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​ = 4 ΩR3​ = 8 Ω+−V1​9 VR2​6 ΩAR4​8 ΩB

Find the equivalent resistance Req seen by the source V1.

V1 = 9 VR1 = 4 ΩR2 = 6 ΩR3 = 8 ΩR4 = 8 Ω
Req = Ω
2.Series & parallel resistorsHard
R1​ = 12 ΩIR1​R3​ = 20 ΩR5​ = 8 Ω+−V1​30 VR2​6 ΩAR4​12 ΩBR6​20 Ω

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

V1 = 30 VR1 = 12 ΩR2 = 6 ΩR3 = 20 ΩR4 = 12 ΩR5 = 8 ΩR6 = 20 Ω
IR1 = A
3.Nodal analysisMedium
R1​ = 2 ΩR3​ = 8 Ω+−V1​18 VR2​3 ΩAR4​12 ΩBIs​2 A

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

V1 = 18 VIs = 2 AR1 = 2 ΩR2 = 3 ΩR3 = 8 ΩR4 = 12 Ω
VA = V
4.Nodal analysisHard
R1​ = 4 ΩR3​ = 25 Ω+−V1​15 VR2​10 ΩAR4​4 ΩB0.1·VA​

The diamond is a voltage-controlled current source that pushes a current of 0.1·VA (in A, with VA in V) up into node B. Use nodal analysis to find the node voltage VA (measured from ground).

V1 = 15 VR1 = 4 ΩR2 = 10 ΩR3 = 25 ΩR4 = 4 Ωg = 0.1 A/V
VA = V
5.Mesh analysisMedium
R1​ = 8 ΩR2​ = 12 Ω+−V1​10 VR3​10 ΩA+−V2​6 VI1​I2​

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

V1 = 10 VV2 = 6 VR1 = 8 ΩR2 = 12 ΩR3 = 10 Ω
I1 = A
6.Mesh analysisHard
R1​ = 20 ΩR2​ = 8 ΩR3​ = 2 Ω+−V1​9 VR4​25 ΩAR5​10 ΩIR5​B+−V2​30 VI1​I2​I3​

Use mesh analysis to find the current IR5 through R5 in the direction of the arrow.

V1 = 9 VV2 = 30 VR1 = 20 ΩR2 = 8 ΩR3 = 2 ΩR4 = 25 ΩR5 = 10 Ω
IR5 = A
7.SuperpositionMedium
R1​ = 12 ΩR3​ = 10 Ω+−V1​30 VR2​10 Ω+−VR2​AR4​20 ΩBIs​4 A

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

V1 = 30 VIs = 4 AR1 = 12 ΩR2 = 10 ΩR3 = 10 ΩR4 = 20 Ω
VR2 = V
8.SuperpositionHard
R1​ = 4 ΩR2​ = 8 ΩIR2​R4​ = 25 Ω+−V1​12 VR3​2 ΩAIs​6 AB+−V2​30 V

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

V1 = 12 VIs = 6 AV2 = 30 VR1 = 4 ΩR2 = 8 ΩR3 = 2 ΩR4 = 25 Ω
IR2 = A
9.Thévenin & NortonMedium
R1​ = 6 ΩR3​ = 15 Ω+−V1​10 VR2​5 ΩAab

Find the Thévenin voltage Vth seen from terminals a–b.

V1 = 10 VR1 = 6 ΩR2 = 5 ΩR3 = 15 Ω
Vth = V
10.Thévenin & NortonHard
R1​ = 3 ΩR3​ = 5 Ω+−V1​36 VR2​6 ΩAIs​1 ABab

Find the Thévenin voltage Vth seen from terminals a–b.

V1 = 36 VIs = 1 AR1 = 3 ΩR2 = 6 ΩR3 = 5 Ω
Vth = V
11.RC & RL transientsMedium
+−Vs​5 Vt = 0R = 50 ΩL47 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 1 ms, in mJ.

