FE Electrical and Computer exam practice: cram pack

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

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1Formulas and methods

Series & parallel resistors

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

Method

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

Nodal analysis

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

Method

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

Mesh analysis

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

Method

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

Superposition

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

Method

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

Thévenin & Norton

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

Method

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

RC & RL transients

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

Method

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

AC impedance & phasors

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

Method

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

AC power & power factor

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

Three-phase power

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

Transformers

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

Op-amp circuits

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

Diodes & transistors

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

Number systems

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

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

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

Method

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

Stability: Routh–Hurwitz & root locus

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

Method

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

Steady-state error & PID control

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

Method

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

2Practice problems

1.Series & parallel resistorsMedium
R1​ = 10 ΩR3​ = 8 Ω+−V1​15 VR2​5 ΩAR4​10 ΩB

Find the equivalent resistance Req seen by the source V1.

V1 = 15 VR1 = 10 ΩR2 = 5 ΩR3 = 8 ΩR4 = 10 Ω
Req = Ω
2.Series & parallel resistorsHard
R1​ = 20 ΩR3​ = 4 ΩR5​ = 10 Ω+−V1​5 VR2​10 ΩAR4​2 ΩBR6​30 Ω

Find the equivalent resistance Req seen by the source V1.

V1 = 5 VR1 = 20 ΩR2 = 10 ΩR3 = 4 ΩR4 = 2 ΩR5 = 10 ΩR6 = 30 Ω
Req = Ω
3.Nodal analysisMedium
R1​ = 12 ΩR3​ = 5 Ω+−V1​30 VR2​20 ΩAR4​20 ΩBIs​3 A

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

V1 = 30 VIs = 3 AR1 = 12 ΩR2 = 20 ΩR3 = 5 ΩR4 = 20 Ω
PIs = W
4.Nodal analysisHard
R1​ = 12 ΩR3​ = 5 Ω+−V1​6 VR2​30 ΩAR4​10 ΩBIs​?

The current source is adjusted until the node voltage at A is VA = 7 V. Use nodal analysis to find the source current Is (direction as shown).

V1 = 6 VR1 = 12 ΩR2 = 30 ΩR3 = 5 ΩR4 = 10 ΩVA = 7 V
Is = A
5.Mesh analysisMedium
R1​ = 5 ΩR2​ = 3 ΩIR2​+−V1​15 VR3​6 ΩA+−V2​24 VI1​I2​

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

V1 = 15 VV2 = 24 VR1 = 5 ΩR2 = 3 ΩR3 = 6 Ω
IR2 = A
6.Mesh analysisHard
R1​ = 20 ΩR2​ = 12 ΩR3​ = 2 Ω+−V1​5 VR4​5 ΩAR5​8 ΩB+−V2​9 VI1​I2​I3​

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

V1 = 5 VV2 = 9 VR1 = 20 ΩR2 = 12 ΩR3 = 2 ΩR4 = 5 ΩR5 = 8 Ω
I1 = A
7.SuperpositionMedium
R1​ = 10 ΩR3​ = 5 ΩIR3​+−V1​6 VR2​3 ΩAR4​5 ΩBIs​?

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

V1 = 6 VR1 = 10 ΩR2 = 3 ΩR3 = 5 ΩR4 = 5 Ω
Is = 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 = 5.5 mA. With V1 doubled and the other two unchanged, I = 0.5 mA. With the direction of Is reversed and the other two unchanged, I = −12.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 5.5 mA2·V1: I = 0.5 mA−Is: I = −12.5 mA
I = mA
9.Thévenin & NortonMedium
R1​ = 5 ΩR3​ = 2 Ω+−V1​18 VR2​2 ΩAab

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

V1 = 18 VR1 = 5 ΩR2 = 2 ΩR3 = 2 Ω
IN = A
10.Thévenin & NortonHard
R1​ = 10 ΩR3​ = 6 Ω+−V1​9 VR2​6 ΩAIs​1 ABRL​2 ΩIL​

A load RL = 2 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 9 VIs = 1 AR1 = 10 ΩR2 = 6 ΩR3 = 6 ΩRL = 2 Ω
IL = A
11.RC & RL transientsMedium
+−Vs​15 Vt = 0R = 20 ΩL47 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the voltage vL across the inductor at t = 5 ms.

