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​ = 2 ΩIR1​R3​ = 15 Ω+−V1​9 VR2​8 ΩAR4​4 ΩB

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

V1 = 9 VR1 = 2 ΩR2 = 8 ΩR3 = 15 ΩR4 = 4 Ω
IR1 = A
2.Series & parallel resistorsHard
R1​ = 5 ΩR3​ = 12 ΩR5​ = 2 Ω+−V1​24 VR2​12 ΩIR2​AR4​3 ΩBR6​20 Ω

Find the current IR2 through R2.

V1 = 24 VR1 = 5 ΩR2 = 12 ΩR3 = 12 ΩR4 = 3 ΩR5 = 2 ΩR6 = 20 Ω
IR2 = A
3.Nodal analysisMedium
R1​ = 10 ΩIR1​R3​ = 3 Ω+−V1​30 VR2​20 ΩAR4​12 ΩBIs​5 A

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

V1 = 30 VIs = 5 AR1 = 10 ΩR2 = 20 ΩR3 = 3 ΩR4 = 12 Ω
IR1 = A
4.Nodal analysisHard
R1​ = 10 ΩIR1​R2​ = 12 ΩR4​ = 2 Ω+−V1​5 VR3​12 ΩAR5​6 ΩBIs​6 A+−V2​36 V

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

V1 = 5 VIs = 6 AV2 = 36 VR1 = 10 ΩR2 = 12 ΩR3 = 12 ΩR4 = 2 ΩR5 = 6 Ω
IR1 = A
5.Mesh analysisMedium
R1​ = 6 ΩR2​ = 15 Ω+−V1​10 VR3​3 Ω+−VR3​A+−V2​15 VI1​I2​

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

V1 = 10 VV2 = 15 VR1 = 6 ΩR2 = 15 ΩR3 = 3 Ω
VR3 = V
6.Mesh analysisHard
R1​ = 5 Ω+−VR1​R2​ = 3 ΩR3​ = 4 Ω+−V1​15 VR4​3 ΩAR5​12 ΩB+−V2​12 VI1​I2​I3​

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

V1 = 15 VV2 = 12 VR1 = 5 ΩR2 = 3 ΩR3 = 4 ΩR4 = 3 ΩR5 = 12 Ω
VR1 = V
7.SuperpositionMedium
R1​ = 8 ΩR3​ = 3 Ω+−V1​12 VR2​4 ΩAR4​5 Ω+−VR4​BIs​2.5 A

Use superposition to find the voltage VR4 across R4 with the polarity shown.

V1 = 12 VIs = 2.5 AR1 = 8 ΩR2 = 4 ΩR3 = 3 ΩR4 = 5 Ω
VR4 = V
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 = 20.5 mA. With V1 doubled and the other two unchanged, I = 28 mA. With the direction of Is reversed and the other two unchanged, I = 6.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 20.5 mA2·V1: I = 28 mA−Is: I = 6.5 mA
I = mA
9.Thévenin & NortonMedium
R1​ = 3 ΩR3​ = 5 Ω+−V1​30 VR2​20 ΩAab

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

V1 = 30 VR1 = 3 ΩR2 = 20 ΩR3 = 5 Ω
Rth = Ω
10.Thévenin & NortonHard
R1​ = 30 ΩR3​ = 2 Ω+−V1​18 VR2​20 ΩAIs​5 ABRL​25 ΩIL​

A load RL = 25 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.

V1 = 18 VIs = 5 AR1 = 30 ΩR2 = 20 ΩR3 = 2 ΩRL = 25 Ω
PL = W
11.RC & RL transientsMedium
+−Vs​12 Vt = 0R = 5 ΩL22 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 5 ms, in mJ.

Vs = 12 VR = 5 ΩL = 22 mHt = 5 ms
W = mJ
12.RC & RL transientsHard
+−Vs​12 Vt = 0R = ?C1 µF+−vC​

A timer starts charging an uncharged capacitor through a resistor from a 12 V supply at t = 0. It must trigger when the capacitor voltage reaches 9 V, exactly 250 ms after the start. With C = 1 µF, find the resistance R needed.