Vs = 5 VR = 50 ΩL = 47 mHt = 1 ms
W = mJ
12.RC & RL transientsHard
+−Vs​6 Vopens at t = 0R1​ = 3.3 kΩR2​ = 4.7 kΩC22 µF+−vC​

The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the capacitor voltage vC at t = 120 ms.

Vs = 6 VR1 = 3.3 kΩR2 = 4.7 kΩC = 22 µFt = 120 ms
vC = V
13.AC impedance & phasorsMedium
100 V200 HzR = 100 ΩL = 100 mHI

A resistor and an inductor are connected in parallel across a 100 V (rms), 200 Hz supply. Find the magnitude of the impedance |Z|.

V = 100 V (rms)f = 200 HzR = 100 ΩL = 100 mH
|Z| = Ω
14.AC impedance & phasorsHard
230 V100 HzR = 15 ΩL = 250 mHC = 4.7 µFI

A series RLC circuit is connected to a 230 V (rms), 100 Hz supply. Find the rms current I.

V = 230 V (rms)f = 100 HzR = 15 ΩL = 250 mHC = 4.7 µF
I = A
15.AC power & power factorMedium
230 V16 kW, 0.7221 kW, 0.8I = ?

Two loads are connected in parallel to a 230 V single-phase supply: a 6 kW load at a power factor of 0.7 lagging and a 21 kW load at a power factor of 0.8 lagging. Find the total current I drawn from the supply.

V = 230 VP1 = 6 kW, pf1 = 0.7 laggingP2 = 21 kW, pf2 = 0.8 lagging
I = A
16.AC power & power factorHard
P1​ = 5 kWQ1​S1​φ

A 10 kVA single-phase transformer supplies a load of 5 kW at a power factor of 0.85 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?

Smax = 10 kVAP1 = 5 kWpf1 = 0.85
ΔP = kW
17.Three-phase powerMedium
ABCVL​ = 480 VNZ = 12 + j9 Ωper phase

A balanced Y-connected load with an impedance of 12 + j9 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.

VL = 480 VZ = 12 + j9 Ω per phase
P = kW
18.Three-phase powerHard
P = 120 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 120 kW at a power factor of 0.78 lagging from a 690 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 120 kWpf1 = 0.78pf2 = 0.92VL = 690 Vf = 50 Hz
Cph = µF
19.TransformersMedium
5 Ω51

A 5 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 5:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 5:1RL = 5 Ω
R1′ = Ω
20.TransformersHard
250 kVA

A single-phase 250 kVA transformer has a core loss of 1500 W and a copper loss of 4500 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 250 kVAPcore = 1500 WPcu,fl = 4500 W
Sη,max = kVA
21.Op-amp circuitsMedium
−+Vout​R2​ = 100 kΩV1​ = 0.5 VR1​ = 47 kΩV2​ = 2 VR3​ = 47 kΩR4​ = 100 kΩ

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

V1 = 0.5 VV2 = 2 VR1 = R3 = 47 kΩR2 = R4 = 100 kΩ
Vout = V
22.Op-amp circuitsHard
−+Vout​R2​ = 1 kΩV1​ = −1.7 VR1​ = 1 kΩV2​ = −0.6 VR3​ = 1 kΩR4​ = 22 kΩ

The op-amp is ideal and runs from ±15 V supplies. The resistor ratios are not matched. Find the output voltage Vout.

V1 = −1.7 VV2 = −0.6 VR1 = 1 kΩR2 = 1 kΩR3 = 1 kΩR4 = 22 kΩ
Vout = V
23.Diodes & transistorsMedium
RC​ = 2.7 kΩβ = 50IC​ = ?VCC​ = 10 VRB​ = 390 kΩ

An NPN transistor with β = 50 is biased by a base resistor RB = 390 kΩ from the supply VCC = 10 V. The collector resistor is RC = 2.7 kΩ and the emitter is grounded. Take VBE = 0.7 V and assume the transistor is in the active region. Find the collector current IC in mA.

