FE Electrical and Computer exam practice: cram pack

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

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1Formulas and methods

Series & parallel resistors

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

Method

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

Nodal analysis

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

Method

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

Mesh analysis

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

Method

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

Superposition

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

Method

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

Thévenin & Norton

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

Method

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

RC & RL transients

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

Method

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

AC impedance & phasors

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

Method

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

AC power & power factor

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

Three-phase power

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

Transformers

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

Op-amp circuits

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

Diodes & transistors

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

Number systems

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

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

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

Method

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

Stability: Routh–Hurwitz & root locus

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

Method

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

Steady-state error & PID control

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

Method

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

2Practice problems

1.Series & parallel resistorsMedium
R1​ = 5 ΩIR1​R3​ = 3 Ω+−V1​9 VR2​2 ΩAR4​10 ΩB

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

V1 = 9 VR1 = 5 ΩR2 = 2 ΩR3 = 3 ΩR4 = 10 Ω
IR1 = A
2.Series & parallel resistorsHard
R1​ = 15 ΩIR1​R3​ = 15 ΩR5​ = 5 Ω+−V1​18 VR2​8 ΩAR4​8 ΩBR6​3 Ω

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

V1 = 18 VR1 = 15 ΩR2 = 8 ΩR3 = 15 ΩR4 = 8 ΩR5 = 5 ΩR6 = 3 Ω
IR1 = A
3.Nodal analysisMedium
R1​ = 5 ΩR3​ = 12 Ω+−V1​10 VR2​5 ΩAR4​3 ΩBIs​5 A

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

V1 = 10 VIs = 5 AR1 = 5 ΩR2 = 5 ΩR3 = 12 ΩR4 = 3 Ω
PIs = W
4.Nodal analysisHard
R1​ = 15 ΩR2​ = 25 ΩR4​ = 12 Ω+−V1​18 VR3​15 ΩAR5​30 ΩBIs​2 A−+V2​10 V

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

V1 = 18 VIs = 2 AV2 = 10 VR1 = 15 ΩR2 = 25 ΩR3 = 15 ΩR4 = 12 ΩR5 = 30 Ω
VB = V
5.Mesh analysisMedium
R1​ = 15 ΩR2​ = 12 ΩIR2​+−V1​10 VR3​6 ΩA+−V2​12 VI1​I2​

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

V1 = 10 VV2 = 12 VR1 = 15 ΩR2 = 12 ΩR3 = 6 Ω
IR2 = A
6.Mesh analysisHard
R1​ = 6 ΩIR1​R2​ = 30 ΩR3​ = 30 Ω+−V1​9 VR4​20 ΩAR5​20 ΩB+−V2​6 VI1​I2​I3​

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

V1 = 9 VV2 = 6 VR1 = 6 ΩR2 = 30 ΩR3 = 30 ΩR4 = 20 ΩR5 = 20 Ω
IR1 = A
7.SuperpositionMedium
R1​ = 3 ΩR3​ = 6 ΩIR3​+−V1​9 VR2​15 ΩAR4​6 ΩBIs​4 A

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

V1 = 9 VIs = 4 AR1 = 3 ΩR2 = 15 ΩR3 = 6 ΩR4 = 6 Ω
IR3 = A
8.SuperpositionHard
linear circuitV1​V2​Is​IRL​

A linear resistive circuit contains three independent sources: V1, V2 and Is. The current I through a load is measured three times. With all sources at their normal values, I = 21.5 mA. With V1 doubled and the other two unchanged, I = 28 mA. With the direction of Is reversed and the other two unchanged, I = 2.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 21.5 mA2·V1: I = 28 mA−Is: I = 2.5 mA
I = mA
9.Thévenin & NortonMedium
R1​ = 8 ΩR3​ = 6 Ω+−V1​18 VR2​12 ΩARL​6 ΩIL​B

