Interpreting a signal block diagram to form transfer funciton

[Evo8]

Homework Statement

Given the following block diagram. Calculate the transfer function H(z) and find $\theta (f=\frac{f_{s}}{6})$

Homework Equations

$H(z)=\frac{Y(z)}{X(z)}$

The Attempt at a Solution

Ok So I know i need to find my y(k) output equation from this. I can then rearange the equation so I have x terms and y terms on opposite sides and then take the Z transform of each side. Then rearange so that I end up with

$\frac{Y(z)}{X(z)}$ Which should be my H(z). Then I can find $\theta(f)$ and evaluate at f= fs/6.

I think that's the correct process anyway.


However I am not 100% on how to interpret the block diagram in this case. I have what looks like a feed back from x(k) i think i can handle that but not sure about the y(k) terms if any.

This is what I have come up with so far.

$3x(k-1)-\frac{1}{3}x(k-1)+y(k)$

However I am not sure what I have here is 100% correct. Any hints would be much appreciated.

Thanks,

 

[I like Serena]

Hi Evo8!

I prefer to work in the z-domain for things like this.
I do not know how you are taught these things, so perhaps the way I do it may be different from the way you are taught.
The result however will be the same.


So if you'll bear with me, then to the left, you have X(z), and to the right you have Y(z).

Your problem is that you have a feedback loop that you need to solve.
Let's give it a name.
Let's say that in the middle you have V(z).

This means that the left summing junction does:
$$X(z) + 3z^{-1}V(z)=V(z)$$

Can you solve V(z) from this?

 

[Evo8]

If I've done my algebra correctly I get

V(z)=3zX(z)

So would $Y(z)=3zX(z)-\frac{1}{3}z^{-1}$ maybe I am forgetting an X(z) in there?

 

[I like Serena]

Your algebra appears to be a little off.
How did you arrive at V(z)=3zX(z)?

Could you say what the right summing junction does in terms of V(z) and Y(z)?

 

[Evo8]

Your algebra appears to be a little off.
How did you arrive at V(z)=3zX(z)?

Could you say what the right summing junction does in terms of V(z) and Y(z)?

I subtracted the x(z) term from either side moving it to the right. then divided both sides by v(z). so I would have

$3z^{-1} V(z)=\frac{V(z)}{V(z)}-\frac{X(z)}{V(z)}$

then errr wait. As I am typing this I realize my mistake. When I canceled out the v(z)/v(z) i didnt leave the 1...Man my algebra skills are killing me here.

In this case i guess I am not really sure how to go about it. If I used the solver on my graphing calculator i get

$V(z)=\frac{X(z)z}{z-3}$ I am not really sure how to come to this conclusion though. Or if it is correct.

Ill have to play around with the algebra again and see if i can get it.

 

[Evo8]

I think the right summing junction would be V(z)-(1/3)z^-1?

 

[I like Serena]

Suppose you had the equation 2+3v=v.
How would you solve it?

 

[I like Serena]

I think the right summing junction would be V(z)-(1/3)z^-1?

You appear to have dropped a V(z).
When going down the first step is that V(z) gets multiplied by z^-1, resulting in z^-1V(z)...

 

[Evo8]

Suppose you had the equation 2+3v=v.
How would you solve it?

v=2/(1-3)

So...

$V(z)=\frac{X(z)}{1-3z^{-1}}$

Man its so easy too. I played with it for a little while and wasnt sure. Once i saw you question above I was like sure i guess I could factor the v out. This is assuming that what I have written now is correct...*fingers crossed* I really should know my algebra better then this. I certainly used to anyway.

 

[I like Serena]

Well you can uncross your fingers again, or push Ronald McDonald down the stairs, or both.
You've got it right now! :)

 

[Evo8]

So would it be

$V(z)- \frac{1}{3}z^{-1}V(z)$?

 

[I like Serena]

Yep. That is Y(z).

Can you find H(z)=Y(z)/X(z) now?

