# Feedback System Help

1. Oct 28, 2008

### NW8800

Hey,

I have the following standard feedback system:

http://img112.imageshack.us/img112/7152/feedbacksystemuq5.jpg [Broken]

This shows the Z-Transform where:

A(z) is the z transform for input a[n] and C(z) is the z-transform of the system's output c[n]

http://img504.imageshack.us/img504/9418/feedbacksystemscanpm3.jpg [Broken]
For this question I assume I would need to get another difference equation with c[n] and b[n] terms then find c[-1] from the above question..?

Any help would be greatly appreciated, just not sure how to get started on this one.

Cheers,

NW

Last edited by a moderator: May 3, 2017
2. Oct 30, 2008

### jhicks

You can find the responses by doing nodal analysis. Take the left node, for instance. The sum of the quantities pointing into the node must equal that going out, so

$$E(z)=A(z)+H(z)C(z)$$, but $$C(z)=E(z)G(z)$$ and so $$E(z)=A(z)+H(z)E(z)G(z)$$, or $$\frac{E(z)}{A(z)}=\frac{1}{1-H(z)G(z)}$$. From here, it's obvious what C(z)/A(z) is in terms of H(z) and G(z). This quantity is the so-called impulse response. Can you do the rest?

3. Oct 30, 2008

### CEL

Although there is no sign indication in the input comparator, the sign for the feedback loop is probably negative, so we should have:
$$E(z)=A(z)-H(z)E(z)G(z)$$
and
$$\frac{E(z)}{A(z)}=\frac{1}{1+H(z)G(z)}$$

4. Oct 30, 2008

### jhicks

Yes I forgot to edit my post to say just that, you are correct.

5. Oct 31, 2008

### NW8800

Hey,

Thanks for the replies... I can’t believe I made that mistake... CEL is right is a negative feedback system, sorry about that, it was late when I was drawing that pic :P

Thanks for showing the steps of how you got that jhicks :)

For the comparator (circle) A(z) is + and the Output from the H(z) block is a -... I.e. A(z) - [Output from the H(z) block], I guess you could call it [D(z)] which equals, $$D(z) = H(z)E(z)G(z)$$

Which gives me $$\frac{C(z)}{A(z)}=\frac{G(z)}{1+H(z)G(z)}$$

Subbing in for G(z) and H(z) to get the total system transfer function I.e. solve for $$C(z)$$ I end up with:

$$\frac{C(z)}{A(z)}=\frac{z(z-8)}{(z-2)(z+4)}$$

What would the unit step response be..? I think I'm starting to get the hang of it... Would I just times the above system transfer function by the standard unit step z-transform or what..?

Also, how do I find c[n] with initial conditions that's my 3rd question... (c[-1]=-3 c[-2]=4)?

Thanks once gain,

NW

Last edited: Oct 31, 2008
6. Oct 31, 2008

### CEL

The transfer function is
$$\frac{C(z)}{A(z)}=\frac{z(z+8)}{(z+2)(z+4)}$$
For the step response you multiply the transfer function by the z-transform of the step and take the inverse z-transform of the result.
For your last question, the inverse z-transform of the transfer function gives you the relationship of c[n] and a[n]. Substitute the initial conditions in the equation and then substitute a[n] by its values for n = 0, 1, 2, ...

7. Oct 31, 2008

### NW8800

Hey CEL,

Can I ask, how you got $$\frac{C(z)}{A(z)}=\frac{z(z+8)}{(z+2)(z+4)}$$

$$\frac{z}{(z+1)} * \frac{(z+1)(z-8)}{(z+1)(z-8)+9z}$$

Simplify to
$$\frac{C(z)}{A(z)}=\frac{z(z-8)}{(z-2)(z+4)}$$

Where H(z) = 9/(z-8)..?

Also would the unit step response for the system be..?

$$\frac{C(z)}{A(z)}=\frac{z^2(z+8)}{(z+2)(z+4)(z-1)}$$

Or

$$\frac{C(z)}{A(z)}=\frac{z^2(z-8)}{(z-2)(z+4)(z-1)}$$???

Cheers,

NW

8. Oct 31, 2008

### CEL

9. Oct 31, 2008

### NW8800

Last edited: Oct 31, 2008
10. Oct 31, 2008

### NW8800

Could someone please show me how to get the difference equation of the transfer function of the system?

11. Nov 1, 2008

### CEL

Multiply the numerator and the denominator by $$z^{-3}$$.
Then, multiply C(z) by the denominator and A(z) by the numerator.
Remember that the inverse transform of $$z^{-1}F(z)$$ is $$f[n-1]$$.

12. Nov 1, 2008

### NW8800

What the inverse z-transform for the transfer function of this system and how do I sub in the initial conditions in this equation and then a[n]?

13. Nov 1, 2008

### CEL

When you make the calculations I suggested you will have c[n] as a function of c[n-1], c[n-2], a[n] and a[n-1].
You have c[-2] and c[-1]. a[-1] = 0, since a[n] = 0.5nu[n].
Now you can calculate c[0].
The other values of c[n] follow.

14. Nov 1, 2008

### NW8800

Thanks for that CEL :) Seem to be getting somewhere :d

For my last question the the 1st post... With B(z) - the output from block H(z)...

would B(z) = C(z) * H(z)

Which equals $$B(z) = H(z)G(z)[R(z)-H(z)C(z)]$$...>?

15. Nov 1, 2008

### NW8800

How would I find the initial conditions terms, like would I take the Z-transform of the difference equation when i sub in the initial condition terms to the difference eqn?

16. Nov 2, 2008

### CEL

Yes, that's it.

17. Nov 2, 2008

### CEL

You use the time domain equation.
You have
c[n] = f{c[n-1], c[n-2], a[n], a[n-1]}
making n = 0, you get
c[0] = f{c[-1], c[-2], a[0], a[-1]}
once you have c[0], you can calculate
c[1] = f{c[0], c[-1], a[1], a[-1]}

and so on