A load RL = 6 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 18 VR1 = 8 ΩR2 = 12 ΩR3 = 6 ΩRL = 6 Ω
IL = A
10.Thévenin & NortonHard
R1​ = 25 ΩR3​ = 20 Ω+−V1​5 VR2​20 ΩAIs​2 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 = 5 VIs = 2 AR1 = 25 ΩR2 = 20 ΩR3 = 20 ΩRL = 12 Ω
PL = W
11.RC & RL transientsMedium
R = 47 Ωt = 0L100 mHi(t)

The inductor carries a steady current of 2 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. Find the current i at t = 5 ms.

i(0) = 2 AR = 47 ΩL = 100 mHt = 5 ms
i = A
12.RC & RL transientsHard
+−Vs​10 Vt = 0R = 6.8 kΩC220 µF+−vC​vC​(0) = 24 V

Just before the switch closes at t = 0, the capacitor holds 24 V with the polarity shown. Find the capacitor voltage vC at t = 1000 ms.

Vs = 10 VR = 6.8 kΩC = 220 µFvC(0) = 24 Vt = 1000 ms
vC = V
13.AC impedance & phasorsMedium
24 V500 HzR = 20 ΩL = 20 mHC = 1 µF

A series RLC circuit is connected to a 24 V (rms), 500 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 24 V (rms)f = 500 HzR = 20 ΩL = 20 mHC = 1 µF
φ = °
14.AC impedance & phasorsHard
230 V60 HzR = 80 ΩL = 150 mHC = 22 µF+−VC​

A series RLC circuit is connected to a 230 V (rms), 60 Hz supply. Find the rms voltage VC across the capacitor.

V = 230 V (rms)f = 60 HzR = 80 ΩL = 150 mHC = 22 µF
VC = V
15.AC power & power factorMedium
P = 13 kWQSφ

A single-phase motor takes 13 kW at a power factor of 0.8 lagging from a 400 V supply. Find the line current I.

P = 13 kWpf = 0.8V = 400 V
I = A
16.AC power & power factorHard
P = 35 kWQ1​S1​φ1​QC​

A single-phase load takes 35 kW at a power factor of 0.65 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the new line current Inew.

P = 35 kWpf1 = 0.65pf2 = 0.95V = 400 Vf = 50 Hz
Inew = A
17.Three-phase powerMedium
ABCVL​ = 415 VIL​ = 25 Acos φ = 0.7lagging

A balanced three-phase load draws a line current of 25 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.7 lagging. Find the total real power P drawn by the load in kW.

VL = 415 VIL = 25 Acos φ = 0.7
P = kW
18.Three-phase powerHard
P = 200 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 200 kW at a power factor of 0.72 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 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 200 kWpf1 = 0.72pf2 = 0.95VL = 400 Vf = 60 Hz
Cph = µF
19.TransformersMedium
RL​ = 15 Ω2014600 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 20:1 has 4600 V applied to its primary winding. A 15 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 20:1V1 = 4600 VRL = 15 Ω
I1 = A
20.TransformersHard
pf 0.850 kVA, η = 98.2%

A single-phase 50 kVA transformer has a core loss of 200 W. At rated load and a power factor of 0.8 lagging its efficiency is 98.2%. Find the full-load copper loss Pcu,fl in watts.

S = 50 kVAPcore = 200 Wpf = 0.8η = 98.2%
Pcu,fl = W
21.Op-amp circuitsMedium
−+Vout​Rf​ = 10 kΩR1​ = 1 kΩVin​ = −0.6 V

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

Vin = −0.6 VR1 = 1 kΩRf = 10 kΩ
Vout = V
22.Op-amp circuitsHard
Vin​V1​Vout​stage 1: invertingR1​ = 4.7 kΩRf1​ = 10 kΩstage 2: non-invertingR2​ = 1 kΩRf2​ = 10 kΩ

Two ideal op-amp stages are connected in cascade: an inverting amplifier (R1, Rf1) followed by a non-inverting amplifier (R2 to ground, Rf2 in the feedback). Both run from ±15 V supplies. Find the overall voltage gain Av = Vout/Vin.

