Finding the input from the impulse response and output

I have $$y,h,x$$ which are all vectors. From $$y[n]=x[n]*h[n]$$ which is basically how I got $$y[n]$$. I also know $$h[n]$$.

I put this through a Fourier transform. Let's assume that the capitalized versions of those are the fourier transforms:

$$Y[k]=X[k]H[k]$$, then $$Y[k] inv(H[k])=X[k]$$

But inverse of $$H$$ doesn't exist because it is a vector.

How else can I solve this?

• you're confusing convolution with a matrix representing convolution, and that doesn't even make much sense, because your time-domain convolution becomes point-wise multiplication in frequency domain. Tidy up your notation! That will solve your confusion; I recommend using $\cdot$ as multiplication operator instead of $*$, which is often used as convolution operator. (Also, matrices can have inverses. So what you claim isn't even true!) – Marcus Müller Mar 19 '19 at 21:34
• Point taken about the notation. Do you mean that according to my notation it would seem like it is pointwise multiplicaiton, or is it the case that I am wrong to assume that it is a matrix multiplicaiton, and it is actually pointwise multiplication? Also, since H is a 1D vector that can't be a square unless it is 1x1, can I really have it's inverse? – canibour Mar 19 '19 at 21:53
• a 1×1 matrix has an inverse, unless it's $0$. Anyway, $H$ is not a matrix, I don't even know how you come to that conclusion! And: your notation simply won't work; even if $H$ is a vector (which it actually is), your notation would imply a inner product or a convolution or something; you probably want to write something like $Y_kH^{-1}_k=X_k, k=1,2,\ldots$ to make it clear you mean point-wise multiplication. Avoid the $*$; and a product of vectors is not a point-wise product, usually. – Marcus Müller Mar 19 '19 at 22:00
• Hi ! First of all, is this a homework, please indicate and please show your effort? Then as Marcus has outlined, you are confusing the basic notation and representation terminology. So please differentiate between 1-) discrete-time sequences $x[n]$ , $h[n]$, $y[n]$ and their relationship through LTI convolution sum as $$y[n] = x[n] \star h[n]$$ and 2-) a set of finite length vectors (matrices) $x,h,y$ that hold samples of the sequences, and relations between matrices (vectors) that provide an alternate linear algebra representation of the convolution sum as $$y = H \cdot x$$. – Fat32 Mar 19 '19 at 22:10
• and 3-) a third relationship by using the Fourler transform convolution theorem where $X,H,Y$ indicate the DTFT and DFT of the corresponding sequences and the DTFT / DFT of the output $y[n]$ is given by $$Y(\omega) = H(\omega) X(\omega)$$ and $$Y[k] = H[k] X[k]$$ where $Y[k],H[k]$ and $X[k]$ are complex valued DFT sequences which can be represented by their vectors in a computer. – Fat32 Mar 19 '19 at 22:13