# Plotting responses to z-plane transfer functions

1. May 20, 2004

### Jimbo

Hi

I am looking for any tutorials in how to plot a time response plot from a z-plane transfer function

For example, if you have G = 4 / z + 2

I understand that plotting this on the z-plane would result in a point at -2 (although I am not sure about how the numerator of 4 might affect this) but I am unsure about the steps to be taken to convert this to a time response plot?

Any help is much appreciated!

2. May 20, 2004

### enigma

Staff Emeritus
If you're talking about what I think you're talking about (control systems... my experience uses 's' instead of 'z'), then to plot the time domain, you need to take the inverse Laplace to get to the time domain.

In this case, you're looking at e^-2t for a time domain response to a step input.

3. May 21, 2004

### jamesrc

You'd only use the inverse Laplace tranform to find the time response if this were a continuous-time tranfer function. Since we're talking about the z-plane, I'll assume you're talking about a discrete-time system. You should specify the form of the input function, since the output function response will depend on the input convolved with the transfer function; since you didn't say, I'll assume you're looking for the unit impulse response (that's what the tranfer function is).

Anyhow, let's see if I can do this without messing up:

$$Y(z) = G(z)X(z) = \frac 4 {z+2} = 4 \frac{z^{-1}}{1+2z^{-1}} = 4z^{-1}\frac 1 {1+2z^{-1}}$$

(X(z) = 1)

which is the same as
$$Y(z) = 4\frac{1}{1+2z^{-1}}$$ time shifted one sample to the right.

so
$$y[n] = 4(-2)^{n-1}u[n-1]$$

where u is the unit step function.

The standard way to plot this would be to use a stem plot
(so it kind of looks like a bunch of lollipops).
y[0] = 0
y[1] = 4
y[2] = -8
y[3] = 16
and so on...

4. May 27, 2004

### turin

If you want the inverse Z-transform, then it involves a contour integration. You should have a table in your textbook that would include Z-transforms of the basic functions. If you want the general rule:

x(n) = (1/2&pi;i) integral of { X(z) zn-1 dz }

where x(n) is the discrete time signal, X(z) is the Z-transform of it, and the integration is around a closed contour (but I can't remember what the rules are for the contour; it probably has to contain all of the poles and the origin). In your case, X(z) is G and x(n) is the response to a Kronecker Delta input of &delta;0n.

5. Jun 1, 2004

### Jimbo

Great!

Thanks for all the help!

Just what I needed!

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