# finding phase shift using i and q signals

Discussion in 'Math' started by ashwini1, Aug 1, 2009.

1. ### ashwini1 Thread Starter New Member

Aug 1, 2009
9
0
Hi,

I have read in one document that the phase of a signal (say sine wave) can be calculated by making use of I and Q signals of it. For ex:
Fs = 10e3;
t=(0:Fs)/Fs';
ph=45;
x=sin(2*pi*50*t+(ph*pi/180));

i=x.*cos(2*pi*50*t);
q=x.*sin(2*pi*50*t);

it was mentioned that by subtracting the max values of i and q, i will get the ph value.

But that is not happening.

what is wrong with this ??
i have to make use of i and q signals to find the ph.

Can anyone help me plz..

Ashwini

2. ### rspuzio Active Member

Jan 19, 2009
77
0

$x = \sin (2 \pi 50 t+ (\phi \pi/180)) =
\sin (2 \pi 50 t) \cos (\phi \pi/180) +
\cos (2 \pi 50 t) \sin (\phi \pi/180)
$

Now that you have x expressed as a superposition of
$\sin (2 \pi 50 t)$ and
$\cos (2 \pi 50 t)$, it should be easy enough
to multiply by a sine or cosine and find the maximum.

3. ### rspuzio Active Member

Jan 19, 2009
77
0
You can also use the trigonometric identities for
product of two sines or a product of a sine and
a cosine:

$i=\sin(2 \pi 50 t+(\phi \pi/180)) \cos(2 \pi 50 t) =
{1 \over 2} \left[ \sin (\phi \pi/180) +
\sin (4 \pi 50 t+(\phi \pi/180)) \right]$

$q=\sin(2 \pi 50 t+(\phi \pi/180)) \sin(2 \pi 50 t) =
{1 \over 2} \left[ \cos (\phi \pi/180) +
\cos (4 \pi 50 t+(\phi \pi/180)) \right]$

Note that in these expressions the first term is constant
whilst the second term is a sine or a cosine so that its
maximum value will be 1. So we have that the
maximum value of i is

$i_{\rm max} = {1 \over 2} \left[ \sin (\phi \pi/180)
+ 1 \right]$

and the maximum value for q is

$q_{\rm max} = {1 \over 2} \left[ \cos (\phi \pi/180)
+ 1 \right]$

So we see that the maximum values of i and q depend
on $\phi$. To recover the value of $\phi$
we can use inverse trig functions. Not quite as simple as
subtracting them, but you can use the maximum values
of i and q to determine the phase.

4. ### ashwini1 Thread Starter New Member

Aug 1, 2009
9
0
Thank you very much ..
It helped me a lot ..
I got the required output.