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​ = 3 ΩR3​ = 15 Ω+−V1​24 VR2​4 ΩIR2​AR4​20 ΩB

Find the current IR2 through R2.

V1 = 24 VR1 = 3 ΩR2 = 4 ΩR3 = 15 ΩR4 = 20 Ω
IR2 = A
2.Series & parallel resistorsHard
R1​ = 8 ΩR3​ = 12 ΩR5​ = 8 Ω+−V1​10 VR2​2 ΩAR4​10 ΩBR6​15 Ω

Find the equivalent resistance Req seen by the source V1.

V1 = 10 VR1 = 8 ΩR2 = 2 ΩR3 = 12 ΩR4 = 10 ΩR5 = 8 ΩR6 = 15 Ω
Req = Ω
3.Nodal analysisMedium
R1​ = 8 ΩR3​ = 6 ΩIR3​+−V1​9 VR2​2 ΩAR4​15 ΩBIs​1 A

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

V1 = 9 VIs = 1 AR1 = 8 ΩR2 = 2 ΩR3 = 6 ΩR4 = 15 Ω
IR3 = A
4.Nodal analysisHard
R1​ = 8 Ω+−Vs​ = 12 V+−V1​12 VR2​2 ΩAR3​8 ΩBIs​2 A

The voltage source Vₛ between A and B has its + terminal at A, so it makes VA − VB = Vₛ. Use nodal analysis with a supernode to find the node voltage VA (measured from ground).

V1 = 12 VIs = 2 AVs = 12 VR1 = 8 ΩR2 = 2 ΩR3 = 8 Ω
VA = V
5.Mesh analysisMedium
R1​ = 15 ΩR2​ = 15 Ω+−V1​9 VR3​8 ΩA+−V2​30 VI1​I2​

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

V1 = 9 VV2 = 30 VR1 = 15 ΩR2 = 15 ΩR3 = 8 Ω
I1 = A
6.Mesh analysisHard
R1​ = 30 ΩR2​ = 20 ΩR3​ = 5 Ω+−V1​36 VR4​15 ΩAR5​10 ΩB+−V2​12 VI1​I2​I3​

Use mesh analysis to find the mesh current I3 (clockwise).

V1 = 36 VV2 = 12 VR1 = 30 ΩR2 = 20 ΩR3 = 5 ΩR4 = 15 ΩR5 = 10 Ω
I3 = A
7.SuperpositionMedium
R1​ = 10 ΩR3​ = 10 Ω+−V1​6 VR2​4 ΩAR4​6 Ω+−VR4​BIs​1.5 A

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

V1 = 6 VIs = 1.5 AR1 = 10 ΩR2 = 4 ΩR3 = 10 ΩR4 = 6 Ω
VR4 = V
8.SuperpositionHard
R1​ = 15 ΩR2​ = 6 ΩR4​ = 8 Ω+−V1​5 VR3​12 ΩIR3​AIs​2 AB+−V2​?

Find the value of the voltage source V2 (polarity as shown) that makes the current IR3 through R3 equal to zero.

V1 = 5 VIs = 2 AR1 = 15 ΩR2 = 6 ΩR3 = 12 ΩR4 = 8 Ω
V2 = V
9.Thévenin & NortonMedium
R1​ = 3 ΩR3​ = 6 Ω+−V1​12 VR2​12 ΩAab

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

V1 = 12 VR1 = 3 ΩR2 = 12 ΩR3 = 6 Ω
Rth = Ω
10.Thévenin & NortonHard
R1​ = 2 ΩR3​ = 30 Ω+−V1​10 VR2​4 ΩAIs​0.5 ABRL​25 ΩIL​

A load RL = 25 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 10 VIs = 0.5 AR1 = 2 ΩR2 = 4 ΩR3 = 30 ΩRL = 25 Ω
IL = A
11.RC & RL transientsMedium
+−Vs​20 Vt = 0R = 20 ΩL33 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the voltage vL across the inductor at t = 2 ms.

Vs = 20 VR = 20 ΩL = 33 mHt = 2 ms
vL = V
12.RC & RL transientsHard
+−Vs​5 Vt = 0R = 1 kΩC100 µF+−vC​vC​(0) = 8 V

Just before the switch closes at t = 0, the capacitor holds 8 V with the polarity shown. Find the capacitor voltage vC at t = 75 ms.

