Five resistors are wired like this. From point P a 10 ohm element reaches point R and a 15 ohm element reaches point S. From R a 20 ohm element continues to point Q, and from S a 30 ohm element also continues to Q. A 25 ohm element joins R directly to S. An ideal 36 V supply is then applied between P and Q. What current passes through the 25 ohm element, and what current does the supply deliver?

Zero through the 25 ohm element, and 2.0 A from the supply

Zero through the 25 ohm element, and 1.44 A from the supply

0.48 A through the 25 ohm element, and 2.0 A from the supply

Zero through the 25 ohm element, and 2.0 A from the supply

Zero through the 25 ohm element, and 3.44 A from the supply

Over its useful band a system has the phase response theta(omega) = -0.004·omega - 0.2 radians, with omega in radians per second. What is its group delay, and does the phase delay equal it?

4 ms, and the phase delay differs from it at every finite frequency because the phase line does not pass through the origin

0.2 ms, and the phase delay is equal to it

4 ms, and the phase delay differs from it at every finite frequency because the phase line does not pass through the origin

4.2 ms, and the phase delay is equal to it

2 ms, and the phase delay is twice as large at every frequency

A causal continuous-time LTI system has the transfer function H(s) = (s - 4)/((s + 2)(s + 5)). Which of the following statements about it are correct?

The system is BIBO stable, because both poles lie strictly in the left half plane

The system is non-minimum-phase, so its step response may at first move opposite to its final direction

The impulse response is a weighted sum of exp(-2t) and exp(-5t) for t greater than zero

The system is BIBO stable, because both poles lie strictly in the left half plane

The system is non-minimum-phase, so its step response may at first move opposite to its final direction

The zero at s = +4 lies in the right half plane and therefore makes the system unstable

The impulse response is a weighted sum of exp(-2t) and exp(-5t) for t greater than zero

Transients, Laplace, Two-Port Networks and LTI Systems

The previous chapter left the sources on and waited for everything to settle. This one is about the waiting itself, and then about the language that makes the waiting unnecessary. A circuit containing an inductor or a capacitor cannot change state instantly, because inductor current and capacitor voltage are the two quantities that cannot jump without infinite voltage or infinite current respectively — and that single continuity statement supplies the initial conditions for every transient problem in the section. A first-order RL or RC circuit then has exactly one number worth knowing, the time constant τ, and one universal form: the response goes from its initial value to its final value along x(t) = x(∞) + [x(0) - x(∞)]e^(-t/τ), so you never integrate anything, you read off three quantities and write the answer. Second-order RLC circuits add one number, the damping ratio ζ, and with it the whole vocabulary of overdamped, critically damped and underdamped behaviour — including the fact that an underdamped capacitor voltage overshoots the source that is charging it, which is not a paradox but the inductor refusing to stop delivering current. Then Laplace replaces all of this with algebra: an inductor becomes sL in series with a voltage source L i(0), a capacitor becomes 1/sC in series with v(0)/s, initial conditions stop being a separate step, and a network becomes a rational function of s. Two structural tools follow. The wye-delta transformation unlocks the bridge circuits that series-parallel reduction cannot touch, and two-port parameters — z, y, h and ABCD — compress a whole network into four numbers so that cascaded stages multiply instead of interacting. Finally the chapter states in general terms what circuits have been doing all along. An LTI system is defined by linearity and time invariance; it is completely described by its impulse response, its output is the convolution of input with that response, its stability is the absolute integrability of that response, and in the frequency domain it acts as a single complex gain at each frequency. The last piece is the one candidates skip: phase delay and group delay are different quantities that agree only when the phase is linear, and knowing which one a question is asking about is usually the whole question.

First-Order RL and RC: Three Numbers and One Exponential

Everything about a single-energy-storage circuit follows from two continuity statements. The current through an inductor cannot change instantaneously, because v = L di/dt would demand infinite voltage; the voltage across a capacitor cannot change instantaneously, because i = C dv/dt would demand infinite current. So at t = 0⁺ an inductor holds whatever current it held at t = 0⁻ and a capacitor holds whatever voltage it held, and those are your initial conditions — obtained by analysing the circuit before the switch operates, with the inductor treated as a short and the capacitor as an open, because in DC steady state nothing is changing and both derivatives are zero. Then at t → ∞ the circuit is back in DC steady state and you use the same shorts and opens again. That leaves only the middle, and the middle is one exponential.

The universal form is worth writing once and then using for every first-order question you ever see: x(t) = x(∞) + [x(0⁺) - x(∞)] e^(-t/τ), where x is any voltage or current in the circuit and τ is RC for a capacitive circuit or L/R for an inductive one, with R being the Thevenin resistance seen by the storage element — which is where the previous chapter earns its keep, because in anything but a two-element loop that resistance has to be found and not guessed. Three numbers, no integration. A 2 kΩ resistor charging a 5 μF capacitor has τ = 10 ms, so switching a 10 V source onto an initially uncharged capacitor gives v_C(t) = 10(1 - e^(-t/0.01)): 6.32 V after one time constant, 8.65 V after two, 9.50 V after three and 9.93 V after five. The engineering convention that a first-order transient is over after 5τ comes from that last figure — 99.3 per cent, which is inside the tolerance of most components.

