Power Signals

Power signals have finite average power over infinite time and model waveforms that run continuously, like AC mains voltage and carrier waves in radio transmitters. Characterising a signal as a power signal tells the designer the sustained load it places on a system.

Core Concept

Average power is computed by integrating |x(t)|^2 over a window of duration 2T, dividing by 2T, and taking the limit as T approaches infinity. If this limit is finite and positive, the signal is a power signal. The signal's energy is then infinite because finite power accumulates indefinitely over all time.

Periodic signals are always power signals. For a periodic signal with period T_0, the average power equals the integral of |x(t)|^2 over exactly one period divided by T_0. You never need to handle the infinite limit directly — periodicity does the heavy lifting.

Discrete-time average power is P = lim(N->inf) [1/(2N+1)] sum(n=-N to N) |x[n]|^2. A periodic sequence with period N has P = (1/N) sum over one period of |x[n]|^2. A unit step u[n] is a power signal with P = 1/2.

Key Formula

P = lim(T->inf) (1/2T) integral(-T to T) |x(t)|^2 dt. For a periodic signal with period T_0: P = (1/T_0) integral(0 to T_0) |x(t)|^2 dt. Condition: 0 < P < infinity, which forces E = infinity.

Given: x(t) = A cos(2 pi f0 t). Find average power.

Formula:
  P = (1/T0) * integral(0 to T0) |x(t)|^2 dt
  where T0 = 1/f0

Step 1: Square the signal
  |A cos(2 pi f0 t)|^2 = A^2 cos^2(2 pi f0 t)

Step 2: Use identity cos^2(theta) = (1 + cos(2 theta)) / 2
  = A^2 * (1 + cos(4 pi f0 t)) / 2

Step 3: Integrate over one period T0
  integral(0 to T0) A^2/2 dt + integral(0 to T0) A^2/2 cos(4 pi f0 t) dt
  = A^2 T0 / 2   +   0   (cosine integrates to zero over full period)

Step 4: Divide by T0
  P = (A^2 T0 / 2) / T0 = A^2 / 2

Final Answer:
  P = A^2 / 2
  x(t) is a power signal (finite P, infinite E)

Exam Tip: Anna University MCQs frequently test whether u(t) is an energy or power signal. Its energy diverges because |u(t)|^2 = 1 for all t >= 0. Its average power over a large window T is approximately T/(2T) = 1/2, which is finite. So u(t) is a power signal with P = 1/2. Do not confuse average power P = A^2/2 with peak power A^2 — the factor of 1/2 comes from the cosine-squared identity and only applies to pure sinusoids.

Properties Summary

• Definition: P = lim(T->inf) (1/2T) integral |x(t)|^2 dt; power signal if 0 < P < inf.
• Energy of a power signal is always infinite: E = inf.
• Periodic signals are power signals; P = (1/T0) * one-period integral of |x|^2.
• For A cos(omega t + phi): P = A^2 / 2, independent of frequency and phase.
• Unit step u(t) is a power signal with P = 1/2.
• A signal cannot simultaneously be both an energy signal and a power signal.
• Some signals (like e^(t)u(t)) are neither — energy diverges and average power diverges.

Quick Revision

• Power signal: 0 < P < infinity; E = infinity.
• P for a periodic signal: integrate |x(t)|^2 over one period, divide by period.
• Sinusoid A cos(omega t): P = A^2 / 2.
• Unit step u(t): P = 1/2, so it is a power signal.
• Exponentially growing signals are neither energy nor power.
• All periodic signals are power signals without exception.
• Exam trap: computing P = A^2 instead of A^2/2 for a cosine. Always apply the cos^2 identity before integrating.