How Is a Simplified DSP Transfer Function Derived?

[Jag1972]

I have taken this example from a Digital Signal Processing book. I have got the simplified transfer function; however do not understand how it was done. I have had a go could someone please look at it and let me know where I am going wrong.

I have attached my working out on a Word document.
Thanks in advance.

 

[Zryn]

You're not doing too badly, everything you need to solve the problem is in that word document, but let's try a different approach, because that denominator is crazy long (but might be right, I didn't check the math)

Instead of converting re^(j*pi/2) and re^(-j*pi/2) by Euler's formula, expand the function first using the FOIL method (First pair, Outer pair, Inner pair, Last pair).

Looking at the 2nd and 3rd terms, they have a common factor, so factorise them, and then convert using Euler's formula and simplify. (At what point does the cosine curve cross the x-axis?)

Now looking at the 4th term, when you multiply two exponential's together, you add their indices right? So simplify this term.

And now, what are you left with, keeping in mind that r = 0.937?

 

[AlephZero]

Remember the IDEA behind what the Euler formula is telling you: the exponential of an imaginary number is very similar to a trig function.

You should know that the trig functions of angles like pi/2 have nice simple values.

Using those facts, you should be able to see that the denominator is just

(z - rj)(z+rj)
= z^2 - (rj)^2
= z^2 + r^2

That was almost as easy as the numerator

 

[Jag1972]

Zryn\AlphaZero:
Thank you very much to both of you. Your helps sorts my issue out completely :)

Jag.

 

[Jag1972]

Zyrn: I know what the answer is as AlephZero’s post solves it however can not work it out the maths as you described. Could you please tell me where I am going wrong?
This is where I am currently:

 

[Zryn]

The maths I describe is the way of showing that AlephZero's shortcuts (based on the things he knows to be true since he has done it the long way before) is correct.

(z-re(j*pi/2))(z-re(-j*pi/2)) --> z^2 - zre(-j*pi/2) - zre(j*pi/2) + zre(j*pi/2 - j*pi/2)

This is not a correct expansion of the original equation, your fourth term is wrong.

Term 2 & 3 can be simplified as: -zre(-j*pi/2 + j*pi/2)

No they can't. You can only add the indices when the terms are multiplied together. Taking into account that you were told to use eulers formula and think about when the cos curve crosses the x-axis, and that this simplification results in e^0 = 1 and has no cos functions in it, it should have clicked that something went wrong.

The fourth term is raised to the power of 0, therefore that is equal to zr as e^0=1.

This is true, but the fourth term is wrong, so it doesn't help.

 

[Jag1972]

Thanks again Zryn: I think I know where I went wrong, whenever you have time could you please let me know if its correct.

Original equation:

\frac{(z-1)(z+1)}{(z-re^{j\frac{\pi}{2}}})(z-re^{-j\frac{\pi}{2}})


The top (Numerator) I think is straight forward.

I think the bottom should look like this:

z^{2}-zre^{(-j\frac{\pi}{2})}-zre^{(j\frac{\pi}{2})}+r^{2}e^{j\frac{\pi}{2})-(-j\frac{\pi}{2}}<br /> <br /> Using Eulers formulae terms 2 and 3 become<br /> <br /> -z*r(cos(^{pi/2})-jsin^{\pi/2})-z*r(cos^{\pi/2}+jsin^{\pi/2})<br /> <br /> trig rules: cos^{\pi/2} is 0 <br /> sin^{\pi/2}=1<br /> therefore:<br /> <br /> -z*r(0-1)-z*r(0+1)<br /> +z*r-z*r<br /> <br /> terms 2 and 3 = 0<br /> <br /> term 4 is: <br /> <br /> r^{2}e^{0} <br /> r^{2}<br /> <br /> therefore final result is:<br /> <br /> H(Z)=\frac{z^2-1}{z^2+r^2}<br /> <br /> I apologise for not using the latex code in the first place as you had rewrite it to reply to me.<br /> Thnaks in advance.

 

[Zryn]

That looks close, though I think your Latex broke somewhere in the middle of the equation.

When I did it, I grouped the 2nd & 3rd term and came up with:

factorised: -zr[ e^(-j*pi/2) + e^(-j*pi/2) ]
eulers: -zr[ cos(-pi/2) + jsin(-pi/2) + cos(pi/2) + jsin(pi/2) ]
simplify: -zr [ - cos(pi/2) - jsin(pi/2) + cos(pi/2) + jsin(pi/2) ] (the 4 terms sum to 0 here)
simplify: -zr [ - 0 - j + 0 + j ] (can see the 4 terms sum to 0 more easily than above)

-z*r(0-1)-z*r(0+1)

Slightly different to yours, as when you went from eulers to simplifying I think you left out the 'j' in front of the sin(-pi/2) and sin(pi/2), but since they sum to 0 in this case you got the right answer.

and the 4th term:

-re^(-j*pi/2) * -re^(j*pi/2)
simplify: [-r*-r] * e^(-j*pi/2 + j*pi/2)
simplify: [r^2] * e^(0)

And you know the answer, so its all good.

 

[Jag1972]

Thanks Zryn: I was thinking about what I wrote and I found a mistake in my solution, however that doesn't quite match yours, my working our is as follows. I am just going to show the denominator:

= (z-re^{-j\pi/2})(z-re^{j\pi/2})

= z^{2} - zre^{j\pi/2} - zre^{-j\pi/2} + r^{2}e^{-j\pi/2 + j\pi/2}

= -zr(e^{j\pi/2} + e^{-j\pi/2})

= -zr((cos\pi/2 + jsin \varpi/2)+(cos \varpi/2 - jsin \pi/2)

= -zr((0+1)+(0-1))

= -zr(1-1) -> -zr(0)

Terms 2 & 3 are 0.

Term 4 is the same as your solution:

r^{2}e(0) -> r^{2}

Thanks in advance Zyrn

 

[Zryn]

= -zr((0+1)+(0-1))

I believe this step is wrong again, for the same reason as before, but everything else looks good.

When you simplify j*sin(pi/2) and -j*sin(pi/2) you get j*1 and -j*1 not just 1 and -1, so that line should be:

-zr((0+j)+(0-j))

Alternatively you could just say j*sin(pi/2) + -j*sin(pi/2) = 0, but to put the 1 and -1 in there is incorrect.

-zr(1-1) -> -zr(0)

As per the previous mistake, it should be -zr(j-j) -> -zr(0), so you get the same answer, but you're just not taking into account the j factor and may be marked down because of it. Unless this is just a typing mistake as you entered all the math into Latex .

 

[Jag1972]

Zryn: Thanks a lot, it was not a typing mistake I was failing to account for j. I can see where I have been going wrong now thanks. I have defo learned something