Vs = 5 VR = 1 kΩC = 100 µFvC(0) = 8 Vt = 75 ms
vC = V
13.AC impedance & phasorsMedium
120 V100 HzR = 80 ΩC = 47 µFI

A resistor and a capacitor are connected in parallel across a 120 V (rms), 100 Hz supply. Find the magnitude of the impedance |Z|.

V = 120 V (rms)f = 100 HzR = 80 ΩC = 47 µF
|Z| = Ω
14.AC impedance & phasorsHard
230 V100 HzR = 50 ΩL = 150 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 = 50 ΩL = 150 mHC = 4.7 µF
I = A
15.AC power & power factorMedium
P = 34 kWQSφ

A single-phase motor takes 34 kW at a power factor of 0.8 lagging from a 230 V supply. Find the line current I.

P = 34 kWpf = 0.8V = 230 V
I = A
16.AC power & power factorHard
230 V60 HzRs​ = 10 ΩXs​ = j10 ΩZL​ = ?

A 230 V (rms), 60 Hz source has an internal impedance Zs = 10 + j10 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.

Vs = 230 V (rms)f = 60 HzZs = 10 + j10 Ω
C = µF
17.Three-phase powerMedium
ABCVL​ = 400 VIL​ = ?M37 kW, η = 95%cos φ = 0.84

A three-phase induction motor delivers 37 kW of mechanical output power at full load. At this load its efficiency is 95% and its power factor is 0.84 lagging. The supply has a line-to-line voltage of 400 V. Find the line current IL.

Pout = 37 kWη = 95%cos φ = 0.84VL = 400 V
IL = A
18.Three-phase powerHard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.65 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. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 75 kWpf1 = 0.65pf2 = 0.98VL = 400 Vf = 60 Hz
Cph = µF
19.TransformersMedium
RL​ = 48 Ω2012400 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 20:1 has 2400 V applied to its primary winding. A 48 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 20:1V1 = 2400 VRL = 48 Ω
I1 = A
20.TransformersHard
pf 0.950 kVA, 50% load

A single-phase 50 kVA transformer has a core loss of 200 W and a copper loss of 300 W at rated load. It supplies 50% of its rated kVA at a power factor of 0.9 lagging. Find the total losses Ploss in watts at this load.

S = 50 kVAPcore = 200 WPcu,fl = 300 Wx = 0.5pf = 0.9
Ploss = W
21.Op-amp circuitsMedium
−+Vout​Rf​ = 47 kΩR1​ = 4.7 kΩVin​ = −4 VRa​ = 22 kΩRb​ = 4.7 kΩ

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

Vin = −4 VRa = 22 kΩRb = 4.7 kΩR1 = 4.7 kΩRf = 47 kΩ
Vout = V
22.Op-amp circuitsHard
−+Vout​Rf​ = 22 kΩV1​ = 0.6 VR1​ = 4.7 kΩV2​ = 0.2 VR2​ = 10 kΩ

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

V1 = 0.6 VV2 = 0.2 VR1 = 4.7 kΩR2 = 10 kΩRf = 22 kΩ
Vout = V
23.Diodes & transistorsMedium
RC​ = 3.3 kΩβ = 75VCC​ = 9 VRB​ = 680 kΩ

An NPN transistor with β = 75 is biased by a base resistor RB = 680 kΩ from the supply VCC = 9 V. The collector resistor is RC = 3.3 kΩ and the emitter is grounded. Take VBE = 0.7 V and assume the transistor is in the active region. Find the collector–emitter voltage VCE.

VCC = 9 VRB = 680 kΩRC = 3.3 kΩβ = 75VBE = 0.7 V
VCE = V
24.Diodes & transistorsHard
RC​ = 680 Ωβ = 150IC​ = ?VCC​ = 9 V5 VRB​ = 10 kΩ

An NPN transistor with β = 150 switches a 680 Ω load resistor connected between VCC = 9 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB = 10 kΩ. Take VBE = 0.7 V and VCE(sat) = 0.2 V. Find the collector current IC in mA when the input is high. Decide first whether the transistor is saturated.

VCC = 9 VRC = 680 Ωβ = 150Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 10 kΩ
IC = mA
25.Number systemsMedium
place values in two's complement0−12806413201618040211

The 8-bit pattern 0010 1001 stores a signed integer in two's complement. What is its value N in decimal?