VCC = 10 VRB = 390 kΩRC = 2.7 kΩβ = 50VBE = 0.7 V
IC = mA
24.Diodes & transistorsHard
V (V)I (mA)0.766Q = ?

A diode is modelled as 0.7 V in series with 20 Ω (it conducts above 0.7 V). It is in series with a 1000 Ω resistor across a 6 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 6 VR = 1000 ΩV0 = 0.7 Vrd = 20 Ω
ID = mA
25.Number systemsMedium
place values in two's complement1−12806413211618140211

The 8-bit pattern 1011 1101 stores a signed integer in two's complement. What is its value N in decimal?

1011 11018 bit
N =
26.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0111 0100 OR 0110 0101. What is the result's value R in decimal?

0111 0100OR0110 0101
R =
27.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010110312141507160121130151140809111010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 1, 2, 4, 5, 6, 11, 13, 14). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 1, 2, 4, 5, 6, 11, 13, 14)
N =
28.Digital logic: gates, K-maps & flip-flopsHard
CLKD12345678

A positive-edge-triggered D 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 = 0D
  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
  8. 8clock edge 8
29.Laplace transformsMedium
t (s)f(t)t = 1.2f(1.2) = ?

Find f(t) for F(s) = 55/(s(s + 1)(s + 3)) by partial fractions, and evaluate f(1.2).

F(s) = 55/(s(s + 1)(s + 3))t = 1.2 s
f(1.2) =
30.Laplace transformsHard
ReIm−2−4−6−8−10−12

A system is described by the differential equation y″ + 3y′ + 2y = u′ + 11u, 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 = 1.2 s. Find g(1.2).

y″ + 3y′ + 2y = u′ + 11ut = 1.2 s
g(1.2) =
31.Block diagramsMedium
R(s)+Ks(s + 17)G(s)C(s)−

The plant G(s) = K/(s(s + 17)) is in a unity negative feedback loop. Find the gain K for which the closed loop is critically damped (two equal real poles).

G(s) = K/(s(s + 17))unity negative feedback
K =
32.Block diagramsHard
R(s)+70s(s + 11)G(s)C(s)2H−

The plant G(s) = 70/(s(s + 11)) is in a negative feedback loop with sensor gain H = 2. The closed-loop poles are complex. Find their real part σ.

G(s) = 70/(s(s + 11))H = 2
σ =
33.Step responseMedium
t (s)y(t)1tp​ = ?

A system has the transfer function G(s) = 97/(s² + 15s + 97) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 97/(s² + 15s + 97)
tp = s
34.Step responseHard
t (s)y(t)1tp​ = 0.25 s12 %

The step response of a second-order system without zeros shows 12 % overshoot, and the first peak occurs at t = 0.25 s. Find the natural frequency ωn.

%OS = 12 %tp = 0.25 s
ωn = rad/s
35.Bode plots & marginsMedium
vin​R = 47 kΩC = 100 nFvout​

The RC low-pass filter shown has R = 47 kΩ, C = 100 nF. A sinusoidal input of 6.8 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.

R = 47 kΩC = 100 nFf = 6.8 Hz
φ = °
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) = 11/(s(s + 1)). Find the phase margin PM in degrees.

L(s) = 11/(s(s + 1))
PM = °
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+201s(s + 1)(s + 5)C(s)+

The loop in the diagram uses positive feedback, with L(s) = 20/(s(s + 1)(s + 5)). How many closed-loop poles N lie in the right half-plane?

L(s) = 20/(s(s + 1)(s + 5))
N =
38.Stability: Routh–Hurwitz & root locusHard
ReImζ = 0.8−2−4−6−8−10j2

The root locus of L(s) = K/(s(s + 8)) is shown. Find the gain K for which the closed-loop poles have the damping ratio ζ = 0.8.

L(s) = K/(s(s + 8))ζ = 0.8
K =
39.Steady-state error & PID controlMedium
R(s)+K(s + 1)(s + 5)C(s)−

A unity-feedback loop has G(s) = K/((s + 1)(s + 5)). Find the gain K that gives a steady-state error of 2 % for a step input.