🎯 Charging a capacitor through a resistor wastes exactly half the energy, whatever the resistor

Charge that 5 μF capacitor from 10 V through the 2 kΩ and account for the energy. The source pushes a total charge Q = CV = 50 μC at a constant terminal voltage of 10 V, so it delivers QV = 500 μJ. The capacitor ends up storing ½CV² = 250 μJ. The other 250 μJ was dissipated in the resistor — and here is the part that catches people: that figure does not depend on the resistor at all. Integrate it and see. The current is i = (V/R)e^(-t/τ), so P_R = i²R = (V²/R)e^(-2t/τ), and ∫₀^∞ P_R dt = (V²/R)(τ/2) = (V²/R)(RC/2) = ½CV². The R cancels. Use a 2 Ω resistor instead of 2 kΩ and the transient finishes a thousand times sooner, the peak current is a thousand times larger, and the total heat is identical. Two consequences worth carrying. First, the efficiency of charging a capacitor through any resistance is exactly 50 per cent, which is why switched-capacitor charge pumps and capacitive touchscreens are analysed for charge rather than for loss, and why real power supplies charge through an inductor — an inductor stores energy instead of dissipating it, and that is the whole idea of a switching regulator. Second, the same algebra run on an inductor energised through a resistor gives the mirror result, ½LI² stored and ½LI² burned. Energy accounting is often the fastest route to a transient answer, because it needs no time-domain solution at all.

Second-Order RLC: One Damping Ratio Decides Everything

A circuit with both an inductor and a capacitor obeys a second-order equation, and the standard form to normalise it to is s² + 2ζω_n s + ω_n² = 0. For a series RLC circuit ω_n = 1/√(LC) and ζ = (R/2)√(C/L); for the parallel case ω_n is the same but ζ = (1/2R)√(L/C), the reciprocal dependence on R, which is the one asymmetry between the two that is examined. The quality factor is Q = 1/(2ζ), so a series circuit has Q = (1/R)√(L/C) and a parallel circuit Q = R√(C/L). Everything else is read from ζ alone.

One series RLC with L = 10 mH and C = 1 μF, so ωn = 10 000 rad/s (1591.5 Hz), driven at four values of R
Series resistance R	Damping ratio and Q	Character	Roots or ringing frequency
R = 20 Ω	`ζ = 0.1`, `Q = 5`	lightly underdamped	`ω_d = 9950 rad/s`, overshoot 72.9%
R = 100 Ω	`ζ = 0.5`, `Q = 1`	underdamped	`ω_d = 8660 rad/s`, overshoot 16.3%
R = 200 Ω	`ζ = 1`, `Q = 0.5`	critically damped	double root at `-10 000`
R = 400 Ω	`ζ = 2`, `Q = 0.25`	overdamped	real roots `-2679` and `-37 321`

Read the last row as the physical picture, because it explains why overdamping is slow. At ζ = 2 the two real roots are -2679 and -37 321 s⁻¹, a ratio of fourteen: the fast root dies almost immediately and the slow one governs, so the circuit behaves like a first-order circuit with a time constant of 1/2679 = 373 μs. Compare that with the critically damped case, whose double root at -10 000 gives τ = 100 μs. Adding resistance past ζ = 1 makes the circuit slower, not faster — the extra resistance impedes the very current that charges the capacitor. Critical damping is the fastest response with no overshoot, and that is exactly why it is the design target for a moving-coil meter, a hard-disk head actuator or a door closer.

⚠️ The capacitor charges to 11.63 V from a 10 V source, and nothing is wrong

Take the ζ = 0.5 row — L = 10 mH, C = 1 μF, R = 100 Ω — and switch a 10 V DC source onto the series loop at t = 0 with everything initially at rest. Ask for the maximum capacitor voltage. The tempting answer is 10 V, on the entirely reasonable ground that a capacitor charged from a 10 V source ends at 10 V and a voltage cannot exceed its source. The correct answer is 11.63 V. The overshoot of an underdamped second-order step response is M_p = exp(-πζ/√(1-ζ²)), which at ζ = 0.5 is exp(-1.8138) = 0.1630, so the peak is 10 × 1.1630 = 11.63 V, reached at t_p = π/ω_d = π/8660 = 363 μs. Then it rings back down and settles at 10 V, within two per cent after roughly 4/(ζω_n) = 800 μs. The mechanism is worth understanding rather than memorising, because it also tells you when it will not happen. As the capacitor voltage approaches 10 V the loop current is still substantial, and an inductor carrying current cannot be told to stop — its current must decay continuously, so that current keeps flowing into the capacitor and pushes its voltage past the source. Only when the current finally reverses does the capacitor start discharging. Two practical readings follow. A first-order RC circuit can never overshoot, because there is nothing to store momentum; overshoot always means a second energy store. And this is precisely why a long PCB trace or a cable is a hazard for a fast logic edge: the trace inductance and the load capacitance form an underdamped RLC whose overshoot can drive a gate input above its supply rail, and the standard cure is a series termination resistor chosen to push ζ towards 1 — which the table above prices exactly, since going from R = 20 Ω to R = 100 Ω drops the overshoot from 72.9 per cent to 16.3 per cent.

