# Using the definition of load voltage reflection coefficient solving for Zt

Discussion in 'Homework Help' started by Kayne, Mar 31, 2011.

1. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Hi,

My question relates to how to multiply out a complex number which is multiplied by a whole number.

where p = rho, Zo = 50ohm

$\frac{Zt}{Zo}= \frac{1+pt}{1-pt}$

$Zt = \frac{1+pt}{1-pt} * Zo$

Where

$pt = 0.5305 angle 69.33deg$
or
$0.18726+j0.49635$

$Zt = \frac{(1+(0.5305 angle 69.33deg))}{(1-(0.5305 angle 69.33deg))}*50$

I would like to know if the following step is correct as this is where I am getting stuck

$Zt = \frac{(1.5305 angle 69.33deg))}{(0.4695 angle 69.33deg))}*50$

Do you have to change the equation from Polar back to Rect then multiply it by 50 or can this be done in the equation above?

What part of the equation do you have to mulitpy by 50 just the real part or does both the real and theta get multiplied.?

Any help would be appricated

Thanks

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
That's incorrect - you have to convert the polar form to rectangular then add the (real) 1 term to the rectangular real part.

If you multiply a rectangular form by a constant (50 in your case) then both the real and imaginary parts are multiplied. If you multiply the polar form by a constant only the magnitude changes by that factor - the angle term remains the same.

Kayne likes this.
3. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Thanks for the help, I will try it now and see how I go

Regards,
Kayne

4. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
I have worked though the question again which you can see below however I the answer I am getting is incorrect as I have worked it out on a TI-84 calculator and got and answer of 67.56angle54.1 or 39.62+j54.73.

So now i am unsure of what I have done wrong. as I thought that I follow the instructions given above, obviously I havent.

$\frac{(1+(0.5305\angle69.33\deg))}{(1-(0.5305\angle69.33\deg))}*50$

So change POL2RECT

$\frac{(1+(0.18726+j0.49635))}{(1-(0.18726+j0.49635))}*50$

Deal with the real 1

$\frac{(1.18726+j0.49635)}{(0.81274+j0.49635)}*50$

Change RECT2POL

$\frac{(1.28818\angle22.6\deg)}{(0.9523\angle31.412\deg)}*50$

$\frac{(64.40\angle22.6\deg)}{(47.6159\angle31.412\deg)}$
Sovle
$\frac{64.40}{47.6159} \angle22.6\deg-\angle31.412\deg$

$1.352\angle -8.722\deg$

5. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,301
338
Above, where you have:

"Deal with the real 1

$\frac{(1.18726+j0.49635)}{(0.81274+j0.49635)}*50$
"

there is a sign error. It should be:

$\frac{(1.18726+j0.49635)}{(0.81274-j0.49635)}*50$

You've also made a mistake in the last operation here:

"Change RECT2POL

$\frac{(1.28818\angle22.6\deg)}{(0.9523\angle31.412\deg)}*50$

$\frac{(64.40\angle22.6\deg)}{(47.6159\angle31.412\deg)}$
"

You shouldn't multiply both the numerator and denominator by 50; only multiply the numerator by 50

Can't the TI-84 do complex arithmetic? It should be able to do these operations directly.

6. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Yes the TI-84 can do the operation and this is how I got the answer but I would still like to know how to do it by hand. So if I dont have a calculator I can work it out.

And I think that it make you better doing it by hand and not having to rely on it.

Thanks for the help will try again and see how i go.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Without a calculator it could prove difficult doing any polar to rectangular conversions and vice versa. Are you suggesting dispensing with the calculator altogether?

You should at least develop the skill in using the calculator reliably. If you can't trust that approach, then the likelihood of success with just pencil and paper will be doubtful.

If you admit the use of the calculator for just the conversions between polar and rectangular then it might allow you to refine your complex number operation skill involving division, multiplication, etc.

I would suggest you adopt the polar form for multiplication & division operations and the polar form for addition & subtraction.

The process is then fairly straightforward. For example with two complex numbers C1 & C2

$C_1=A\angle{\theta}$
$C_2=B\angle{\phi}$
$C_1.C_2=AB\angle{(\theta+\phi)}$
$\frac{C_1}{C_2}=\frac{A}{B}\angle{(\theta-\phi)}$