Is it true or fale:

Given LTI, $H(z)$, with impulse response $h[n]$ that is not zero for all $n$. The following system is not LTI:

$$x[n]\longrightarrow\boxed{\downarrow2}\longrightarrow H(z)\longrightarrow\boxed{\uparrow 2}\longrightarrow y[n]$$


Why? If I take delta to be the input, and then take another input: delta shifted by one...how I see that it will support the statement?


1 Answer 1


Lets say that $x_1[n]=\delta[n]$ and $x_2[n] = \delta[n-1]$, where $\delta[n]$ is Kronecker delta and that $y_1[n],y_2[n]$ are the output signals for $x_1[n],x_2[n]$.
The downsampling operator discards every other sample. Without the loss of generality, lets assume that we are keeping the even samples and throwing the odd ones. Therefore, after downsampling $x_2[n]$, it is all zeros, which means that $y_2[n] = 0$ for each $n$. Now since you say that $h[n]$ is not all zeros, it is guarantied that $y_1[n]$ will have some non-zeros values.
In other words, $y_1[n]\ne y_2[n]$ for some $n$, and you system is not TI.


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