In the frequency domain the same circuit is a resonator. A series RLC driven by a sinusoid has minimum impedance R at ω = ω_n, where the inductive and capacitive reactances cancel exactly; the current peaks there, and the half-power bandwidth is BW = ω_n/Q. At R = 20 Ω in the table, Q = 5 and the bandwidth is 10 000/5 = 2000 rad/s. Note that Q does two jobs at once and they are the same job: it is the sharpness of the resonance in the frequency domain and the persistence of the ringing in the time domain, related by the fact that a narrow band cannot support a fast decay. A high-Q circuit is a good filter and a badly-behaved switch, and one number tells you both.

Laplace: Initial Conditions Stop Being a Separate Step

The syllabus asks specifically for the solution of network equations using Laplace transform, and the reason it is worth a separate mention is that Laplace does something the phasor method cannot: it handles the transient and the steady state in one calculation, with the initial conditions built in rather than fitted afterwards. Transform the element laws with their initial values included and you get s-domain equivalents you can then treat exactly like resistors:

A resistor is R, unchanged, with no initial condition to carry.
An inductor is an impedance sL in series with a voltage source L i(0⁻) whose polarity aids the initial current. Equivalently, sL in parallel with a current source i(0⁻)/s.
A capacitor is an impedance 1/sC in series with a voltage source v(0⁻)/s opposing the charging direction. Equivalently, 1/sC in parallel with a current source C v(0⁻).

🧠 One circuit, one line of algebra, and the answer already contains the initial condition

Take a series RL circuit with R = 4 Ω and L = 1 H that has been sitting on an 8 V supply long enough to settle, so the inductor carries 8/4 = 2 A. At t = 0 the supply steps up to 12 V. In the s domain the loop contains 12/s from the source, L i(0) = 2 V from the inductor's initial condition, and an impedance 4 + s, so I(s) = (12/s + 2)/(s + 4) = (2s + 12)/(s(s + 4)). Partial fractions give the residue at s = 0 as 12/4 = 3 and at s = -4 as (2(-4) + 12)/(-4) = -1, so i(t) = 3 - e^(-4t) A. Check all three things it must satisfy. At t = 0 it gives 3 - 1 = 2 A, the pre-switch current, which is continuity ✓. As t → ∞ it gives 3 A = 12/4, the new DC steady state ✓. And the exponent is -4 = -R/L, so τ = 0.25 s, which is a property of the circuit and not of the source ✓. Compare the classical route: assume a general solution, apply the final value, then apply the initial value to fix the arbitrary constant, then verify. The Laplace route did all of that as one algebraic step, because the L i(0) source is the initial condition. Two further habits make this the fastest method rather than merely the most systematic. The initial value theorem x(0⁺) = lim_(s→∞) sX(s) and the final value theorem x(∞) = lim_(s→0) sX(s) let you check an s-domain answer without inverting it at all — here lim sI(s) = 2 and lim_(s→0) sI(s) = 12/4 = 3, both correct. And the final value theorem is only valid when the poles of sX(s) are strictly in the left half plane, which is a warning and not a footnote: applying it to an oscillator returns a confident, meaningless number.

One more reason Laplace matters here rather than in a mathematics chapter: the ratio of the transformed output to the transformed input, with all initial conditions set to zero, is the transfer function H(s). That object is the bridge to the rest of this chapter and the whole of the next one. Its poles are the natural frequencies of the network — the exponents that appear in the transient, whatever the input — and its zeros are the frequencies the network refuses to transmit. Setting s = jω in H(s) recovers the phasor result of the previous chapter, so the sinusoidal steady state is not a separate theory at all but the transfer function evaluated on the imaginary axis.

Wye-Delta: The Only Tool for a Bridge

Series and parallel reduction handles almost every network, and fails on exactly one common shape: the bridge, in which no two elements are in series (they have a third element joined at their common node) and no two are in parallel (they do not share both nodes). The wye-delta transformation is what unlocks it, by replacing a three-terminal star of three elements with an electrically identical triangle, or the reverse. Going delta to wye, each wye arm is the product of the two adjacent delta arms divided by the sum of all three: R_a = R_ab R_ca/(R_ab + R_bc + R_ca). Going wye to delta, each delta arm is the sum of the pairwise products of the wye arms divided by the opposite wye arm: R_ab = (R_a R_b + R_b R_c + R_c R_a)/R_c. The balanced case is the one to carry in memory as a sanity check: a delta of three equal 3R is a wye of three equal R, so R_Δ = 3R_Y — and if a transformation ever gives you a delta smaller than its wye, you have the formula upside down.

