Class 5 (Continuous-time linear, time-invariant (LTI) systems)

Representing any signal with impulses

First, we consider a simple property. If we multiply a signal $x(t)$ by shifted impulse $\delta(t-T)$, we notice that $$ x(t)\delta(t-T) = x(T)\delta(t-T) .$$ We get this result because $\delta(t-T)$ is zero everywhere except at time $t = T$. Therefore, only the value at $x(T)$ is expressed in the signal.

We can now expand this idea to the entire signal. Intuitively, we show that we can represent any signal as an infinite, continuous sum (i.e., an integral) of shifted impulses with different amplitudes. Mathematically, this is written as $$ \begin{eqnarray} x(t) &=& \int_{-\infty}^{\infty} x(t) \delta(t - \tau) d\tau \\ &=& \int_{-\infty}^{\infty} x(\tau) \delta(t - \tau) d\tau \; . \end{eqnarray} $$ That is, at every $\tau$, we sample $x(t)$ at $t = \tau$. In the end, we get back $x(t)$. This is known as the sifting property or the sampling property of an impulse function.

At first glance, this may seem like an exercise in tautology. However, this property is key to understanding linear, time-invariant (LTI) systems.

Step 1: Start with a time-invariant system

Let the input to a time-invariant system $\mathcal{H}\{ \cdot \}$ be an impulse $\delta(t)$. Due to time-invariance, we know that delaying or advancing the impulse will delay or advance the output $h(t)$. Hence, we can say $$\mathcal{H}\{ \delta(t + \tau) \} = h(t + \tau).$$

Step 2: Add linearity

Continue to consider the time-invariant system $\mathcal{H}\{ \cdot \}$ discussed in the previous subsection. If this system is also linear, we will know what output to expect from a sum of two or more different $\delta$ inputs. For example, for the input signal $x(t) = x(\tau_1) \delta(t + \tau_1) + x(\tau_2) \delta(t + \tau_2)$, the output response would be $$\begin{eqnarray} \mathcal{H}\{ x(t) \} &=& \mathcal{H}\{ x(\tau_1) \delta(t + \tau_1) + x(\tau_2) \delta(t + \tau_2) \} \\ &=& \mathcal{H}\{ x(\tau_1) \delta(t + \tau_1) \} + \mathcal{H}\{ x(\tau_2) \delta(t + \tau_2) \} \quad (\textrm{due to linearity}) \\ &=& x(\tau_1) \mathcal{H}\{ \delta(t + \tau_1) \} + x(\tau_2) \mathcal{H}\{ \delta(t + \tau_2) \} \quad (\textrm{due to linearity}) \\ &=& x(\tau_1) h(t + \tau_1) + x(\tau_2) h(t + \tau_2). \end{eqnarray}$$

Step 3: Extend with the sampling property

In the previous subsection, we showed that linearity allowed us to determine the output from a sum of two different $\delta$ functions. We can extend this idea to a sum of any number of delta functions. Furthermore, the sampling property tells us that we can represent any signal $x(t)$ as an infinite sum of amplified delta functions. So if we now let $x(t)$ be the input to the linear, time-invariant (LTI) system $\mathcal{H}\{ \cdot \}$, we get $$\begin{eqnarray} \mathcal{H} \left\{ x(t) \right\} &=& \mathcal{H} \left\{ \int_{-\infty}^{\infty} x(t) \delta(t - \tau) d\tau \right\} \\ &=& \mathcal{H} \left\{ \int_{-\infty}^{\infty} x(\tau) \delta(t - \tau) d\tau \right\} \quad (\textrm{due to sifting property}) \\ &=& \int_{-\infty}^{\infty} x(\tau) \mathcal{H} \left\{ \delta(t - \tau) \right\} d\tau \quad (\textrm{due to linearity}) \\ &=& \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\tau \\ &=& x(t) * h(t) . \end{eqnarray}$$ The result of these 3 steps is known as the convolution integral.

This result shows us that we can describe the output of any linear, time-invariant system through the convolution integral. To find the output, we need to know the input signal $x(t)$ and the impulse response $h(t)$. The function $h(t)$ is the response of the system to an single impulse at time t=0, i.e., $\delta(t)$.

If an impulse response of a system is infinite in duration, we refer to it as an infinite impulse response system. If an impulse response of a system is finite in duration, we refer to it as an finite impulse response system.

RC circuit (impulse response)

The impulse response of an RC circuit (across the capacitor) is $$h(t) = \frac{1}{RC} e^{-t/(RC)} u(t).$$

If our input signal $x(t)$ is a unit impulse $x(t) = \delta(t)$, then the output of an RC circuit (across the capacitor) is the impulse response $$ \begin{eqnarray} y(t) &=& h(t) * \delta(t) \\ &=& h(t) \\ &=& \frac{1}{RC} e^{-t/(RC)} u(t). \end{eqnarray} $$

RC circuit (step response)

The impulse response of an RC circuit (across the capacitor) is $$h(t) = \frac{1}{RC} e^{-t/(RC)} u(t).$$

If our input signal $x(t)$ is a unit step function $x(t) = u(t)$, then the output of an RC circuit (across the capacitor) is the called the step response $$ \begin{eqnarray} y(t) &=& h(t) * u(t) \\ &=& (1 - e^{-t/(RC)}) u(t). \end{eqnarray} $$

A box system (step response)

The impulse response of a "box" system is $$ \begin{eqnarray} h(t) &=& u(t) - u(t - t_0) \quad \text{where} \quad t_0 > 0 \end{eqnarray} $$

If our input signal $x(t)$ is a unit step function $x(t) = u(t)$, then the output of a "box" system is $$ \begin{eqnarray} y(t) &=& h(t) * u(t) \\ &=& (u(t) - u(t-t_0)) * u(t). \end{eqnarray} $$

Convolution of two boxes

The impulse response of a "box" system is $$h(t) = u(t) - u(t - t_0) \quad \text{where} \quad t_0 > 0$$

If our input signal $x(t)$ is a box or rectangular function $x(t) = u(t) - u(t-t_0)$, then the output of an "box" system is $$ \begin{eqnarray} y(t) &=& h(t) * u(t) \\ &=& t u(t) - 2(t - t_0)u(t - t_0) + (t - 2 t_0)u(t - 2t_0). \end{eqnarray} $$