0010 10018 bit
N =
26.Number systemsHard
16 bit1 px640 px480 pxM = ? KiB

An uncompressed image is 640 × 480 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

640 × 480 px16 bit/px1 KiB = 1024 B
M = KiB
27.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000110302140507060120130151140819011010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(1, 4, 9, 14). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(1, 4, 9, 14)
N =
28.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKDOWN

A 4-bit binary down counter starts at 6 (decimal, Q = 0110). Which count Q (in decimal) does it show after 18 clock pulses?

Q0 = 618n = 4
Q =
29.Laplace transformsMedium
ReIm−1−2−3−4−5−6−71

F(s) = (6s + 39)/(s² + 10s + 24). Its inverse transform has the form f(t) = A·e−4t + B·e−6t. Find the coefficient A by partial fractions.

F(s) = (6s + 39)/(s² + 10s + 24)
A =
30.Laplace transformsHard
t (s)y(t)t = 1.2y(1.2) = ?

A system obeys y″ + 4y′ + 3y = 6·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 1 is applied. Use the Laplace transform to find y(1.2).

y″ + 4y′ + 3y = 6·u(t)y(0) = 2y′(0) = 0u(t) = 1t = 1.2 s
y(1.2) =
31.Block diagramsMedium
R(s)+K(s + 4)(s + 11)G(s)C(s)−

The plant G(s) = K/((s + 4)(s + 11)) 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 + 4)(s + 11))unity negative feedback
K =
32.Block diagramsHard
R(s)+7controller+D(s)+7s + 5plantC(s)−

A proportional controller K = 7 drives the plant G(s) = 7/(s + 5) 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 1 are applied together. Find the steady-state output css.

K = 7G(s) = 7/(s + 5)D = 1R = 2
css =
33.Step responseMedium
t (s)y(t)3?

A system has the transfer function G(s) = 132/(s² + 3s + 44) and receives a unit step input. Find the percent overshoot %OS of the step response.

G(s) = 132/(s² + 3s + 44)
%OS = %
34.Step responseHard
R(s)+170s(s + 10)G(s)C(s)−

A unity negative feedback loop has G(s) = 170/(s(s + 10)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 170/(s(s + 10))
ts = s
35.Bode plots & marginsMedium
|H| (dB)0.11101001000−80−60−40−2002040ω (rad/s)ω = 30

Find the exact magnitude |G(jω)| of G(s) = 882/((s + 7)(s + 9)) at ω = 30 rad/s, in dB.

G(s) = 882/((s + 7)(s + 9))ω = 30 rad/s
|G(jω)| = dB
36.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​

A unity-feedback loop has the open-loop transfer function L(s) = 800/(s(s + 7)(s + 9)). Find the gain margin GM in dB.

L(s) = 800/(s(s + 7)(s + 9))
GM = dB
37.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1(s + 1)(s + 2)(s + 4)C(s)−

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

L(s) = K/((s + 1)(s + 2)(s + 4))
Kmax =
38.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2−3

For the loop L(s) = K/(s(s + 6)(s + 10)), find the gain K that puts a closed-loop pole at s = −3.

L(s) = K/(s(s + 6)(s + 10))s = −3
K =
39.Steady-state error & PID controlMedium
r(t) = 5t+2s(s + 4)C(s)−

A ramp r(t) = 5t is applied to a unity-feedback loop with G(s) = 2/(s(s + 4)). Find the steady-state error ess.

G(s) = 2/(s(s + 4))r(t) = 5t
ess =
40.Steady-state error & PID controlHard
R(s)+K = 55s(s + 4)C(s)−

A proportional controller K = 5 drives the plant G(s) = 5/(s(s + 4)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 5/(s(s + 4))K = 5
%OS = %

3Solutions

1.Series & parallel resistors

Answer: IR2 = 3.268 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​ = 3 ΩR3​ = 15 Ω+−V1​24 VR2​4 ΩAR4​20 ΩB

Combine resistors from the far end

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

R3 + R4 = 15 + 20 = 35 Ω
R2 ∥ 35 = 4·35/(4+35) = 3.59 Ω
Req = R1 + 3.59 = 3 + 3.59 = 6.59 Ω