Vs = 15 VR = 20 ΩL = 47 mHt = 5 ms
vL = V
12.RC & RL transientsHard
+−Vs​5 Vt = 0R = ?C22 µF+−vC​

A timer starts charging an uncharged capacitor through a resistor from a 5 V supply at t = 0. It must trigger when the capacitor voltage reaches 4 V, exactly 250 ms after the start. With C = 22 µF, find the resistance R needed.

Vs = 5 VVth = 4 VT = 250 msC = 22 µF
R = kΩ
13.AC impedance & phasorsMedium
100 V1000 HzR = 50 ΩL = 50 mHC = 22 µF

A series RLC circuit is connected to a 100 V (rms), 1000 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 100 V (rms)f = 1000 HzR = 50 ΩL = 50 mHC = 22 µF
φ = °
14.AC impedance & phasorsHard
VfR = 50 ΩL = 5 mHC = 0.1 µF

A series RLC circuit has R = 50 Ω, L = 5 mH and C = 0.1 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.

R = 50 ΩL = 5 mHC = 0.1 µF
B = Hz
15.AC power & power factorMedium
400 V118 kW, 0.6220 kW, 0.85I = ?

Two loads are connected in parallel to a 400 V single-phase supply: an 18 kW load at a power factor of 0.6 lagging and a 20 kW load at a power factor of 0.85 leading. Find the total current I drawn from the supply.

V = 400 VP1 = 18 kW, pf1 = 0.6 laggingP2 = 20 kW, pf2 = 0.85 leading
I = A
16.AC power & power factorHard
P = 20 kWQ1​S1​φ1​QC​

A single-phase load takes 20 kW at a power factor of 0.65 lagging from a 240 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the capacitance C needed.

P = 20 kWpf1 = 0.65pf2 = 0.98V = 240 Vf = 50 Hz
C = µF
17.Three-phase powerMedium
ABCVL​ = 208 VZ = 16 + j12 Ωper phase

A balanced Δ-connected load with an impedance of 16 + j12 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.

VL = 208 VZ = 16 + j12 Ω per phase
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 0.78 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 0.98 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 200 kWpf1 = 0.78pf2 = 0.98VL = 400 Vf = 60 Hz
QC = kvar
19.TransformersMedium
2 ΩN1​N2​128 Ω

An amplifier with an output resistance of 128 Ω drives a 2 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 128 Ω from the amplifier. Find the turns ratio a = N₁/N₂.

Rs = 128 ΩRL = 2 Ω
a =
20.TransformersHard
50 kVA

A single-phase 50 kVA transformer has a core loss of 250 W and a copper loss of 1000 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 50 kVAPcore = 250 WPcu,fl = 1000 W
Sη,max = kVA
21.Op-amp circuitsMedium
−+Vout​Rf​ = 47 kΩVin​R1​ = 4.7 kΩ

The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±12 V. The 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 = 4.7 kΩRf = 47 kΩVsat = ±12 V
Vin,max = V
22.Op-amp circuitsHard
−+Vout​Rf​ = 4.7 kΩV1​ = 1.9 VR1​ = 47 kΩV2​ = 1 VR2​ = 10 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 = 1.9 VV2 = 1 VR1 = 47 kΩR2 = 10 kΩRf = 4.7 kΩ
If = mA
23.Diodes & transistorsMedium
RC​ = 5.6 kΩβ = 100RE​ = 1.2 kΩVCC​ = 15 VR1​ = 82 kΩR2​ = 15 kΩ

An NPN transistor with β = 100 has voltage-divider bias from VCC = 15 V: R₁ = 82 kΩ from VCC to the base and R₂ = 15 kΩ from the base to ground. The collector resistor is RC = 5.6 kΩ and the emitter resistor is RE = 1.2 kΩ. Take VBE = 0.7 V and use the Thévenin equivalent of the divider (do not neglect the base current). Find the collector–emitter voltage VCE.