R1 = 4.7 kΩRf1 = 10 kΩR2 = 1 kΩRf2 = 10 kΩ
Av =
23.Diodes & transistorsMedium
RC​ = 2.2 kΩβ = 120RE​ = 1 kΩVCC​ = 18 VR1​ = 47 kΩR2​ = 8.2 kΩ

An NPN transistor with β = 120 has voltage-divider bias from VCC = 18 V: R₁ = 47 kΩ from VCC to the base and R₂ = 8.2 kΩ from the base to ground. The collector resistor is RC = 2.2 kΩ and the emitter resistor is RE = 1 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 = 47 kΩR2 = 8.2 kΩRC = 2.2 kΩRE = 1 kΩβ = 120
VCE = V
24.Diodes & transistorsHard
RC​ = 220 Ωβ = 200VCC​ = 12 V5 VRB​ = ?

An NPN transistor with β = 200 switches a 220 Ω load resistor connected between VCC = 12 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB. Take VBE = 0.7 V and VCE(sat) = 0.2 V. Find the largest base resistor RB, in kΩ, that still drives the transistor just into saturation.

VCC = 12 VRC = 220 Ωβ = 200Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 V
RB = kΩ
25.Number systemsMedium
8-bit two's complement?−128?64?32?16?8?4?2?1

Write −41 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −418 bit
U =
26.Number systemsHard
24 bit1 px256 px256 pxM = ? KiB

An uncompressed image is 256 × 256 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

256 × 256 px24 bit/px1 KiB = 1024 B
M = KiB
27.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011001131204050716

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(0, 2, 3, 6). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(0, 2, 3, 6)
N =
28.Digital logic: gates, K-maps & flip-flopsHard
CLKJK12345678

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?

Q₀ = 0JK
N =
29.Laplace transformsMedium
t (s)f(t)t = 0.3f(0.3) = ?

Find f(t) for F(s) = 60/(s(s + 1)(s + 4)) by partial fractions, and evaluate f(0.3).

F(s) = 60/(s(s + 1)(s + 4))t = 0.3 s
f(0.3) =
30.Laplace transformsHard
ReIm−1−2−3−4−5−61

A system is described by the differential equation y″ + 9y′ + 18y = 2u′ + 10u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 0.1 s. Find g(0.1).

y″ + 9y′ + 18y = 2u′ + 10ut = 0.1 s
g(0.1) =
31.Block diagramsMedium
R(s)+Ks(s + 15)G(s)C(s)−

The plant G(s) = K/(s(s + 15)) 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 + 15))unity negative feedback
K =
32.Block diagramsHard
R(s)+12controller+D(s)+4s + 4plantC(s)−

A proportional controller K = 12 drives the plant G(s) = 4/(s + 4) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). A reference step of size 10 and a disturbance step of size 5 are applied together. Find the steady-state output css.

K = 12G(s) = 4/(s + 4)D = 5R = 10
css =
33.Step responseMedium
t (s)y(t)2

A system has the transfer function G(s) = 124/(s² + 8s + 62) and receives a unit step input. Find the natural frequency ωn in rad/s.

G(s) = 124/(s² + 8s + 62)
ωn = rad/s
34.Step responseHard
R(s)+Ks(s + 6)G(s)C(s)−

A unity negative feedback loop has G(s) = K/(s(s + 6)). Find the gain K that gives a step response with 5 % overshoot.

G(s) = K/(s(s + 6))%OS = 5 %
K =
35.Bode plots & marginsMedium
vin​R = 47 kΩC = 4.7 nFvout​

The RC low-pass filter shown has R = 47 kΩ, C = 4.7 nF. A sinusoidal input of 2.2 kHz is applied. Find the gain |H| = |vout/vin| in dB.

R = 47 kΩC = 4.7 nFf = 2.2 kHz
|H| = dB
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 + 5)(s + 9)). Find the gain K that gives a gain margin of 10 dB.