Current from the source

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

IR1 = V1/Req = 24/6.59 = 3.642 A

Current through R₂

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

VA = V1 − IR1·R1 = 24 − 3.642·3 = 13.07 V
IR2 = VA/R2 = 13.07/4 = 3.268 A

Answer

IR2 = 3.268 A
2.Series & parallel resistors

Answer: Req = 9.809 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​ = 8 ΩR3​ = 12 ΩR5​ = 8 Ω+−V1​10 VR2​2 ΩAR4​10 ΩBR6​15 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 + 15 = 23 Ω
R4 ∥ 23 = 6.97 Ω
R3 + 6.97 = 18.97 Ω
R2 ∥ 18.97 = 1.809 Ω
Req = R1 + 1.809 = 8 + 1.809 = 9.809 Ω

Answer

Req = 9.809 Ω
3.Nodal analysis

Answer: IR3 = −0.5841 A

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

R1​ = 8 ΩR3​ = 6 Ω+−V1​9 VR2​2 ΩAR4​15 ΩBIs​1 A

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 9 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/2 + (VA − 9)/8 + (VA − VB)/6 = 0
0.7917·VA − 0.1667·VB = 1.125

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/15 + (VB − VA)/6 = 1
−0.1667·VA + 0.2333·VB = 1

Solve

Δ = (0.7917)(0.2333) − (−0.1667)(−0.1667) = 0.1569
VA = Δ1/Δ = 0.4292/0.1569 = 2.735 V
VB = Δ2/Δ = 0.9792/0.1569 = 6.239 V

Answer

IR3 = (VA − VB)/R3 = (2.735 − 6.239)/6 = −0.5841 A
4.Nodal analysis

Answer: VA = 6.667 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.

Form a supernode

The voltage source between A and B carries an unknown current, so KCL cannot be written at A or B separately. Enclose A, B and the source in one supernode: the current entering it equals the current leaving it.

VB = VA − 12
(VA − V1)/R1 + VA/R2 + VB/R3 = Is

Solve

(VA − 12)/8 + VA/2 + (VA − 12)/8 = 2
VA·(1/8 + 1/2 + 1/8) = 2 + 12/8 + 12/8 = 5
VA = 5/0.75 = 6.667 V
VB = −5.333 V

Answer

VA = 6.667 V
5.Mesh analysis

Answer: I1 = −0.07097 A

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

R1​ = 15 ΩR2​ = 15 Ω+−V1​9 VR3​8 ΩA+−V2​30 VI1​I2​

Assign mesh currents

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

KVL around mesh 1

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

R1·I1 + R3(I1 − I2) = V1
15·I1 + 8(I1 − I2) = 9
23·I1 − 8·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 + R3(I2 − I1) = −V2
15·I2 + 8(I2 − I1) = (−30)
−8·I1 + 23·I2 = −30

Solve

Δ = (23)(23) − (−8)(−8) = 465
I1 = Δ1/Δ = −33/465 = −0.07097 A
I2 = Δ2/Δ = −618/465 = −1.329 A

Answer

I1 = −0.07097 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: I3 = −0.72 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​ = 30 ΩR2​ = 20 ΩR3​ = 5 Ω+−V1​36 VR4​15 ΩAR5​10 ΩB+−V2​12 VI1​I2​I3​

Assign mesh currents

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

KVL around mesh 1

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

R1·I1 + R4(I1 − I2) = V1
30·I1 + 15(I1 − I2) = 36
45·I1 − 15·I2 = 36

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
20·I2 + 15(I2 − I1) + 10(I2 − I3) = 0
−15·I1 + 45·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
5·I3 + 10(I3 − I2) = (−12)
−10·I2 + 15·I3 = −12

Solve

Δ = det of the coefficient matrix = 22500
I1 = Δ1/Δ = 18900/22500 = 0.84 A
I2 = Δ2/Δ = 2700/22500 = 0.12 A
I3 = Δ3/Δ = −16200/22500 = −0.72 A

Answer

I3 = −0.72 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: VR4 = 6.682 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​ = 10 ΩR3​ = 10 Ω+−V1​6 VR2​4 ΩAR4​6 ΩBOPEN

Case 1 · V1 acting alone

Iₛ becomes an open circuit, so R₃ and R₄ form one series branch from A to ground, in parallel with R₂.