Vs = 12 VVth = 9 VT = 250 msC = 1 µF
R = kΩ
13.AC impedance & phasorsMedium
230 V50 HzR = 25 ΩL = 50 mHI

A resistor and an inductor are connected in parallel across a 230 V (rms), 50 Hz supply. Find the magnitude of the impedance |Z|.

V = 230 V (rms)f = 50 HzR = 25 ΩL = 50 mH
|Z| = Ω
14.AC impedance & phasorsHard
VfR = 10 ΩL = 50 mHC = 2.2 µF

A series RLC circuit has R = 10 Ω, L = 50 mH and C = 2.2 µF. Find its quality factor Q at resonance.

R = 10 ΩL = 50 mHC = 2.2 µF
Q =
15.AC power & power factorMedium
P = 8 kWQSφ

A single-phase motor takes 8 kW at a power factor of 0.7 lagging from a 400 V supply. Find the line current I.

P = 8 kWpf = 0.7V = 400 V
I = A
16.AC power & power factorHard
P = 30 kWQ1​S1​φ1​QC​

A single-phase load takes 30 kW at a power factor of 0.75 lagging from a 240 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.

P = 30 kWpf1 = 0.75pf2 = 0.98V = 240 Vf = 50 Hz
Inew = A
17.Three-phase powerMedium
ABCVL​ = 400 VIL​ = 50 Acos φ = 0.85lagging

A balanced three-phase load draws a line current of 50 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.

VL = 400 VIL = 50 Acos φ = 0.85
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 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 250 kWpf1 = 0.7pf2 = 0.98VL = 415 Vf = 50 Hz
Cph = µF
19.TransformersMedium
7200 V240 VIHV​ILV​ = ?250 kVA

A single-phase 250 kVA, 7200/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 250 kVAV1 = 7200 VV2 = 240 V
ILV = A
20.TransformersHard
pf 0.95500 kVA, η = 98%

A single-phase 500 kVA transformer has a core loss of 2500 W. At rated load and a power factor of 0.95 lagging its efficiency is 98%. Find the full-load copper loss Pcu,fl in watts.

S = 500 kVAPcore = 2500 Wpf = 0.95η = 98%
Pcu,fl = W
21.Op-amp circuitsMedium
−+Vout​Rf​ = 47 kΩR1​ = 10 kΩVin​

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

R1 = 10 kΩRf = 47 kΩVsat = ±12 V
Vin,max = V
22.Op-amp circuitsHard
−+Vout​Rf​ = 100 kΩV1​ = 1.7 VR1​ = 47 kΩV2​ = 0.5 VR2​ = 22 kΩIf​

The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the current If through the feedback resistor, flowing towards the output.

V1 = 1.7 VV2 = 0.5 VR1 = 47 kΩR2 = 22 kΩRf = 100 kΩ
If = mA
23.Diodes & transistorsMedium
+−24 V330 Ω5.6 V470 ΩIZ​ = ?

A 5.6 V Zener regulator is fed from 24 V through 330 Ω and supplies a 470 Ω load. Find the Zener current IZ.

Vin = 24 VVZ = 5.6 VR = 330 ΩRL = 470 Ω
IZ = mA
24.Diodes & transistorsHard
RC​ = 330 Ωβ = 80VCC​ = 12 V3.3 VRB​ = ?

An NPN transistor with β = 80 switches a 330 Ω load resistor connected between VCC = 12 V and its collector. The emitter is grounded. The base is driven from a 3.3 V logic output through a base resistor RB. Take VBE = 0.7 V and VCE(sat) = 0.2 V. For reliable switching, design for a base current 4 times the minimum needed for saturation. Find RB in kΩ.

VCC = 12 VRC = 330 Ωβ = 80Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 Voverdrive = 4
RB = kΩ
25.Number systemsMedium
place values in two's complement1−12816403201618140211

The 8-bit pattern 1100 1101 stores a signed integer in two's complement. What is its value N in decimal?

1100 11018 bit
N =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0110 1011 XOR 1010 1110. What is the result's value R in decimal?

