system diagram

Hi, I am really confused with the system in the diagram.

The input-output relation of the system is given by $y[n]=\exp(x^2[n])$

I need to find the transfer function of this system $Y(z)$ in $z$-domain. My question is: is it really the simple exponential system I am missing out or something complex. If otherwise what would be the system transfer function.


1 Answer 1


This system doesn't have a transfer function, so there's no way to find it. Only linear time-invariant (LTI) systems can be described by a transfer function. In discrete-time, the $\mathcal{Z}$-transforms of input and output sequences of an LTI system are related by


where $H(z)$ is the system's transfer function, which is the $\mathcal{Z}$-transform of its impulse response. Equation $(1)$ is another way to state that the output sequence is given by the convolution of the input sequence with the system's impulse response. If the system is not LTI there is no function $H(z)$ which relates $X(z)$ and $Y(z)$ as in Eq. $(1)$.

Your system is definitely non-linear so whoever asked you to find its transfer function is either clueless or wanted to fool you.

  • $\begingroup$ Damn, exactly, I completely ignored the fact the system is not even a LSI system. Great, thanks a lot. Actually the question was a bit bigger. This system was the last part of a cascade of three systems. Where first two parts were perfectly fine: causal LSI systems. The actual question was to comment about the overall system. Stable? Shift-Invariant? $\endgroup$ Commented Mar 5, 2016 at 8:48
  • $\begingroup$ @OptimusPrime: If the first two systems are stable (which doesn't necessarily follow from the fact that they are causal and LSI), stability and SI only depends on that last system. Shift-invariance can be tested very easily, and if you define stability as BIBO-stability, you can also directly draw your conclusions. $\endgroup$
    – Matt L.
    Commented Mar 5, 2016 at 9:11

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