If I have two functions in a convolution like



then it means (trivially) $Y=Z$.

Is this correct or are there subtleties in the convolution theorem where $Y=Z$ isn't always true?


Here is a counter example. The functions $X,Y,Z$ below will satisfy that $X * Y = X * Z = 1$ but $ Y \not= Z$:

$$ X(t) := 1, \forall \,t \in \mathbb{R},$$

$$ Y(t) := \begin{cases} \frac{1}{C_Y} e^{-\frac{1}{1-t^2}} \quad \text{if $-1 \leq t\leq 1$}\\ 0 \quad \text{otherwise,} \end{cases} $$ where $C_Y$ is a (normalization) constant such that the integral of $Y$ over $[-1,1]$ equals 1, and $$ Z(t) := \begin{cases} \frac{1}{C_Z} e^{-\frac{1}{2^2-t^2}} \quad \text{if $-2 \leq t\leq 2$}\\ 0 \quad \text{otherwise,} \end{cases} $$ where, similar to $C_Y$, $C_Z$ is a (normalization) constant such that the integral of $Z$ over $[-2,2]$ equals 1.

Those are concrete functions as an explicit example. Any normalized (integral equals 1) function $f$ that has compact support will satisfy that $X * f = 1$. From there you can choose any $Y$ and $Z$ you want as long as they are different.


Good Question :

Here we can use the concept of Laplace Transformation

$$X(s) Y(s) = X(s) Z(s)$$

$$Y(s) = Z(s)$$

We know that "Laplace Transform of two signals can be same but their ROC will be definitely different"

Here is a example :

$$e^{-at} u(t) \rightleftharpoons \frac{1}{s+a}$$

$$-e^{-at} u(-t) \rightleftharpoons \frac{1}{s+a}$$

HENCE now it is proved that it is not necessary that

$$Y = Z$$


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