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Impulse Response of a System

A $discrete$ signal $s (n)$ can be written as a linear combination of shifted impulses.

s (n) = \sum_{k = 0}^{k = N - 1} s (k) δ (n-k)

$s (n) = s (0) δ (n) + s (1) δ (n - 1) + \dots + s (N - 1) δ (n-(N - 1))$

If the system is $linear$ we can compute the response of the system to the signal $s (n)$ if we know the $impulse response$ of the system to each $impulse$ in the signal.

For a $time invariant$ system, this is easy to do because the characteristics of its $impulse response$ does not change with time: The system has only one impulse response.

For a $time variant$ system, however, this is not so easy.

Time Variant System

time variant

system's impulse response changes with time.

$δ (n - k) \to h_{k} (n) o r T [δ (n - k)] = h_{k} (n)$

That is, the system produces the output $h_{k} (n)$ in response to the impulse input $δ (n - k)$ occurring at time $k$ . The characteristics of the system's impulse response changes as a function of time: The system can potentially have many impulse responses (at least two.)

Time Invariant System

time invariant

system's impulse response does not change with time.

$δ (n - k) \to h (n - k) o r T [δ (n - k)] = h (n - k)$

Namely, the system produces the output $h (n - k)$ in response to the impulse input $δ (n - k)$ occurring at time $k$ . The characteristics of the system's impulse response does not change as a function of time, but the response is shifted in time by $k$ units of time.

Response of a Linear System

$x_{1} (n) \to y_{1} (n) o r T [x_{1} (n)] = y_{1} (n)$

$x_{2} (n) \to y_{2} (n) o r T [x_{2} (n)] = y_{2} (n)$

$x_{1} (n) + x_{2} (n) \to y_{1} (n) + y_{2} (n) o r T [x_{1} (n) + x_{2} (n)] = y_{1} (n) + y_{2} (n)$

The response of the system to the sum of individual inputs is identical to the sum of system's response to each individual input.

We can represent a discrete signal $s (n)$ as a linear combination of shifted impulses.

$s (n) = \sum_{k = 0}^{k = N - 1} s (k) δ (n-k)$

We can then compute the response of the system to each individual component of the linear combination and add up all the responses.

$T [s (n)] = \sum_{k = 0}^{k = N - 1} T [s (k) δ (n-k)] = \sum_{k = 0}^{k = N - 1} s (k) T [δ (n-k)]$

The response of the system to each individual component of the linear combination will be the product of the value of the signal $s (n)$ at time $k$ , $s (k)$ , and the response of the system to the impulse occurring at time $k$ , $T [δ (n - k)]$ .

$s (n) = s (0) δ (n) + s (1) δ (n - 1) + \dots + s (N - 1) δ (n-(N - 1))$

$T [s (n)] = s (0) T [δ (n)] + s (1) T [δ (n - 1)] + \dots + s (N - 1) T [δ (n-(N - 1))]$

Then the response of a $time variant$ system is given by:

$y (n) = T [s (n)] = \sum_{k = 0}^{k = N - 1} s (k) h_{k} (n)$

$y (n) = T [s (n)] = s (0) h_{0} (n) + s (1) h_{1} (n - 1) + \dots + s (N - 1) h_{N - 1} (n)$

And the response of a $time invariant$ system is given by:

$y (n) = T [s (n)] = \sum_{k = 0}^{k = N - 1} s (k) h (n-k)$

$y (n) = T [s (n)] = s (0) h (n) + s (1) h (n - 1) + \dots + s (N - 1) h (n-(N - 1))$

An Example

We compute the $response$ $y (n)$ of a $linear system$ system to the input signal $s (n)$ .

Input Signal

Given the signal

s (n) = {2, 3, 5, 7}

We can write it in terms of shifted impulses as

s (n) = s (0) δ (n) + s (1) δ (n - 1) + s (2) δ (n - 2) + s (3) δ (n - 3)

s (n) = 2 δ (n) + 3 δ (n - 1) + 5 δ (n - 2) + 7 δ (n - 3)

The input signal consists of $4$ impulses.

If we know the

impulse response

of the system for each

impulse

, we can compute its

response

y (n)

to the input signal

s (n)

Time Variant System

Since a $time variant$ system can potentially have more than one impulse response, we write its response $y (n)$ as:

y (n) = s (0) h_{0} (n) + s (1) h_{1} (n) + s (2) h_{2} (n) + s (3) h_{3} (n)

y (n) = 2 h_{0} (n) + 3 h_{1} (n) + 5 h_{2} (n) + 7 h_{3} (n)

Notice the subscripts. They account for the fact that the system can respond differently to each impulse in the input.

Time Invariant System

Since a $time invariant$ system has only one impulse response, we write its response $y (n)$ as:

y (n) = s (0) h (n) + s (1) h (n - 1) + s (2) h (n - 2) + s (3) h (n - 3)

y (n) = 2 h (n) + 3 h (n - 1) + 5 h (n - 2) + 7 h (n - 3)

Notice the time delays. They account for the fact that the system responds identically to each impulse in the input but the response is shifted in time by the same amount as the corresponding impulse is shifted in time.

[This content was created with SL, available on the App Store.]