Let's assume that there are two convolutions:

$y_1 = (h_2[n]\cdot x[n])*(\bar{h_1}[n]\cdot\bar{x}[n])$

$y_2 = (h_3[n]\cdot x[n])*(\bar{h_2}[n]\cdot\bar{x}[n])$

where "$\bar{x}$" is complex conjugate, $\cdot$ is element-by-element multiplication, and $*$ is linear convolution, that is $(f*g)[n]=\sum_mf[m]g[n-m]$.

I am looking to simplify $y = y_1+y_2$.

It would not be difficult to show that $$y\neq ((h_2[n]+h_3[n])\cdot x[n])*((\bar{h_1}[n]+\bar{h_2}[n])\cdot\bar{x}[n])$$ But is there any other way to algebraically simplify $y$?

This can also be seen as the sum of two products $Y_1$ and $Y_2$, (the FFTs), where in each FFT there are two convolutions:

$Y_1 = (H_2*X)\cdot(\tilde{H_1}*\tilde{X})$

$Y_2 = (H_3*X)\cdot(\tilde{H_2}*\tilde{X})$


1 Answer 1


Here is the simplest approach I could think of:

$$x_2[n]= h_2[n]x[n]$$

$$\bar{x_2}[n] = \bar{h_2}[n]\bar{x}[n]$$

Thus we eliminate one product as we can get $\bar{x_2}[n]$ from $x_2[n]$. This leaves two more products before going to the frequency domain:

$$\bar{x}_1[n]= \bar{h}_1[n]\bar{x}[n]$$


With the three products use the FFT to compute the convolution in the frequency domain (zero pad to twice the length if a linear convolution result is desired):

$$y_1+y_2 = \text{ifft}(X_2[k]\bar{X_1}[N-k]+X_3[k]\bar{X_2}[N-k])$$

So in summary I simplified the operations by using the same results whereever possible if already computed, and doing all convolution operations as products in the frequency domain. Given $x_1[n]$, $x_2[n]$ and $x_3[n]$ are completely independent given no relationship between complex $h_1[n]$, $h_2[n]$ and $h_3[n]$, I do not see the opportunity for any further algebraic simplifications.

  • $\begingroup$ accepting. It is already known though. $\endgroup$
    – user67236
    Apr 11 at 7:46

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