🎯 An unbalanced bridge reduced to 7.2 Ω, then checked by node analysis

Take a bridge with terminals A at the top and D at the bottom and nodes B and C across the middle. The arms are AB = 6 Ω, AC = 18 Ω, BC = 12 Ω, BD = 4 Ω and CD = 8 Ω, and the question is the resistance between A and D. Nothing here is in series or parallel with anything, and the bridge is not balanced (6/18 ≠ 4/8), so the 12 Ω cannot be ignored. Convert the delta A-B-C, whose arms are R_ab = 6, R_bc = 12 and R_ca = 18 summing to 36, into a wye with an internal node N: R_AN = 6(18)/36 = 3 Ω, R_BN = 6(12)/36 = 2 Ω, R_CN = 12(18)/36 = 6 Ω. All three come out whole, which is not luck — the delta values were chosen so the transformation is checkable by eye, and running it backwards confirms it: the pairwise product sum is 6 + 12 + 18 = 36, so 36/6 = 6, 36/3 = 12 and 36/2 = 18, returning the original delta exactly ✓. Now the circuit is a simple tree. From A there is 3 Ω to N; from N two paths run to D, one of 2 + 4 = 6 Ω and one of 6 + 8 = 14 Ω, in parallel giving 6(14)/20 = 4.2 Ω. So R_AD = 3 + 4.2 = 7.2 Ω. Because a wrong transformation still produces a plausible number, check it by a method that shares no arithmetic with it: put 1 V across A-D and solve for the node voltages. Node B gives 6V_B - V_C = 2 after clearing, node C gives -6V_B + 19V_C = 4, and adding them gives 18V_C = 6, so V_C = 1/3 and V_B = 7/18. The current out of A is (1 - 7/18)/6 + (1 - 1/3)/18 = 11/108 + 4/108 = 15/108 = 5/36 A, so the resistance is 36/5 = 7.2 Ω ✓. Two independent routes, one answer. The reason to know both is that the transformation is far faster when the bridge is one part of a larger circuit, while node analysis is faster when you want the branch currents anyway.

Two extensions are worth a sentence each. The transformation is not restricted to resistors — replace every R by an impedance Z and both formulae hold unchanged in the s domain or the phasor domain, which is how three-phase problems move between star and delta connections. And if the bridge is balanced, meaning the ratio of the two arms on one side equals the ratio on the other, the bridge element carries no current at all and can be either removed or short-circuited, whichever makes the remaining reduction easier. Recognising balance first saves the transformation entirely, and it is the first thing to test.

Two-Port Parameters: A Network in Four Numbers

A two-port is any network reached through exactly two pairs of terminals, with four variables at its boundary: V1, I1, V2, I2, conventionally with both currents defined as flowing into the network. Any two of the four can be treated as independent and the other two written as a linear combination, which is why there are several parameter sets rather than one, and why each is convenient for a different situation. The z parameters express the voltages in terms of the currents, so they are measured with ports open-circuited. The y parameters express currents in terms of voltages, measured with ports short-circuited. The h parameters mix the two — V1 and I2 in terms of I1 and V2 — which is precisely the pairing a bipolar transistor makes natural, and is why h_fe is the number printed on a transistor datasheet. The transmission parameters A, B, C and D express the input pair in terms of the output pair, and they exist for one overwhelming reason: cascaded two-ports multiply their transmission matrices, in order, so a chain of stages is one matrix product rather than a simultaneous solution.

Work all four sets out for one network, so that the relations between them are visible rather than looked up. Use the T-network from the previous chapter: R1 = 4 Ω from port 1 to internal node x, R2 = 6 Ω from x to port 2, and R3 = 12 Ω from x to the common reference. The z parameters come straight off the topology, because with a port open no current flows in its series arm: z11 = R1 + R3 = 16 Ω, z22 = R2 + R3 = 18 Ω, and z12 = z21 = R3 = 12 Ω, the shunt arm being the only path common to both loops. The determinant is Δz = 16(18) - 144 = 144.

All four parameter sets for the same 4-6-12 Ω T-network
Parameter set	Diagonal entries	Off-diagonal entries	How each is measured
`z`, open-circuit impedances	`z11 = 16 Ω`, `z22 = 18 Ω`	`z12 = z21 = 12 Ω`	far port left open
`y`, short-circuit admittances	`y11 = 1/8 S`, `y22 = 1/9 S`	`y12 = y21 = -1/12 S`	far port shorted
`h`, hybrid or transistor form	`h11 = 8 Ω`, `h22 = 1/18 S`	`h12 = 2/3`, `h21 = -2/3`, so `h21 = -h12`	output shorted, then input opened
`ABCD`, transmission form	`A = 4/3`, `D = 3/2`, both dimensionless	`B = 12 Ω`, `C = 1/12 S`	output opened, then output shorted

🧠 Every entry in that table can be measured directly, and two of them you already have