VCC = 15 VR1 = 82 kΩR2 = 15 kΩRC = 5.6 kΩRE = 1.2 kΩβ = 100
VCE = V
24.Diodes & transistorsHard
+−≤ 24 V100 Ω12 V

A 12 V Zener regulator has a 100 Ω series resistor. The input can rise to 24 V, and the load may be disconnected. Find the largest power PZ,max the Zener must dissipate.

Vin,max = 24 VVZ = 12 VR = 100 Ω
PZ,max = mW
25.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −48 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −488 bit
U =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0101 1010 >> 2. What is the result's value R in decimal?

0101 1010>> 2
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011001031204051716

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 2, 6, 7). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(0, 2, 6, 7)
L =
28.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKMOD-7

A synchronous mod-7 counter counts 0, 1, …, 6 and then returns to 0. It starts at 6. Which count Q (in decimal) does it show after 25 clock pulses?

Q0 = 625N = 7
Q =
29.Laplace transformsMedium
t (s)f(t)t = 0.2f(0.2) = ?

Find f(t) for F(s) = (4s + 14)/(s² + 8s + 16) and evaluate f(0.2).

F(s) = (4s + 14)/(s² + 8s + 16)t = 0.2 s
f(0.2) =
30.Laplace transformsHard
t (s)y(t)t = 0.1y(0.1) = ?

Solve y″ + 10y′ + 24y = 0 with y(0) = −4 and y′(0) = −15 using the Laplace transform, and find y(0.1).

y″ + 10y′ + 24y = 0y(0) = −4y′(0) = −15t = 0.1 s
y(0.1) =
31.Block diagramsMedium
R(s)+26s + 6G(s)C(s)0.25H−

A plant G(s) = 26/(s + 6) is controlled in a negative feedback loop whose sensor has the gain H = 0.25. A step of size 1 is applied at R. Find the steady-state output css.

G(s) = 26/(s + 6)H = 0.25R = 1
css =
32.Block diagramsHard
R(s)+14controller+D(s)+1s + 7plantC(s)−

A proportional controller K = 14 drives the plant G(s) = 1/(s + 7) 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 2 and a disturbance step of size 0.5 are applied together. Find the steady-state output css.

K = 14G(s) = 1/(s + 7)D = 0.5R = 2
css =
33.Step responseMedium
t (s)y(t)3?

A system has the transfer function G(s) = 300/(s² + 12s + 100) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 300/(s² + 12s + 100)
%OS = %
34.Step responseHard
t (s)y(t)140 %

A second-order system without zeros must have a step response with 40 % overshoot. Find the damping ratio ζ.

%OS = 40 %
ζ =
35.Bode plots & marginsMedium
vin​C = 100 nFR = 3.3 kΩvout​

The RC high-pass filter shown has R = 3.3 kΩ, C = 100 nF. A sinusoidal input of 960 Hz is applied. Find the gain |H| = |vout/vin| in dB.

R = 3.3 kΩC = 100 nFf = 960 Hz
|H| = dB
36.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​

A unity-feedback loop has the open-loop transfer function L(s) = 8.8/(s(s + 4)). Find the phase margin PM in degrees.

L(s) = 8.8/(s(s + 4))
PM = °
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 1)(s + 5)(s + 6)C(s)−

A unity-feedback loop has L(s) = K/((s + 1)(s + 5)(s + 6)), with K > 0. Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/((s + 1)(s + 5)(s + 6))
Kmax =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−1−2−3−4−5−6j1

The root locus of L(s) = K/(s(s + 1)(s + 5)) leaves the real axis between s = 0 and s = −1. Find the break-away point sb.

L(s) = K/(s(s + 1)(s + 5))
sb =
39.Steady-state error & PID controlMedium
r(t) = 1t+50s(s + 2)(s + 5)C(s)−

A ramp r(t) = 1t is applied to a unity-feedback loop with G(s) = 50/(s(s + 2)(s + 5)). Find the steady-state error ess.

G(s) = 50/(s(s + 2)(s + 5))r(t) = 1t
ess =
40.Steady-state error & PID controlHard
R(s)+2 + Kd​s1s(s + 1)C(s)−

A PD controller C(s) = 2 + Kd·s drives the plant G(s) = 1/(s(s + 1)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.707.

