Let $f(t)$ and $g(t)$ be signals.

I can't figure out what happens to the convolution of $f(t)$ and $g(t)$ if

  1. both $f(t)$ and $g(t)$ are reversed

  2. one of $f(t)$ and $g(t)$ is reversed

I tried the change of variable in integral, but couldn't find out the answer.

  • $\begingroup$ @EdV Sure I searched a lot but I can't find the answer that I want. Actually I know if both input signals are reversed, then convolution doesn't change. But I just wrote a whole, and expected there would be a fundamental explanation that can explain both situation in the same manner. Anyway, first of all, I couldn't find the answer that explains situation 2 with integral method. $\endgroup$
    – achilles
    Apr 7 '20 at 21:20
  • $\begingroup$ Show your work and we may be able to help you you where you got stuck. I suggest writing out the equations and also doing it graphically so that you intuitively see what is happening. $\endgroup$ Apr 8 '20 at 4:01
  • $\begingroup$ When I tried to do some calculation with some simple discrete signals that I just made for experiment, my conclusion is that output signal when one of f, g is time-reversed just gets totally different from the original output signal. $\endgroup$
    – achilles
    Apr 8 '20 at 9:05
  • $\begingroup$ Also in continuous signal, I wrote a convolution integral of f and g in two terms, which means I wrote two integral terms which have range of -inf~0 and 0~+inf respectively. Then I compared the original convolution of f, g with the convolution of time-reversed f and g by assuming t = 3. Then the difference between these two output couldn't be disappeared even though I shifted or time-reversed the original output signal. $\endgroup$
    – achilles
    Apr 8 '20 at 9:10

This question appears to involve confusion about the relationship between convolution and cross-correlation. To illustrate the relationship, consider the two rectangular pulses shown below:

Two rectangular pulses

The shorter pulse has width = 1 and the longer pulse has width = 2. For simplicity, assume unity heights for the pulses.

In my answer to a previous question, I used a figure from a textbook 1 to illustrate the convolution of $\mathrm {x_1(\lambda)}$ and $\mathrm {x_2(\lambda)}$, denoted by $\mathrm {x_1*x_2}$.

The convolution integral was given by

$$\mathrm {x_1*x_2 =\int_{-\infty}^{\infty}x_1(\lambda)x_2(t-\lambda)\,\mathrm d\lambda} \tag{1}$$

so $\mathrm {x_2(\lambda)}$ was reflected around the ordinate and is the function that gets shifted. This is clearly shown in the figure in my linked answer.

But it is also possible to reflect $\mathrm {x_1(\lambda)}$ about the ordinate and shift it. Then the convolution would be denoted by $\mathrm {x_2*x_1}$ and the convolution integral would be

$$\mathrm {x_2*x_1 =\int_{-\infty}^{\infty}x_1(t-\lambda)x_2(\lambda)\,\mathrm d\lambda} \tag{2}$$

This is illustrated the following figure that I drew in the same fashion as the figure in my linked answer:

Convolution of rectangular pulses

Thus $\mathrm {x_1*x_2 = x_2*x_1}$, i.e., it does not matter which function is reversed for the convolution. Of course, if a function is symmetric about its center, then reversing becomes rather trivial. In this example, both rectangular pulses are symmetric about their centers, so the situation is especially simple.

In corroboration of the above, the following (through their equation 1.5-3) is directly quoted from Blinchikoff and Zverev [2, p. 12]:

Let p(t) and q(t) each be a piecewise continuous function for $-\infty < t < \infty$. The convolution of p and q, denoted by $p*q$, is defined to be the third function

$$\mathrm {v(t) = p*q =\int_{-\infty}^{\infty}p(t-\tau)g(\tau)\,\mathrm d\tau} \tag{1.5-1}$$

and the integral on the right is called the convolution integral. Furthermore, the convolution of q and p, $q*p$, is identical to p*q, that is,

$$q*p = p*q \tag{1.5-2}$$

This is easily shown by letting $\mathrm {x = t -\tau}$, $\mathrm {dx = -d\tau}$ in (1.5-1), yielding

$$\mathrm {p*q =\int_{-\infty}^{\infty}p(x)g(t-x)\,\mathrm dx} = q*p \tag{1.5-3}$$

Therefore, in answer to the OP's second question, it does not matter mathematically if f(t) or g(t) is the reversed function, but the calculation may be simplified, in practice, if either function is symmetric about its center.

For cross-correlation, all that is necessary is to shift one function by $\tau$, as per this photo image from [2, p. 305]:

B and K Table 7-2

So cross-correlation calculations are quite similar to convolution calculations, but note the important considerations discussed directly below Blinchikoff and Zverev's Table 7-2. I suggest that the OP think about their first question, i.e., what happens when both f(t) and g(t) are reversed and the overlap integrals are calculated.

1 C.D. McGillem, G.R. Cooper, "Continuous and Discrete Signal and System Analysis", 2nd Ed., Holt, Rinehart and Winston, NY, ©1984, pp. 58-59.

2 H.J. Blinchikoff, A.I. Zverev, "Filtering in the Time and Frequency Domains", Wiley-Interscience, John Wiley & Sons, NY, ©1976.


If you view convolution as a symmetric operation with the rather non-formal but symmetric definition $$f(t)\star g(t)=\int_{t_1+t_2 = t} f(t_1)\,g(t_2)\,d?$$ (a more formal version would be $$f(t)\star g(t)=\int_{t_1}\int_{t_2} f(t_1)\,g(t_2)\,\delta(t_1+t_2-t)\,dt_1\,dt_2$$)

then it becomes clear that flipping both functions does not create the same result as before but actually does flip the result in consequence.

Flipping just one function replaces the convolution with cross-correlation. However, the results for flipping one or the other are not identical but again differ by one cross-correlation being the mirrored version of the other cross-correlation.


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