Do not convert between parameter sets with formulae you half-remember; go back to the definitions, because each one is a one-line measurement on the circuit. y11 = I1/V1 with port 2 shorted is the input conductance with the output shorted, so 1/y11 = R1 + (R3 ∥ R2) = 4 + (12 ∥ 6) = 4 + 4 = 8 Ω, giving y11 = 1/8 S — and that agrees with the matrix inverse z22/Δz = 18/144 = 1/8 ✓. y22 is the same measurement from the other end: 1/y22 = R2 + (R3 ∥ R1) = 6 + 3 = 9 Ω, so y22 = 1/9 S, matching z11/Δz = 16/144 ✓. Now the entry you already computed. y21 = I2/V1 with port 2 shorted is exactly the reciprocity experiment of the previous chapter: driving port 1 with 12 V and shorting port 2 gave 1 A through the 6 Ω out of port 2, and since the convention has I2 flowing into the port, I2 = -1 A and y21 = -1/12 S ✓. The minus sign is not an error to be corrected; it is the sign convention doing its job, and a candidate who normalises it away will get the wrong sign in every cascade. Finally use the table to read off the two conditions that GATE asks about most. Reciprocity shows as z12 = z21 (12 = 12 ✓), equivalently y12 = y21, equivalently h12 = -h21 (2/3 and −2/3 ✓), equivalently AD - BC = (4/3)(3/2) - 12(1/12) = 2 - 1 = 1 ✓ — four different-looking statements, one property. Symmetry would additionally need z11 = z22, and 16 ≠ 18, so this network is reciprocal but not symmetric, exactly as the previous chapter's asymmetric measurements showed. Note also that B = 12 Ω and 1/C = 12 Ω both equal the shunt arm; for a T-network the transfer impedance is the shunt element, and knowing that lets you write three of the eight parameters without any algebra at all.

Two-port parameters also fix how networks combine, and the rule is different for each connection. Series connection adds the z matrices; parallel connection adds the y matrices; cascade multiplies the transmission matrices. The cascade rule is the one used most, and it comes with a caution worth stating: the transmission-matrix product is valid only when connecting the ports does not change the internal behaviour of either stage, which for grounded two-ports is automatic and for floating ones needs checking. That is also why the transmission set is the standard description of a transmission line, a matching section and a filter section — the whole point of those objects is to be cascaded.

LTI Systems: Definition, Impulse Response, Convolution, Causality, Stability

Everything above has been a special case of one abstraction. A system is linear if scaling the input scales the output and superposing inputs superposes outputs — the same additivity and homogeneity that let the previous chapter switch sources on one at a time. It is time-invariant if delaying the input delays the output by the same amount and changes nothing else. A system with both properties is LTI, and it has a remarkable consequence: it is completely characterised by its response to a single input. Decompose any input into a sum of scaled, shifted impulses; linearity says the output is the sum of the responses to each; time invariance says each of those responses is the same shape shifted. The result is convolution: y(t) = ∫ x(τ)h(t - τ) dτ in continuous time and y[n] = Σ x[k]h[n - k] in discrete time, where h is the impulse response. Every filter, every channel, every circuit in this chapter is a convolution wearing different clothes.

Convolution in the exam is almost always short and discrete, and it has a check that costs nothing. Let x[n] = {1, 2, 3} for n = 0, 1, 2 and h[n] = {1, 1, 1}. The output runs from n = 0 to n = 4, a length of 3 + 3 - 1 = 5, and evaluating term by term gives y[n] = {1, 3, 6, 5, 3}. Now check: the sum of a convolution equals the product of the sums, because Σy = (Σx)(Σh). Here 1 + 3 + 6 + 5 + 3 = 18 and (1 + 2 + 3)(1 + 1 + 1) = 6 × 3 = 18 ✓. That identity is the DC gain statement in disguise — Σh is the response to a constant input — and it catches almost every arithmetic slip in a hand convolution. Change h to {1, -1}, a first difference, and the same x gives y[n] = {1, 1, 1, -3} summing to zero, exactly as Σh = 0 demands: a differencing filter annihilates DC, which is what makes it a highpass.

Two properties are then read directly off h, and reading them off is the whole examinable skill. Causality: the output cannot depend on future input, which means h(t) = 0 for t < 0, or h[n] = 0 for n < 0. Any physical circuit is causal; a symmetric FIR filter used offline on a stored file need not be. BIBO stability: every bounded input gives a bounded output if and only if h is absolutely integrable (or absolutely summable), ∫|h(t)| dt < ∞ or Σ|h[n]| < ∞. Test three examples. h(t) = e^(-2t)u(t) integrates to 1/2, so stable. h(t) = u(t), a pure integrator, integrates to infinity, so it is not stable — and note that its response to a bounded input, a step, is a ramp, which is the counterexample made explicit. h[n] = (0.5)^n u[n] sums to 2, stable; h[n] = 2^n u[n] diverges, unstable. Notice that absolute summability is stronger than merely decaying: h[n] = 1/n for n ≥ 1 decays to zero and is still unstable, because the harmonic series diverges.

⚠️ Poles, zeros, and the cancellation that hides an unstable mode

For a system described by a rational H(s), the poles are the roots of the denominator and the zeros the roots of the numerator, and each carries a different piece of information. The poles are the natural frequencies — the exponents that appear in the transient no matter what you excite the system with — so a causal system is stable exactly when every pole lies strictly in the left half plane, Re(s) < 0. Poles on the imaginary axis are marginally stable and poles to the right of it are unstable. The zeros do not affect stability at all; they shape the response, and a zero in the right half plane produces a non-minimum-phase system whose step response may initially move the wrong way. Now the trap, which is a favourite because it looks like ordinary algebra. Suppose a feedback arrangement gives H(s) = (s - 2)/((s - 2)(s + 3)). Cancel and you have 1/(s + 3), one pole at -3, apparently a perfectly stable first-order lowpass. It is not stable. The pole at s = +2 is a genuine natural frequency of the physical system, and cancelling it on paper does not remove the e^(2t) mode from the circuit; it only removes it from the input-output description. Any initial energy in that mode, any component tolerance that makes the cancellation inexact, and the response grows without bound. The formal statement is that pole-zero cancellation costs you observability or controllability of that mode, and a mode you cannot see is not a mode you have fixed. The examinable form of this is usually simpler and just as sharp: given H(s), do not cancel before you have identified the poles, and if a question gives you the differential equation rather than the transfer function, get the natural frequencies from the equation's own characteristic polynomial.

