Sinusoidal Signals

Sinusoidal signals are the fundamental test waveforms for every linear system because their frequency content is concentrated at a single point. Frequency response measurements in audio amplifier design and vibration testing both rely on injecting known sinusoids and observing the output.

Core Concept

A sinusoidal signal is x(t) = A cos(omega_0 t + phi) where A is amplitude in volts or any unit, omega_0 = 2 pi f_0 is angular frequency in radians per second, and phi is initial phase in radians. The signal repeats with period T_0 = 1/f_0 = 2 pi / omega_0. Cosine and sine are related by a 90-degree phase shift: sin(omega t) = cos(omega t - pi/2).

Sinusoids are power signals because they repeat forever. Their average power is A^2/2 regardless of frequency or phase. Their Fourier transform consists of two impulses at +f_0 and -f_0, with weight A/2 each. This two-sided spectrum is a direct consequence of Euler's formula.

In discrete time, x[n] = A cos(omega_0 n + phi). A key difference from continuous time: discrete sinusoids are periodic only if omega_0 / (2 pi) is rational. If omega_0 = 2 pi k / N for integers k and N, the period is N samples. Otherwise the sequence never repeats exactly.

Key Formula

x(t) = A cos(omega_0 t + phi). Parameters: A = amplitude, omega_0 = 2 pi f_0 rad/s, f_0 = 1/T_0 Hz, phi = phase in radians. Fourier transform: X(f) = (A/2)[delta(f - f_0) + delta(f + f_0)] e^(j phi). Average power: P = A^2 / 2. RMS value: A / sqrt(2).

Given: x(t) = 3 cos(20 pi t + pi/4). Find: frequency, period, power, Fourier transform.

Step 1: Identify parameters
  A = 3
  omega_0 = 20 pi  rad/s
  phi = pi/4 rad

Step 2: Find frequency and period
  f_0 = omega_0 / (2 pi) = 20 pi / (2 pi) = 10 Hz
  T_0 = 1 / f_0 = 0.1 s

Step 3: Average power
  P = A^2 / 2 = 9 / 2 = 4.5 W

Step 4: Fourier transform
  X(f) = (3/2) e^(j pi/4) delta(f - 10) + (3/2) e^(-j pi/4) delta(f + 10)

Final Answer:
  f0 = 10 Hz,  T0 = 0.1 s
  P = 4.5 W
  Spectrum: two impulses at f = +10 Hz and f = -10 Hz

Exam Tip: In Anna University problems, the phrase "find the spectrum" means plot |X(f)| and angle(X(f)) versus f. For a cosine with phase phi, the magnitude spectrum has two equal spikes at +/-f0, each of height A/2. The phase spectrum has +phi at +f0 and -phi at -f0. This sign flip in phase is a direct result of conjugate symmetry of the Fourier transform of any real signal. Forgetting this sign flip is a common error in GATE numerical questions.

Properties Summary

• General form: A cos(omega_0 t + phi); period T_0 = 2 pi / omega_0.
• Average power: P = A^2/2; RMS = A/sqrt(2).
• Fourier transform: two impulses at +/-f_0 weighted A/2.
• Phase spectrum: +phi at +f_0 and -phi at -f_0 for a real cosine.
• Discrete-time periodicity: periodic only if omega_0/(2 pi) is rational.
• Euler relation: cos(x) = [e^(jx) + e^(-jx)]/2; sin(x) = [e^(jx) - e^(-jx)]/(2j).
• Energy: infinite (power signal); use P = A^2/2 not energy for characterisation.

Quick Revision

• x(t) = A cos(omega t + phi); omega = 2 pi f, T = 1/f.
• Power = A^2/2; RMS = A/sqrt(2).
• Fourier spectrum: impulses at +/-f0, amplitude A/2 each.
• Discrete sinusoid is periodic only when omega0/(2pi) is rational.
• cos(omega t) = [e^(j omega t) + e^(-j omega t)] / 2.
• Sinusoid is a power signal; energy is infinite.
• Exam trap: computing power as A^2 instead of A^2/2. Always remember the factor of 1/2 from integrating cosine-squared over one period.