# Combined impulse responses

I'm looking for a simple way to show that you can form a single impulse response that is the equivalent $M$ other impulse responses, ie:

$$h\left(t\right)=h_0\left(t\right)\star h_1\left(t\right)...\star h_{M-1}\left(t\right)$$

I know convolution is multiplication in the frequency domain, so I cooked up this "equation":

$$h\left(t\right)=\mathcal F^{-1}\left\{\prod_{i=0}^{M-1} \mathcal F\left\{h_{i}\right\}\right\}(t)$$

With $\mathcal F^{-1}$ being the inverse Fourier transform operator, and $\mathcal F$ being the Fourier transform operator. Expressing myself in math notation isn't really my strong side, what I intended to show was that the combined impulse response $h(t)$ equals the sequences multiplied together in the frequency domain.

• Does it make any sense?
• Or is there perhaps a simple way to show this?
• What about circular convolution in the frequency domain, and infinite impulse responses in continuous time?
• Your equations are correct and you can easily show that using the properties of the Fourier transform as pointed out by Maximilan. The same applies in the discrete domain. – MBaz Nov 30 '16 at 23:59

## 1 Answer

You can readily use the associativity of the convolution to explain this fact. Assume you send a signal $x(t)$ through a chain of convolutions $h_i(t)$:

$$y(t) = (((x(t)*h_1(t))*h_2(t))*h_3(t))$$

By associativity, this becomes

$$y(t) = x(t) * ((h_1(t)*h_2(t))*h_3(t))$$

I.e. you can see this operation as if sending your signal $x(t)$ through a single filter with impulse response $h(t)=h_1(t)*h_2(t)*h_3(t)$.