Frequency Response, Phase Delay and Group Delay

Feed an LTI system a complex exponential e^(jωt) and the output is H(jω)e^(jωt) — the same exponential, scaled by one complex number. That is the frequency response, and it says an LTI system can do exactly two things to a sinusoid: change its amplitude by |H(jω)| and shift its phase by ∠H(jω) = θ(ω). It cannot create a new frequency, which is why any new frequency appearing at the output of a supposedly linear amplifier is proof that it is not linear. Take H(s) = 1000/((s + 10)(s + 100)), a two-pole lowpass with unity DC gain. At ω = 10 rad/s, on the first pole, |H| = 0.7036 (−3.05 dB) and θ = -50.71°. At ω = 100, on the second pole, |H| = 0.0704 (−23.05 dB) and θ = -129.29°. The two phases sum to exactly -180°, and the two gains differ by exactly a factor of ten, both of which are the symmetry of the response about the geometric mean of the poles — a useful sanity check when sketching a Bode plot by hand.

Now the part of the syllabus that is stated in three words and carries a disproportionate number of marks. From the phase θ(ω) two different delays are defined. The phase delay τ_p = -θ(ω)/ω is the delay experienced by a single sinusoid at that frequency: it is the phase shift converted into a time. The group delay τ_g = -dθ/dω is the delay experienced by the envelope of a narrowband group of frequencies — by the information, in other words, since a modulated signal carries its message in its envelope and not in its carrier. They are different quantities with the same units, and confusing them is the single most common error in this topic.

First-order lowpass with τ = 1 ms: the two delays part company as soon as the phase bends
Normalised frequency ωτ	Phase θ	Phase delay τ_p	Group delay τ_g
`ωτ = 0`, at DC	0°, no shift at all	1.000 ms	1.000 ms
`ωτ = 0.5`	−26.57°	0.927 ms	0.800 ms
`ωτ = 1`, the cutoff	−45.00°	0.785 ms	0.500 ms
`ωτ = 2`	−63.43°	0.554 ms	0.200 ms
`ωτ = 5`	−78.69°	0.275 ms	0.038 ms

🎯 Only a straight line makes the two delays agree, and that is what linear phase is for

For H(s) = 1/(1 + sτ) the phase is θ(ω) = -arctan(ωτ), so τ_p = arctan(ωτ)/ω and τ_g = τ/(1 + (ωτ)²). At DC both equal τ, exactly — the two definitions always agree in the limit ω → 0, because a curve and its tangent agree at the origin. Away from DC they diverge fast. At the cutoff, ωτ = 1, the phase delay is 0.785 ms while the group delay is 0.500 ms, a ratio of π/2; by ωτ = 5 the ratio is over seven. Now ask what makes them equal at every frequency. Set -θ(ω)/ω = -dθ/dω and the only solution is θ(ω) = -ω t_0, a straight line through the origin: phase delay equals group delay if and only if the phase is linear in frequency, and then both equal the slope t_0. That is the whole reason linear phase is a design objective rather than an aesthetic preference. A system with linear phase delays every frequency component by the same time t_0, so the output waveform is the input waveform shifted — undistorted in shape. A system with non-linear phase delays different components by different times, and the waveform arrives smeared even if the amplitude response is perfectly flat. Three consequences worth carrying into the next chapter. First, flat amplitude is not enough: an all-pass filter has |H| = 1 at every frequency and can still wreck a pulse. Second, this is why a symmetric FIR filter is prized — symmetry forces exactly linear phase, and a length-N symmetric FIR has constant group delay (N - 1)/2 samples, which is a delay and not a distortion. Third, group delay is where dispersion lives: an optical fibre or a long copper pair whose group delay varies across the band spreads a transmitted pulse until adjacent symbols overlap, which is intersymbol interference arriving from the physical layer rather than from the filter.

