Continuous-Time Linear Time-Invariant Systems

In this lesson, we will discuss linear time-invariant (LTI) systems - these are systems that are both linear and time-invariant. We will see that an LTI system has an input-output relationship described by convolution.

Impulse Representation of Continuous-Time Signals

Using the sifting property, we can write a signal x(t) as:

which is writing a general signal x(t) as a function of an impulse. This expresses the input x(t) as an integral (continuum sum) of shifted impulses that are weighted by weights x(τ).

Continuous Time Convolution

Again, we write:

Now take a system and define the impulse response h(t) of the system as the response of the system to an unit impulse input:

h(t)  =  S[δ(t)]

Next, define the response of the system to a shifted unit impulse as:

h(t,τ)  =  S[δ (t - τ)]

If the system is linear, then

S[αx1(t) +  βx2(t)]  =  αy1(t) +  βy2(t)

where of course S[x1(t)] = y1(t) and S[x2(t)] = y2(t).



But what if the system is also Time-Invariant?

Then S[δ(t - τ)] =  h(t , τ)  =  h(t - τ), since we had S[δ(t)] = h(t). Therefore,


We have seen that for a linear time-invariant system, the output is the input convolved with the system's impulse response h(t). In other words, we can completely characterize an LTI system by its impulse response. This is a very important result!

Again, we write the convolution integral as:

The notation h(t-τ) means that the function h(τ) is flipped and shifted across the function x(τ).

Convolution is a tough concept to get at first. Here are two rules that will greatly simplify doing convolutions:

  1. 1) DRAW A PICTURE of the functions x(τ) and h(t  - τ)


Why can we pick which function to flip?

Because convolution is commutative:

Change variables: λ = t - τ  → τ = t - λ,  = -.

(minus signs cancel)

Let's examine convolution formula:

  1. First, we time-reverse h(τ) and shift it to form h(t  - τ).
    Note: the independent variable of h(t - τ) is τ, not t! The variable t is the shift parameter, i.e. the function h(τ) is shifted by an amount t.
  2. Next, fix t and multiply x(τ) with h(t - τ) for all values of τ.
  3. Then integrate x(τ)h(t - τ) over all τ to get y(t) which is a single value that depends on t. Remember that τ is the integration variable and that t is treated like a constant when doing the integral.
  4. Finally, repeat for all values of t.

Fortunately, it usually falls out that there are only several regions of interest and the rest of y(t) is zero.

Example 1 Find y(t) = x(t)*h(t).

Form x(τ) and h(t - τ) (to shift the time-reversed function h(- τ) by t, just add t to all points).

When you finish notice:

  1. (a) The nonzero "width" of x(t) = 4
    (b) The nonzero "width" of h(t) = 3
    (c) The nonzero "width" of y(t) = 4+3 = 7

  2. y(t) is "smoother" than x(t) or h(t)