An example is presented in this page. Both the functions, h(t) and f(t), are pulses.
Functions h(t) and f(t) are defined as shown below. Obtain the convolution integral of h(t) and f(t). Let y(t) = f(t) * h(t).
SOLUTION :
From equations (7.1) and (7.2), the functions h(t) and f(t) can be sketched, as shown below.
Convolution Integral Approach: Analytical Technique
Given h(t) and f(t), the convolution integral of h(t) and f(t) can be obtained as presented below. From Fig. 27,
Function f(t) can be defined as follows.
Function f(t - λ) is a waveform that is obtained by reflecting f(t) and shifting to the right as the value of t increases from zero onwards. That is,
For t < 0, f(t - λ) lies on the left side of vertical axis and there is no overlap of h( λ) and f(t - λ) and the convolution integral has zero value. The overlap of h (λ) and f(t - λ) starts when t equals zero, as shown above in Fig. 28. Hence we get the equation, presented below.
The overlap of of h (λ) and f(t - λ) over the range 0 < t < 1 s, has been presented in Fig. 29. The product and the convolution integral are obtained as shown below. At any instant t , the convolution integral is the area enclosed by the plot of the product over the range from 0 to t .
The values of convolution integral, evaluated at the boundary values of t, are presented above.
The overlap of h (λ) and f(t - λ), obtained when t equals unity, is presented in Fig. 30. The plots of the product of h (λ) and f(t - λ) and the convolution integral have to be interpreted carefully. The plot of the product is a moving image, very much like the plot of f(t - λ). For the plots in Fig. 29, the value of t = 0.67 s, whereas for the plots in Fig. 30, the value of t = 1 s. In Fig. 29, the product has a value of 0.67 at τ = 0, where it has a value of 1 at τ = 0 in Fig. 30. On the other the value of the convolution integral equals the area of the product curve at any t. It is a cumulative curve. For τ = 0.67, the convolution integral has a value of 0.22 for its plot in Fig. 29 and in Fig. 30.
The plots of h (λ) and f(t - λ), their product and their convolution integral are presented in Fig. 31.
It can be seen from equations (7.10) and (7.12) that y(1) has the same value whether it is computed from equation (7.9) or equation (7.11). It means that the convolution integral is a continuous function at t = 1 s, and we are verifying that the solutions presented by equations (7.9) and (7.11). The plots obtained at t = 2 s, are presented in Fig. 32.
For values of t in the range 2 < t < 3 s, the trailing edge of f(t - λ) occurs after the trailing edge of h(λ) and there is no overlap from τ = 0 s till τ = (t - 2) s. From equations (7.12) and (7.14), it is seen that y(2) has the same value, validating the solutions presented by equations (7.11) and (7.13).
For the range 3 < t < 4 s, equation (7.15) presents the convolution integral, and the relevant plots are shown in Fig. 33.
The plot of the convolution integral over the period 0 < t < 4 s is shown above.
Laplace Transforms Approach
The Laplace transform of a time-shifted pulse is presented below.
Hence for this problem,
The inverse transform obtained is presented below. The inverse transform of Y(s) yields the convolution integral.
We can verify the solution by by evaluating y(t) at specific instants.
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