Key takeaways

Charging a capacitor through a resistor is exactly 50 per cent efficient whatever the resistor — 500 μJ delivered, 250 μJ stored, 250 μJ burned, because the R cancels out of the integral. That is why real supplies charge through an inductor instead.
One ζ decides a second-order circuit. At ζ = 0.5 a 10 V step drives the capacitor to 11.63 V at 363 μs — a first-order circuit can never overshoot, so overshoot always means a second energy store. And past ζ = 1 more resistance makes the circuit slower, not faster.
Inductor current and capacitor voltage cannot jump, and in the s domain that fact is already built in: an inductor is sL in series with L i(0) and a capacitor is 1/sC in series with v(0)/s, so initial conditions stop being a separate step. i(t) = 3 - e^(-4t) came out of one line of partial fractions and satisfies continuity, final value and τ = L/R all at once.
Wye-delta is the only tool for an unbalanced bridge: the 6-18-12-4-8 Ω bridge reduces to 7.2 Ω, confirmed independently by node analysis. Test for balance first, because a balanced bridge lets you delete the bridging element instead.
Get two-port parameters from their defining measurement, not from a conversion table: 1/y11 = 4 + (12 ∥ 6) = 8 Ω on the T-network, and y21 = -1/12 S is the reciprocity experiment with the sign convention applied. Reciprocity is z12 = z21, h12 = -h21, AD - BC = 1; symmetry additionally needs z11 = z22.
An LTI system is its impulse response. Convolution's cheapest check is Σy = (Σx)(Σh); stability is absolute summability of h, not mere decay; a causal system is stable if and only if every pole is strictly in the left half plane — and never cancel a pole with a zero before you have listed the poles.
τ_p = -θ/ω and τ_g = -dθ/dω agree at DC and nowhere else unless the phase is a straight line. On a first-order lowpass at cutoff they are 0.785 ms and 0.500 ms. Linear phase is exactly the condition for undistorted delay, which is why a symmetric FIR of length N is worth its (N-1)/2 samples of lag.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A 5 microfarad capacitor is the only energy-storage element in a DC circuit. With the capacitor lifted out and every independent source deactivated, the resistance between the two terminals it occupied is a 12 kilo-ohm resistor in parallel with a 4 kilo-ohm and a 2 kilo-ohm resistor joined in series. What is the time constant of the natural response?
1. 60 ms
2. 90 ms
3. 30 ms
4. 20 ms
Show answer
Answer: D — 20 ms
The resistance the capacitor sees is the Thevenin resistance at its own terminals, which here is 12 in parallel with 4 + 2 = 6, that is 12 times 6 over 18 = 4 kilo-ohm. The time constant is then RC = 4000 times 5 microfarad = 20 ms. Taking the 12 kilo-ohm branch alone gives 60 ms and is the commonest error, because it ignores the parallel path the discharge current can also take; adding all three resistors in series gives 18 kilo-ohm and 90 ms, which supposes a series loop that the topology does not contain. The whole skill in a first-order problem is finding the correct resistance, since the exponential form itself is fixed.
A series loop of L = 20 mH, C = 0.5 microfarad and R = 120 ohm is completely at rest when a 24 V DC source is switched across it at t = 0. What is the largest voltage that appears across the capacitor at any time afterwards?
1. 32.9 V
2. 24.0 V
3. 41.5 V
4. 27.9 V
Show answer
Answer: A — 32.9 V
First find the damping ratio, because nothing else can be decided without it: zeta = (R/2) times the square root of C/L = 60 times the square root of 0.5e-6/0.02 = 60 times 0.005 = 0.3, so the loop is underdamped. The step-response overshoot is exp(-pi·zeta divided by the square root of 1 - zeta squared) = exp(-0.9425/0.9539) = 0.372, so the peak is 24 times 1.372 = 32.9 V. The answer 24.0 V is the intuitive one, on the ground that a capacitor charged from a 24 V source cannot exceed it, and it is wrong because the inductor current cannot stop the instant the capacitor reaches 24 V and keeps pushing charge in. Using zeta = 0.1 by mistake would give 41.5 V and zeta = 0.5 would give 27.9 V.
A branch current in a network has the Laplace transform I(s) = (3s + 20)/(s(s + 5)) amperes. Using the initial-value and final-value theorems, what are i(0+) and i(infinity)?
1. 4 A and 3 A
2. 3 A and 4 A
3. 0 A and 4 A
4. 3 A and 20 A
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Answer: B — 3 A and 4 A
The initial value is the limit of s·I(s) as s tends to infinity, which is (3s + 20)/(s + 5) tending to 3 A. The final value is the limit of s·I(s) as s tends to zero, which is 20/5 = 4 A. Reporting 4 A and 3 A is the standard slip of applying the two limits the wrong way round, and it is worth a sanity check: the transform has a pole at the origin, meaning a constant term, and its residue there is the steady value 4 A, so the current must end at 4 A and start below it. The value 0 A would follow from evaluating I(s) itself rather than s·I(s) at large s, which drops the factor the theorems require.
Five resistors are wired like this. From point P a 10 ohm element reaches point R and a 15 ohm element reaches point S. From R a 20 ohm element continues to point Q, and from S a 30 ohm element also continues to Q. A 25 ohm element joins R directly to S. An ideal 36 V supply is then applied between P and Q. What current passes through the 25 ohm element, and what current does the supply deliver?
1. Zero through the 25 ohm element, and 1.44 A from the supply