0110 1011XOR1010 1110
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000110302141517161121131151141809011110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(1, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(1, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15)
N =
28.Digital logic: gates, K-maps & flip-flopsHard
CLKJK1234567

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0JK
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
29.Laplace transformsMedium
ReIm−5−10−15−205

Find f(t) for F(s) = (2s + 38)/(s² + 6s + 45) by completing the square, and evaluate f(0.25).

F(s) = (2s + 38)/(s² + 6s + 45)t = 0.25 s
f(0.25) =
30.Laplace transformsHard
ReIm−0.5−10.51

A system is described by the differential equation y″ + 2y′ + 2y = 7u, 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 = 0.6 s. Find g(0.6).

y″ + 2y′ + 2y = 7ut = 0.6 s
g(0.6) =
31.Block diagramsMedium
R(s)+8(s + 2)(s + 1)(s + 4)G(s)C(s)−

A unity negative feedback loop has G(s) = 8(s + 2)/((s + 1)(s + 4)). A step of size 10 is applied at the reference. Find the steady-state error ess.

G(s) = 8(s + 2)/((s + 1)(s + 4))R = 10
ess =
32.Block diagramsHard
R(s)+20s + 5G₁+8s + 7G₂C(s)1H₁−0.25H₂−

Reduce the block diagram with G₁ = 20/(s + 5), G₂ = 8/(s + 7), an inner negative feedback H₁ = 1 around G₂ and an outer negative feedback H₂ = 0.25. Find the DC gain T(0) of C(s)/R(s).

G₁ = 20/(s + 5)G₂ = 8/(s + 7)H₁ = 1H₂ = 0.25
T(0) =
33.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 4/(s² + 2s + 4) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 4/(s² + 2s + 4)
ts = s
34.Step responseHard
t (s)y(t)1tp​ = 0.2 s8 %

The step response of a second-order system without zeros shows 8 % overshoot, and the first peak occurs at t = 0.2 s. Find the natural frequency ωn.

%OS = 8 %tp = 0.2 s
ωn = rad/s
35.Bode plots & marginsMedium
|H| (dB)1101001000−60−40−2002040ω (rad/s)ω = 10

Find the exact magnitude |G(jω)| of G(s) = 4080/((s + 16)(s + 17)) at ω = 10 rad/s, in dB.

G(s) = 4080/((s + 16)(s + 17))ω = 10 rad/s
|G(jω)| = dB
36.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 82.5/(s(s + 5)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 82.5/(s(s + 5))
ωc = rad/s
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 3)(s + 4)(s + 6)C(s)−

A unity-feedback loop has L(s) = K/((s + 3)(s + 4)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 3)(s + 4)(s + 6))
Kmax =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6−7−81j1j2

The root locus of L(s) = K(s + 7)/(s(s + 6)) leaves the real axis between s = 0 and s = −6, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.

L(s) = K(s + 7)/(s(s + 6))
sb =
39.Steady-state error & PID controlMedium
r(t) = 5t+Ks(s + 2)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 2)) must follow the ramp r(t) = 5t with a steady-state error of 0.05. Find the gain K.

G(s) = K/(s(s + 2))r(t) = 5tess = 0.05
K =
40.Steady-state error & PID controlHard
r(t) = 1t+5 + 10/s20(s + 1)(s + 5)C(s)−

A PI controller C(s) = 5 + 10/s drives the plant G(s) = 20/((s + 1)(s + 5)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 1t.

C(s) = 5 + 10/sG(s) = 20/((s + 1)(s + 5))r(t) = 1t
ess =

3Solutions

1.Series & parallel resistors

Answer: IR1 = 1.18 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​ = 2 ΩR3​ = 15 Ω+−V1​9 VR2​8 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 = 15 + 4 = 19 Ω
R2 ∥ 19 = 8·19/(8+19) = 5.63 Ω
Req = R1 + 5.63 = 2 + 5.63 = 7.63 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 = 9/7.63 = 1.18 A

Answer

IR1 = 1.18 A
2.Series & parallel resistors

Answer: IR2 = 1.138 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​ = 12 ΩR5​ = 2 Ω+−V1​24 VR2​12 ΩAR4​3 ΩBR6​20 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 = 2 + 20 = 22 Ω
R4 ∥ 22 = 2.64 Ω
R3 + 2.64 = 14.64 Ω
R2 ∥ 14.64 = 6.595 Ω
Req = R1 + 6.595 = 5 + 6.595 = 11.59 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 = 24/11.59 = 2.07 A

Current through R₂

Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.

