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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
Start at the end of the network furthest from the terminals.
Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
Redraw the circuit after each step.
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
Choose a reference node (ground) and label the other node voltages.
Write Kirchhoff’s current law at every unknown node: currents leaving through each branch equal the currents entering.
Express every branch current as (Va − Vb)/R.
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
Draw a clockwise current in every mesh (window) of the circuit.
Apply Kirchhoff’s voltage law around each mesh: the sum of voltage drops is zero.
In a resistor shared by two meshes the current is the difference of the two mesh currents.
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
Count the independent sources. You will solve one simpler circuit per source.
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).
Solve the simpler circuit for the voltage or current you need, keeping track of its direction.
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
Remove the load and mark the two terminals a and b.
Find the open-circuit voltage between a and b: this is Vth.
Switch off all independent sources and find the resistance seen from a–b: this is Rth.
Put the load back on the equivalent circuit: Vth in series with Rth.
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
Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
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
Convert the frequency: ω = 2πf.
Find each reactance: XL = ωL and XC = 1/(ωC).
Add impedances like resistors, with Z = R + jX.
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
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.
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.
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.
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
Write the closed-loop characteristic polynomial: the denominator of 1 + K·G(s) = 0, multiplied out.
Put the coefficients alternately into the first two rows of the Routh array and build the remaining rows.
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.
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
Count the poles of the open loop C(s)·G(s) at s = 0: that is the system type.
Work out the error constant that matches the input: Kp for a step, Kv for a ramp, Ka for a parabola.
Apply ess = A/(1 + Kp), A/Kv or A/Ka, or solve it for the gain you need.
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
Find the current IR1 delivered by the source, which flows through R1.
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
V1 = 15 VR1 = 6 ΩR2 = 5 ΩR3 = 10 Ω
IN = A
10.Thévenin & NortonHard
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
V1 = 30 VIs = 4 AR1 = 30 ΩR2 = 20 ΩR3 = 2 Ω
IN = A
11.RC & RL transientsMedium
The inductor carries a steady current of 1.5 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. How long does it take for the current to fall to 0.15 A? Find t.
i(0) = 1.5 AR = 68 ΩL = 22 mH
t = ms
12.RC & RL transientsHard
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 = 300 ms.
Vs = 24 VR1 = 1 kΩR2 = 3.3 kΩC = 68 µFt = 300 ms
vC = V
13.AC impedance & phasorsMedium
A resistor and a capacitor are connected in parallel across a 12 V (rms), 100 Hz supply. Find the magnitude of the impedance |Z|.
V = 12 V (rms)f = 100 HzR = 10 ΩC = 100 µF
|Z| = Ω
14.AC impedance & phasorsHard
A series RLC circuit is connected to a 50 V (rms), 200 Hz supply. Find the resonant frequency f0 of the circuit.
V = 50 V (rms)f = 200 HzR = 10 ΩL = 50 mHC = 4.7 µF
f0 = Hz
15.AC power & power factorMedium
A load with impedance Z = 7 + j24 Ω is connected to a 120 V (rms) single-phase supply. Find the apparent power S.
V = 120 VZ = 7 + j24 Ω
S = VA
16.AC power & power factorHard
A 12 V (rms), 50 Hz source has an internal impedance Zs = 8 + j15 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 = 12 V (rms)f = 50 HzZs = 8 + j15 Ω
C = µF
17.Three-phase powerMedium
A balanced Δ-connected load with an impedance of 8 + j6 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 400 V. Find the total real power P absorbed by the load in kW.
VL = 400 VZ = 8 + j6 Ω per phase
P = kW
18.Three-phase powerHard
A balanced three-phase load takes 300 kW at a power factor of 0.8 lagging from a 480 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 = 300 kWpf1 = 0.8pf2 = 0.98VL = 480 Vf = 60 Hz
Cph = µF
19.TransformersMedium
A single-phase 500 kVA, 6600/230 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.
S = 500 kVAV1 = 6600 VV2 = 230 V
ILV = A
20.TransformersHard
A single-phase 200 kVA transformer has a core loss of 600 W and a copper loss of 1500 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
S = 200 kVAPcore = 600 WPcu,fl = 1500 W
Sη,max = kVA
21.Op-amp circuitsMedium
The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±14 V. The non-inverting amplifier is driven by a sine wave. Find the largest input amplitude Vin,max (peak value) for which the output does not clip.
