Periodic and non-periodic signals; LTI output is convolution

Periodic signal

A signal is periodic if it repeats exactly after a fixed time $T_0$:

$$g(t)=g(t+T_0)\quad\text{for all }t$$

The smallest such $T_0$ is the fundamental period and $f_0=1/T_0$ is the fundamental frequency. Periodic signals are power signals and have line (discrete) spectra given by the Fourier series. Examples: $A\cos(2\pi f_ct)$ with $T_0=1/f_c$; a square wave clock; the 50 Hz mains voltage.

Non-periodic (aperiodic) signal

A signal that does not repeat for any $T_0$ is non-periodic. It has a continuous spectrum given by the Fourier transform. Examples: a single rectangular pulse $\text{rect}(t/T)$; the decaying exponential $e^{-at}u(t)$; a speech sentence; noise.

 Periodic (square wave)        Non-periodic (single pulse)
  _   _   _   _                      ____
 | |_| |_| |_| |_ ...         ______|    |______
  <-T0->

Output of an LTI system is convolution of input and impulse response

Let the system be LTI with impulse response $h(t)$, i.e. $\delta(t)\to h(t)$.

Step 1 – write the input as a sum of impulses. By the sifting property:

$$x(t)=\int_{-\infty}^{\infty}x(\tau)\,\delta(t-\tau)\,d\tau$$

This expresses $x(t)$ as a continuous sum of impulses located at $t=\tau$ with strengths $x(\tau)\,d\tau$.

Step 2 – time invariance: $\delta(t-\tau)\to h(t-\tau)$.

Step 3 – homogeneity (scaling): $x(\tau)\delta(t-\tau)\to x(\tau)h(t-\tau)$, because $x(\tau)$ is a constant for each $\tau$.

Step 4 – additivity (superposition): since the input is the integral of these terms, the output is the integral of their responses:

$$y(t)=\mathcal{T}\{x(t)\}=\int_{-\infty}^{\infty}x(\tau)\,\mathcal{T}\{\delta(t-\tau)\}\,d\tau=\int_{-\infty}^{\infty}x(\tau)\,h(t-\tau)\,d\tau$$ $$y(t)=x(t)*h(t)$$

Hence the output of an LTI system is the convolution of the input with the impulse response. Putting $\lambda=t-\tau$ shows that convolution is commutative: $y(t)=\int h(\lambda)x(t-\lambda)d\lambda=h(t)*x(t)$.

Frequency domain: by the convolution property, $Y(f)=X(f)H(f)$, where $H(f)$ is the transfer function.

Example: input $x(t)=u(t)$ to $h(t)=e^{-at}u(t)$:

$$y(t)=\int_0^t e^{-a(t-\tau)}d\tau=\frac{1}{a}\left(1-e^{-at}\right),\quad t\ge0$$