VA = V1 − IR1·R1 = 24 − 2.07·5 = 13.65 V
IR2 = VA/R2 = 13.65/12 = 1.138 A

Answer

IR2 = 1.138 A
3.Nodal analysis

Answer: IR1 = −0.2308 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​ = 10 ΩR3​ = 3 Ω+−V1​30 VR2​20 ΩAR4​12 ΩBIs​5 A

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 30 V). Unknowns: VA, VB.

KCL at node A

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

VA/R2 + (VA − V1)/R1 + (VA − VB)/R3 = 0
VA/20 + (VA − 30)/10 + (VA − VB)/3 = 0
0.4833·VA − 0.3333·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/12 + (VB − VA)/3 = 5
−0.3333·VA + 0.4167·VB = 5

Solve

Δ = (0.4833)(0.4167) − (−0.3333)(−0.3333) = 0.09028
VA = Δ1/Δ = 2.917/0.09028 = 32.31 V
VB = Δ2/Δ = 3.417/0.09028 = 37.85 V

Answer

IR1 = (V1 − VA)/R1 = (30 − 32.31)/10 = −0.2308 A
4.Nodal analysis

Answer: IR1 = −0.7302 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​ = 10 ΩR2​ = 12 ΩR4​ = 2 Ω+−V1​5 VR3​12 ΩAR5​6 ΩBIs​6 A+−V2​36 V

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 5 V, V2 = 36 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/12 + (VA − 5)/10 + (VA − VB)/12 = 0
0.2667·VA − 0.08333·VB = 0.5

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/6 + (VB − VA)/12 + (VB − 36)/2 = 6
−0.08333·VA + 0.75·VB = 24

Solve

Δ = (0.2667)(0.75) − (−0.08333)(−0.08333) = 0.1931
VA = Δ1/Δ = 2.375/0.1931 = 12.3 V
VB = Δ2/Δ = 6.442/0.1931 = 33.37 V

Answer

IR1 = (V1 − VA)/R1 = (5 − 12.3)/10 = −0.7302 A
5.Mesh analysis

Answer: VR3 = 4.706 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​ = 6 ΩR2​ = 15 Ω+−V1​10 VR3​3 ΩA+−V2​15 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
6·I1 + 3(I1 − I2) = 10
9·I1 − 3·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
15·I2 + 3(I2 − I1) = (−15)
−3·I1 + 18·I2 = −15

Solve

Δ = (9)(18) − (−3)(−3) = 153
I1 = Δ1/Δ = 135/153 = 0.8824 A
I2 = Δ2/Δ = −105/153 = −0.6863 A

Answer

IR3 = I1 − I2 = 1.569 A
VR3 = IR3·R3 = 1.569·3 = 4.706 V

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: VR1 = 8.571 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​ = 5 ΩR2​ = 3 ΩR3​ = 4 Ω+−V1​15 VR4​3 ΩAR5​12 ΩB+−V2​12 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 + 3(I1 − I2) = 15
8·I1 − 3·I2 = 15

KVL around mesh 2

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

R2·I2 + R4(I2 − I1) + R5(I2 − I3) = 0
3·I2 + 3(I2 − I1) + 12(I2 − I3) = 0
−3·I1 + 18·I2 − 12·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
4·I3 + 12(I3 − I2) = (−12)
−12·I2 + 16·I3 = −12

Solve

Δ = det of the coefficient matrix = 1008
I1 = Δ1/Δ = 1728/1008 = 1.714 A
I2 = Δ2/Δ = −432/1008 = −0.4286 A
I3 = Δ3/Δ = −1080/1008 = −1.071 A

Answer

IR1 = I1 = 1.714 A
VR1 = IR1·R1 = 1.714·5 = 8.571 V

A negative mesh current flows anticlockwise.

