Time Dependent Properties of Signal

A measurement signal is the physical or electrical quantity carrying information about the measurand. Signals are classified in several ways.

Static and dynamic

• Static signal: does not change with time or changes very slowly. Example: weight of an object on a scale, dead-load pressure.
• Dynamic signal: varies with time. It may be steady-periodic, aperiodic or transient. Example: engine vibration, pressure in a cylinder.

Deterministic and random

• Deterministic: can be described by a mathematical function of time and predicted exactly. Example: $x(t) = A\sin\omega t$.
• Random (nondeterministic): cannot be predicted; described only statistically by mean, variance and probability density. Example: turbulent wind velocity, noise.

Periodic signals

• Simple periodic (harmonic): a single sine or cosine at one frequency, $x = A\sin(\omega t + \phi)$.
• Complex periodic: repeats after a period $T$ but contains many harmonics, $x(t) = x(t+T)$. Example: a square wave from a rotating machine, which has a Fourier series.

Aperiodic and transient

• Almost periodic: sum of harmonics whose frequency ratios are not rational, so the signal never exactly repeats.
• Transient: exists for a short time then dies away. Example: step input, impulse from a hammer blow, decaying vibration after impact.

 Signals
  |-- Static
  |-- Dynamic
        |-- Deterministic
        |     |-- Periodic (simple, complex)
        |     |-- Aperiodic (almost periodic, transient)
        |-- Random (stationary, non-stationary)

(a) Single harmonic

Compare $y(t) = 10\sin(100\pi t + \pi/6)$ with $y = A\sin(\omega t + \phi)$.

• Amplitude $A = 10$ V
• Angular frequency $\omega = 100\pi$ rad/s, so $f = \omega/2\pi = 50$ Hz
• Period $T = 1/f = 0.02$ s = 20 ms
• RMS value $= A/\sqrt{2} = 10/\sqrt{2} = 7.07$ V
• Phase $\phi = \pi/6 = 30^\circ$. Time lead:

$$t_d = \frac{\phi}{\omega} = \frac{\pi/6}{100\pi} = \frac{1}{600}\ \text{s} = 1.667\ \text{ms}$$

Answer: $f = 50$ Hz, $T = 20$ ms, $A = 10$ V, RMS $= 7.07$ V, lead $= 1.667$ ms.

(b) Two harmonics

The components have frequencies 50 Hz and 150 Hz. The 150 Hz is the third harmonic of 50 Hz, so the fundamental frequency is 50 Hz (period 20 ms).

For components of different frequency, mean squares add:

$$x_{rms} = \sqrt{\frac{A_1^2}{2} + \frac{A_2^2}{2}} = \sqrt{\frac{16}{2} + \frac{4}{2}} = \sqrt{10} = 3.16\ \text{V}$$

Answer: fundamental 50 Hz, RMS $= 3.16$ V.

(a) Fourier series of the square wave

A periodic signal of period $T$ and $\omega_0 = 2\pi/T$ is written as

$$f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos n\omega_0 t + b_n\sin n\omega_0 t\right)$$

with $a_0 = \frac{2}{T}\int_0^T f\,dt$, $a_n = \frac{2}{T}\int_0^T f\cos n\omega_0 t\,dt$, $b_n = \frac{2}{T}\int_0^T f\sin n\omega_0 t\,dt$.

   +A  ____        ____
        |  |      |  |
   0 ---+--+------+--+----> t
        0  T/2    T
   -A      |______|

The wave has zero mean over a period, so $a_0 = 0$. Taking the origin at the rising edge it is an odd function (about $t=0$ over $-T/2 < t < T/2$), so all $a_n = 0$.

$$b_n = \frac{2}{T}\left[\int_0^{T/2} A\sin n\omega_0 t\,dt - \int_{T/2}^{T} A\sin n\omega_0 t\,dt\right]$$ $$\int_0^{T/2}\sin n\omega_0 t\,dt = \frac{1-\cos n\pi}{n\omega_0}, \qquad \int_{T/2}^{T}\sin n\omega_0 t\,dt = \frac{\cos n\pi - 1}{n\omega_0}$$ $$b_n = \frac{2A}{T}\cdot\frac{2(1-\cos n\pi)}{n\omega_0} = \frac{2A(1-\cos n\pi)}{n\pi}$$

So $b_n = 4A/(n\pi)$ for odd $n$ and $b_n = 0$ for even $n$. Therefore

$$f(t) = \frac{4A}{\pi}\left[\sin\omega_0 t + \frac{1}{3}\sin 3\omega_0 t + \frac{1}{5}\sin 5\omega_0 t + \cdots\right]$$

Only odd harmonics exist and their amplitudes fall as $1/n$.

(b) 4-point DFT

$$X_k = \sum_{n=0}^{3} x_n e^{-j2\pi kn/4}, \qquad x = \{1, 3, 2, 0\}$$

Using $e^{-j\pi k n/2}$ = $1, -j, -1, j$ for $k n = 0,1,2,3$:

• $X_0 = 1+3+2+0 = 6$
• $X_1 = 1 + 3(-j) + 2(-1) + 0 = -1 - j3$
• $X_2 = 1 + 3(-1) + 2(1) + 0 = 0$
• $X_3 = 1 + 3(j) + 2(-1) + 0 = -1 + j3$

Answer: $X = \{6,\ -1-j3,\ 0,\ -1+j3\}$.

Why FFT

A direct DFT needs about $N^2$ complex multiplications. The FFT (Cooley-Tukey) splits the sum into even and odd samples and needs only $(N/2)\log_2 N$. For $N = 1024$ this is about 5120 instead of about 1,000,000, so spectra are obtained quickly in real time.