram with G₁ = 13/(s + 4), G₂ = 5/(s + 1), an inner negative feedback H₁ = 4 around G₂ and an outer negative feedback H₂ = 0.2. Find the DC gain T(0) of C(s)/R(s).

G₁ = 13/(s + 4)G₂ = 5/(s + 1)H₁ = 4H₂ = 0.2
T(0) =
33.Step responseMedium
t (s)y(t)1

A system has the transfer function G(s) = 69/(s² + 7s + 69) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.

G(s) = 69/(s² + 7s + 69)
σ =
34.Step responseHard
t (s)y(t)1tp​ = 0.8 s12 %

The step response of a second-order system without zeros shows 12 % overshoot, and the first peak occurs at t = 0.8 s. Find the natural frequency ωn.

%OS = 12 %tp = 0.8 s
ωn = rad/s
35.Bode plots & marginsMedium
vin​C = 10 nFR = 22 kΩvout​

The RC high-pass filter shown has R = 22 kΩ, C = 10 nF. A sinusoidal input of 2.9 kHz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 22 kΩC = 10 nFf = 2.9 kHz
φ = °
36.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​

A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 2)(s + 3)). Find the gain K that gives a gain margin of 12 dB.

L(s) = K/(s(s + 2)(s + 3))GM = 12 dB
K =
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+Ks + 10(s − 3)(s + 3)C(s)−

The plant G(s) = (s + 10)/((s − 3)(s + 3)) has an unstable pole at s = 3. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.

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

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

L(s) = K/(s(s + 10)(s + 4))
σa =
39.Steady-state error & PID controlMedium
r(t) = 2t+5s(s + 5)(s + 10)C(s)−

A ramp r(t) = 2t is applied to a unity-feedback loop with G(s) = 5/(s(s + 5)(s + 10)). Find the steady-state error ess.

G(s) = 5/(s(s + 5)(s + 10))r(t) = 2t
ess =
40.Steady-state error & PID controlHard
R(s)+K5s(s + 6)C(s)−

A proportional controller K drives the plant G(s) = 5/(s(s + 6)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.707.

G(s) = 5/(s(s + 6))ζ = 0.707
K =

3Solutions

1.Series & parallel resistors

Answer: IR1 = 2.172 A

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

R1​ = 10 ΩR3​ = 12 Ω+−V1​30 VR2​5 ΩAR4​4 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 + 4 = 16 Ω
R2 ∥ 16 = 5·16/(5+16) = 3.81 Ω
Req = R1 + 3.81 = 10 + 3.81 = 13.81 Ω

Current from the source

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

IR1 = V1/Req = 30/13.81 = 2.172 A

Answer

IR1 = 2.172 A
2.Series & parallel resistors

Answer: Req = 7.745 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​ = 2 ΩR3​ = 3 ΩR5​ = 10 Ω+−V1​18 VR2​30 ΩAR4​6 ΩBR6​3 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 = 10 + 3 = 13 Ω
R4 ∥ 13 = 4.105 Ω
R3 + 4.105 = 7.105 Ω
R2 ∥ 7.105 = 5.745 Ω
Req = R1 + 5.745 = 2 + 5.745 = 7.745 Ω

Answer

Req = 7.745 Ω
3.Nodal analysis

Answer: PIs = 2.632 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/10 + (VA − 15)/5 + (VA − VB)/20 = 0
0.35·VA − 0.05·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/2 + (VB − VA)/20 = 1
−0.05·VA + 0.55·VB = 1

Solve

Δ = (0.35)(0.55) − (−0.05)(−0.05) = 0.19
VA = Δ1/Δ = 1.7/0.19 = 8.947 V
VB = Δ2/Δ = 0.5/0.19 = 2.632 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 = 2.632·1 = 2.632 W

Answer

PIs = 2.632 W
4.Nodal analysis

Answer: IR2 = −1.632 A

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

R1​ = 30 ΩR2​ = 10 ΩR4​ = 6 Ω+−V1​9 VR3​10 ΩAR5​25 ΩBIs​4 A+−V2​24 V

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 9 V, V2 = 24 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/10 + (VA − 9)/30 + (VA − VB)/10 = 0
0.2333·VA − 0.1·VB = 0.3