7.Superposition

Answer: VR4 = 8.516 V

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

R1​ = 8 ΩR3​ = 3 Ω+−V1​12 VR2​4 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 = 3+5 = 8 Ω
R2 ∥ (R3+R4) = 4·8/(4+8) = 2.667 Ω
VA = 12·2.667/(8+2.667) = 3 V
VB = VA·R4/(R3+R4) = 3·5/8 = 1.875 V
VR4′ = VB − 0 = 1.875 − 0 = 1.875 V
R1​ = 8 ΩR3​ = 3 ΩSHORTR2​4 ΩAR4​5 ΩBIs​2.5 A

Case 2 · Is acting alone

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

R1 ∥ R2 = 8·4/(8+4) = 2.667 Ω
Req = R4 ∥ (R3+R1∥R2) = 5 ∥ 5.667 = 2.656 Ω
VB = Is·Req = 2.5·2.656 = 6.641 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 6.641·2.667/5.667 = 3.125 V
VR4″ = VB − 0 = 6.641 − 0 = 6.641 V

Add the contributions

VR4 = VR4′ + VR4″
VR4 = 1.875 + 6.641 = 8.516 V

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

8.Superposition

Answer: I = 14.5 mA

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

Write the measurements with contributions

Each contribution is proportional to its own source.

I₁ + I₂ + I₃ = 20.5
2·I₁ + I₂ + I₃ = 28
I₁ + I₂ − I₃ = 6.5

Solve for the contributions

I₁ = 28 − 20.5 = 7.5 mA
I₃ = (20.5 − 6.5)/2 = 7 mA
I₂ = 20.5 − 7.5 − 7 = 6 mA

Switch V₂ off

I = I₁ + I₃ = 7.5 + 7 = 14.5 mA

Answer

I = 14.5 mA
9.Thévenin & Norton

Answer: Rth = 7.609 Ω

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

R1​ = 3 ΩR3​ = 5 ΩSHORTR2​20 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 = 3·20/(3+20) = 2.609 Ω
Rth = (R1 ∥ R2) + R3 = 2.609 + 5 = 7.609 Ω
10.Thévenin & Norton

Answer: PL = 97.96 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​ = 30 ΩR3​ = 2 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 = 30·20/(30+20) = 12 Ω
Rth = (R1 ∥ R2) + R3 = 12 + 2 = 14 Ω
R1​ = 30 ΩR3​ = 2 Ω+−V1​18 VR2​20 ΩAIs​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) = 7.2 + 5·12 = 67.2 V
Vth = VB = VA + Is·R3 = 67.2 + 5·2 = 77.2 V
Rth​ = 14 Ω+−Vth​77.2 VRL​25 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) = 77.2/(14 + 25) = 1.979 A
PL = IL²·RL = 1.979²·25 = 97.96 W
11.RC & RL transients

Answer: W = 29.21 mJ

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

Time constant and final current

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

τ = L/R = 22 mH / 5 Ω = 4.4 ms
i(∞) = Vs/R = 12/5 = 2.4 A

Current at the given time

i(t) = 2.4·(1 − e−5/4.4) = 1.63 A

Stored energy

W = ½·L·i² = ½·22·1.63² = 29.21 mJ

Answer

W = 29.21 mJ
12.RC & RL transients

Answer: R = 180.3 kΩ

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.

Required time constant

9 = 12·(1 − e−250/τ) → τ = 250/ln(12/(12 − 9))
τ = 250/1.386 = 180.3 ms

Resistance

kΩ times µF gives milliseconds.

R = τ/C = 180.3/1 = 180.3 kΩ

Answer

R = 180.3 kΩ
13.AC impedance & phasors

Answer: |Z| = 13.3 Ω

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

Reactance

XL = ωL = 2π·50·0.05 = 15.71 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

IR = V/R = 230/25 = 9.2 A
IX = V/XL = 230/15.71 = 14.64 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(IR² + IX²) = √(9.2² + 14.64²) = 17.29 A

Impedance

|Z| = V/I = 230/17.29 = 13.3 Ω

Answer

|Z| = 13.3 Ω
14.AC impedance & phasors

Answer: Q = 15.08

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·√(0.05·2.2·10⁻⁶)) = 479.9 Hz