L(s) = K/(s(s + 5)(s + 9))GM = 10 dB
K =
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 1)(s + 2)(s + 4)C(s)−

A unity-feedback loop has L(s) = K/((s + 1)(s + 2)(s + 4)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 1)(s + 2)(s + 4))
Kmax =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−8−9−101j1j2

For L(s) = K/((s + 8)(s² + 4s + 5)), find the angle of departure θd of the root locus from the complex pole in the upper half-plane, measured from the positive real axis.

L(s) = K/((s + 8)(s² + 4s + 5))
θd = °
39.Steady-state error & PID controlMedium
R(s)+K = 210.5s + 1C(s)−

A process G(s) = 1/(0.5s + 1) is controlled by a proportional gain K = 2 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 1/(0.5s + 1)K = 2
ess = %
40.Steady-state error & PID controlHard
r(t) = 5t+1 + 0.5/s4(s + 1)(s + 4)C(s)−

A PI controller C(s) = 1 + 0.5/s drives the plant G(s) = 4/((s + 1)(s + 4)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 5t.

C(s) = 1 + 0.5/sG(s) = 4/((s + 1)(s + 4))r(t) = 5t
ess =

3Solutions

1.Series & parallel resistors

Answer: IR1 = 1.337 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​ = 5 ΩR3​ = 3 Ω+−V1​9 VR2​2 ΩAR4​10 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 = 3 + 10 = 13 Ω
R2 ∥ 13 = 2·13/(2+13) = 1.733 Ω
Req = R1 + 1.733 = 5 + 1.733 = 6.733 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 = 9/6.733 = 1.337 A

Answer

IR1 = 1.337 A
2.Series & parallel resistors

Answer: IR1 = 0.8725 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​ = 15 ΩR3​ = 15 ΩR5​ = 5 Ω+−V1​18 VR2​8 ΩAR4​8 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 = 5 + 3 = 8 Ω
R4 ∥ 8 = 4 Ω
R3 + 4 = 19 Ω
R2 ∥ 19 = 5.63 Ω
Req = R1 + 5.63 = 15 + 5.63 = 20.63 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 = 18/20.63 = 0.8725 A

Answer

IR1 = 0.8725 A
3.Nodal analysis

Answer: PIs = 66.43 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/5 + (VA − 10)/5 + (VA − VB)/12 = 0
0.4833·VA − 0.08333·VB = 2

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/3 + (VB − VA)/12 = 5
−0.08333·VA + 0.4167·VB = 5

Solve

Δ = (0.4833)(0.4167) − (−0.08333)(−0.08333) = 0.1944
VA = Δ1/Δ = 1.25/0.1944 = 6.429 V
VB = Δ2/Δ = 2.583/0.1944 = 13.29 V

Power delivered by the current source

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

PIs = VB·Is = 13.29·5 = 66.43 W

Answer

PIs = 66.43 W
4.Nodal analysis

Answer: VB = 9.791 V

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

R1​ = 15 ΩR2​ = 25 ΩR4​ = 12 Ω+−V1​18 VR3​15 ΩAR5​30 ΩBIs​2 A−+V2​10 V

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 18 V, V2 = −10 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/15 + (VA − 18)/15 + (VA − VB)/25 = 0
0.1733·VA − 0.04·VB = 1.2

KCL at node B

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

VB/R5 + (VB − VA)/R2 + (VB − V2)/R4 = Is
VB/30 + (VB − VA)/25 + (VB − (−10))/12 = 2
−0.04·VA + 0.1567·VB = 1.167

Solve

Δ = (0.1733)(0.1567) − (−0.04)(−0.04) = 0.02556
VA = Δ1/Δ = 0.2347/0.02556 = 9.183 V
VB = Δ2/Δ = 0.2502/0.02556 = 9.791 V