2. 0.48 A through the 25 ohm element, and 2.0 A from the supply
3. Zero through the 25 ohm element, and 2.0 A from the supply
4. Zero through the 25 ohm element, and 3.44 A from the supply
Show answer
Answer: C — Zero through the 25 ohm element, and 2.0 A from the supply
Check the ratios of the two chains before attempting any reduction: 10/20 = 0.5 along one path and 15/30 = 0.5 along the other, so R and S sit at exactly the same potential whatever the supply is. The 25 ohm element therefore carries nothing at all and can be struck out, which leaves 10 + 20 = 30 ohm in parallel with 15 + 30 = 45 ohm, that is 1350/75 = 18 ohm, and the supply delivers 36/18 = 2.0 A. Answering 3.44 A comes from keeping the 25 ohm element and putting it in parallel with the reduced 18 ohm, which would only be right if it joined P to Q instead of R to S; 1.44 A treats it as the whole load. Testing for equal ratios first is what makes the wye-delta transformation unnecessary here.
A two-port network is described by the open-circuit parameters z11 = 30 ohm, z12 = z21 = 18 ohm and z22 = 24 ohm. What is the value of its transmission parameter A?
1. 1.67
2. 1.33
3. 0.60
4. 0.75
Show answer
Answer: A — 1.67
A is defined as V1/V2 with the output port open, so no current flows at port 2 and the z-equations reduce to V1 = z11·I1 and V2 = z21·I1. Dividing gives A = z11/z21 = 30/18 = 1.67, a dimensionless number as it must be. The value 1.33 is z22/z21, which is the parameter D and belongs to the other diagonal of the transmission matrix; a candidate who remembers only that A and D are the two dimensionless entries can easily hand back the wrong one. The ratios 0.60 and 0.75 are z21/z11 and z12/z22, reciprocals of the two correct forms, and neither is a transmission parameter.
Two finite sequences are x[n] = {2, 1, 3} for n = 0, 1, 2 and h[n] = {1, -1} for n = 0, 1. What is their linear convolution?
1. {2, 1, 3, -3}
2. {2, 3, 4, 3}
3. {2, -1, -2, -3}
4. {2, -1, 2, -3}
Show answer
Answer: D — {2, -1, 2, -3}
The output runs from n = 0 to n = 3, a length of 3 + 2 - 1 = 4. Term by term: y[0] = 2(1) = 2; y[1] = 1(1) + 2(-1) = -1; y[2] = 3(1) + 1(-1) = 2; y[3] = 3(-1) = -3. The cheapest check costs nothing: the samples of a convolution sum to the product of the two input sums, and here 2 + 1 + 3 = 6 multiplied by 1 - 1 = 0 gives zero, which the answer 2 - 1 + 2 - 3 = 0 satisfies. The sequence {2, 1, 3, -3} fails that check because it sums to 3; it comes from sliding only the first tap of h and never folding in the -1 except at the last sample, which is the usual slip in a hand convolution.
Over its useful band a system has the phase response theta(omega) = -0.004·omega - 0.2 radians, with omega in radians per second. What is its group delay, and does the phase delay equal it?
1. 0.2 ms, and the phase delay is equal to it
2. 4 ms, and the phase delay differs from it at every finite frequency because the phase line does not pass through the origin
3. 4.2 ms, and the phase delay is equal to it
4. 2 ms, and the phase delay is twice as large at every frequency
Show answer
Answer: B — 4 ms, and the phase delay differs from it at every finite frequency because the phase line does not pass through the origin
Group delay is minus the derivative of the phase with respect to omega, and the derivative of -0.004·omega - 0.2 is -0.004, so the group delay is 0.004 s = 4 ms at every frequency in the band. Phase delay is minus theta divided by omega, which here is 0.004 + 0.2/omega and therefore changes with frequency, exceeding 4 ms everywhere and approaching it only as omega grows large. The two delays agree at all frequencies only when the phase is a straight line through the origin; a constant offset of -0.2 rad makes the phase affine rather than linear, and that offset is exactly what the tempting answer overlooks. The 0.2 ms answer treats the constant phase offset in radians as though it were already a time, and 4.2 ms adds that offset to the slope instead of differentiating it away.
A causal continuous-time LTI system has the transfer function H(s) = (s - 4)/((s + 2)(s + 5)). Which of the following statements about it are correct?
1. The system is BIBO stable, because both poles lie strictly in the left half plane
2. The system is non-minimum-phase, so its step response may at first move opposite to its final direction
3. The zero at s = +4 lies in the right half plane and therefore makes the system unstable
4. The impulse response is a weighted sum of exp(-2t) and exp(-5t) for t greater than zero
Show answer
Answer: A — The system is BIBO stable, because both poles lie strictly in the left half plane; B — The system is non-minimum-phase, so its step response may at first move opposite to its final direction; D — The impulse response is a weighted sum of exp(-2t) and exp(-5t) for t greater than zero
Stability of a causal system depends on the poles alone, and both sit at -2 and -5, strictly in the left half plane, so the system is BIBO stable and its impulse response is a combination of exp(-2t) and exp(-5t) decaying to zero. A zero in the right half plane changes the shape of the response and not its boundedness: it makes the system non-minimum-phase, whose classic signature is a step response that starts off in the wrong direction before turning round. Treating that zero as a stability hazard is the standard confusion, and it is the mirror of the equally common error of cancelling a right-half-plane pole against a numerator factor and declaring the result stable.