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The question asks to compute the convolution of x(t) and h(t). While I know how to do this mathematically, using a combination of derivatives and integrals, I don't know how to convolve the two using the graphical approach.

Either method should work, but I want to know the specifics of the graphical approach.

enter image description here

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2 Answers 2

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Doing this graphically, or conceptually, is quite easy. This problem seems to have been designed to make it easy.

Imagine the top signal sliding along, with the bottom signal staying fixed. For each position of the top signal, produce the product of the two signals, then add up the resulting area under the curve.

For example, at T = 0, the top signal is as shown. Note that only the part of the top signal between t=1 and t=2 matters. The bottom signal is 1 there, so basically the product of the two is always the small window of the top signal from t=1 to t=2.

At T = 0, that product is just -1 from t = 1 to 2, for a total area of -1. You should be able to see from inspection that with the top signal advanced by 1 to the right (T = 1), that same product will be a positive rectangle with area +1. The values in between will obviously be linear from -1 to +1. Likewise, at T = 2 you also get +1, then at T = 3 and greater, 0.

So basically you solve the convolution by inspection at three points, then also see from inspection that it will vary linearly between these point.

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Major edit to improve clarity and show link between graphical and mathematial methods:

Mathematically, the convolution integral is defined as: $$ \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau $$

Graphically, what you are doing is taking your second function \$h(\tau)\$, reflecting it about the y-axis (\$h(-\tau)\$), and shifting it (to the right) by some value \$t\$: $$ h(-(\tau-t))=h(-\tau+t))=h(t-\tau). $$

The convolution of the two signals at \$t\$ is equal to the area overlapped by the two signals when \$h(\tau)\$ is shifted by \$t\$, mathematically described as: $$ y(t) = \int_{-\infty}^{t}x(\tau)h(t-\tau)d\tau $$

In our case, \$h(-\tau)\$ is valued on the interval \$[-2, -1]\$ when \$t=0\$, and \$x(\tau)\$ is valued on the interval \$[-1, 2]\$. Therefore, overlap only occurs when \$t\geq0\$, which redefines our limits:

$$ y(t) = \int_{0}^{t}x(\tau)h(t-\tau)d\tau $$

Graphically, then, the convolution of the two functions can be thought of as the area of the overlap of the two functions as you move \$h(t-\tau)\$ along the \$\tau\$-axis. At \$\tau=0\$ in your case, the convolution \$y(t)=0\$, as there is no intersecting area. Shift \$h(\tau)\$ to \$t=0.5\$ and \$y(t)=0.5\$ (change in t times the height of the two signals).

\$y(t)\$ will continue to increase until the "trailing" edge of \$h(\tau)\$ passes the "leading" edge of \$x(\tau)\$ at \$t=1\$. Now the area (and therefore \$y(t)\$) will remain constant until \$t=2\$, when the pulse in \$h(\tau)\$ begins to overlap the negative portion of \$x(\tau)\$. Negative function values have negative areas, and so as \$h(\tau)\$ continues, \$y(t)\$ drops to zero at \$t=2.5\$ (when the sums of the positive and negative areas encompassed within the \$h(\tau)\$ pulse cancel out) and then continues to decrease to -1 when \$t=3\$. At this point, the non-zero portion of \$h(\tau)\$ is fully enclosing the negative portion of \$h(\tau)\$. Finally, \$y(t)\$ will begin to return to zero as the negative area diminishes, ending at \$t=4\$.

In summary, graphic convolution describes the area overlapped by a reflected \$h(\tau)\$ as it passes "through" \$x(\tau)\$ along the \$\tau\$-axis. The value of the convolution at any \$t\$-value is equal to the overlapped area when \$h(t-\tau)\$ is at that \$t\$-value.

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  • \$\begingroup\$ While I understand your logic and follow it very well, what particularly concerns me is that when I apply the mathematical method, and even when I perform the convolution in Mathematica, I get a result that begins at t = 0, instead of t = -1. To me, it seems as if any method I choose should end in the same result, so I'm a bit confused as to why the graphical method does not begin at t = 0. \$\endgroup\$
    – Wei
    Jun 6, 2017 at 19:31
  • \$\begingroup\$ The limits of the convolution integral are \$ (-\infty, \infty) \$. Is Mathematica assuming x(t) starts at zero, and integrating from zero? Are you integrating from zero? \$\endgroup\$
    – Chris M.
    Jun 6, 2017 at 20:30
  • \$\begingroup\$ I see the problem - the mistake is mine. I will update the answer to correct it. \$\endgroup\$
    – Chris M.
    Jun 6, 2017 at 21:51
  • \$\begingroup\$ I don't quite understand the concept of t being the leading edge + 1. I see what you're saying in regards to having to reflect h(t) over the y-axis, in which case it is a square wave that has value 1 from {-2,-1}, and 0 elsewhere. There is still overlapping area at time t = -1 still, is there not? \$\endgroup\$
    – Wei
    Jun 7, 2017 at 1:55
  • \$\begingroup\$ Remember that when you do the convolution integral, you reflect \$h(t)\$ about y and shift it by a value \$\tau\$. When you're sliding along the axis, you're actually changing the value of \$\tau\$. At \$\tau=0\$ we have just the reflection as it is, with \$h(t)\$ valued at [-2, -1], where it starts to intersect. \$\endgroup\$
    – Chris M.
    Jun 7, 2017 at 2:24

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