Quality factor

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

Q = (1/R)·√(L/C) = (1/10)·√(0.05/(2.2·10⁻⁶)) = 15.08

Answer

Q = 15.08
15.AC power & power factor

Answer: I = 28.57 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) = 8000/(400·0.7) = 28.57 A

Answer

I = 28.57 A
16.AC power & power factor

Answer: Inew = 127.6 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 = 30·tan(41.41°) = 26.46 kvar
Q2 = P·tan φ2 = 30·tan(11.48°) = 6.092 kvar
QC = Q1 − Q2 = 20.37 kvar

New line current

Inew = P/(V·cos φ2) = 30000/(240·0.98) = 127.6 A

Answer

Inew = 127.6 A
17.Three-phase power

Answer: P = 29.44 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 = √3·400·50 = 34.64 kVA

Real power

P = S·cos φ = 34.64·0.85 = 29.44 kW

Answer

P = 29.44 kW
18.Three-phase power

Answer: Cph = 3776 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 0.98) = 250·0.2031 = 50.76 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 204.3 kvar

Capacitance per phase

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

V = 415/√3 = 239.6 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 68095/(314.2·239.6²) = 3776 µF

Answer

Cph = 3776 µF
19.Transformers

Answer: ILV = 1042 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.

Rated current

Both windings carry the same apparent power.

ILV = S/V = 250000/240 = 1042 A

Answer

ILV = 1042 A
20.Transformers

Answer: Pcu,fl = 7194 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 = 500·0.95 = 475 kW
Pin = 475000/0.98 = 484694 W

Total and copper losses

Ploss = 484694 − 475000 = 9694 W
Pcu,fl = 9694 − 2500 = 7194 W

Answer

Pcu,fl = 7194 W
21.Op-amp circuits

Answer: Vin,max = 2.105 V

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

Gain

Av = 1 + Rf/R1 = 1 + 47/10 = 5.7

Largest input before clipping

Vin,max = Vsat/|Av| = 12/5.7 = 2.105 V

Answer

Vin,max = 2.105 V
22.Op-amp circuits

Answer: If = 0.0589 mA

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 = 1.7/47 = 0.03617 mA
I2 = V2/R2 = 0.5/22 = 0.02273 mA
If = I1 + I2 = 0.0589 mA

Answer

If = 0.0589 mA
23.Diodes & transistors

Answer: IZ = 43.84 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 = 18.4/330 = 55.76 mA

Load current

IL = 5.6/470 = 11.91 mA

Zener current

IZ = 55.76 − 11.91 = 43.84 mA

Answer

IZ = 43.84 mA
24.Diodes & transistors

Answer: RB = 1.454 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)/330 = 35.76 mA

Minimum base current

IB(min) = IC(sat)/β = 35.76/80 = 447 µA

Design base current

IB = 4·447 µA = 1788 µA

Base resistor

RB = (3.3 − 0.7)/IB = 2.6 V/1.788 mA = 1.454 kΩ

Answer

RB = 1.454 kΩ
25.Number systems

Answer: N = −51

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

Add the place values

The other bits keep their normal positive values.

−128 + 64 + 8 + 4 + 1 = −51

Answer

N = −51
26.Number systems

Answer: R = 197

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

Bit by bit

XOR gives 1 where the two bits differ.

0110 1011
XOR 1010 1110
= 1100 0101

Value of the result

1100 0101₂ = 197

Answer

R = 197
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.

ABCD000111100001111000110302141517161121131151141809011110

Group the 0s

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

(B + C′ + D′)
(A + B + D)
(A′ + B + D′)

Minimal POS form

F = (B + C′ + D′)(A + B + D)(A′ + B + D′)
N = 3

Answer

N = 3

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

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

Answer: clock edge 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 = 0, K = 1 → Q = 0
4: J = 0, K = 0 → Q = 0
5: J = 1, K = 0 → Q = 1
6: J = 0, K = 1 → Q = 0
7: J = 0, K = 1 → Q = 0

First change from 1 to 0

clock edge 3

Answer

clock edge 3
29.Laplace transforms

Answer: f(0.25) = 2.58

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

Complete the square

s² + 6s + 45 = (s + 3)² + 36 = (s + 3)² + 6²

Match the numerator

F(s) = 2·(s + 3)/((s + 3)² + 6²) + (32/6)·6/((s + 3)² + 6²)