R1 = 10 kΩRf = 47 kΩVsat = ±14 V
Vin,max = V
22.Op-amp circuitsHard
The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±13 V. The capacitor is uncharged when a constant input Vin = 2.3 V is applied at t = 0. Find the output voltage Vout at t = 50 ms.
A 9.1 V Zener regulator is fed from 24 V through 270 Ω and supplies a 680 Ω load. Find the Zener current IZ.
Vin = 24 VVZ = 9.1 VR = 270 ΩRL = 680 Ω
IZ = mA
24.Diodes & transistorsHard
A 5.6 V Zener regulator must supply up to 50 mA to a load. The input can drop to 18 V, and the Zener needs at least 2 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.
A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 0, I₁ = C, I₂ = C′, I₃ = 1. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.
I₀ = 0I₁ = CI₂ = C′I₃ = 1
N =
28.Digital logic: gates, K-maps & flip-flopsHard
A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 7 clock edges.
Q₃Q₂Q₁Q₀ = 10007
Q =
29.Laplace transformsMedium
Find f(t) for F(s) = (2s + 33)/(s² + 4s + 4) and evaluate f(0.3).
F(s) = (2s + 33)/(s² + 4s + 4)t = 0.3 s
f(0.3) =
30.Laplace transformsHard
Solve y″ + 11y′ + 30y = 0 with y(0) = 1 and y′(0) = 11 using the Laplace transform, and find y(0.1).
y″ + 11y′ + 30y = 0y(0) = 1y′(0) = 11t = 0.1 s
y(0.1) =
31.Block diagramsMedium
A plant G(s) = 25/(s + 8) is controlled in a negative feedback loop whose sensor has the gain H = 0.1. Find the time constant τ of the closed loop.
G(s) = 25/(s + 8)H = 0.1
τ = s
32.Block diagramsHard
The plant G(s) = 40/(s(s + 7)) is in a negative feedback loop with sensor gain H = 0.5. The closed-loop poles are complex. Find their imaginary part ωd (the positive value), in rad/s.
G(s) = 40/(s(s + 7))H = 0.5
ωd = rad/s
33.Step responseMedium
A system has the transfer function G(s) = 43.5/(s² + 13s + 87) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 43.5/(s² + 13s + 87)
%OS = %
34.Step responseHard
A unity negative feedback loop has G(s) = K/(s(s + 8)). Find the gain K that gives a step response with 30 % overshoot.
G(s) = K/(s(s + 8))%OS = 30 %
K =
35.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 42(s + 12)/(s(s + 9)) at ω = 2 rad/s, in degrees.
G(s) = 42(s + 12)/(s(s + 9))ω = 2 rad/s
∠G(jω) = °
36.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 80/(s(s + 8)). Find the phase margin PM in degrees.
L(s) = 80/(s(s + 8))
PM = °
37.Stability: Routh–Hurwitz & root locusMedium
The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 1)(s + 6)). The gain is raised until the closed loop is on the edge of stability. Find the frequency ω of the sustained oscillation.
L(s) = K/(s(s + 1)(s + 6))
ω = rad/s
38.Stability: Routh–Hurwitz & root locusHard
The root locus of L(s) = K(s + 14)/(s(s + 7)) leaves the real axis between s = 0 and s = −7, circles round and joins the real axis again to the left of the zero. Find the break-away point sb.
L(s) = K(s + 14)/(s(s + 7))
sb =
39.Steady-state error & PID controlMedium
A process G(s) = 5/(2s + 1) is controlled by a proportional gain K = 20 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.
G(s) = 5/(2s + 1)K = 20
ess = %
40.Steady-state error & PID controlHard
A proportional controller K = 16 drives the plant G(s) = 4/(s(s + 10)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.
G(s) = 4/(s(s + 10))K = 16
%OS = %
3Solutions
1.Series & parallel resistors
Answer: IR1 = 0.8242 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.