G(s) = 1/(s(s + 1))Kp = 2ζ = 0.707
Kd = s

3Solutions

1.Series & parallel resistors

Answer: Req = 13.91 Ω

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

R1​ = 10 ΩR3​ = 8 Ω+−V1​15 VR2​5 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 = 8 + 10 = 18 Ω
R2 ∥ 18 = 5·18/(5+18) = 3.913 Ω
Req = R1 + 3.913 = 10 + 3.913 = 13.91 Ω

Answer

Req = 13.91 Ω
2.Series & parallel resistors

Answer: Req = 23.71 Ω

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

R1​ = 20 ΩR3​ = 4 ΩR5​ = 10 Ω+−V1​5 VR2​10 ΩAR4​2 ΩBR6​30 Ω

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R5 + R6 = 10 + 30 = 40 Ω
R4 ∥ 40 = 1.905 Ω
R3 + 1.905 = 5.905 Ω
R2 ∥ 5.905 = 3.713 Ω
Req = R1 + 3.713 = 20 + 3.713 = 23.71 Ω

Answer

Req = 23.71 Ω
3.Nodal analysis

Answer: PIs = 103.8 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/20 + (VA − 30)/12 + (VA − VB)/5 = 0
0.3333·VA − 0.2·VB = 2.5

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/20 + (VB − VA)/5 = 3
−0.2·VA + 0.25·VB = 3

Solve

Δ = (0.3333)(0.25) − (−0.2)(−0.2) = 0.04333
VA = Δ1/Δ = 1.225/0.04333 = 28.27 V
VB = Δ2/Δ = 1.5/0.04333 = 34.62 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 = 34.62·3 = 103.8 W

Answer

PIs = 103.8 W
4.Nodal analysis

Answer: Is = 1.175 A

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

KCL at node A

VA is known, so this equation gives VB.

(VA − V1)/R1 + VA/R2 + (VA − VB)/R3 = 0
(7 − 6)/12 + 7/30 = (VB − 7)/5
0.3167 = (VB − 7)/5 → VB = 8.583 V

KCL at node B

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

Is = (VB − VA)/R3 + VB/R4 = (8.583 − 7)/5 + 8.583/10 = 1.175 A

Answer

Is = 1.175 A
5.Mesh analysis

Answer: IR2 = −2.762 A

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

R1​ = 5 ΩR2​ = 3 Ω+−V1​15 VR3​6 ΩA+−V2​24 VI1​I2​

Assign mesh currents

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

KVL around mesh 1

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

R1·I1 + R3(I1 − I2) = V1
5·I1 + 6(I1 − I2) = 15
11·I1 − 6·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 + R3(I2 − I1) = −V2
3·I2 + 6(I2 − I1) = (−24)
−6·I1 + 9·I2 = −24

Solve

Δ = (11)(9) − (−6)(−6) = 63
I1 = Δ1/Δ = −9/63 = −0.1429 A
I2 = Δ2/Δ = −174/63 = −2.762 A

Answer

IR2 = I2 = −2.762 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: I1 = 0.1295 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​ = 20 ΩR2​ = 12 ΩR3​ = 2 Ω+−V1​5 VR4​5 ΩAR5​8 ΩB+−V2​9 VI1​I2​I3​

Assign mesh currents

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

KVL around mesh 1

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

R1·I1 + R4(I1 − I2) = V1
20·I1 + 5(I1 − I2) = 5
25·I1 − 5·I2 = 5

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
12·I2 + 5(I2 − I1) + 8(I2 − I3) = 0
−5·I1 + 25·I2 − 8·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
2·I3 + 8(I3 − I2) = (−9)
−8·I2 + 10·I3 = −9

Solve

Δ = det of the coefficient matrix = 4400
I1 = Δ1/Δ = 570/4400 = 0.1295 A
I2 = Δ2/Δ = −1550/4400 = −0.3523 A
I3 = Δ3/Δ = −5200/4400 = −1.182 A

Answer

I1 = 0.1295 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: Is = 0.2769 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​ = 10 ΩR3​ = 5 Ω+−V1​6 VR2​3 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 = 5+5 = 10 Ω
R2 ∥ (R3+R4) = 3·10/(3+10) = 2.308 Ω
VA = 6·2.308/(10+2.308) = 1.125 V
VB = VA·R4/(R3+R4) = 1.125·5/10 = 0.5625 V
IR3′ = (VA − VB)/R3 = (1.125 − 0.5625)/5 = 0.1125 A

