SIGNALS AND SYSTEMS-III PROPERTIES OF SYSTEMS CONVOLUTION-1

EDITOR: B. SOMANATHAN NAIR

1. CONVOLUTION

Convolution is a mathematical operation by which we obtain the output of a system as the sum of the products of the inputs (present and past) and respective impulse responses of the system appropriately delayed in time.

            We can express convolution in continuous time domain using the integral

where x(t) represents input of a system, y(t) represents its output, and h(t) represents its impulse response. The thick dot symbol between x(t) and y(t) represents the operation of convolution. In discrete time domain, (1) gets modified as a summation given by

We can prove that the convolution in time domain becomes a multiplication in frequency domain. For example, convolution defined in (1) can be expressed in terms of z transform components as

                                                          Y(z) = X(z) H(z)                                                                                                                       

where Y(z) is the z transform of y(n) and so on.

            There are several methods for performing convolution. Some of these include the direct, graphical, and the transformation methods. In the following section, we discuss the graphical method for performing convolution.

2. GRAPHICAL CONVOLUTION

Graphical convolution is one of the convenient methods of solving convolution problems involving continuous-time functions. In this Section, we make use of a simplified approach for solving convolution problems using graphical methods. The graphical convolution explained here makes use of the piece-wise linear approach for performing convolution. In this approach we integrate the product x(τ)h(t –τ) in between finite limits to yield the convolution sum. It can be seen that this method gives sufficiently accurate results using straight-line segments. This method is illustrated using numerical examples as given below.

ILLUSTRATIVE EXAMPLE 1: Figures 1 and 2 show pulse waveforms x(t) and h(t), respectively. Obtain their convolution using graphical method.

Solution: As stated earlier, convolution of the functions x(t) and h(t) is defined mathematically as:

In (1), the term x(t) is represented using the dummy variable τ (tau) as x(τ). The second function h(t – τ) is moved from the left (i.e., from –∞) to the right (to +∞) in regular steps indicated by the time intervals of x(τ) and h(t – τ). In each interval, we multiply the areas common to both these functions through integration. These operations are illustrated as given below.

Step 1: Invert h(τ) to obtain h(–τ). This operation is shown in Fig. 3.

Step 2: Now start moving h(–τ) in steps of 1 unit time interval to obtain the functions h(1–τ), h(2–τ),  h(3–τ), and  h(4–τ). We now observe that the functions h(4–τ) and x(τ) just touch each other as shown in Fig. 4. There is no common area of the two functions in this position (or interval) and hence convolution sum at this point is zero.

We designate this as y(4), which indicates that the convolution sum in this interval t = 4 is 0.

Step 3: In this step, we move the function h(–τ) again by one unit time interval to get the function h(5–τ) as shown in Fig. 5. This function now extends from t = 1 to 5. In this figure, we notice that x(τ) and h(5–τ) have some common area (indicated by shading)  in the interval between t = 4 and t = 5.  The limits of the convolution integral in this case are thus t = 4 and t = 5, respectively. Hence, the magnitudes of x(τ) and h(5–τ) in this interval are given by

                                                          x(t) = 2,   4≤ t ≤ 5  

h(5 ‒ τ) = 1,   4≤ t ≤ 5

Using these values, we now use the convolution integral to get the convolution sum as:

This result shows that the convolution sum in the interval 4 to 5 is 2 and is designated as y(5).

Step 4: In this step, we move the function h(–τ) again by one unit interval of time to get the function h(6–τ) as shown in Fig. 6. This function now extends from t = 2 to 6. In this figure, we notice that x(τ) and h(6–τ) have some common area (indicated by shading) in the interval between t = 4 and t = 6.  The limits of the convolution integral in this case are thus t = 4 and t = 6, respectively. Hence the magnitudes of x(τ) and h(6–τ) in this interval are given by

                                                                  x(t) = 2,   4≤ t ≤ 6  

             h(6 ‒ τ) = 1,   4≤ t ≤ 6

Using these values, we now use the convolution integral to get the convolution sum as:

This result shows that the convolution sum in the interval 4 to 6 is 4 and is designated as y(6).

Step 5: In this step, we move the function h(–τ) again by one unit interval of time to get the function h(7–τ) as shown in Fig. 7. This function now extends from t = 3 to 7. In this figure, we notice that x(τ) and h(7–τ) have again the same common area (indicated by shading) in the interval between t = 4 and t = 6.  The limits of the convolution integral in this case also are t = 4 and t = 6, respectively. Hence the magnitudes of x(τ) and h(7–τ) in this interval are given by

                                                                x(t) = 2,   4≤ t ≤ 6  

             h(7 ‒ τ) = 1,   4≤ t ≤ 6

Using these values, we now use the convolution integral to get the convolution sum as:

This result shows that the convolution sum in the interval 4 to 6 in this case also is 4 and is designated as y(7).

Step 6: In this step, we move the function h(–τ) again by one unit interval of time to get the function h(8–τ) as shown in Fig. 8. This function now extends from t = 4 to 8. In this figure, we notice that x(τ) and h(8–τ) have again the same common area (indicated by shading) in the interval between t = 4 and t = 6.  The limits of the convolution integral in this case also are t = 4 and t = 6, respectively. Hence the magnitudes of x(τ) and h(8–τ) in this interval are given by

                                                                x(t) = 2,   4≤ t ≤ 6  

             h(8 ‒ τ) = 1,   4≤ t ≤ 6

Using these values, we now use the convolution integral to get the convolution sum as:

This result shows that the convolution sum in the interval 4 to 6 in this case also is 4 and is designated as y(8).

Step 7: In this step, we move the function h(–τ) again by one unit interval of time to get the function h(9–τ) as shown in Fig. 9. This function now extends from t = 5 to 9. In this figure, we notice that x(τ) and h(9–τ) have again the same common area (indicated by shading) in the interval between t = 5 and t = 6.  The limits of the convolution integral in this case also are t = 5 and t = 6, respectively. Hence the magnitudes of x(τ) and h(9–τ) in this interval are given by

                                                             x(t) = 2,   5≤ t ≤ 6  

                                                            h(8 ‒ τ) = 1,   5≤ t ≤ 6

Using these values, we now use the convolution integral to get the convolution sum as:

This result shows that the convolution sum in the interval 4 to 6 in this case gets reduced again to 2 and is designated as y(9).

Step 8: In this final step, we move the function h(–τ) is again moved by one unit interval of time to get the function h(10–τ) as shown in Fig. 10 This function now extends from t = 6 to 10. and hence we notice that x(τ) and h(10–τ) have no common area at all. Hence the convolution sum y(10) = 0 and we stop our computation.

  

            Figure 11 shows the result of convolution sum as a plot. To plot this trapezoidal waveform, we use the values of y(0) to y(10). It can be seen that the convolution sum has values only from t = 4 to t = 10. This is because only in between these two intervals that x(τ) and h(–τ) have common area and hence can be integrated to obtain the convolution sum.

   In the next blog we shall discuss discrete convolution.