G(s) = K/((s + 1)(s + 5))ess = 2 %
K =
40.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s5s + 1C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 5/(s + 1) in a unity-feedback loop. The closed loop must have ωₙ = 3 rad/s and ζ = 0.8. Find the integral gain Ki.

G(s) = 5/(s + 1)ωₙ = 3 rad/sζ = 0.8
Ki =

3Solutions

1.Series & parallel resistors

Answer: Req = 8.364 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​ = 4 ΩR3​ = 8 Ω+−V1​9 VR2​6 ΩAR4​8 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 + 8 = 16 Ω
R2 ∥ 16 = 6·16/(6+16) = 4.364 Ω
Req = R1 + 4.364 = 4 + 4.364 = 8.364 Ω

Answer

Req = 8.364 Ω
2.Series & parallel resistors

Answer: IR1 = 1.77 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​ = 12 ΩR3​ = 20 ΩR5​ = 8 Ω+−V1​30 VR2​6 ΩAR4​12 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 = 8 + 20 = 28 Ω
R4 ∥ 28 = 8.4 Ω
R3 + 8.4 = 28.4 Ω
R2 ∥ 28.4 = 4.953 Ω
Req = R1 + 4.953 = 12 + 4.953 = 16.95 Ω

Current from the source

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

IR1 = V1/Req = 30/16.95 = 1.77 A

Answer

IR1 = 1.77 A
3.Nodal analysis

Answer: VA = 8.83 V

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

R1​ = 2 ΩR3​ = 8 Ω+−V1​18 VR2​3 ΩAR4​12 ΩBIs​2 A

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 18 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/3 + (VA − 18)/2 + (VA − VB)/8 = 0
0.9583·VA − 0.125·VB = 9

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)/8 = (−2)
−0.125·VA + 0.2083·VB = −2

Solve

Δ = (0.9583)(0.2083) − (−0.125)(−0.125) = 0.184
VA = Δ1/Δ = 1.625/0.184 = 8.83 V
VB = Δ2/Δ = −0.7917/0.184 = −4.302 V

Answer

VA = 8.83 V
4.Nodal analysis

Answer: VA = 10.12 V

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

KCL at node A

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

(VA − V1)/R1 + VA/R2 + (VA − VB)/R3 = 0
0.39·VA − 0.04·VB = 3.75

KCL at node B

The dependent source pushes g·VA into node B. Treat it like an independent source, then move the VA term to the left-hand side.

(VB − VA)/R3 + VB/R4 = 0.1·VA
−0.14·VA + 0.29·VB = 0

Solve

Δ = (0.39)(0.29) − (−0.04)(−0.14) = 0.1075
VA = Δ1/Δ = 1.087/0.1075 = 10.12 V
VB = Δ2/Δ = 0.525/0.1075 = 4.884 V

Answer

VA = 10.12 V
5.Mesh analysis

Answer: I1 = 0.5405 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​ = 8 ΩR2​ = 12 Ω+−V1​10 VR3​10 ΩA+−V2​6 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
8·I1 + 10(I1 − I2) = 10
18·I1 − 10·I2 = 10

KVL around mesh 2

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

R2·I2 + R3(I2 − I1) = −V2
12·I2 + 10(I2 − I1) = (−6)
−10·I1 + 22·I2 = −6

Solve

Δ = (18)(22) − (−10)(−10) = 296
I1 = Δ1/Δ = 160/296 = 0.5405 A
I2 = Δ2/Δ = −8/296 = −0.02703 A

Answer

I1 = 0.5405 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: IR5 = 2.34 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​ = 8 ΩR3​ = 2 Ω+−V1​9 VR4​25 ΩAR5​10 ΩB+−V2​30 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 + 25(I1 − I2) = 9
45·I1 − 25·I2 = 9