Answer

VB = 9.791 V
5.Mesh analysis

Answer: IR2 = −0.5614 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​ = 12 Ω+−V1​10 VR3​6 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 + 6(I1 − I2) = 10
21·I1 − 6·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 + R3(I2 − I1) = −V2
12·I2 + 6(I2 − I1) = (−12)
−6·I1 + 18·I2 = −12

Solve

Δ = (21)(18) − (−6)(−6) = 342
I1 = Δ1/Δ = 108/342 = 0.3158 A
I2 = Δ2/Δ = −192/342 = −0.5614 A

Answer

IR2 = I2 = −0.5614 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: IR1 = 0.4208 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​ = 6 ΩR2​ = 30 ΩR3​ = 30 Ω+−V1​9 VR4​20 ΩAR5​20 ΩB+−V2​6 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
6·I1 + 20(I1 − I2) = 9
26·I1 − 20·I2 = 9

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 + 20(I2 − I1) + 20(I2 − I3) = 0
−20·I1 + 70·I2 − 20·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
30·I3 + 20(I3 − I2) = (−6)
−20·I2 + 50·I3 = −6

Solve

Δ = det of the coefficient matrix = 60600
I1 = Δ1/Δ = 25500/60600 = 0.4208 A
I2 = Δ2/Δ = 5880/60600 = 0.09703 A
I3 = Δ3/Δ = −4920/60600 = −0.08119 A

Answer

IR1 = I1 = 0.4208 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: IR3 = −1.138 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​ = 3 ΩR3​ = 6 Ω+−V1​9 VR2​15 ΩAR4​6 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 = 6+6 = 12 Ω
R2 ∥ (R3+R4) = 15·12/(15+12) = 6.667 Ω
VA = 9·6.667/(3+6.667) = 6.207 V
VB = VA·R4/(R3+R4) = 6.207·6/12 = 3.103 V
IR3′ = (VA − VB)/R3 = (6.207 − 3.103)/6 = 0.5172 A
R1​ = 3 ΩR3​ = 6 ΩSHORTR2​15 ΩAR4​6 ΩBIs​4 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 = 3·15/(3+15) = 2.5 Ω
Req = R4 ∥ (R3+R1∥R2) = 6 ∥ 8.5 = 3.517 Ω
VB = Is·Req = 4·3.517 = 14.07 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 14.07·2.5/8.5 = 4.138 V
IR3″ = (VA − VB)/R3 = (4.138 − 14.07)/6 = −1.655 A

Add the contributions

IR3 = IR3′ + IR3″
IR3 = 0.5172 + (−1.655) = −1.138 A

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

8.Superposition

Answer: I = 16 mA

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

Write the measurements with contributions

Each contribution is proportional to its own source.

I₁ + I₂ + I₃ = 21.5
2·I₁ + I₂ + I₃ = 28
I₁ + I₂ − I₃ = 2.5

Solve for the contributions

I₁ = 28 − 21.5 = 6.5 mA
I₃ = (21.5 − 2.5)/2 = 9.5 mA
I₂ = 21.5 − 6.5 − 9.5 = 5.5 mA

Switch V₂ off

I = I₁ + I₃ = 6.5 + 9.5 = 16 mA

Answer

I = 16 mA
9.Thévenin & Norton

Answer: IL = 0.6429 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​ = 6 ΩSHORTR2​12 ΩAabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 8·12/(8+12) = 4.8 Ω
Rth = (R1 ∥ R2) + R3 = 4.8 + 6 = 10.8 Ω
R1​ = 8 ΩR3​ = 6 Ω+−V1​18 VR2​12 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) = 18·12/(8+12) = 10.8 V
Rth​ = 10.8 Ω+−Vth​10.8 VRL​6 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) = 10.8/(10.8 + 6) = 0.6429 A
10.Thévenin & Norton

