Continuous Time Convolution by Delayed Impulse
Signals & Systems
Continuous-Time Convolution Example
A linear time-invariant system is described by the impulse response h(t) = exp(-t)u(t). An input x(t) is applied to the system, and convolution will be used to determine the expression for the output y(t). The input has a value of 0.6 for -1 < t < 0.5, a value of 0.3 for 0.5 < t < 3, and a value of 0 everywhere else.
Impulse response h(t) and Input x(t)
To perform the convolution, one of the signals must be reversed in time; in this example, it will be x(t). Time-reversing x(t) makes it x(-t), so the signal is just a mirror image about t = 0. Note that plotting h(t) vs. t and x(-t) vs. t is the same as plotting h(tau) vs. tau and x(-tau) vs. tau. It doesn't matter what symbol is used for the time argument. Also note that plotting x(-tau) vs. tau is the same as plotting x(t-tau) vs. tau when t takes on the value t=0.
Impulse Response and Time-Reversed Input
The next step in the convolution is to shift the time-reversed input signal. Since the plots are made vs. the time variable tau, the shifted and reversed input is x(t-tau). The value of t is the amount by which x(-tau) is shifted, and it is the value of time at which the value of the output signal y(t) is being computed: y(t) = Integral[x(t-tau)h(tau)dtau]. Negative values of t shift the signal left; positive values shift it to the right.
Since the value of the output at a particular value of t is the area under the product of h(tau) and x(t-tau), the limits on the integration depend on the defining expressions for h(t) and x(t). Since x(t) is defined over segments of time, the overall integral will be broken into smaller intervals of time. The first segment is when the input is shifted far enough to the left that there is no overlap between x(t-tau) and h(tau). Therefore, the area under the curve is 0, and that is the value of y(t). The time interval for which the output is 0 can be determined by examining the curves. Input x(tau) is 0 for tau < -1, so x(-tau) is 0 for tau > 1. Realizing that x(-tau) = x(t-tau) for t = 0, and seeing that the transition from 0.6 to 0 in x(-tau) occurs at tau = 1 second, it can be seen that that transition occurs at a value of tau = t+1. This relation will always hold for this input, regardless of the value of t (amount of shift in the signal). The transition from 0.6 to 0 will always be at a value tau = t+1. In a like manner, the transition from 0.3 to 0.6 occurs at tau = t-0.5, and the transition from 0 to 0.3 occurs at tau = t-3. Therefore, the time interval for no overlap between the signals is all time where the leading edge of x(t-tau) is to the left of tau = 0, or t+1 < 0. Therefore, y(t) = 0 for all t < -1.
Input Shifted Left, No Overlap
If t is increased above -1, there is overlap between the two signals. The first situation is for there to be overlap with x(t) = 0.6, but no overlap where x(t) = 0.3. This occurs for t+1 > 0 and for t-0.5 < 0. The limits of integration will be the range of overlap, which is from tau = 0 to tau = t+1. The range of time for defining y(t) by this integral is -1 < t < 0.5. In this interval, y(t) = 0.6*[1 - 0.3679*exp(-t)].
Partial Overlap with 1st Segment
The next situation is for the x(t-tau) = 0.6 segment to be completely overlapped by h(tau) and the x(t-tau) = 0.3 segment to be partially overlapped. This occurs when t-0.5 > 0 and t-3 < 0. The limits of integration are from 0 to t-0.5 with x(t-tau) = 0.3 and from t-0.5 to t+1 with x(t-tau) = 0.6. The expression for y(t) in this interval is y(t) = 0.3*[1 + 0.9130*exp(-t)]. The range of time which defines this interval is 0.5 < t < 3.
Partial Overlap with 2nd Segment
The next, and last, situation to consider is full overlap for both segments. This occurs for t-3 > 0. Since h(t) is defined and non-zero for all positive finite time, the type of overlap does not change any more (even though it looks like h(tau) = 0 in the plot by tau = 10). The limits of integration are from t-3 to t-0.5 for x(t-tau) = 0.3 and from t-0.5 to t+1 for x(t-tau) = 0.6. Note the difference between these limits and those for the previous situation. The range of t which defines this interval is t > 3, and the expression for the output is y(t) = 6.2995*exp(-t).
Full Overlap with Both Segments
Here is the output signal produced by the convolution of the input signal x(t) with the system's impulse response h(t). Each expression can be plotted over its corresponding time interval.
Output Signal y(t)
MATLAB Code
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Latest revision on Thursday, May 18, 2006 10:50 PM
Source: https://people-ece.vse.gmu.edu/~gbeale/ece_220/convolution_01.html
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