KVL around mesh 2

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

R2·I2 + R4(I2 − I1) + R5(I2 − I3) = 0
8·I2 + 25(I2 − I1) + 10(I2 − I3) = 0
−25·I1 + 43·I2 − 10·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 + 10(I3 − I2) = (−30)
−10·I2 + 12·I3 = −30

Solve

Δ = det of the coefficient matrix = 11220
I1 = Δ1/Δ = −3756/11220 = −0.3348 A
I2 = Δ2/Δ = −10800/11220 = −0.9626 A
I3 = Δ3/Δ = −37050/11220 = −3.302 A

Answer

IR5 = I2 − I3 = 2.34 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: VR2 = 23.85 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​ = 12 ΩR3​ = 10 Ω+−V1​30 VR2​10 ΩAR4​20 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 = 10+20 = 30 Ω
R2 ∥ (R3+R4) = 10·30/(10+30) = 7.5 Ω
VA = 30·7.5/(12+7.5) = 11.54 V
VB = VA·R4/(R3+R4) = 11.54·20/30 = 7.692 V
VR2′ = VA − 0 = 11.54 − 0 = 11.54 V
R1​ = 12 ΩR3​ = 10 ΩSHORTR2​10 ΩAR4​20 ΩBIs​4 A

Case 2 · Is acting alone

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

R1 ∥ R2 = 12·10/(12+10) = 5.455 Ω
Req = R4 ∥ (R3+R1∥R2) = 20 ∥ 15.45 = 8.718 Ω
VB = Is·Req = 4·8.718 = 34.87 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 34.87·5.455/15.45 = 12.31 V
VR2″ = VA − 0 = 12.31 − 0 = 12.31 V

Add the contributions

VR2 = VR2′ + VR2″
VR2 = 11.54 + 12.31 = 23.85 V

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

8.Superposition

Answer: IR2 = −5.126 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​ = 4 ΩR2​ = 8 ΩR4​ = 25 Ω+−V1​12 VR3​2 ΩAOPENBSHORT

Case 1 · V1 acting alone

V₂ becomes a short and Iₛ an open circuit. Seen from A, R₃ is in parallel with the series pair R₂+R₄.

R3 ∥ (R2+R4) = 2 ∥ 33 = 1.886 Ω
VA = 12·1.886/(4+1.886) = 3.845 V
VB = VA·R4/(R2+R4) = 3.845·25/33 = 2.913 V
IR2′ = (VA − VB)/R2 = (3.845 − 2.913)/8 = 0.1165 A
R1​ = 4 ΩR2​ = 8 ΩR4​ = 25 ΩSHORTR3​2 ΩAIs​6 ABSHORT

Case 2 · Is acting alone

Both voltage sources become shorts. Iₛ splits at B between R₄ (now to ground) and R₂ in series with R₁ ∥ R₃.

R1 ∥ R3 = 1.333 Ω, so the left path is R2+1.333 = 9.333 Ω
Req = R4 ∥ 9.333 = 6.796 Ω
VB = Is·Req = 6·6.796 = 40.78 V
VA = VB·1.333/9.333 = 5.825 V
IR2″ = (VA − VB)/R2 = (5.825 − 40.78)/8 = −4.369 A
R1​ = 4 ΩR2​ = 8 ΩR4​ = 25 ΩSHORTR3​2 ΩAOPENB+−V2​30 V

Case 3 · V2 acting alone

V₁ becomes a short and Iₛ an open circuit. R₁ ∥ R₃ hangs from A, in series with R₂ back to B.