Answer: PL = 26.81 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​ = 25 ΩR3​ = 20 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 = 25·20/(25+20) = 11.11 Ω
Rth = (R1 ∥ R2) + R3 = 11.11 + 20 = 31.11 Ω
R1​ = 25 ΩR3​ = 20 Ω+−V1​5 VR2​20 ΩAIs​2 ABab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 2.222 + 2·11.11 = 24.44 V
Vth = VB = VA + Is·R3 = 24.44 + 2·20 = 64.44 V
Rth​ = 31.11 Ω+−Vth​64.44 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) = 64.44/(31.11 + 12) = 1.495 A
PL = IL²·RL = 1.495²·12 = 26.81 W
11.RC & RL transients

Answer: i = 0.1907 A

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

Time constant

τ = L/R = 100 mH / 47 Ω = 2.128 ms

Decay of the current

i(t) = 2·e−5/2.128 = 0.1907 A

Answer

i = 0.1907 A
12.RC & RL transients

Answer: vC = 17.18 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) = 24 V, vC(∞) = Vs = 10 V
τ = R·C = 6.8 kΩ · 220 µF = 1496 ms
vC(t) = 10 + (24 − 10)·e−t/τ

Voltage at the given time

vC(1000) = 10 + (14)·e−1000/1496 = 17.18 V

Answer

vC = 17.18 V
13.AC impedance & phasors

Answer: φ = −85.52 °

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·500 = 3142 rad/s
XL = ωL = 3142·0.02 = 62.83 Ω
XC = 1/(ωC) = 1/(3142·1·10⁻⁶) = 318.3 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(20² + (−255.5)²) = 256.3 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(−255.5/20) = −85.52°

Answer

φ = −85.52°
14.AC impedance & phasors

Answer: VC = 270.6 V

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

Reactances

ω = 2πf = 2π·60 = 377 rad/s
XL = ωL = 377·0.15 = 56.55 Ω
XC = 1/(ωC) = 1/(377·22·10⁻⁶) = 120.6 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(80² + (−64.02)²) = 102.5 Ω

Current

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

I = V/|Z| = 230/102.5 = 2.245 A

Voltage across the capacitor

VC = I·XC = 2.245·120.6 = 270.6 V

Answer

VC = 270.6 V
15.AC power & power factor

Answer: I = 40.63 A

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

Line current

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

I = P/(V·cos φ) = 13000/(400·0.8) = 40.63 A

Answer

I = 40.63 A
16.AC power & power factor

Answer: Inew = 92.11 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 = 35·tan(49.46°) = 40.92 kvar
Q2 = P·tan φ2 = 35·tan(18.19°) = 11.5 kvar
QC = Q1 − Q2 = 29.42 kvar

New line current

Inew = P/(V·cos φ2) = 35000/(400·0.95) = 92.11 A

Answer

Inew = 92.11 A
17.Three-phase power

Answer: P = 12.58 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.

Apparent power

S = √3·VL·IL = √3·415·25 = 17.97 kVA

Real power

P = S·cos φ = 17.97·0.7 = 12.58 kW

Answer

P = 12.58 kW
18.Three-phase power

Answer: Cph = 702 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 = 200·tan(acos 0.72) = 200·0.9639 = 192.8 kvar
Q2 = 200·tan(acos 0.95) = 200·0.3287 = 65.74 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 127 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) = 42345/(377·400²) = 702 µF

Answer

Cph = 702 µF
19.Transformers

Answer: I1 = 0.7667 A

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

Secondary voltage and current

V2 = 4600/20 = 230 V
I2 = 230/15 = 15.33 A

Primary current

An ideal transformer has no losses: V₁·I₁ = V₂·I₂.

I1 = I2/20 = 0.7667 A

Answer

I1 = 0.7667 A

Check: the load seen from the primary is a²·RL = 6000 Ω, and 4600/6000 gives the same current.