R3+R4 = 10+6 = 16 Ω
R2 ∥ (R3+R4) = 4·16/(4+16) = 3.2 Ω
VA = 6·3.2/(10+3.2) = 1.455 V
VB = VA·R4/(R3+R4) = 1.455·6/16 = 0.5455 V
VR4′ = VB − 0 = 0.5455 − 0 = 0.5455 V
R1​ = 10 ΩR3​ = 10 ΩSHORTR2​4 ΩAR4​6 ΩBIs​1.5 A

Case 2 · Is acting alone

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

R1 ∥ R2 = 10·4/(10+4) = 2.857 Ω
Req = R4 ∥ (R3+R1∥R2) = 6 ∥ 12.86 = 4.091 Ω
VB = Is·Req = 1.5·4.091 = 6.136 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 6.136·2.857/12.86 = 1.364 V
VR4″ = VB − 0 = 6.136 − 0 = 6.136 V

Add the contributions

VR4 = VR4′ + VR4″
VR4 = 0.5455 + 6.136 = 6.682 V

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

8.Superposition

Answer: V2 = 11.33 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​ = 15 ΩR2​ = 6 ΩR4​ = 8 Ω+−V1​5 VR3​12 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) = 12 ∥ 14 = 6.462 Ω
VA = 5·6.462/(15+6.462) = 1.505 V
VB = VA·R4/(R2+R4) = 1.505·8/14 = 0.8602 V
R1​ = 15 ΩR2​ = 6 ΩR4​ = 8 ΩSHORTR3​12 ΩAIs​2 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. The arrow of Iₛ points down, away from its node, so Iₛ draws current out of that node: its term enters with a minus sign.

R1 ∥ R3 = 6.667 Ω, so the left path is R2+6.667 = 12.67 Ω
Req = R4 ∥ 12.67 = 4.903 Ω
VB = −Is·Req = −2·4.903 = −9.806 V
VA = VB·6.667/12.67 = −5.161 V

Case 3 · V2 acting alone, per volt

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

R1 ∥ R3 = 6.667 Ω
VA‴ = V2·6.667/(8 + 6 + 6.667) = 0.3226·V2

Set VA to zero

With VA = 0 there is no voltage across R₃, so its current is zero.

VA′ + VA″ + VA‴ = 1.505 + (−5.161) + 0.3226·V2 = 0
V2 = 3.656/0.3226 = 11.33 V

Answer

V2 = 11.33 V
9.Thévenin & Norton

Answer: Rth = 8.4 Ω

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

R1​ = 3 ΩR3​ = 6 ΩSHORTR2​12 ΩAabRth​

Find Rth with every source switched off

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

R1 ∥ R2 = 3·12/(3+12) = 2.4 Ω
Rth = (R1 ∥ R2) + R3 = 2.4 + 6 = 8.4 Ω
10.Thévenin & Norton

Answer: IL = 0.3964 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​ = 2 ΩR3​ = 30 ΩSHORTR2​4 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 = 2·4/(2+4) = 1.333 Ω
Rth = (R1 ∥ R2) + R3 = 1.333 + 30 = 31.33 Ω
R1​ = 2 ΩR3​ = 30 Ω+−V1​10 VR2​4 ΩAIs​0.5 ABab+−Vth​

Find Vth, the open-circuit voltage

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

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 6.667 + 0.5·1.333 = 7.333 V
Vth = VB = VA + Is·R3 = 7.333 + 0.5·30 = 22.33 V
Rth​ = 31.33 Ω+−Vth​22.33 VRL​25 ΩIL​

Connect the load to the Thévenin equivalent

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

IL = Vth/(Rth + RL) = 22.33/(31.33 + 25) = 0.3964 A
11.RC & RL transients

Answer: vL = 5.951 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 = 33 mH / 20 Ω = 1.65 ms
i(∞) = Vs/R = 20/20 = 1 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 = 20·e−2/1.65 = 5.951 V

Answer

vL = 5.951 V
12.RC & RL transients

Answer: vC = 6.417 V

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

Initial value, final value, time constant

The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.

vC(0) = 8 V, vC(∞) = Vs = 5 V
τ = R·C = 1 kΩ · 100 µF = 100 ms
vC(t) = 5 + (8 − 5)·e−t/τ

Voltage at the given time

vC(75) = 5 + (3)·e−75/100 = 6.417 V

Answer

vC = 6.417 V
13.AC impedance & phasors

Answer: |Z| = 31.18 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

XC = 1/(ωC) = 1/(2π·100·47·10⁻⁶) = 33.86 Ω

Branch currents

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

IR = V/R = 120/80 = 1.5 A
IX = V/XC = 120/33.86 = 3.544 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) = √(1.5² + 3.544²) = 3.848 A