R1 ∥ R3 = 4·2/(4+2) = 1.333 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 30·9.333/34.33 = 8.155 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 8.155·1.333/9.333 = 1.165 V
IR2‴ = (VA − VB)/R2 = (1.165 − 8.155)/8 = −0.8738 A

Add the contributions

IR2 = IR2′ + IR2″ + IR2‴
IR2 = 0.1165 + (−4.369) + (−0.8738) = −5.126 A

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

9.Thévenin & Norton

Answer: Vth = 4.545 V

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

R1​ = 6 ΩR3​ = 15 ΩSHORTR2​5 ΩAabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 6·5/(6+5) = 2.727 Ω
Rth = (R1 ∥ R2) + R3 = 2.727 + 15 = 17.73 Ω
R1​ = 6 ΩR3​ = 15 Ω+−V1​10 VR2​5 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) = 10·5/(6+5) = 4.545 V
10.Thévenin & Norton

Answer: Vth = 31 V

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​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 = 3·6/(3+6) = 2 Ω
Rth = (R1 ∥ R2) + R3 = 2 + 5 = 7 Ω
R1​ = 3 ΩR3​ = 5 Ω+−V1​36 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) = 24 + 1·2 = 26 V
Vth = VB = VA + Is·R3 = 26 + 1·5 = 31 V
11.RC & RL transients

Answer: W = 0.1008 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 = 47 mH / 50 Ω = 0.94 ms
i(∞) = Vs/R = 5/50 = 0.1 A

Current at the given time

i(t) = 0.1·(1 − e−1/0.94) = 0.06549 A

Stored energy

W = ½·L·i² = ½·47·0.06549² = 0.1008 mJ

Answer

W = 0.1008 mJ
12.RC & RL transients

Answer: vC = 1.104 V

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

Initial capacitor voltage

In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.

vC(0) = 6·4.7/(3.3 + 4.7) = 3.525 V

Time constant after the switch opens

τ = R2·C = 4.7·22 = 103.4 ms

Discharge

vC(t) = 3.525·e−120/103.4 = 1.104 V

Answer

vC = 1.104 V
13.AC impedance & phasors

Answer: |Z| = 78.25 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·200·0.1 = 125.7 Ω

Branch currents

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

IR = V/R = 100/100 = 1 A
IX = V/XL = 100/125.7 = 0.7958 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) = √(1² + 0.7958²) = 1.278 A

Impedance

|Z| = V/I = 100/1.278 = 78.25 Ω

Answer

|Z| = 78.25 Ω
14.AC impedance & phasors

Answer: I = 1.263 A

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·100 = 628.3 rad/s
XL = ωL = 628.3·0.25 = 157.1 Ω
XC = 1/(ωC) = 1/(628.3·4.7·10⁻⁶) = 338.6 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(15² + (−181.5)²) = 182.2 Ω

Current

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

I = V/|Z| = 230/182.2 = 1.263 A

Answer

I = 1.263 A
15.AC power & power factor

Answer: I = 151.1 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 = 6·tan(arccos 0.7) = 6.121 kvar
Q2 = 21·tan(arccos 0.8) = 15.75 kvar

Totals

P = 6 + 21 = 27 kW
Q = 6.121 + 15.75 = 21.87 kvar
S = √(27² + 21.87²) = 34.75 kVA

Total current

I = S/V = 34747/230 = 151.1 A

Answer

I = 151.1 A
16.AC power & power factor

Answer: ΔP = 4.508 kW

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 the existing load

Q1 = 5·tan(arccos 0.85) = 3.099 kvar

Largest total real power

At full rating S = Smax, and Q is unchanged by the resistive load.

P = √(Smax² − Q1²) = √(10² − 3.099²) = 9.508 kW

Extra real power

ΔP = 9.508 − 5 = 4.508 kW

Answer

ΔP = 4.508 kW
17.Three-phase power

Answer: P = 12.29 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 = 480/√3 = 277.1 V

Phase current

|Z| = √(12² + 9²) = 15 Ω
Iph = 277.1/15 = 18.48 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·18.48²·12 = 12.29 kW

Answer

P = 12.29 kW

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

18.Three-phase power

Answer: Cph = 301.9 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 = 120·tan(acos 0.78) = 120·0.8023 = 96.27 kvar
Q2 = 120·tan(acos 0.92) = 120·0.426 = 51.12 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 45.15 kvar

Capacitance per phase

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

V = 690/√3 = 398.4 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 15051/(314.2·398.4²) = 301.9 µF

Answer

Cph = 301.9 µF
19.Transformers

Answer: R1′ = 125 Ω

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

Referred resistance

The voltage goes up by a and the current down by a, so the ratio V/I goes up by a².