Case 2 · Is acting alone

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

R1 ∥ R2 = 2.308 Ω
IR3″ = −Is·R4/(R4 + R3 + R1∥R2) = −Is·5/12.31 = −0.4063·Is

Set the total to zero

IR3′ + IR3″ = 0.1125 − 0.4063·Is = 0
Is = 0.1125/0.4063 = 0.2769 A

Answer

Is = 0.2769 A
8.Superposition

Answer: I = 4 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 = 5.5
2·I₁ + I₂ + I₃ = 0.5
I₁ + I₂ − I₃ = −12.5

Solve for the contributions

I₁ = 0.5 − 5.5 = −5 mA
I₃ = (5.5 − (−12.5))/2 = 9 mA
I₂ = 5.5 − (−5) − 9 = 1.5 mA

Switch V₂ off

I = I₁ + I₃ = (−5) + 9 = 4 mA

Answer

I = 4 mA
9.Thévenin & Norton

Answer: IN = 1.5 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​ = 5 ΩR3​ = 2 ΩSHORTR2​2 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 = 5·2/(5+2) = 1.429 Ω
Rth = (R1 ∥ R2) + R3 = 1.429 + 2 = 3.429 Ω
R1​ = 5 ΩR3​ = 2 Ω+−V1​18 VR2​2 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) = 18·2/(5+2) = 5.143 V
IN​1.5 ARth​3.429 Ωab

Norton equivalent

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

IN = Vth/Rth = 5.143/3.429 = 1.5 A
10.Thévenin & Norton

Answer: IL = −0.5426 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​ = 10 ΩR3​ = 6 ΩSHORTR2​6 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 = 10·6/(10+6) = 3.75 Ω
Rth = (R1 ∥ R2) + R3 = 3.75 + 6 = 9.75 Ω
R1​ = 10 ΩR3​ = 6 Ω+−V1​9 VR2​6 ΩAIs​1 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) = 3.375 + (−1)·3.75 = −0.375 V
Vth = VB = VA + Is·R3 = −0.375 + (−1)·6 = −6.375 V
Rth​ = 9.75 Ω−+Vth​6.375 VRL​2 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) = −6.375/(9.75 + 2) = −0.5426 A
11.RC & RL transients

Answer: vL = 1.787 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.

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 = 47 mH / 20 Ω = 2.35 ms
i(∞) = Vs/R = 15/20 = 0.75 A

Inductor voltage

At t = 0 the whole supply voltage appears across the inductor; it then decays to zero.

vL(t) = Vs·e−t/τ = 15·e−5/2.35 = 1.787 V

Answer

vL = 1.787 V
12.RC & RL transients

Answer: R = 7.061 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

4 = 5·(1 − e−250/τ) → τ = 250/ln(5/(5 − 4))
τ = 250/1.609 = 155.3 ms

Resistance

kΩ times µF gives milliseconds.

R = τ/C = 155.3/22 = 7.061 kΩ

Answer

R = 7.061 kΩ
13.AC impedance & phasors

Answer: φ = 80.75 °

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·1000 = 6283 rad/s
XL = ωL = 6283·0.05 = 314.2 Ω
XC = 1/(ωC) = 1/(6283·22·10⁻⁶) = 7.234 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(50² + (306.9)²) = 311 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(306.9/50) = 80.75°

Answer

φ = 80.75°
14.AC impedance & phasors

Answer: B = 1592 Hz

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

Resonant frequency

f0 = 1/(2π·√(0.005·0.1·10⁻⁶)) = 7118 Hz

Quality factor

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

Q = (1/R)·√(L/C) = (1/50)·√(0.005/(0.1·10⁻⁶)) = 4.472

Bandwidth

B = f0/Q = 7118/4.472 = 1592 Hz

Answer

B = 1592 Hz
15.AC power & power factor

Answer: I = 99.33 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 of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 18·tan(arccos 0.6) = 24 kvar
Q2 = −20·tan(arccos 0.85) = −12.39 kvar

Totals

P = 18 + 20 = 38 kW
Q = 24 + (−12.39) = 11.61 kvar
S = √(38² + 11.61²) = 39.73 kVA

Total current

I = S/V = 39733/400 = 99.33 A

Answer

I = 99.33 A
16.AC power & power factor

Answer: C = 1068 µF

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

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 = 20·tan(49.46°) = 23.38 kvar
Q2 = P·tan φ2 = 20·tan(11.48°) = 4.061 kvar
QC = Q1 − Q2 = 19.32 kvar

Capacitance

A capacitor across the supply delivers Q = V²·ωC.