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)/10 + (VB − 24)/6 = 4
−0.1·VA + 0.3067·VB = 8

Solve

Δ = (0.2333)(0.3067) − (−0.1)(−0.1) = 0.06156
VA = Δ1/Δ = 0.892/0.06156 = 14.49 V
VB = Δ2/Δ = 1.897/0.06156 = 30.81 V

Answer

IR2 = (VA − VB)/R2 = (14.49 − 30.81)/10 = −1.632 A
5.Mesh analysis

Answer: VR1 = 5.538 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​ = 3 ΩR2​ = 2 Ω+−V1​24 VR3​4 ΩA+−V2​24 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
3·I1 + 4(I1 − I2) = 24
7·I1 − 4·I2 = 24

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
2·I2 + 4(I2 − I1) = (−24)
−4·I1 + 6·I2 = −24

Solve

Δ = (7)(6) − (−4)(−4) = 26
I1 = Δ1/Δ = 48/26 = 1.846 A
I2 = Δ2/Δ = −72/26 = −2.769 A

Answer

IR1 = I1 = 1.846 A
VR1 = IR1·R1 = 1.846·3 = 5.538 V

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: I2 = 0.1406 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​ = 5 ΩR2​ = 30 ΩR3​ = 20 Ω+−V1​18 VR4​4 ΩAR5​4 ΩB+−V2​18 VI1​I2​I3​

Assign mesh currents

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

KVL around mesh 1

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

R1·I1 + R4(I1 − I2) = V1
5·I1 + 4(I1 − I2) = 18
9·I1 − 4·I2 = 18

KVL around mesh 2

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

R2·I2 + R4(I2 − I1) + R5(I2 − I3) = 0
30·I2 + 4(I2 − I1) + 4(I2 − I3) = 0
−4·I1 + 38·I2 − 4·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 + 4(I3 − I2) = (−18)
−4·I2 + 24·I3 = −18

Solve

Δ = det of the coefficient matrix = 7680
I1 = Δ1/Δ = 15840/7680 = 2.063 A
I2 = Δ2/Δ = 1080/7680 = 0.1406 A
I3 = Δ3/Δ = −5580/7680 = −0.7266 A

Answer

I2 = 0.1406 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: Is = 2.7 A

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

R1​ = 5 ΩR3​ = 2 Ω+−V1​18 VR2​15 ΩAR4​5 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 = 2+5 = 7 Ω
R2 ∥ (R3+R4) = 15·7/(15+7) = 4.773 Ω
VA = 18·4.773/(5+4.773) = 8.791 V
VB = VA·R4/(R3+R4) = 8.791·5/7 = 6.279 V
IR3′ = (VA − VB)/R3 = (8.791 − 6.279)/2 = 1.256 A

Case 2 · Is acting alone

V₁ becomes a short, so R₁ ∥ R₂ hangs from A to ground. Iₛ splits at B between R₄ and the path R₃ + R₁∥R₂; the part through R₃ flows from B to A, against the arrow.

