Basic signal analysis (system invertible)

FrogPad
Messages
801
Reaction score
0
Ok I have this really simple question that is bugging me.

Lets say you have the signal:

y(t) = x(t-4)

where y(t) corresponds to the output, and x(t) the input.

I know this system is invertible, but I don't really know how to show that this is the case. I see that the output is x(t) with an independet variable transformation such that the input shifted by 4 units to the right. So if we shift the output four units to the left then we get the input without the independent variable transformation. I just don't know how to express what is going on here mathematically.


Maybe I don't understand invertibility well enough to apply it.
From what I gather it can be shown by,

x(t) --> [system] --> y(t) = T{x(t)}
y(t) --> [invert] --> T{y(t)} = x(t)

I'm getting confused since the problem has x(t-4) in this case. I'm guessing I can show it with some type of function composition, but I need some help.

thanks
 
Physics news on Phys.org
You can make the substitution
t' = t - 4
so that
x(t') = y(t'+4)
The inverted function is non-causal, since you need to know future values of y to find the present value of x.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top