# Z - Transform of a non recursive block diagramm

i am currently struggling to find the Transfer Function of the following Block Diagram, since i have never done it for a non recursive System, could somebody please walk me through the process of it? My first step was setting up the Difference Equation:

$$g(n) = a*s(n)+s(n)*(a+1)*z^{-1}+a*s(n)*z^{-2}$$

Now I went on to divide $$s(n)$$ so I have the Transfer Function in the z Domain

$$H(z) = a+(a+1)*z^{-1}+a*z^{-2}$$

this is basically as far as I got, since I am unsure on how to continue this problem without the poles you would get from a recursive System

• Hi Ricardo. This looks like homework. Can you edit your question with your attempts at solving this? Those blocks should read z^{-1}, not "z-^1"
– Jdip
Jan 24 at 14:08
• Your time domain notation seems odd (of not wrong ) I would write the difference equation as $g[n] = a \cdot s[n] + (a+1)\cdot s[n-1] + a \cdot s[n-2]$ . Your transfer function looks fine though, what's your concern ? Jan 24 at 15:21
• Sorry, I wrote this before I saw your answer Jan 24 at 15:24

1. The standard way to represent the convolution operator is to use the "$$*$$" sign. In general it's preferable not to use it to represent multiplication like you did.
2. Your difference equation is wrong. The thing is, you don't even need it to get the correct transfer function (straight from the block diagram which is already in the transfer function domain, i.e. the $$\mathcal{Z}$$-domain), which you actually got right: $$H(z) = \frac{g(z)}{s(z)} = a + (a+1)z^{-1} + az^{-2}$$ If you really want to get the difference equation (in the difference equation domain, i.e. the time domain), use the inverse $$\mathcal{Z}$$-transform: $$g[n] = as[n] + (a+1)s[n-1] + as[n-2]$$
1. If you want to go further, this is a Finite Impulse Response (FIR), symmetric, Type I, Low-Pass filter of odd length $$N=3$$, and here is its frequency response for different values of $$a$$: $$H(z) = \frac{g(z)}{s(z)} = a + (a+1)z^{-1} + az^{-2} = \frac{az^{2} + (a+1)z^{1} + a }{z^2}$$
Now we can see that the transfer function has two real zeros and also two poles at $$z = 0$$. That's the case for all FIR filters: an FIR filter of order N has N poles at $$z=0$$