Impedance

|Z| = V/I = 120/3.848 = 31.18 Ω

Answer

|Z| = 31.18 Ω
14.AC impedance & phasors

Answer: I = 0.9221 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 = 628.3·0.15 = 94.25 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) = √(50² + (−244.4)²) = 249.4 Ω

Current

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

I = V/|Z| = 230/249.4 = 0.9221 A

Answer

I = 0.9221 A
15.AC power & power factor

Answer: I = 184.8 A

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

Line current

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

I = P/(V·cos φ) = 34000/(230·0.8) = 184.8 A

Answer

I = 184.8 A
16.AC power & power factor

Answer: C = 265.3 µF

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

Conjugate match

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

ZL = Zs* = 10 − j10 Ω

Capacitance

XC = 10 Ω → C = 1/(2π·60·10) = 265.3 µF

Answer

C = 265.3 µF
17.Three-phase power

Answer: IL = 66.92 A

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.

Electrical input power

Pin = Pout/η = 37/0.95 = 38.95 kW

Line current

IL = Pin/(√3·VL·cos φ) = 38947/(√3·400·0.84) = 66.92 A

Answer

IL = 66.92 A
18.Three-phase power

Answer: Cph = 1201 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 = 75·tan(acos 0.65) = 75·1.169 = 87.68 kvar
Q2 = 75·tan(acos 0.98) = 75·0.2031 = 15.23 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 72.46 kvar

Capacitance per phase

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

V = 400/√3 = 230.9 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 24152/(377·230.9²) = 1201 µF

Answer

Cph = 1201 µF
19.Transformers

Answer: I1 = 0.125 A

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

Secondary voltage and current

V2 = 2400/20 = 120 V
I2 = 120/48 = 2.5 A

Primary current

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

I1 = I2/20 = 0.125 A

Answer

I1 = 0.125 A

Check: the load seen from the primary is a²·RL = 19200 Ω, and 2400/19200 gives the same current.

20.Transformers

Answer: Ploss = 275 W

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

Copper loss at this load

Pcu = x²·Pcu,fl = 0.25·300 = 75 W

Total losses

The core loss is the same at any load.

Ploss = 200 + 75 = 275 W

Answer

Ploss = 275 W

The efficiency at this load is η = 98.79%.

21.Op-amp circuits

Answer: Vout = −7.745 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₊ = Vin·Rb/(Ra + Rb) = −4·4.7/(22 + 4.7) = −0.7041 V

Non-inverting gain

V₋ = V₊, and the current through R₁ also flows through Rf.

Vout = V₊·(1 + Rf/R1) = −0.7041·(1 + 47/4.7) = −7.745 V

Answer

Vout = −7.745 V
22.Op-amp circuits

Answer: Vout = −3.249 V

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

Input currents

V₋ = 0, so each input resistor has its full input voltage across it.

I1 = V1/R1 = 0.6/4.7 = 0.1277 mA
I2 = V2/R2 = 0.2/10 = 0.02 mA
If = I1 + I2 = 0.1477 mA

Output voltage

Vout = −Rf·If = −22·0.1477 = −3.249 V

Answer

Vout = −3.249 V
23.Diodes & transistors

Answer: VCE = 5.979 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.

Base current

IB = (VCC − VBE)/RB = (9 − 0.7)/680 kΩ = 12.21 µA

Collector current

IC = β·IB = 75·12.21 µA = 0.9154 mA

Collector–emitter voltage

VCE = 9 − 0.9154 mA·3.3 kΩ = 5.979 V

Answer

VCE = 5.979 V
24.Diodes & transistors

Answer: IC = 12.94 mA

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

Base current

IB = (5 − 0.7)/10 kΩ = 430 µA

Saturated or active?

β·IB is larger than IC(sat), so the transistor saturates.

β·IB = 64.5 mA
IC(sat) = (9 − 0.2)/680 = 12.94 mA

Collector current

IC = IC(sat) = 12.94 mA

Answer

IC = 12.94 mA

The transistor is saturated.

25.Number systems

Answer: N = 41

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.

32 + 8 + 1 = 41

Answer

N = 41
26.Number systems

Answer: M = 600 KiB

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

Bits and bytes

640·480·16 = 4915200 bit = 614400 B

Convert to KiB

614400/1024 = 600 KiB

Answer

M = 600 KiB
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.