20.Transformers

Answer: Pcu,fl = 533.2 W

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

Output and input power

Pout = 50·0.8 = 40 kW
Pin = 40000/0.982 = 40733 W

Total and copper losses

Ploss = 40733 − 40000 = 733.2 W
Pcu,fl = 733.2 − 200 = 533.2 W

Answer

Pcu,fl = 533.2 W
21.Op-amp circuits

Answer: Vout = −6.6 V

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

Inverting input follows the input

V₋ = V₊ = −0.6 V
I = V₋/R1 = −0.6/1 = −0.6 mA

Output voltage

The current through R₁ also flows through Rf.

Vout = V₋ + I·Rf = −0.6 + (−0.6)·10 = −6.6 V

Answer

Vout = −6.6 V
22.Op-amp circuits

Answer: Av = −23.4

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

Gain of each stage

A1 = −Rf1/R1 = −10/4.7 = −2.128
A2 = 1 + Rf2/R2 = 1 + 10/1 = 11

Overall gain

Av = A1·A2 = −2.128·11 = −23.4

Answer

Av = −23.4
23.Diodes & transistors

Answer: VCE = 12.06 V

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

Thévenin equivalent of the divider

VTH = 18·8.2/(47 + 8.2) = 2.674 V
RTH = 47·8.2/(47 + 8.2) = 6.982 kΩ

Base current

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

IB = (2.674 − 0.7)/(6.982 + 121·1) = 15.42 µA

Collector current

IC = 120·15.42 µA = 1.851 mA
IE = 121·15.42 µA = 1.866 mA

Collector–emitter voltage

VCE = 18 − 1.851·2.2 − 1.866·1 = 12.06 V

Answer

VCE = 12.06 V
24.Diodes & transistors

Answer: RB = 16.03 kΩ

BJT as a switch: to switch a load fully on, the transistor is driven into saturation, where VCE = VCE(sat) and the collector current is set by the load: IC(sat) = (VCC − VCE(sat))/RC. Saturation needs at least IB = IC(sat)/β. If the base current is smaller, the transistor stays active and IC = β·IB.

Collector current in saturation

IC(sat) = (12 − 0.2)/220 = 53.64 mA

Minimum base current

IB(min) = IC(sat)/β = 53.64/200 = 268.2 µA

Base resistor

RB = (5 − 0.7)/IB = 4.3 V/0.2682 mA = 16.03 kΩ

Answer

RB = 16.03 kΩ

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

25.Number systems

Answer: U = 215

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

Invert and add one

41 = 0010 1001
inverted: 1101 0110
+1 → 1101 0111

Read it as unsigned

1101 0111₂ = 256 − 41 = 215

Answer

U = 215
26.Number systems

Answer: M = 192 KiB

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.

Bits and bytes

256·256·24 = 1572864 bit = 196608 B

Convert to KiB

196608/1024 = 192 KiB

Answer

M = 192 KiB
27.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

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.

ABC00011110011001131204050716

Group the 0s

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

(B + C′)
(A′ + C′)
(A′ + B)

Minimal POS form

F = (B + C′)(A′ + C′)(A′ + B)
N = 3

Answer

N = 3
28.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

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 JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 1, K = 0 → Q = 1
2: J = 0, K = 0 → Q = 1
3: J = 1, K = 0 → Q = 1
4: J = 1, K = 0 → Q = 1
5: J = 1, K = 0 → Q = 1
6: J = 1, K = 0 → Q = 1
7: J = 1, K = 1 → Q = 0
8: J = 1, K = 0 → Q = 1

Count the changes

0 → 1 → 1 → 1 → 1 → 1 → 1 → 0 → 1
N = 3

Answer

N = 3
29.Laplace transforms

Answer: f(0.3) = 1.69

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 = 60/(1·4) = 15
B = 60/((−1)·3) = −20
C = 60/((−4)·(−3)) = 5

Inverse transform

f(t) = 15 − 20·e−t + 5·e−4t

Evaluate

f(0.3) = 15 − 20·e−0.3 + 5·e−4·0.3 = 1.69

Answer

f(0.3) = 1.69
30.Laplace transforms

Answer: g(0.1) = 1.354

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

Transfer function

G(s) = Y(s)/U(s) = (2s + 10)/(s² + 9s + 18)