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 = 4 + 12 = 16 Ω
R2 ∥ 16 = 4·16/(4+16) = 3.2 Ω
Req = R1 + 3.2 = 15 + 3.2 = 18.2 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 = 15/18.2 = 0.8242 A
Answer
IR1 = 0.8242 A
2.Series & parallel resistors
Answer: IR2 = 1.66 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.
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 = 30 + 15 = 45 Ω
R4 ∥ 45 = 6.792 Ω
R3 + 6.792 = 11.79 Ω
R2 ∥ 11.79 = 4.766 Ω
Req = R1 + 4.766 = 6 + 4.766 = 10.77 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 = 30/10.77 = 2.786 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 = 30 − 2.786·6 = 13.28 V
IR2 = VA/R2 = 13.28/8 = 1.66 A
Answer
IR2 = 1.66 A
3.Nodal analysis
Answer: IR3 = −1.066 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.
Set up the nodes
Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 30 V). Unknowns: VA, VB.
KCL at node A
Currents leaving A through the resistors equal the current pushed into A by sources.
VA/R2 + (VA − V1)/R1 + (VA − VB)/R3 = 0
VA/2 + (VA − 30)/15 + (VA − VB)/6 = 0
0.7333·VA − 0.1667·VB = 2
KCL at node B
Currents leaving B through the resistors equal the current pushed into B by sources.
VB/R4 + (VB − VA)/R3 = Is
VB/3 + (VB − VA)/6 = 5
−0.1667·VA + 0.5·VB = 5
Solve
Δ = (0.7333)(0.5) − (−0.1667)(−0.1667) = 0.3389
VA = Δ1/Δ = 1.833/0.3389 = 5.41 V
VB = Δ2/Δ = 4/0.3389 = 11.8 V
Answer
IR3 = (VA − VB)/R3 = (5.41 − 11.8)/6 = −1.066 A
4.Nodal analysis
Answer: Is = 0.6 A
Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.
KCL at node A
VA is known, so this equation gives VB.
(VA − V1)/R1 + VA/R2 + (VA − VB)/R3 = 0
(14 − 36)/15 + 14/15 = (VB − 14)/5
−0.5333 = (VB − 14)/5 → VB = 11.33 V
KCL at node B
Currents leaving B through the resistors equal the current pushed into B by sources.
Is = (VB − VA)/R3 + VB/R4 = (11.33 − 14)/5 + 11.33/10 = 0.6 A
Answer
Is = 0.6 A
5.Mesh analysis
Answer: I1 = −0.3448 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.
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
12·I1 + 5(I1 − I2) = 10
17·I1 − 5·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
5·I2 + 5(I2 − I1) = (−30)
−5·I1 + 10·I2 = −30
Solve
Δ = (17)(10) − (−5)(−5) = 145
I1 = Δ1/Δ = −50/145 = −0.3448 A
I2 = Δ2/Δ = −460/145 = −3.172 A
Answer
I1 = −0.3448 A
A negative mesh current flows anticlockwise.
6.Mesh analysis
Answer: VR1 = 25.45 V
Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.
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
12·I1 + 6(I1 − I2) = 36
18·I1 − 6·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
12·I2 + 6(I2 − I1) + 10(I2 − I3) = 0
−6·I1 + 28·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
15·I3 + 10(I3 − I2) = (−10)
−10·I2 + 25·I3 = −10
Solve
Δ = det of the coefficient matrix = 9900
I1 = Δ1/Δ = 21000/9900 = 2.121 A
I2 = Δ2/Δ = 3600/9900 = 0.3636 A
I3 = Δ3/Δ = −2520/9900 = −0.2545 A
Answer
IR1 = I1 = 2.121 A
VR1 = IR1·R1 = 2.121·12 = 25.45 V
A negative mesh current flows anticlockwise.
7.Superposition
Answer: IR1 = −1.415 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.
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 = 4+12 = 16 Ω
R2 ∥ (R3+R4) = 10·16/(10+16) = 6.154 Ω
VA = 30·6.154/(2+6.154) = 22.64 V
VB = VA·R4/(R3+R4) = 22.64·12/16 = 16.98 V
IR1′ = (VA − V1)/R1 = (22.64 − 30)/2 = −3.679 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 = 2·10/(2+10) = 1.667 Ω
Req = R4 ∥ (R3+R1∥R2) = 12 ∥ 5.667 = 3.849 Ω
VB = Is·Req = 4·3.849 = 15.4 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 15.4·1.667/5.667 = 4.528 V
IR1″ = (VA − V1)/R1 = (4.528 − 0)/2 = 2.264 A
Add the contributions
IR1 = IR1′ + IR1″
IR1 = (−3.679) + 2.264 = −1.415 A
A negative result means the real current flows against the arrow.
