2
$\begingroup$

I got this answer from here. But i don't know how this method is correct. Until now to prove shift invariance what i did was :
step 1: delay output and write it in terms of input.
step 2: delay input and write the modified output
step 3: if output from step 1 and 2 are same then system is shift invariant.

what this guy has done is "different". the oppenheim book and proakis manokalis book don't follow this method. which book follows this method?

the method.

The test for this is $$x_{1}[n]=x[n-n_{0}] $$ $$y_{1}[n]=y[n-n_{0}] $$

So let's first express in terms of just $y[n]$. $$ y[n]=\frac{x[n-3]-y[n-3]-y[n-1]}{x[n]} $$ Next, we go through the test. $$y[n-n_{0}]=\frac{x[n-3-n_{0}]-y[n-3-n_{0}]-y[n-1-n_{0}]}{x[n-n_{0}]} $$ Substituting $y_{1}[n] $ $$y_{1}[n]=\frac{x[n-3-n_{0}]-y[n-3-n_{0}]-y[n-1-n_{0}]}{x[n-n_{0}]} $$ $$y_{1}[n]=\frac{x[n-3-n_{0}]-y_{1}[n-3]-y_{1}[n-1]}{x[n-n_{0}]} $$ And finally substituting $x_{1}[n] $ $$y_{1}[n]=\frac{x_{1}[n-3]-y_{1}[n-3]-y_{1}[n-1]}{x_{1}[n]} $$ Because the shifted sequence has the exact same relationship, it is said to be time-invariant. Most systems that don't alter $n$ or $t$ meet this. Be wary of anything multiplying or otherwise messing with the vector arguments beyond simple delays.

$\endgroup$
2
$\begingroup$

The so called example system is nonlinear and recursive and those books generally exclude the treatment of nonlinear & recursive systems. Hence you may not find example solutions.

The method relies on the fact that the relationship between the output $y[n]$ and input $x[n]$ remains the same between $y[n-d]$ and $x[n-d]$.

As an absurd example, think of the following. Consider the equation of a circle at origin: $x^2 + y^2 = r^2$. Now, shift this circle to the point $(a,b)$ and re-write its equation: $(x-a)^2 + (y-b)^2 = r^2$. Finally, redefine the variables $x' = x-a$ and $y' = y-b$ and rewrite the equation: $(x')^2 + (y')^2 = r^2$.

It's seen that the circle is shift invariant (not in the sense of the signal processing context though)

$\endgroup$
  • $\begingroup$ do you know a book that follows this method? $\endgroup$ – abhishek Jan 26 at 12:23
  • $\begingroup$ you should look at every question and answer... no I don't remember. $\endgroup$ – Fat32 Jan 26 at 12:59
1
$\begingroup$

"Invariant" here can be misleading, see for instance: What is the difference between “equivariant to translation” and “invariant to translation”.

For the system perspective (which could be called shift-equivariant, see above), a shift-invariant (sometimes called translation-invariant or time-invariant) system is a system (with action denoted by $S$) does not "vary" with time delays $D_\tau$ in the input, in some sense. More formally, if you defines input $x(t)$, with $D_\tau(x(t))=x(t-\tau)$ , you expect that the system and any delay commute, for any input:

$$ S(D_\tau(x(t))) = D_\tau(S(x(t))\,.$$

This might sound a bit abstract. Simple examples are given here. An informal version would be: does my system explicitly depends on something else than a time-shift? You can think there exists some absolute time, and that you can only change either 1) the inner time of the input signal or 2) the inner time of the output signal.

First, suppose that I look at $nx[n]$: a $x[n-n_0]$ will be inner, but the multiplications by $n$ is "outer". Second, suppose that I change my inner time in a non-shift way: $x[n^2-3]$. In both cases, you have a strong hint that something "non-shift invariant" is happening. In such cases, it is clever to try examples, and hopefully find a counter-example showing that the system is not shift-invariant.

As @Fat32 already said: I am not aware of a generic method to detect time-invariant systems. However, you can find inspiration in differentiating with respect to the time variable. If the system is $x(t)\to tx(t)$, the derivative with respect to $t$ is $x(t)+tx'(t)$. If the system is $x(t)\to x(t^2-3)$, the derivative with respect to $t$ is $2tx'(t^2-3)$, both showing a lack of shift-invariance, as $t$ becomes explicit in an "outer" way.

Honestly, I am not sure (as for today) this can transfer totally on discrete systems. Yet, replacing $[n]$ by $(t)$ followed by differentiation in your ratio won't show explicit outer dependence on $t$.

In the given example, in a formal way, the system would be shift-invariant for non-vanishing signals, but this is a bit unfair, as the output is recursive in the past only.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.