# Flipping the impulse response in convolution

During convolution on a signal, why do we need to flip the impulse response during the process?

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The latter half of this answer might help you understand. –  Dilip Sarwate Nov 14 '12 at 23:16
In addition to reading @DilipSarwate's great answer it's a good exercise to take a sheet of paper and to calculate the output of an LTI system graphically by adding up time-shifted and scaled versions of the impulse response. –  Deve Nov 15 '12 at 9:15
Note that you can flip either argument - the result is the same. –  wakjah Nov 28 '12 at 22:23

Adapted from an answer to a different question (as mentioned in a comment) in the hope that this question will not get thrown up repeatedly by Community Wiki as one of the Top Questions....

There is no "flipping" of the impulse response by a linear (time-invariant) system. The output of a linear time-invariant system is the sum of scaled and time-delayed versions of the impulse response, not the "flipped" impulse response.

We break down the input signal $x$ into a sum of scaled unit pulse signals. The system response to the unit pulse signal $\cdots, ~0, ~0, ~1, ~0, ~0, \cdots$ is the impulse response or pulse response $$h[0], ~h[1], \cdots, ~h[n], \cdots$$ and so by the scaling property the single input value $x[0]$, or, if you prefer $$x[0](\cdots, ~0, ~0, ~1, ~0,~ 0, \cdots) = \cdots ~0, ~0, ~x[0], ~0, ~0, \cdots$$ creates a response $$x[0]h[0], ~~x[0]h[1], \cdots, ~~x[0]h[n], \cdots$$

Similarly, the single input value $x[1]$ or creates $$x[1](\cdots, ~0, ~0, ~0, ~1,~ 0, \cdots) = \cdots ~0, ~0, ~0, ~x[1], ~0, \cdots$$ creates a response $$0, x[1]h[0], ~~x[1]h[1], \cdots, ~~x[1]h[n-1], x[1]h[n] \cdots$$ Notice the delay in the response to $x[1]$. We can continue further in this vein, but it is best to switch to a more tabular form and show the various outputs aligned properly in time. We have $$\begin{array}{l|l|l|l|l|l|l|l} \text{time} \to & 0 &1 &2 & \cdots & n & n+1 & \cdots \\ \hline x[0] & x[0]h[0] &x[0]h[1] &x[0]h[2] & \cdots &x[0]h[n] & x[0]h[n+1] & \cdots\\ \hline x[1] & 0 & x[1]h[0] &x[1]h[1] & \cdots &x[1]h[n-1] & x[1]h[n] & \cdots\\ \hline x[2] & 0 & 0 &x[2]h[0] & \cdots &x[2]h[n-2] & x[2]h[n-1] & \cdots\\ \hline \vdots & \vdots & \vdots & \vdots & \ddots & \\ \hline x[m] & 0 &0 & 0 & \cdots & x[m]h[n-m] & x[m]h[n-m+1] & \cdots \\ \hline \vdots & \vdots & \vdots & \vdots & \ddots \end{array}$$ The rows in the above array are precisely the scaled and delayed versions of the impulse response that add up to the response $y$ to input signal $x$. But if you ask a more specific question such as

What is the output at time $n$?

then you can get the answer by summing the $n$-th column to get \begin{align*} y[n] &= x[0]h[n] + x[1]h[n-1] + x[2]h[n-2] + \cdots + x[m]h[n-m] + \cdots\\ &= \sum_{m=0}^{\infty} x[m]h[n-m], \end{align*} the beloved convolution formula that befuddles generations of students because the impulse response seems to be "flipped over" or running backwards in time. But, what people seem to forget is that instead we could have written \begin{align*} y[n] &= x[n]h[0] + x[n-1]h[1] + x[n-2]h[2] + \cdots + x[0]h[n] + \cdots\\ &= \sum_{m=0}^{\infty} x[n-m]h[m], \end{align*} so that it is the input that seems "flipped over" or running backwards in time! In other words, it is human beings who flip the impulse response (or the input) over when computing the response at time $n$ using the convolution formula, but the system itself does nothing of the sort.