8.Superposition
Answer: IR3 = −0.4356 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.
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 ∥ 9 = 5.143 Ω
VA = 30·5.143/(20+5.143) = 6.136 V
VB = VA·R4/(R2+R4) = 6.136·4/9 = 2.727 V
IR3′ = (VA − 0)/R3 = (6.136 − 0)/12 = 0.5114 A
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 = 7.5 Ω, so the left path is R2+7.5 = 12.5 Ω
Req = R4 ∥ 12.5 = 3.03 Ω
VB = Is·Req = (−2.5)·3.03 = −7.576 V
VA = VB·7.5/12.5 = −4.545 V
IR3″ = (VA − 0)/R3 = (−4.545 − 0)/12 = −0.3788 A
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 = 20·12/(20+12) = 7.5 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = (−15)·12.5/16.5 = −11.36 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = −11.36·7.5/12.5 = −6.818 V
IR3‴ = (VA − 0)/R3 = (−6.818 − 0)/12 = −0.5682 A
Add the contributions
IR3 = IR3′ + IR3″ + IR3‴
IR3 = 0.5114 + (−0.3788) + (−0.5682) = −0.4356 A
A negative result means the real current flows against the arrow.
9.Thévenin & Norton
Answer: IN = 0.5357 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.
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 + 10 = 12.73 Ω
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) = 15·5/(6+5) = 6.818 V
Norton equivalent
A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.
IN = Vth/Rth = 6.818/12.73 = 0.5357 A
10.Thévenin & Norton
Answer: IN = 4.857 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.
Find Rth with every source switched off
Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.
R1 ∥ R2 = 30·20/(30+20) = 12 Ω
Rth = (R1 ∥ R2) + R3 = 12 + 2 = 14 Ω
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) = 12 + 4·12 = 60 V
Vth = VB = VA + Is·R3 = 60 + 4·2 = 68 V
Norton equivalent
A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.
IN = Vth/Rth = 68/14 = 4.857 A
11.RC & RL transients
Answer: t = 0.745 ms
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Time constant
τ = L/R = 22 mH / 68 Ω = 0.3235 ms
Solve for the time
0.15 = 1.5·e−t/τ → t = τ·ln(1.5/0.15)
t = 0.3235·ln(10) = 0.745 ms
Answer
t = 0.745 ms
12.RC & RL transients
Answer: vC = 4.838 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) = 24·3.3/(1 + 3.3) = 18.42 V
Time constant after the switch opens
τ = R2·C = 3.3·68 = 224.4 ms
Discharge
vC(t) = 18.42·e−300/224.4 = 4.838 V
Answer
vC = 4.838 V
13.AC impedance & phasors
Answer: |Z| = 8.467 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·100·100·10⁻⁶) = 15.92 Ω
Branch currents
Both branches are connected straight across the supply, so each sees the full voltage V.
IR = V/R = 12/10 = 1.2 A
IX = V/XC = 12/15.92 = 0.754 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) = √(1.2² + 0.754²) = 1.417 A
Impedance
|Z| = V/I = 12/1.417 = 8.467 Ω
Answer
|Z| = 8.467 Ω
14.AC impedance & phasors
Answer: f0 = 328.3 Hz
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
Resonance
At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.
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.
Current
|Z| = √(7² + 24²) = 25 Ω
I = V/|Z| = 120/25 = 4.8 A
Apparent power
S = V·I = 120·4.8 = 576 VA
Answer
S = 576 VA
16.AC power & power factor
Answer: C = 212.2 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* = 8 − j15 Ω
Capacitance
XC = 15 Ω → C = 1/(2π·50·15) = 212.2 µF
Answer
C = 212.2 µF
17.Three-phase power
Answer: P = 38.4 kW
Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.