R1′ = a²·RL = 5²·5 = 125 Ω

Answer

R1′ = 125 Ω
20.Transformers

Answer: Sη,max = 144.3 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

x²·4500 = 1500 → x = √(1500/4500) = 0.5774

Load at maximum efficiency

Sη,max = x·S = 0.5774·250 = 144.3 kVA

Answer

Sη,max = 144.3 kVA
21.Op-amp circuits

Answer: Vout = 3.191 V

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

Voltage at the inputs

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

V₊ = V2·R4/(R3 + R4) = 2·100/(47 + 100) = 1.361 V
V₋ = V₊

KCL at the inverting input

(0.5 − 1.361)/47 = (1.361 − Vout)/100
Vout = (R2/R1)·(V2 − V1) = (100/47)·(2 − 0.5) = 3.191 V

Answer

Vout = 3.191 V
22.Op-amp circuits

Answer: Vout = 0.5522 V

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

Voltage at the + input

V₊ = V2·R4/(R3 + R4) = −0.6·22/(1 + 22) = −0.5739 V

Superpose the two inputs

With V₂ = 0, V₁ sees an inverting amplifier. With V₁ = 0, V₊ is amplified by the non-inverting gain 1 + R₂/R₁.

Vout = −(1/1)·(−1.7) + (1 + 1/1)·(−0.5739) = 1.7 + (−1.148) = 0.5522 V

Answer

Vout = 0.5522 V
23.Diodes & transistors

Answer: IC = 1.192 mA

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.

Base current

IB = (VCC − VBE)/RB = (10 − 0.7)/390 kΩ = 23.85 µA

Collector current

IC = β·IB = 50·23.85 µA = 1.192 mA

Answer

IC = 1.192 mA

Check: VCE = 6.781 V is well above 0.2 V, so the transistor is indeed active.

24.Diodes & transistors

Answer: ID = 5.196 mA

LED and diode circuits: a forward-biased LED drops roughly its forward voltage VF (about 2 V for red, 3 V for blue or white), whatever the current. A series resistor sets the current: R = (Vs − n·VF)/I for n LEDs in series. With a more exact diode model V = V₀ + I·rd, the operating point is where the diode line meets the load line I = (Vs − V)/R, so I = (Vs − V₀)/(R + rd).

Combine the two equations

6 = 0.7 + I·(1000 + 20)

Operating current

ID = 5.3/1020 = 5.196 mA
VD = 0.7 + 0.005196·20 = 0.8039 V

Answer

ID = 5.196 mA
25.Number systems

Answer: N = −67

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 + 32 + 16 + 8 + 4 + 1 = −67

Answer

N = −67
26.Number systems

Answer: R = 117

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

OR gives 1 where at least one bit is 1.

0111 0100
OR 0110 0101
= 0111 0101

Value of the result

0111 0101₂ = 117

Answer

R = 117
27.Digital logic: gates, K-maps & flip-flops

Answer: N = 5

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.

ABCD000111100001111010110312141507160121130151140809111010

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

Minimal POS form

F = (B′ + C′ + D′)(A + C′ + D′)(A′ + C + D)(A′ + B + D)(A′ + B + C)
N = 5

Answer

N = 5
28.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 4

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

Q after every edge

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

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

First change from 1 to 0

clock edge 4

Answer

clock edge 4
29.Laplace transforms

Answer: f(1.2) = 10.3

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

Partial fractions

F(s) = A/s + B/(s + 1) + C/(s + 3)
A = 55/(1·3) = 18.33
B = 55/((−1)·2) = −27.5
C = 55/((−3)·(−2)) = 9.167

