# Simulation diagram from a Transfer function/difference equation

#### josh007

Joined Sep 20, 2015
43
Greetings,

I require assistance.

Calculate the difference equation and then Draw the simulation diagram of the below transfer function.

H(z) = Y(z)/X(z) = (0,4142 + 0.4142(z^-1)) /(1,4142 - 05858(z^-1))

I performed the normal procedure to find the difference equation, by cross multiplying and using the delay property of the z-transforms, I finally ended up with:

y(n) = 0,2929x(n) + 0.2929x(n-1) + 0.4137y(n-1)

How do I Draw the simulation diagram?

I don't require the answer but guidance on how to derive the diagram and final answer.

Thanks.

Regards
Joshy

#### MrAl

Joined Jun 17, 2014
10,599
Hi,

You can start by making n=0 and then increment it, 1,2,3, etc., up to the required last number where you see at least one cycle or whatever it shows.

You might set y(0)=0 unless you have initial conditions and x(-n) equal to zero in most cases unless you have values for x(-n) and then you would have to run n lower than zero also. You also probably set y(-1)=0 too and all other y with n negative unless those have initial conditions also.

The simulation could be a plot of all y(n) and x(n) just as you would with y=2*t where you plot y vs t, or you could plot y(n) vs n and perhaps y(n) and x(n) vs n (two plots with n running horizontal).

#### MrAl

Joined Jun 17, 2014
10,599
Hi,

That's not bad, but how are you getting 0.4137 when you divide 0.5858 by 1.4142 ?
That comes out just a little different than what you show there right?

#### josh007

Joined Sep 20, 2015
43
Hi MrAI,

you right sir thats wrong it should be y(n) = 0.2929x(n) + 0.2929x(n-1) + 0.4142y(n-1),

I am going to take a pic of my block diagram now.

Gracias.
Joshy

#### josh007

Joined Sep 20, 2015
43

#### MrAl

Joined Jun 17, 2014
10,599
Hi again,

That looks reasonable 