Phase voltage
Vph = VL = 400 V
Phase current
|Z| = √(8² + 6²) = 10 Ω
Iph = 400/10 = 40 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·40²·8 = 38.4 kW
Answer
P = 38.4 kW
Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.
18.Three-phase power
Answer: Cph = 1889 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 = 300·tan(acos 0.8) = 300·0.75 = 225 kvar
Q2 = 300·tan(acos 0.98) = 300·0.2031 = 60.92 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 164.1 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.
Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.
Rated current
Both windings carry the same apparent power.
ILV = S/V = 500000/230 = 2174 A
Answer
ILV = 2174 A
20.Transformers
Answer: Sη,max = 126.5 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
x²·1500 = 600 → x = √(600/1500) = 0.6325
Load at maximum efficiency
Sη,max = x·S = 0.6325·200 = 126.5 kVA
Answer
Sη,max = 126.5 kVA
21.Op-amp circuits
Answer: Vin,max = 2.456 V
Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.
Gain
Av = 1 + Rf/R1 = 1 + 47/10 = 5.7
Largest input before clipping
Vin,max = Vsat/|Av| = 14/5.7 = 2.456 V
Answer
Vin,max = 2.456 V
22.Op-amp circuits
Answer: Vout = −1.112 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.
The ideal value lies inside ±Vsat, so the integrator has not clipped yet.
Vout = −Vin·t/(RC) = −2.3·50/103.4 = −1.112 V
Answer
Vout = −1.112 V
23.Diodes & transistors
Answer: IZ = 41.8 mA
Zener regulator: a reverse-biased Zener diode holds its voltage VZ while its current stays above a minimum IZ,min. The series resistor R takes the difference between input and output: IR = (Vin − VZ)/R. That current splits between the load, IL = VZ/RL, and the Zener: IZ = IR − IL. Design for the worst cases: the lowest input with the largest load must still leave IZ,min, and the highest input with no load gives the largest Zener power PZ = VZ·IZ.
Resistor current
IR = 14.9/270 = 55.19 mA
Load current
IL = 9.1/680 = 13.38 mA
Zener current
IZ = 55.19 − 13.38 = 41.8 mA
Answer
IZ = 41.8 mA
24.Diodes & transistors
Answer: Rmax = 238.5 Ω
Zener regulator: a reverse-biased Zener diode holds its voltage VZ while its current stays above a minimum IZ,min. The series resistor R takes the difference between input and output: IR = (Vin − VZ)/R. That current splits between the load, IL = VZ/RL, and the Zener: IZ = IR − IL. Design for the worst cases: the lowest input with the largest load must still leave IZ,min, and the highest input with no load gives the largest Zener power PZ = VZ·IZ.
Worst-case current
IR = 50 + 2 = 52 mA
Largest resistance
Rmax = 12.4/0.052 = 238.5 Ω
Answer
Rmax = 238.5 Ω
25.Number systems
Answer: U = 183
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Invert and add one
73 = 0100 1001
inverted: 1011 0110
+1 → 1011 0111
Read it as unsigned
1011 0111₂ = 256 − 73 = 183
Answer
U = 183
26.Number systems
Answer: M = 1200 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
640·480·32 = 9830400 bit = 1228800 B
Convert to KiB
1228800/1024 = 1200 KiB
Answer
M = 1200 KiB
27.Digital logic: gates, K-maps & flip-flops
Answer: N = 4
Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.
Output expression
F = A′BC + AB′C′ + AB
Rows with F = 1
ABC = 011, 100, 110, 111
N = 4
Answer
N = 4
28.Digital logic: gates, K-maps & flip-flops
Answer: Q = 1
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.
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 + 4s + 4 = (s + 2)²
Partial fractions
2s + 33 = 2(s + 2) + 29
F(s) = 2/(s + 2) + 29/(s + 2)²
Inverse transform
f(t) = 2·e−2t + 29·t·e−2t
f(0.3) = (2 + 29·0.3)·e−2·0.3 = 5.872
Answer
f(0.3) = 5.872
30.Laplace transforms
Answer: y(0.1) = 1.53
Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).
