Q:

Deconvolution by Convolution

This is now a second time I am attempting to ask this very important but simple question here. What I want to know is can you do deconvolution by convolving a signal. It is often stated that, for example by cutting and boosting the same frequency on an equalizer the result is the original signal. Is that the case? Can convolution be removed by convolving? That would certainly brake the identity that the convolved signal must be n+m-1 in length. I tried this with an equalizer and the results seem to be close to perfect, just some quantization distortion.
If the above can be done, I can not arrive at this conclusion. For example If I convolve (1) with (1, 1) the result is (1, 1). I can not find any impulse response that if convolved with (1, 1) would result in (1).
So what is the truth here?

A:

Since convolution describes the operation of a linear time-invariant (LTI) system, the question is if the effect of an LTI system can be compensated by another LTI system. In the discrete-time domain you can use the $\mathcal{Z}$-transform to analyze LTI systems. If a signal $x(n)$ (with $\mathcal{Z}$-transform $X(z)$) is filtered by a system with impulse response $h(n)$ (with $\mathcal{Z}$-transform $H(z)$), then the $\mathcal{Z}$-transform of the resulting output signal $y(n)$ is given by
$$Y(z)=X(z)H(z)$$
Recovering $X(z)$ from $Y(z)$ can theoretically be done by
$$X(z)=\frac{Y(z)}{H(z)}$$
except at points $z$ where $H(z)=0$. The above equation corresponds to convolution with an impulse response $g(n)$ with $\mathcal{Z}$-transform
$$G(z)=\frac{1}{H(z)}\tag{1}$$
The question is if this system can be realized by a causal and stable filter. This is indeed the case if $H(z)$ is a minimum-phase system. Assuming rational transfer functions (which we always get when discrete-time systems consist of adders, multipliers, and delay elements), this means that $H(z)$ is a causal and stable system with all its zeros inside the unit circle of the complex plane (its poles are of course also inside the unit circle because of stability). Due the inversion in (1), the zeros of $H(z)$ are the poles of $G(z)$ and vice versa, i.e. if the zeros of $H(z)$ are inside the unit circle, then $G(z)$ is stable. One more thing that can be seen from (1) is that if $H(z)$ is an FIR system, then $G(z)$ is IIR. Nevertheless, in practice equalization if often done by FIR filters, which approximate the inverse filter in some optimal sense.
Let's look at a simple example. Assume an FIR system
$$h(n)=\delta(n)-a\delta(n-1),\quad |a|<1$$
Then we have
$$H(z)=1-az^{-1}$$
$H(z)$ is a minimum-phase system because it has one pole at $z=0$ and a zero at $z=a$, which is inside the unit circle if $|a|<1$.
The inverse system is given by
$$G(z)=\frac{1}{1-az^{-1}}=\sum_{n=0}^{\infty}a^nz^{-n}$$
the impulse response of which is
$$g(n)=a^n,\quad n\ge 0$$
which is of course IIR.
So in this case the convolution
$$y(n)=(x*h)(n)=x(n)-ax(n-1)$$
can be perfectly compensated for by the convolution
$$x(n)=(y*g)(n)=\sum_{k=0}^{\infty}g(k)y(n-k)=\sum_{k=0}^{\infty}a^ky(n-k)$$
Note that exact deconvolution of a minimum-phase FIR system can only be achieved by an IIR system. The only systems that can be inverted (exactly) by FIR systems are all-pole IIR systems, i.e. systems with all their zeros at the origin of the complex plane (like $g(n)$ in the example above). Note however that in practice exact compensation is usually not the goal. E.g., in digital communications, almost all (linear) equalizers are implemented by transversal (FIR) filters, because they can be adapted more easily and because they are always stable.

A:

What you want is 
$$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$
where $\otimes$ denotes convolution. Taking $Z$ transform,
$$X(z) = X(z) \times H(z) \times H'(z)$$
or
$$H'(z) = \frac{1}{H(z)} \tag{1}$$
So you can deconvolve a convolution sum if you you have inverse transfer function as expressed in (1).
For causal and stable system, the ROC of $H'(z)$ must extend to infinity and ROC should contain the unit circle. 