 

[Evo8]

Ok not so bad. I think

This is what I get for

$\frac{Y(z)}{X(z)}=H(z)=1-3z-\frac{1}{3}z^{-1}(1-3z)$



Simplifies to $H(z)=\frac{(-3z-1)^{2}}{3z}$

I think anyway. The simplification is where I make the simple algebra mistakes. Which you already know. I found this last simplificaiton with my graphing calculator. I tried a few times to simplify, by distributing that -1/3z^-1 into the parenthesis and then clean things up a little. It never came out as clean as this though...:(

 

[I like Serena]

Hmm, I get a different result... so I suspect your algebra needs some more polishing... (or mine does :shy:).

You had $Y(z)=V(z)- \frac{1}{3}z^{-1}V(z)$.
Can you substitute your result for V(z) into this?

 

[Evo8]

Looks like I've jumped ahead too fast again.

$Y(z)=(\frac{X(z)}{1-3z^{-1}})-\frac{1}{3}z^{-1}(\frac{X(z)}{1-3z^{-1}})$

Looking back agian I think i see where i went wrong.

I wanted to get rid of the division in the V(z) term so i thought (I can see that it is wrong now) that I could do this.

$V(z)=\frac{X(z)}{1-3z^{-1}} = X(z)(1+3z).$

Now I am playing around with it to see if there is another way. Or if I can take the above Y(z) and just simplify it all together.

 

[I like Serena]

You have 2 fractions with the same denominator.
Perhaps you could add them?

 

[Evo8]

Indeed I do.

So adding the fractions I would get.

$Y(z)=\frac{X(z)-\frac{1}{3}z^{-1}X(z)}{1-3z^{-1}}$

If I factor out the X(z) I get

$Y(z)=\frac{X(z)(1-\frac{1}{3}z^{-1})}{1-3z^{-1}}$

This is where i get nervous because I am unsure if I am breaking any rules or not

I need to break that fraction up so I can isolate the Y(z) and X(z) on one side. My first thought is to rearange like this.

$Y(z)=X(z)(1-\frac{1}{3}z^{-1})(1-3z)$ Is this ok? If so I can obviously bring my X(z) over to the other side and get my H(z)

 

[I like Serena]

You added the fractions correctly and you also factored out X(z) correctly.
But rearranging like that is not ok.

Suppose you had the equation:

y=3x/4

How would you find y/x?

 

[Evo8]

You know its weird. I ask the same questions but with simple integers and variables and I mess with the algebra again and I usually find a method pretty quickly. I thought I tried that this time. I knew there was a way I just wasnt sure how to get there i guess. I will need to remember in the future to actually write down and replace sections wtih variables and try to simplify or rearrange that way to find a method.

Now i get something like this

$\frac{Y(z)}{X(z)}=\frac{1-\frac{1}{3}z^{-1}}{1-3z^{-1}}$

or I guess I could write it like this.

$\frac{Y(z)}{X(z)}=\frac{1-\frac{z^{-1}}{3}}{1-3z^{-1}}$

Right...?

 

[I like Serena]

Right! Btw, you're not the only one making mistakes with complex expressions.
I know it helps to simplify them, and then suddenly the trees resolve into a small forest.

 

[Evo8]

Awesome! So at this point I think I should be able to evaluate along the unit circle by substituting in e^2pifT and then breaking out the real and imaginary parts to find my Phase response. Then evaluate that at the value given?

Thanks for the help "I like Serena!" As always!

 

[I like Serena]

I think so, although TBH I don't know what the question for $\theta(f={f_s \over 6})$ means.
I suspect $H(e^{j 2\pi fT})$ is intended, but you should know that better than I do.
Note the extra "j" btw...

Aaaaand... you're welcome! :)

 

[Evo8]

What do you mean the extra j?

 

[I like Serena]

z is supposed to be a complex number.

e^2pifT is not complex but real.
e^j2pifT is the correct representation of a complex number on the unit circle.

 

[Evo8]

Ahhh yes I see. I forgot the j in the my exponential. I caused my own confusion. Thanks for mentioning it!