A question about time-shifted convolution

Question

I have 2 time-shifted signals,$f(t-t_0)$ and $g(t-t_0)$. Assuming that $y=f(t)*g(t)$, which is the convolution of the signals.

By definition, I have $f(t-t_0)*g(t-t_0)=\int_{-\infty}^{+\infty}{f(\tau-t_0)g(t-t_0-\tau)d\tau}$

Substitute $\tau-t_0$ with $t_1$ then I have

$f(t-t_0)*g(t-t_0)=\int_{-\infty}^{+\infty}{f(t-2t_0-t_1)dt_1}$ which gives $f(t-t_0)*g(t-t_0)=y(t-2t_0)$

But when I considered substituting the independent variable by $T=t-t_0$, I have

$f(t-t_0)*g(t-t_0)=f(T)*g(T)=y(T)=y(t-t_0)$, which is different from the result above.

Can anyone point out where my mistake is? Thanks in advance.

Answer 1

The mistake lies in the use of bad notation for convolution:

$$y(t)=f(t)*g(t)\qquad \text{(bad!)}\tag{1}$$

The correct way to write convolution is

$$y(t)=(f*g)(t)\tag{2}$$

As long as you don't play around with the variable $t$ everything is fine, so sometimes the sloppy use of notation isn't problematic, but as soon as you shift signals or modify $t$ in some other way, the notation in $(1)$ will get you into trouble, as you've seen yourself.

The correct way to write down what you did would be to define new functions for the shifted versions of the original functions:

$$\tilde{f}(t)=f(t-t_0),\qquad \tilde{g}(t)=g(t-t_0)$$

and write the convolution as $(\tilde{f}*\tilde{g})(t)$.