Inverse transform

f(t) = 18.33 − 27.5·e−t + 9.167·e−3t

Evaluate

f(1.2) = 18.33 − 27.5·e−1.2 + 9.167·e−3·1.2 = 10.3

Answer

f(1.2) = 10.3
30.Laplace transforms

Answer: g(1.2) = 2.195

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) = (s + 11)/(s² + 3s + 2)

Partial fractions

s² + 3s + 2 = (s + 1)(s + 2)
A = 10, B = −9

Impulse response

g(t) = 10·e−t − 9·e−2t
g(1.2) = 2.195

Answer

g(1.2) = 2.195
31.Block diagrams

Answer: K = 72.25

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

Characteristic equation

1 + K/(s(s + 17)) = 0 → s² + 17s + K = 0

Repeated root

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

17² = 4K → K = 289/4 = 72.25

Answer

K = 72.25
Both closed-loop poles are then at s = −8.5.
32.Block diagrams

Answer: σ = −5.5

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

Characteristic equation

s(s + 11) + 70·2 = 0 → s² + 11s + 140 = 0

Roots

s = −5.5 ± j·√(140 − 30.25) = −5.5 ± j10.48
σ = −5.5

Answer

σ = −5.5
33.Step response

Answer: tp = 0.4921 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 = 97 → ωₙ = 9.849 rad/s
2ζωₙ = 15 → ζ = 15/(2·9.849) = 0.7615

Result

ωd = 9.849·√(1 − 0.7615²) = 6.384 rad/s → tp = π/ωd = 0.4921 s

Answer

tp = 0.4921 s
34.Step response

Answer: ωn = 15.16 rad/s

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

Damping ratio

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

Natural frequency

ωd = π/tp = π/0.25 = 12.57 rad/s
ωₙ = ωd/√(1 − ζ²) = 12.57/0.8289 = 15.16 rad/s

Answer

ωn = 15.16 rad/s
35.Bode plots & margins

Answer: φ = −11.35 °

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

Corner frequency

ωc = 1/(RC) = 1/(47000·1 × 10⁻⁷) = 212.8 rad/s
fc = 33.86 Hz → f/fc = 6.8/33.86 = 0.2008

Phase

φ = − atan(0.2008) = −11.35°

Answer

φ = −11.35°
36.Bode plots & margins

Answer: PM = 17.14 °

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

Gain crossover

ω⁴ + 1ω² − 121 = 0
ω² = (−1 + √(1 + 4·121))/2 = 10.51
ωc = 3.242 rad/s

Phase margin

∠L(jωc) = −90° − atan(3.242/1) = −162.9°
PM = 180° + ∠L = 17.14°

Answer

PM = 17.14°
37.Stability: Routh–Hurwitz & root locus

Answer: N = 1

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 + 6s² + 5s − 20 = 0

Routh array

s³: 1, 5
s²: 6, −20
s¹: 8.333
s⁰: −20

Sign changes in the first column

1 → 6 → 8.333 → −20
N = 1

Answer

N = 1

Positive feedback makes the constant coefficient negative: one real pole always moves into the right half-plane.

38.Stability: Routh–Hurwitz & root locus

Answer: K = 25

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.

Compare with the standard form

s² + 8s + K = s² + 2ζωₙs + ωₙ²
2·0.8·ωₙ = 8 → ωₙ = 5 rad/s

Gain

K = ωₙ² = 5² = 25

Answer

K = 25

The closed-loop poles sit where the ζ line crosses the vertical branch at s = −4.

39.Steady-state error & PID control

Answer: K = 245

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 error constant

1/(1 + Kp) = 0.02 → Kp = 49

Gain

K/5 = 49 → K = 245

Answer

K = 245

Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.

40.Steady-state error & PID control

Answer: Ki = 1.8

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

1 + C(s)G(s) = 0, multiplied by s(s + 1).

s² + (1 + 5Kp)s + 5Ki

Target polynomial

s² + 2·0.8·3s + 3² = s² + 4.8s + 9

Match the coefficients

5Ki = 9 → Ki = 1.8

Answer

Ki = 1.8

The integral term makes the loop type 1, so a step is followed without steady-state error.

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