Transform the equation
s²Y − 1s − 11 + 11(sY − 1) + 30Y = 0
Y(s) = (s + 22)/((s + 5)(s + 6))
Partial fractions
A = 17/1 = 17
B = 16/(−1) = −16
Result
y(t) = 17·e−5t − 16·e−6t
y(0.1) = 1.53
Answer
y(0.1) = 1.53
Check: y(0) = A + B = 1.
31.Block diagrams
Answer: τ = 0.09524 s
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).
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 + 7) + 40·0.5 = 0 → s² + 7s + 20 = 0
Roots
s = −3.5 ± j·√(20 − 12.25) = −3.5 ± j2.784
ωd = 2.784 rad/s
Answer
ωd = 2.784 rad/s
33.Step response
Answer: %OS = 4.724 %
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 = 87 → ωₙ = 9.327 rad/s
2ζωₙ = 13 → ζ = 13/(2·9.327) = 0.6969
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−3.053 = 4.724 %
Answer
%OS = 4.724 %
34.Step response
Answer: K = 124.9
Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).
Closed-loop transfer function
T(s) = K/(s² + 8s + K)
Damping ratio from the overshoot
ζ = −ln(0.3)/√(π² + ln²(0.3)) = 0.3579
Gain
ωₙ = 8/(2ζ) = 11.18 rad/s
K = ωₙ² = 124.9
Answer
K = 124.9
35.Bode plots & margins
Answer: ∠G(jω) = −93.07 °
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.
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 + 64ω² − 6400 = 0
ω² = (−64 + √(4096 + 4·6400))/2 = 54.16
ωc = 7.36 rad/s
Phase margin
∠L(jωc) = −90° − atan(7.36/8) = −132.6°
PM = 180° + ∠L = 47.39°
Answer
PM = 47.39°
37.Stability: Routh–Hurwitz & root locus
Answer: ω = 2.449 rad/s
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.
Routh array
s³ + 7s² + 6s + K = 0
s³: 1, 6
s²: 7, K
s¹: (7·6 − K)/7
s⁰: K
Critical gain
K = 7·6 = 42
Auxiliary equation
At the critical gain the s¹ row is zero; the s² row gives the poles on the imaginary axis.
7s² + 42 = 0 → s = ±j√6
ω = 2.449 rad/s
Answer
ω = 2.449 rad/s
38.Stability: Routh–Hurwitz & root locus
Answer: sb = −4.101
Root locus: as the gain K goes from 0 to ∞, the closed-loop poles, the roots of 1 + K·G(s) = 0, move from the open-loop poles to the open-loop zeros or to infinity. A point on the real axis belongs to the locus when the number of real poles and zeros to its right is odd. The n − m branches that go to infinity follow asymptotes at the angles (2q + 1)·180°/(n − m), which start from the centroid σₐ = (Σ poles − Σ zeros)/(n − m). Branches leave or join the real axis where dK/ds = 0. The angle condition ∠G(s) = 180° fixes the shape, and the magnitude condition K = 1/|G(s)| gives the gain at any point of the locus.
Gain along the real axis
K(s) = −(s² + 7s)/(s + 14)
Stationary points
Differentiate with the quotient rule and set the numerator to zero.
(2s + 7)(s + 14) − (s² + 7s) = s² + 28s + 98 = 0
s = −14 ± √98 = −4.101, −23.9
Break-away point
sb = −4.101
Answer
sb = −4.101
Between the two break points the locus is a circle centred on the zero, with radius √98 = 9.899.
39.Steady-state error & PID control
Answer: ess = 0.9901 %
Steady-state error: in a unity-feedback loop with open-loop transfer function G(s) the error is E(s) = R(s)/(1 + G(s)), and by the final value theorem ess = lims→0 s·E(s). The system type is the number of poles of G(s) at s = 0. The error constants are Kp = lim G(s), Kv = lim s·G(s) and Ka = lim s²·G(s), all for s → 0. A step of size A gives ess = A/(1 + Kp), a ramp A·t gives A/Kv and a parabola A·t²/2 gives A/Ka. A type-0 loop follows a step with a finite error, a type-1 loop follows it exactly and a ramp with a finite error.
Position constant
Kp = 20·5 = 100
Step error
ess = 100/(1 + 100) = 0.9901 %
Answer
ess = 0.9901 %
40.Steady-state error & PID control
Answer: %OS = 8.084 %
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).