# Simplification of convolution

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:

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

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

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:

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

$$x_3[n]=h_3[n]x[n]$$

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.

• accepting. It is already known though. Apr 11 at 7:46