Partial fractions

s² + 9s + 18 = (s + 3)(s + 6)
A = 1.333, B = 0.6667

Impulse response

g(t) = 1.333·e−3t + 0.6667·e−6t
g(0.1) = 1.354

Answer

g(0.1) = 1.354
31.Block diagrams

Answer: K = 56.25

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 + 15)) = 0 → s² + 15s + K = 0

Repeated root

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

15² = 4K → K = 225/4 = 56.25

Answer

K = 56.25
Both closed-loop poles are then at s = −7.5.
32.Block diagrams

Answer: css = 9.615

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).

Transfer functions

C/D = G/(1 + K·G) = 4/(s + 4 + 48)
C/R = K·G/(1 + K·G) = 48/(s + 4 + 48)

Steady state

css = 10·48/52 + 5·4/52 = 9.615

Answer

css = 9.615

A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.

33.Step response

Answer: ωn = 7.874 rad/s

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

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 62 → ωₙ = 7.874 rad/s
2ζωₙ = 8 → ζ = 8/(2·7.874) = 0.508

Result

ωₙ = 7.874 rad/s

Answer

ωn = 7.874 rad/s
34.Step response

Answer: K = 18.9

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) = K/(s² + 6s + K)

Damping ratio from the overshoot

ζ = −ln(0.05)/√(π² + ln²(0.05)) = 0.6901

Gain

ωₙ = 6/(2ζ) = 4.347 rad/s
K = ωₙ² = 18.9

Answer

K = 18.9
35.Bode plots & margins

Answer: |H| = −10.14 dB

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

Corner frequency

ωc = 1/(RC) = 1/(47000·4.7 × 10⁻⁹) = 4527 rad/s
fc = 720.5 Hz → f/fc = 2200/720.5 = 3.054

Gain

|H| = 1/√(1 + 3.054²) = 0.3112
20·log₁₀(0.3112) = −10.14 dB

Answer

|H| = −10.14 dB
36.Bode plots & margins

Answer: K = 199.2

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/5)·(ω/9) = 1 → ω180 = √(5·9) = 6.708 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 25)·√(ω² + 81)) = K/(5·9·(5 + 9)) = K/630

Required gain

630/K = 1010/20 = 3.162
K = 630/3.162 = 199.2

Answer

K = 199.2
37.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 90

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

Characteristic polynomial

s³ + 7s² + 14s + 8 + K = 0

Routh array

s³: 1, 14
s²: 7, 8 + K
s¹: (7·14 − 8 − K)/7
s⁰: 8 + K

Condition

The s¹ entry must stay positive.

K < 7·14 − 8 = 90

Answer

Kmax = 90

At the limit the loop oscillates at ω = √14 = 3.742 rad/s.

38.Stability: Routh–Hurwitz & root locus

Answer: θd = 80.54 °

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.

Pole locations

s² + 4s + 5 = 0 → s = −2 ± j1
s = −8

Angles from the other poles

∠(−2 + j1 + 8) = atan2(1, 6) = 9.462°
∠(−2 + j1 − (−2 − j1)) = ∠(j2) = 90°

Angle of departure

θd = 180° − 9.462° − 90° = 80.54°

Answer

θd = 80.54°
39.Steady-state error & PID control

Answer: ess = 33.33 %

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

Position constant

Kp = 2·1 = 2

Step error

ess = 100/(1 + 2) = 33.33 %

Answer

ess = 33.33 %
40.Steady-state error & PID control

Answer: ess = 10

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).

Velocity constant

s·C(s) = 1s + 0.5 → 0.5 as s → 0.

Kv = 0.5·4/4 = 0.5

Ramp error

ess = 5/0.5 = 10

Answer

ess = 10

The PI controller removes the step error completely; a ramp is followed with a constant lag.

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