C = QC/(2πf·V²) = 19321/(2π·50·240²) = 1068 µF

Answer

C = 1068 µF
17.Three-phase power

Answer: P = 5.192 kW

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Phase voltage

Vph = VL = 208 V

Phase current

|Z| = √(16² + 12²) = 20 Ω
Iph = 208/20 = 10.4 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·10.4²·16 = 5.192 kW

Answer

P = 5.192 kW

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

18.Three-phase power

Answer: QC = 119.8 kvar

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 0.78) = 200·0.8023 = 160.5 kvar
Q2 = 200·tan(acos 0.98) = 200·0.2031 = 40.61 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 119.8 kvar

Answer

QC = 119.8 kvar
19.Transformers

Answer: a = 8

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

Match the referred load

a²·2 = 128
a = √(128/2) = √64 = 8

Answer

a = 8
20.Transformers

Answer: Sη,max = 25 kVA

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

Copper loss equals core loss

x²·1000 = 250 → x = √(250/1000) = 0.5

Load at maximum efficiency

Sη,max = x·S = 0.5·50 = 25 kVA

Answer

Sη,max = 25 kVA
21.Op-amp circuits

Answer: Vin,max = 1.2 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| = Rf/R1 = 47/4.7 = 10

Largest input before clipping

Vin,max = Vsat/|Av| = 12/10 = 1.2 V

Answer

Vin,max = 1.2 V
22.Op-amp circuits

Answer: If = 0.1404 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 = 1.9/47 = 0.04043 mA
I2 = V2/R2 = 1/10 = 0.1 mA
If = I1 + I2 = 0.1404 mA

Answer

If = 0.1404 mA
23.Diodes & transistors

Answer: VCE = 6.759 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 = 15·15/(82 + 15) = 2.32 V
RTH = 82·15/(82 + 15) = 12.68 kΩ

Base current

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

IB = (2.32 − 0.7)/(12.68 + 101·1.2) = 12.1 µA

Collector current

IC = 100·12.1 µA = 1.21 mA
IE = 101·12.1 µA = 1.222 mA

Collector–emitter voltage

VCE = 15 − 1.21·5.6 − 1.222·1.2 = 6.759 V

Answer

VCE = 6.759 V
24.Diodes & transistors

Answer: PZ,max = 1440 mW

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.

Worst-case Zener current

IZ = 12/100 = 120 mA

Zener power

PZ = 12·120 = 1440 mW

Answer

PZ,max = 1440 mW
25.Number systems

Answer: U = 208

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

Invert and add one

48 = 0011 0000
inverted: 1100 1111
+1 → 1101 0000

Read it as unsigned

1101 0000₂ = 256 − 48 = 208

Answer

U = 208
26.Number systems

Answer: R = 22

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

Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.

0101 1010 >> 2
= 0001 0110

Value of the result

0001 0110₂ = 22

Answer

R = 22
27.Digital logic: gates, K-maps & flip-flops

Answer: L = 4

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.

ABC00011110011001031204051716

Group the 1s

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

A′C′
AB

Minimal SOP form

F = A′C′ + AB
L = 4

Answer

L = 4
28.Digital logic: gates, K-maps & flip-flops

Answer: Q = 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.

Full cycles

Every 7 pulses bring the counter back to the same state.

