Continuous and discrete-time signals and their classification

Classifies continuous- and discrete-time signals as periodic or aperiodic, even or odd, deterministic or random, and energy or power, with the formulas to compute each.

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Why it matters

Every sensor output you will meet in instrumentation — a thermocouple voltage, a strain-gauge bridge output, the samples coming out of an ADC — is a signal, and the first thing you must decide is what kind of signal it is. The classification (continuous or discrete, periodic or not, energy or power, even or odd) decides which transform you use, which formulas are valid and whether a quantity such as "average power" even exists.

Key ideas

Continuous-time (CT) and discrete-time (DT) signals

  • A CT signal x(t) is defined for every real value of the independent variable t. Example: the output voltage of an LVDT.
  • A DT signal x[n] is defined only at integer n. It often comes from sampling, x[n] = x(nTs), but n itself is just an index with no unit.
  • Discrete-time is not the same as digital. A digital signal is discrete in time and quantised in amplitude; a DT signal can still take any real amplitude.

Periodic and aperiodic signals

  • CT: x(t) is periodic if x(t + T) = x(t) for all t, for some T > 0. The smallest such T is the fundamental period T0.
  • DT: x[n] is periodic if x[n + N] = x[n] for all n, for some positive integer N.
  • A sum of CT periodic signals is periodic only if the ratio of their periods is a rational number; the period is then the LCM of the individual periods.
  • The DT sinusoid cos(Ω0 n) is periodic only if Ω0 / 2π is a rational number k/N (in lowest terms); then the fundamental period is N. So cos(n) (Ω0 = 1 rad/sample) is not periodic, even though cos(t) is.

Even and odd signals

  • Even: x(−t) = x(t); odd: x(−t) = −x(t) (same for x[n]).
  • Any signal splits uniquely into an even part and an odd part (formulas below). An odd signal must be zero at t = 0.

Deterministic and random signals

  • Deterministic: completely specified by a formula or table, e.g. 5 sin(100πt).
  • Random: only statistical properties (mean, autocorrelation, PSD) are known, e.g. thermal noise in a resistor. These are treated in the random-signals topic.

Energy and power signals

  • Energy signal: total energy E is finite and non-zero, so average power P = 0. Typical: time-limited pulses, decaying exponentials.
  • Power signal: P is finite and non-zero, so E = ∞. Typical: periodic signals, the unit step, DC.
  • Some signals are neither (e.g. x(t) = t or e^(2t) for all t): both E and P are infinite.
  • "Energy" and "power" here are normalised quantities — the energy or power that would be dissipated if x(t) were a voltage across (or current through) a 1 Ω resistor. They carry the unit (unit of x)²·s and (unit of x)², not joules and watts, unless that 1 Ω interpretation is stated.

Elementary signals used everywhere later: unit step u(t), unit impulse δ(t) (area 1, ∫x(t)δ(t − t0)dt = x(t0)), ramp r(t) = t·u(t), real and complex exponentials, and their DT counterparts u[n], δ[n].

Operations on the independent variable — shifting x(t − t0) (delay if t0 > 0), scaling x(at) (compression if |a| > 1), reversal x(−t). For x(at − b), shift first by b, then scale by a — or, safer, substitute values of t.

Formulas

  • E = ∫ |x(t)|² dt (limits −∞ to ∞) — energy of a CT signal.
  • E = Σ |x[n]|² (n from −∞ to ∞) — energy of a DT signal.
  • P = lim(T→∞) (1/2T) ∫ from −T to T of |x(t)|² dt — average power of a CT signal.
  • P = lim(N→∞) [1/(2N + 1)] Σ from −N to N of |x[n]|² — average power of a DT signal.
  • P = (1/T0) ∫ over one period of |x(t)|² dt — power of a periodic CT signal.
  • P = A²/2 — power of A cos(ω0 t + φ); for a sum of sinusoids at different frequencies, powers add: P = Σ Ak²/2 (plus C² for a DC term C).
  • xe(t) = [x(t) + x(−t)]/2, xo(t) = [x(t) − x(−t)]/2 — even and odd parts.
  • N = 2πk / Ω0 with the smallest integer k that makes N an integer — period of a DT sinusoid.

Symbols: t time (s); n sample index (dimensionless); T, T0 period (s); N period in samples; Ω0 digital frequency (rad/sample); ω0 angular frequency (rad/s); A amplitude (unit of x); E normalised energy ((unit of x)²·s for CT, (unit of x)² for DT); P normalised power ((unit of x)²).

Worked examples

Example 1 (standard). Classify x[n] = (0.5)ⁿ u[n] and y(t) = 3 cos(200πt) + 4 sin(300πt) as energy or power signals and find E or P.

  1. For x[n]: E = Σ from n=0 to ∞ of (0.5)^(2n) = Σ (0.25)ⁿ.
  2. Geometric series: E = 1/(1 − 0.25) = 1/0.75.
  3. E = 1.333 — finite, so x[n] is an energy signal with E = 4/3 ≈ 1.333 and P = 0.
  4. For y(t): two sinusoids at 100 Hz and 150 Hz (different frequencies), so the powers add: P = 3²/2 + 4²/2 = 4.5 + 8.
  5. P = 12.5 (in (unit of y)²); y(t) is a power signal with infinite energy.

Example 2 (GATE level). Find the fundamental period of x[n] = cos(3πn/8) + sin(πn/6), and decide whether z(t) = cos(4t) + sin(4πt) is periodic.

