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I am a student learning dsp. I like the subject. I could understand the discrete time signals. When I move into z transform. I could not understand it.

Z transform is the mapping from discrete signal to zplane to make frequency analysis easy. It would be great if users give some tips that could help me to clarify various views on z-tranform on discrete signals.

Text is here to explain about broad answers on z-tranform. I am just looking forward in some tips that can help me understand and approach z-transform in text. Just some lines about what you think about z-tranform will surely help. Thank you mate.

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  • $\begingroup$ I would really recommend that you read the chapter on the Z-transform in Introduction to Signal Processing by Orfanidis and come back with a more concrete question. Actually, if you're interested in DSP you should probably read all chapters of that book. $\endgroup$ – Matt L. Oct 31 '15 at 9:54
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Discrete-time signals are generally written in the time domain as $x[n]$ where $n$ is an integer.

We want to be able to see how discrete-time systems can be used to modify these signals: $$ y[n] = \sum_{m=1}^M a_k y[n-m] + \sum_{p=0}^P b_p x[n-p] $$

The above equation is a linear, constant coefficient difference equation.

Such equations can be hard to manipulate for a specific $x$ to figure out what $y$ is.

The $z$ operator is used to attempt to simplify the analysis of such systems. $z$ is the forward-shift operator: it shifts a signal forward by one time instant. $z^{-1}$ is the backward-shift operator: it shifts a signal backward in time by one time instant.

Using this operator, we can rewrite the equation as: $$ Y(z) = Y(z) \sum_{m=1}^M a_k z^{-m} + X(z) \sum_{p=0}^P b_p z^{-p} $$ which, upon rearrangement, we can get:

$$ Y(z)/X(z) = \frac{\displaystyle\sum_{p=0}^P b_p z^{-p}}{\displaystyle1 - \sum_{m=1}^M a_k z^{-m}} $$

which is now a discrete-time system that we can analyze in the $z$ domain, rather than the time domain.

One thing this buys us is the ability to figure out what happens when many discrete-time systems are put together: we just multiply the discrete-time system descriptions (transfer functions). Doing the same thing in the time domain is doable, but very tedious.

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  • $\begingroup$ Thank you so much Peter. Your answer clarified my various views on z-transform. $\endgroup$ – Kishore Vignesh Nov 1 '15 at 11:27
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Z is a complex exponential, a complex exponential is a compact form of representing a sinusoid.

What essentially happens in z transform is we take a discrete signal and try to project it on various sinusoids - various frequency.

We multiply our samples with exp(z,-n). Sum up the values., essentially that's a measure of finding if our sampled data has a frequency that corresponds to that sinusoid.

I will stop here and let your thought process flow on z as a delay operator., transfer function - the compact notation of sampled data's periodic characteristics.

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To add my two cents to Peter K's good answer: Continuous (analog) signal processing systems are described by their time-domain "differential equations." For all but the simplest systems those equations are very difficult to solve in order to predict a system's behavior as a function of input signal frequency. Then unsung hero Oliver Heaviside proposed the "Laplace transform" that converted those differential equations into simpler algebra equations that were much easier to solve.

Likewise, discrete systems are described by their time-domain "difference equations" such as Peter K's $y[n]$ expression. But as he said, such time-domain expressions are difficult to solve in order to predict a discrete system's behavior as a function of input signal frequency. Lucky for us, in the early 1950s Lofti Zadeh and John Ragazzini (in their efforts to analyze discrete-time control systems) developed the discrete-time equivalent to the Laplace transform, called the "z-transform." The z-transform converted those discrete-time difference equations into algebra equations that were much easier to solve in order to predict a discrete-time system's behavior as a function of input signal frequency.

Upon learning of this z-transform, DSP guru James Kaiser (Bell Labs) realized it could be used simplify the analysis of discrete-time signal processing systems, such as digital filters. Kaiser then went on a campaign to convince DSP practitioners to use the z-transform. He knew that the z-transform would produce $Y(z)/X(z)$ algebra expressions enabling regular guys like us to, by way of algebra manipulation, determine both the frequency-domain behavior and stability of discrete-time signal processing systems.

Is it important for you to learn the z-transform? The answer is, the z-transform will not allow you "to see things people wouldn't believe. Attack ships on fire off the shoulder of Orion. Or watch C-beams glitter in the dark near the Tannhauser gate. All those moments are lost in time... like tears in rain." But the z-transform will enable you to completely analyze digital filters.

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  • $\begingroup$ These are points I wanted to know. Great work Richard Lyons. Thank you so much. $\endgroup$ – Kishore Vignesh Nov 2 '15 at 16:45

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