25 = 3·7 + 4

Count after the remaining pulses

Q = (6 + 4) mod 7 = 3
Q = 011₂

Answer

Q = 3
29.Laplace transforms

Answer: f(0.2) = 1.618

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

Factor the denominator

s² + 8s + 16 = (s + 4)²

Partial fractions

4s + 14 = 4(s + 4) + −2
F(s) = 4/(s + 4) − 2/(s + 4)²

Inverse transform

f(t) = 4·e−4t − 2·t·e−4t
f(0.2) = (4 − 2·0.2)·e−4·0.2 = 1.618

Answer

f(0.2) = 1.618
30.Laplace transforms

Answer: y(0.1) = −4.565

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

Transform the equation

s²Y − (−4)s − (−15) + 10(sY − (−4)) + 24Y = 0
Y(s) = (−4s − 55)/((s + 4)(s + 6))

Partial fractions

A = −39/2 = −19.5
B = −31/(−2) = 15.5

Result

y(t) = −19.5·e−4t + 15.5·e−6t
y(0.1) = −4.565

Answer

y(0.1) = −4.565

Check: y(0) = A + B = −4.

31.Block diagrams

Answer: css = 2.08

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).

Closed-loop transfer function

T(s) = G/(1 + G·H) = 26/(s + 6 + 6.5) = 26/(s + 12.5)

Result

css = R·T(0) = 1·26/12.5 = 2.08

Answer

css = 2.08
32.Block diagrams

Answer: css = 1.357

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) = 1/(s + 7 + 14)
C/R = K·G/(1 + K·G) = 14/(s + 7 + 14)

Steady state

css = 2·14/21 + 0.5·1/21 = 1.357

Answer

css = 1.357

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

33.Step response

Answer: %OS = 9.478 %

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 = 100 → ωₙ = 10 rad/s
2ζωₙ = 12 → ζ = 12/(2·10) = 0.6

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−2.356 = 9.478 %

Answer

%OS = 9.478 %
34.Step response

Answer: ζ = 0.28

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(0.4)/√(π² + ln²(0.4)) = 0.9163/√(9.87 + 0.8396) = 0.28

Answer

ζ = 0.28
35.Bode plots & margins

Answer: |H| = −0.9774 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/(3300·1 × 10⁻⁷) = 3030 rad/s
fc = 482.3 Hz → f/fc = 960/482.3 = 1.991

Gain

|H| = 1.991/√(1 + 1.991²) = 0.8936
20·log₁₀(0.8936) = −0.9774 dB

Answer

|H| = −0.9774 dB
36.Bode plots & margins

Answer: PM = 63.74 °

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 + 16ω² − 77.44 = 0
ω² = (−16 + √(256 + 4·77.44))/2 = 3.893
ωc = 1.973 rad/s

Phase margin

∠L(jωc) = −90° − atan(1.973/4) = −116.3°
PM = 180° + ∠L = 63.74°

Answer

PM = 63.74°
37.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 462

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 + 12s² + 41s + 30 + K = 0

Routh array

s³: 1, 41
s²: 12, 30 + K
s¹: (12·41 − 30 − K)/12
s⁰: 30 + K

Condition

The s¹ entry must stay positive.

K < 12·41 − 30 = 462

Answer

Kmax = 462

At the limit the loop oscillates at ω = √41 = 6.403 rad/s.

38.Stability: Routh–Hurwitz & root locus

Answer: sb = −0.4725

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) = −s(s + 1)(s + 5) = −(s³ + 6s² + 5s)

Stationary points

dK/ds = −(3s² + 12s + 5) = 0
s = (−12 ± √(144 − 60))/6 = −0.4725, −3.528

Pick the point on the locus

The locus lies on the real axis between 0 and −1; the other root is not on the locus.

sb = −0.4725

Answer

sb = −0.4725
39.Steady-state error & PID control

Answer: ess = 0.2

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

Velocity constant

Kv = lim s·G(s) = 5

Ramp error

ess = 1/5 = 0.2

Answer

ess = 0.2
40.Steady-state error & PID control

Answer: Kd = 0.9997 s

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

Characteristic polynomial

s(s + 1) + 1(2 + Kds) = s² + (1 + 1Kd)s + 2

Natural frequency

ωₙ = √2 = 1.414 rad/s

Derivative gain

1 + 1Kd = 2·0.707·1.414 = 2
Kd = 0.9997 s

Answer

Kd = 0.9997 s

The derivative term adds damping without changing ωₙ.

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