  1. First term: Ω1 = 3π/8, Ω1/2π = 3/16 (rational, lowest terms), so N1 = 16.
  2. Second term: Ω2 = π/6, Ω2/2π = 1/12, so N2 = 12.
  3. N = LCM(16, 12) = 48. Fundamental period N = 48 samples.
  4. For z(t): T1 = 2π/4 = π/2 s, T2 = 2π/4π = 0.5 s.
  5. T1/T2 = π, which is irrational, so no common period exists: z(t) is not periodic (it is still a power signal, with P = 1/2 + 1/2 = 1).

Common mistakes

  • Assuming every DT sinusoid is periodic. cos(0.5n) is not, because 0.5/2π is irrational.
  • Taking the period of a DT sinusoid as 2π/Ω0 without checking it is an integer: cos(3πn/8) has period 16, not 16/3.
  • Adding the powers of two sinusoids at the same frequency. Same-frequency terms must first be combined into one phasor; only different-frequency terms add in power.
  • Writing energy in joules and power in watts without the 1 Ω assumption.
  • Calling a sampled signal "digital" — quantisation is a separate step.
  • Forgetting the 1/(2N + 1) (not 1/2N) in the DT power formula.
  • Reading x(2t − 4) as "compress by 2, then delay by 4". It is x(2(t − 2)): compress, then delay by 2 — or delay by 4 first, then compress. Check by substituting one point.

For GATE IN

Expect one-mark questions on periodicity (especially DT sinusoids and sums of sinusoids), energy/power classification and even–odd parts, and NAT questions asking for the energy of an exponential or pulse, the power of a sum of sinusoids, or a fundamental period in samples. Time shifting/scaling of a sketched signal also appears, usually as an MCQ with sketches. Practise computing E and P quickly with geometric series and the A²/2 rule.

Quick check

  1. Is x[n] = cos(2n) periodic?
  2. Find the energy of x(t) = e^(−3|t|).
  3. What is the average power of x[n] = u[n]?
  4. Is x(t) = sin(t)·u(t) even, odd or neither?
  5. What is the fundamental period of cos(4πt) + sin(6πt)? Answers: 1. No — 2/2π = 1/π is irrational. 2. E = 2 × 1/6 = 1/3. 3. P = lim (N + 1)/(2N + 1) = 1/2. 4. Neither. 5. T1 = 0.5 s, T2 = 1/3 s, so T0 = 1 s.

Try answering each one aloud before you open it.

  1. 1.What is a continuous-time signal, and how does it differ from a discrete-time signal?Concept

    A continuous-time signal x(t) is defined for every real value of time, like the voltage from a thermocouple. A discrete-time signal x[n] is defined only at integer indices n, typically obtained by sampling x(nTs). Discrete-time is not the same as digital: a digital signal is also quantised in amplitude, while a discrete-time signal can take any real value at each sample.

  2. 2.Explain the classification of signals based on periodicity.Concept

    Signals can be classified as periodic or aperiodic based on periodicity. A periodic signal repeats itself after a fixed interval of time, known as the period. Mathematically, a signal x(t) is periodic if there exists a positive constant T such that x(t) = x(t + T) for all t. Aperiodic signals do not repeat over time.

  3. 3.What are deterministic and random signals?Concept

    Deterministic signals are those whose values are completely specified for any given time. They can be described by mathematical functions. Random signals, on the other hand, have some level of uncertainty and cannot be precisely predicted. They are often described using statistical properties.

  4. 4.Why are discrete-time signals used in digital systems?Application

    Discrete-time signals are used in digital systems because they can be easily processed, stored, and transmitted using digital technology. Digital systems are less susceptible to noise and distortion compared to analog systems, and they allow for more complex signal processing algorithms to be implemented.

  5. 5.What happens if a continuous-time signal is not sampled properly?Application

    If a continuous-time signal is not sampled properly, it can lead to aliasing. Aliasing occurs when different continuous-time signals become indistinguishable from each other after sampling. To avoid aliasing, the sampling frequency must be at least twice the highest frequency present in the signal, as per the Nyquist-Shannon sampling theorem.

  6. 6.Explain the concept of signal energy and power.Concept

    Signal energy is a measure of the total magnitude of a signal over time, calculated as the integral of the square of the signal's amplitude. For a continuous-time signal x(t), energy E is given by E = ∫ |x(t)|² dt over all time. Signal power is the average energy per unit time, especially relevant for periodic signals. For a periodic signal, power P is given by P = (1/T) ∫ |x(t)|² dt over one period T.

  7. 7.Calculate the energy of a continuous-time signal x(t) = e^(-2t) for t ≥ 0.Numerical

    To calculate the energy of the signal x(t) = e^(-2t), we use the formula for energy: E = ∫ |x(t)|² dt from 0 to ∞. Substituting x(t), we get E = ∫ (e^(-2t))² dt = ∫ e^(-4t) dt from 0 to ∞. Evaluating this integral, E = [-1/4 e^(-4t)] from 0 to ∞ = 1/4.

  8. 8.Determine if the discrete-time signal x[n] = cos(πn/4) is periodic.Numerical

    A discrete-time signal x[n] is periodic if there exists a positive integer N such that x[n] = x[n + N] for all n. For x[n] = cos(πn/4), the period N is the smallest integer for which (π(n + N)/4) - (πn/4) = 2πk, where k is an integer. Solving, N = 8, so the signal is periodic with period 8.

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