Inverse transform

f(t) = e−3t·(2·cos 6t + 5.333·sin 6t)
f(0.25) = e−3·0.25·(2·cos(6·0.25) + 5.333·sin(6·0.25)) = 2.58

Answer

f(0.25) = 2.58
30.Laplace transforms

Answer: g(0.6) = 2.169

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) = 7/(s² + 2s + 2)

Partial fractions

s² + 2s + 2 = (s + 1)² + 1²
G(s) = (7/1)·1/((s + 1)² + 1²)

Impulse response

g(t) = e−t·(7·sin 1t)
g(0.6) = 2.169

Answer

g(0.6) = 2.169
31.Block diagrams

Answer: ess = 2

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

Position error constant

Kₚ = G(0) = 8·2/(1·4) = 4

Steady-state error

ess = A/(1 + Kₚ) = 10/(1 + 4) = 2

Answer

ess = 2
32.Block diagrams

Answer: T(0) = 1.391

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) = 8/(s + 15)

Outer loop

T(s) = G₁Gᵢ/(1 + G₁GᵢH₂) = 160/((s + 5)(s + 15) + 40)

DC gain

T(0) = 160/(5·15 + 40) = 160/115 = 1.391

Answer

T(0) = 1.391
33.Step response

Answer: ts = 4 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 = 4 → ωₙ = 2 rad/s
2ζωₙ = 2 → ζ = 2/(2·2) = 0.5

Result

ζωₙ = 1 → ts ≈ 4/1 = 4 s

Answer

ts = 4 s
34.Step response

Answer: ωn = 20.15 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(0.08)/√(π² + ln²(0.08)) = 2.526/√(9.87 + 6.379) = 0.6266

Natural frequency

ωd = π/tp = π/0.2 = 15.71 rad/s
ωₙ = ωd/√(1 − ζ²) = 15.71/0.7794 = 20.15 rad/s

Answer

ωn = 20.15 rad/s
35.Bode plots & margins

Answer: |G(jω)| = 20.8 dB

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

Multiply the magnitudes

|G(j10)| = 4080/(√(10² + 16²)·√(10² + 17²)) = 4080/(18.87·19.72) = 10.96

Result

20·log₁₀(10.96) = 20.8 dB

Answer

|G(jω)| = 20.8 dB
36.Bode plots & margins

Answer: ωc = 8.423 rad/s

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

Gain crossover

ω⁴ + 25ω² − 6806 = 0
ω² = (−25 + √(625 + 4·6806))/2 = 70.94
ωc = 8.423 rad/s

Answer

ωc = 8.423 rad/s
37.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 630

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 + 13s² + 54s + 72 + K = 0

Routh array

s³: 1, 54
s²: 13, 72 + K
s¹: (13·54 − 72 − K)/13
s⁰: 72 + K

Condition

The s¹ entry must stay positive.

K < 13·54 − 72 = 630

Answer

Kmax = 630

At the limit the loop oscillates at ω = √54 = 7.348 rad/s.

38.Stability: Routh–Hurwitz & root locus

Answer: sb = −4.354

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

Gain along the real axis

K(s) = −(s² + 6s)/(s + 7)

Stationary points

Differentiate with the quotient rule and set the numerator to zero.

(2s + 6)(s + 7) − (s² + 6s) = s² + 14s + 42 = 0
s = −7 ± √7 = −4.354, −9.646

Break-away point

sb = −4.354

Answer

sb = −4.354

Between the two break points the locus is a circle centred on the zero, with radius √7 = 2.646.

39.Steady-state error & PID control

Answer: K = 200

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

Required velocity constant

Kv = A/ess = 5/0.05 = 100

Gain

K = 2·Kv = 200

Answer

K = 200

A large gain makes this loop very lightly damped: ζ = 0.07071. That conflict is why compensators are used.

40.Steady-state error & PID control

Answer: ess = 0.025

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) = 5s + 10 → 10 as s → 0.

Kv = 10·20/5 = 40

Ramp error

ess = 1/40 = 0.025

Answer

ess = 0.025

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

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