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At index c[n], the convolution of a[n] and b[n], is such that:

"c[n] is a summation of all products (a[k]b[m]) such that m+k=n," so m = n - k or k = n - m, which means that one of the sequences has to be flipped.

Now why does convolution behave this way in the first place? Because of its connection with multiplying polynomials.

Multiplying two polynomials results in a new polynomial with co-efficients. The co-efficients of the product polynomial define the operation of convolution. Now, in signal processing, transfer functions- Laplace transforms or z-transforms are these polynomials, with each co-efficient corresponding to a different time-delay. Matching the co-efficients of the product and the multiplicands results in the fact that 'multiplication in one representation corresponds to convolution in the transformed representation'.

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During convolution, no "flip" of the impulse response needs to occur at all...

However, if you want to prevent any phase alteration, you can convolute a signal with an impulse response and then reverse the impulse response and re-convolute to cancel phase effects.

In offline processing, you could just as easily reverse the signal after the first convolution to get to the same conclusion (as the comments suggest).

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He's referring to the "time reversal" on the impulse response in the convolution integral: $y(t) = \int_{-\infty}^{\infty}x(\tau) h(t-\tau) d\tau$. As someone else already pointed out, you don't have to flip the impulse response $h(t)$; you can flip either term (i.e. $x(t) * h(t) = h(t) * x(t)$). I think he's trying to figure out what the qualitative interpretation of that "flip-and-slide" action is. –  Jason R Nov 29 '12 at 0:55
@JasonR Ah, whoops! Sometimes hard to see what the question is getting at. Izhak, once you understand the answer you were looking for, you will understand where I was going. Ignore me for now! –  learnvst Nov 29 '12 at 3:08

It's only 'flipped' for pointwise computation.

@Dilip explains what the convolution integral/summation represents, but to explain why one of the two input functions (often h(t)) is flipped for computation purposes, consider a discrete-time system with input x[n] and impulse response h[n]:

• You could take your input function x[n], and for each non-zero* sample x[n] calculate the scaled impulse response from sample n and on until the time-shifted h[n] dies down to zero (assuming a causal h[n]). This would involve no 'flipping' (or more accurately 'time-reversal') of either x[n] or h[n]. However, at the end you would have to add/superimpose all these scaled+shifted 'echos' of the impulse response for each non-zero x[n].

• Or, for convenience you could time-reverse one of the functions about the time origin (usually 0), making your computation {multiply, add, multiply, add, ...} instead of {multiply, multiply, ..., add, add, ...}. This results in the same output signal because it will perform the exact same multiply and add operations. For example, think about the output contribution from a non-zero input signal at time 0 x[0]. When k = 0 for the equation $$\sum_{k=-\infty}^{\infty} x[k]h[n-k]$$ the impulse reponse h[n] will only be time-reversed but not shifted, giving us the first sample response for x[n], which is x[0]h[0]. Then, incrementing k by one will shift h[n] to the right one time step, such that the time-reversed h[n]s second entry (h[1]) will now be laying on top of x[0], waiting to be multiplied. This will yield the desired contribution x[0]h[1] at time n=1, just as would have been done in the previous method.

*I say non-zero x[n] because $$\forall x[n] = 0$$ the impulse response h[n] is scaled to zero, thus contributing nothing to the final output y[n].

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"You could take your input function x[n], and for each non-zero* sample x[n] calculate the scaled impulse response from sample n and on until the time-shifted h[n] dies down to zero (assuming a causal h[n])" Are the five $n$ that occur in this sentence all the same number or are they different? –  Dilip Sarwate Jan 28 '13 at 4:21
@Dilip. All n are the same, except for 'the time-shifted h[n]', which implies 'h[n-k]', where 'k' is a constant used to shift the impulse response to the desired point of signal x[n]. ie: h[n-2] for calculating response to signal at x[2]. –  sixstring91 Feb 24 '13 at 17:41