ABCD000111100001111000110302140507060120130151140819011010

Group the 0s

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

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

Minimal POS form

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

Answer

N = 5

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

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

Answer: Q = 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.

Full cycles

Every 16 pulses bring the counter back to the same state.

18 = 1·16 + 2

Count after the remaining pulses

Q = (6 − 2) mod 16 = 4
Q = 0100₂

Answer

Q = 4
29.Laplace transforms

Answer: A = 7.5

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 + 10s + 24 = (s + 4)(s + 6)
F(s) = A/(s + 4) + B/(s + 6)

Cover-up rule

Multiply by (s + 4) and set s = −4: the B-term vanishes.

A = (6s + 39)/(s + 6) |s = −4 = 15/2 = 7.5

Answer

A = 7.5
(B = −1.5, so f(t) = 7.5·e−4t − 1.5·e−6t)
30.Laplace transforms

Answer: y(1.2) = 2

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

(s² + 4s + 3)·Y(s) = 2s + 8 + 6/s
Y(s) = (2s² + 8s + 6)/(s(s + 1)(s + 3))

Partial fractions

A = 6/(1·3) = 2
B = 0/((−1)·2) = 0
C = 0/((−3)·(−2)) = 0

Result

y(t) = 2
y(1.2) = 2

Answer

y(1.2) = 2

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

31.Block diagrams

Answer: K = 12.25

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 + 4)(s + 11)) = 0 → s² + 15s + 44 + K = 0

Repeated root

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

15² = 4(44 + K) → K = 225/4 − 44 = 12.25

Answer

K = 12.25
Both closed-loop poles are then at s = −7.5.
32.Block diagrams

Answer: css = 1.944

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) = 7/(s + 5 + 49)
C/R = K·G/(1 + K·G) = 49/(s + 5 + 49)

Steady state

css = 2·49/54 + 1·7/54 = 1.944

Answer

css = 1.944

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

33.Step response

Answer: %OS = 48.22 %

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 = 44 → ωₙ = 6.633 rad/s
2ζωₙ = 3 → ζ = 3/(2·6.633) = 0.2261

Result

%OS = 100·e−ζπ/√(1−ζ²) = 100·e−0.7293 = 48.22 %

Answer

%OS = 48.22 %
34.Step response

Answer: ts = 0.8 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).

Closed-loop transfer function

T(s) = 170/(s² + 10s + 170): ωₙ = 13.04 rad/s, ζ = 0.3835

Settling time

ζωₙ = 10/2 = 5
ts ≈ 4/(ζωₙ) = 0.8 s

Answer

ts = 0.8 s

For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.

35.Bode plots & margins

Answer: |G(jω)| = −0.78 dB

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

Multiply the magnitudes

|G(j30)| = 882/(√(30² + 7²)·√(30² + 9²)) = 882/(30.81·31.32) = 0.9141

Result

20·log₁₀(0.9141) = −0.78 dB

Answer

|G(jω)| = −0.78 dB
36.Bode plots & margins

Answer: GM = 2.007 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.

Phase crossover

(ω/7)·(ω/9) = 1 → ω180 = √(7·9) = 7.937 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 49)·√(ω² + 81)) = K/(7·9·(7 + 9)) = K/1008

Gain margin

GM = 20·log₁₀(1008/800) = 2.007 dB

Answer

GM = 2.007 dB
37.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 90

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

Characteristic polynomial

s³ + 7s² + 14s + 8 + K = 0

Routh array

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

Condition

The s¹ entry must stay positive.

K < 7·14 − 8 = 90

Answer

Kmax = 90

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

38.Stability: Routh–Hurwitz & root locus

Answer: K = 63

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.

Magnitude condition

K = |s|·|s + 6|·|s + 10|

Evaluate at the point

K = 3·3·7 = 63

Answer

K = 63

Check: s(s + 6)(s + 10) + K is zero at s = −3.

39.Steady-state error & PID control

Answer: ess = 10

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

Velocity constant

Kv = lim s·G(s) = 0.5

Ramp error

ess = 5/0.5 = 10

Answer

ess = 10
40.Steady-state error & PID control

Answer: %OS = 25.38 %

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 + 4) + 5·5 = s² + 4s + 25

ωₙ and ζ

ωₙ = √25 = 5 rad/s
ζ = 4/(2·5) = 0.4

Overshoot

%OS = 100·e−0.4π/√(1 − 0.4²) = 25.38 %

Answer

%OS = 25.38 %

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