R1 ∥ R2 = 3.75 Ω
IR3″ = −Is·R4/(R4 + R3 + R1∥R2) = −Is·5/10.75 = −0.4651·Is

Set the total to zero

IR3′ + IR3″ = 1.256 − 0.4651·Is = 0
Is = 1.256/0.4651 = 2.7 A

Answer

Is = 2.7 A
8.Superposition

Answer: PR4 = 60.01 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​ = 20 ΩR2​ = 4 ΩR4​ = 6 Ω+−V1​5 VR3​4 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) = 4 ∥ 10 = 2.857 Ω
VA = 5·2.857/(20+2.857) = 0.625 V
VB = VA·R4/(R2+R4) = 0.625·6/10 = 0.375 V
IR4′ = (VB − V2)/R4 = (0.375 − 0)/6 = 0.0625 A
R1​ = 20 ΩR2​ = 4 ΩR4​ = 6 ΩSHORTR3​4 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 = 3.333 Ω, so the left path is R2+3.333 = 7.333 Ω
Req = R4 ∥ 7.333 = 3.3 Ω
VB = Is·Req = 4·3.3 = 13.2 V
VA = VB·3.333/7.333 = 6 V
IR4″ = (VB − V2)/R4 = (13.2 − 0)/6 = 2.2 A
R1​ = 20 ΩR2​ = 4 ΩR4​ = 6 ΩSHORTR3​4 ΩAOPENB−+V2​12 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 = 20·4/(20+4) = 3.333 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = (−12)·7.333/13.33 = −6.6 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = −6.6·3.333/7.333 = −3 V
IR4‴ = (VB − V2)/R4 = (−6.6 − −12)/6 = 0.9 A

Add the current contributions

IR4 = IR4′ + IR4″ + IR4‴ = 0.0625 + 2.2 + 0.9 = 3.163 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 = 3.163²·6 = 60.01 W

Answer

PR4 = 60.01 W

Adding the powers of the separate cases would give 33.92 W, which is wrong.

9.Thévenin & Norton

Answer: IN = 0.3636 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​ = 3 ΩR3​ = 15 ΩSHORTR2​3 ΩAabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 3·3/(3+3) = 1.5 Ω
Rth = (R1 ∥ R2) + R3 = 1.5 + 15 = 16.5 Ω
R1​ = 3 ΩR3​ = 15 Ω+−V1​12 VR2​3 ΩAab+−Vth​

Find Vth, the open-circuit voltage

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

Vth = V1·R2/(R1+R2) = 12·3/(3+3) = 6 V
IN​0.3636 ARth​16.5 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 = 6/16.5 = 0.3636 A
10.Thévenin & Norton

Answer: PL = 19.91 W

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

R1​ = 5 ΩR3​ = 8 ΩSHORTR2​30 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 = 5·30/(5+30) = 4.286 Ω
Rth = (R1 ∥ R2) + R3 = 4.286 + 8 = 12.29 Ω
R1​ = 5 ΩR3​ = 8 Ω+−V1​15 VR2​30 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) = 12.86 + 1.5·4.286 = 19.29 V
Vth = VB = VA + Is·R3 = 19.29 + 1.5·8 = 31.29 V
Rth​ = 12.29 Ω+−Vth​31.29 VRL​12 ΩIL​

Connect the load to the Thévenin equivalent

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

IL = Vth/(Rth + RL) = 31.29/(12.29 + 12) = 1.288 A
PL = IL²·RL = 1.288²·12 = 19.91 W
11.RC & RL transients

Answer: vC = 1.671 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.

Thévenin equivalent seen by the capacitor

Short the source to find Rth; open the capacitor to find the voltage it will settle at.

Rth = R1 ∥ R2 = 2.2·1/(2.2 + 1) = 0.6875 kΩ
vC(∞) = 15·1/(2.2 + 1) = 4.688 V

Time constant

τ = Rth·C = 0.6875·33 = 22.69 ms

Charging curve

vC(t) = 4.688·(1 − e−10/22.69) = 1.671 V

Answer

vC = 1.671 V
12.RC & RL transients

Answer: vC = 8.767 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 = 1 kΩ · 68 µF = 68 ms
vC(t) = 10 + (6 − 10)·e−t/τ

Voltage at the given time

vC(80) = 10 + (−4)·e−80/68 = 8.767 V

Answer

vC = 8.767 V
13.AC impedance & phasors

Answer: φ = −62.27 °

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 = 1257·0.25 = 314.2 Ω
XC = 1/(ωC) = 1/(1257·2.2·10⁻⁶) = 361.7 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(25² + (−47.56)²) = 53.73 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(−47.56/25) = −62.27°

Answer

φ = −62.27°
14.AC impedance & phasors

Answer: B = 397.9 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|.

Resonant frequency

f0 = 1/(2π·√(0.002·10·10⁻⁶)) = 1125 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/5)·√(0.002/(10·10⁻⁶)) = 2.828

Bandwidth

B = f0/Q = 1125/2.828 = 397.9 Hz

Answer

B = 397.9 Hz
15.AC power & power factor

Answer: S = 1440 VA

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

Current

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

Apparent power

S = V·I = 120·12 = 1440 VA

Answer

S = 1440 VA
16.AC power & power factor

Answer: C = 530.5 µF

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

Conjugate match

The load reactance cancels the source reactance, and the load resistance equals the source resistance.

