Introduction to time-domain representation, convolution sum, and evaluation procedure.
The convolution sum
To begin explaining the convolution sum we start by assuming a discrete-time case.
At first, an arbitrary signal is expressed as a weighted superposition of shifted impulses. Then, the convolution sum is found by applying a signal represented in an LTI system. A similar procedure is used to find convolution integral later on.
Let the signal x[n] be multiplied by the impulse sequence δ[n].
Implying: x[n]δ[n] = x[0]δ[n].
The relationship is generalized to the product of x[n] and a time-shifted impulse sequence to obtain:
x[n]δ[n-k] = x[k]δ[n-k].
Where n represents the time index; therefore x[n] denotes the entire signal, while x[k] represents a specific value of x[n] at time k.
The multiplication of a signal by a time-shifted impulse results in a time-shifted impulse with amplitude given by the value of the signal at the time the impulse occurs.
It allows us to express x[n] as the following weighted sum of time-shifted impulses:
x[n]=........+x[-2]δ[n+2] +x[-1]δ[n+1]+ x[0]δ[n]......
This equation can be rewritten as:
x[n]=Σ x[k]δ[n-k].
The graphical illustration of convolution sum is given below:
Image is taken from electrical academia.
Let us now use the H operator to write the equation:
y[n]=H{x[n]} =H{Σ x[k]δ[n-k]}.
Now using the linearity property to interchange the system operator H with the summation, we obtain:
y[n]=Σ H{x[k]δ[n-k]}.
As n is the time index, the quantity x[k] is a constant with respect to the system operator H. Using linearity again, we can interchange H with x[k] to obtain:
y[n]=Σ x[k]H{δ[n-k]}.
It indicates that the system output is a weighted sum of the response of the system to time-shifted impulses. This response completely characterizes the system’s input-output behavior and is a fundamental property of the linear system.
Moving ahead, if we assume that the system is time-invariant, then a time shift in the input results in a time shift in the output.
It implies that the output due to a time-shifted impulse is a time-shifted version of the output.
The impulse is given by:
H{δ[n-k]}=h[n-k]
Therefore, the output of an LTI system is given by a weighted sum of time-shifted impulse responses.
It is a direct consequence of expressing the input as a weighted sum of time-shifted basis functions.
The convolution sum equation is given below:
x[n]*h[n]=Σ x[k]h[n-k].
The input will be decomposed as a sum of weighted and time-shifted unit impulses, with the kth impulse input represented in the right half as:
H{x[k]δ[n-k]}=x[k]h[n-k].
The output component is obtained by shifting the impulse response k units in time and multiplying by x[k]. The total output y[n] in response to the input x[n] is obtained by finding the summation of all the individual outputs:
y[n]=Σ x[k]h[n-k].
For each value of n, we find the sum of the outputs associated with each weighted and time-shifted impulse input from k=-infinity to +inifnity.
Convolution sum evaluation procedure
As mentioned earlier convolution sum is expressed as:
y[n]=Σ x[k]h[n-k].
Let us suppose, we define the intermediate signal as:
wn[k]=x[k]h[n-k].
as the product of x[k] and h[n-k]. In this definition, k is the independent variable and we explicitly indicate that n is treated as a constant by writing n as a subscript.
Now as we know,
h[n-k] = h[-(k-n)], it is reflected and time shifted version of h[k].
Therefore, if n is negative, then h[n-k] is obtained by time shifting h[-k] to the left.
While if the n is positive then h[-k] is shifted to the right. This time shift determines the time at which we evaluate the output of the system, as:
y[n]=Σ wn[k].
Though, we only need to determine one signal wn[k], for each time n at which we desire to evaluate the output.
Procedure 1:
Reflect and shift convolution sum evaluation
Step 1: Graph both x[k] and h[n-k] as a function of the independent variable k. To determineh[n-k], first reflect h[k] about k=0 to obtain h[-k], then shift by -n.
Step 2: Start with n being large and negative.
Step 3: Write the mathematical expression to represent the intermediate signal i.e wn[k].
Step 4: Increase shift n toward the right until the mathematical representation for wn[k] changes.
Step 5: Let n, be the new interval.
Step 6: At every interval of time shifts, sum all the values of the corresponding wn[k] to obtain y[n] on that interval.
Reference
Introduction to time-domain representation, convolution sum, and evaluation procedure.