ZL = Zs* = 10 − j6 Ω

Capacitance

XC = 6 Ω → C = 1/(2π·50·6) = 530.5 µF

Answer

C = 530.5 µF
17.Three-phase power

Answer: P = 20.67 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| = √(9² + 12²) = 15 Ω
Iph = 415/15 = 27.67 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·27.67²·9 = 20.67 kW

Answer

P = 20.67 kW

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

18.Three-phase power

Answer: Cph = 1065 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 = 250·tan(acos 0.7) = 250·1.02 = 255.1 kvar
Q2 = 250·tan(acos 0.95) = 250·0.3287 = 82.17 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 172.9 kvar

Capacitance per phase

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

V = 415 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 57627/(314.2·415²) = 1065 µF

Answer

Cph = 1065 µF
19.Transformers

Answer: a = 4

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.

Match the referred load

a²·75 = 1200
a = √(1200/75) = √16 = 4

Answer

a = 4
20.Transformers

Answer: Ploss = 203.1 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 = 0.25·312.5 = 78.13 W

Total losses

The core loss is the same at any load.

Ploss = 125 + 78.13 = 203.1 W

Answer

Ploss = 203.1 W

The efficiency at this load is η = 98.32%.

21.Op-amp circuits

Answer: V− = 0.0625 V

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

Voltage at the inputs

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

V₊ = V2·R4/(R3 + R4) = 0.2·10/(22 + 10) = 0.0625 V
V₋ = V₊

Answer

V− = 0.0625 V
22.Op-amp circuits

Answer: Vout = 7.819 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 = 0.3/22 = 0.01364 mA
I2 = V2/R2 = −1.8/10 = −0.18 mA
If = I1 + I2 = −0.1664 mA

Output voltage

Vout = −Rf·If = −47·(−0.1664) = 7.819 V

Answer

Vout = 7.819 V
23.Diodes & transistors

Answer: IZ = 30.91 mA

Zener regulator: a reverse-biased Zener diode holds its voltage VZ while its current stays above a minimum IZ,min. The series resistor R takes the difference between input and output: IR = (Vin − VZ)/R. That current splits between the load, IL = VZ/RL, and the Zener: IZ = IR − IL. Design for the worst cases: the lowest input with the largest load must still leave IZ,min, and the highest input with no load gives the largest Zener power PZ = VZ·IZ.

Resistor current

IR = 12/330 = 36.36 mA

Load current

IL = 12/2200 = 5.455 mA

Zener current

IZ = 36.36 − 5.455 = 30.91 mA

Answer

IZ = 30.91 mA
24.Diodes & transistors

Answer: RB = 2.105 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) = (5 − 0.2)/470 = 10.21 mA

Minimum base current

IB(min) = IC(sat)/β = 10.21/50 = 204.3 µA

Design base current

IB = 10·204.3 µA = 2043 µA

Base resistor

RB = (5 − 0.7)/IB = 4.3 V/2.043 mA = 2.105 kΩ

Answer

RB = 2.105 kΩ
25.Number systems

Answer: U = 157

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

Invert and add one

99 = 0110 0011
inverted: 1001 1100
+1 → 1001 1101

Read it as unsigned

1001 1101₂ = 256 − 99 = 157

Answer

U = 157
26.Number systems

Answer: R = 4

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

Bit by bit

AND gives 1 only where both bits are 1.

0010 0110
AND 0100 0101
= 0000 0100

Value of the result

0000 0100₂ = 4

Answer

R = 4
27.Digital logic: gates, K-maps & flip-flops

Answer: N = 2

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′C′ + ABC′

Rows with F = 1

ABC = 000, 110
N = 2

Answer

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

Answer: Q = 12

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

State after every edge

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

Decimal value

1100₂ = 12

Answer

Q = 12
29.Laplace transforms

Answer: f(0.3) = 0.2816

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

Partial fractions

F(s) = A/s + B/(s + 1) + C/(s + 4)
A = 10/(1·4) = 2.5
B = 10/((−1)·3) = −3.333
C = 10/((−4)·(−3)) = 0.8333

Inverse transform

f(t) = 2.5 − 3.333·e−t + 0.8333·e−4t

Evaluate

f(0.3) = 2.5 − 3.333·e−0.3 + 0.8333·e−4·0.3 = 0.2816

Answer

f(0.3) = 0.2816
30.Laplace transforms

Answer: y(1) = −0.6957

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 − (−11) + 4(sY − 2) + 3Y = 0
Y(s) = (2s − 3)/((s + 1)(s + 3))

Partial fractions

A = −5/2 = −2.5
B = −9/(−2) = 4.5

Result

y(t) = −2.5·e−t + 4.5·e−3t
y(1) = −0.6957

Answer

y(1) = −0.6957

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

31.Block diagrams

Answer: K = 25

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

1 + K/(s(s + 10)) = 0 → s² + 10s + K = 0

Repeated root

A quadratic s² + βs + γ has a double root when β² = 4γ.

10² = 4K → K = 100/4 = 25

Answer

K = 25
Both closed-loop poles are then at s = −5.
32.Block diagrams

Answer: T(0) = 0.6701

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

Inner loop

Gᵢ = G₂/(1 + G₂H₁) = 5/(s + 21)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 65/((s + 4)(s + 21) + 13)

DC gain

T(0) = 65/(4·21 + 13) = 65/97 = 0.6701

Answer

T(0) = 0.6701
33.Step response

Answer: σ = −3.5

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 = 69 → ωₙ = 8.307 rad/s
2ζωₙ = 7 → ζ = 7/(2·8.307) = 0.4214

Result

σ = −ζωₙ = −7/2 = −3.5

Answer

σ = −3.5
34.Step response

Answer: ωn = 4.738 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).

Damping ratio

ζ = −ln(0.12)/√(π² + ln²(0.12)) = 2.12/√(9.87 + 4.496) = 0.5594

Natural frequency

ωd = π/tp = π/0.8 = 3.927 rad/s
ωₙ = ωd/√(1 − ζ²) = 3.927/0.8289 = 4.738 rad/s

Answer

ωn = 4.738 rad/s
35.Bode plots & margins

Answer: φ = 14.01 °

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 = 1/(RC) = 1/(22000·1 × 10⁻⁸) = 4545 rad/s
fc = 723.4 Hz → f/fc = 2900/723.4 = 4.009

Phase

φ = 90° − atan(4.009) = 14.01°

Answer

φ = 14.01°
36.Bode plots & margins

Answer: K = 7.536

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

Phase crossover

(ω/2)·(ω/3) = 1 → ω180 = √(2·3) = 2.449 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 4)·√(ω² + 9)) = K/(2·3·(2 + 3)) = K/30

Required gain

30/K = 1012/20 = 3.981
K = 30/3.981 = 7.536

Answer

K = 7.536
37.Stability: Routh–Hurwitz & root locus

Answer: Kmin = 0.9

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

Characteristic polynomial

(s − 3)(s + 3) + K(s + 10) = s² + (0 + K)s + (10K − 9)

Conditions

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

0 + K > 0 → K > 0
10K − 9 > 0 → K > 0.9
Kmin = 0.9

Answer

Kmin = 0.9
38.Stability: Routh–Hurwitz & root locus

Answer: σa = −4.667

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: 0, −10, −4
n − m = 3 − 0 = 3

Centroid

σa = (−14)/3 = −4.667

Answer

σa = −4.667

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

39.Steady-state error & PID control

Answer: ess = 20

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

Velocity constant

Kv = lim s·G(s) = 0.1

Ramp error

ess = 2/0.1 = 20

Answer

ess = 20
40.Steady-state error & PID control

Answer: K = 3.601

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 + 6) + 5K = s² + 6s + 5K

Natural frequency

2·0.707·ωₙ = 6 → ωₙ = 4.243 rad/s

Gain

5K = ωₙ² = 18.01 → K = 3.601

Answer

K = 3.601

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