# Magnitude & Phase

1. Jan 31, 2013

### JeeebeZ

1. The problem statement, all variables and given/known data

H(e^jw) = (1-1.25e^(-jw))/(1-0.8e^(-jw))

Prove |H(e^jw)|^2 = G^2, and what is G
Find Magnitude & Phase

2. Relevant equations

H(e^jw) = (e^(jw)-1.25)/(e^(jw)-0.8)

H(e^jw) = 1 - (0.45e^(-jw))/(1-0.8e^(-jw))

H(e^jw) = 1 - 0.45/(e^(jw)-0.8)

3. The attempt at a solution

I don't know how to approch this question. I can rewrite H(e^jw) in 4 different formats, but none of which make me understand how to attempt to get the magnitude in the first place.

2. Jan 31, 2013

### Simon Bridge

H(e^jw) = (1-1.25e^(-jw))/(1-0.8e^(-jw))
this would be:$$H(e^{j\omega})=\frac{1-(1.25)e^{-j\omega}}{1-(0.8)e^{-j\omega}}$$

What is the definition of the magnitude of a complex exponential/number?

3. Jan 31, 2013

### JeeebeZ

$$Magnitude = \sqrt{Re(z)^2 + Im(z)^2}$$

4. Jan 31, 2013

### Simon Bridge

Great - so all you need to do is identify the real and imaginary parts of $H(e^{j\omega})$ ... how do you do that?
Hint: the exponential describes a phasor.

5. Jan 31, 2013

### JeeebeZ

Because |Z|^2 = z * zbar

we have
$$|H(e^{j\omega})|^2=(\frac{1-(1.25)e^{-j\omega}}{1-(0.8)e^{-j\omega}})(\frac{1-(1.25)e^{j\omega}}{1-(0.8)e^{j\omega}})$$

substituting
$$x=e^{j\omega}$$

$$|H(e^{j\omega})|^2=(\frac{1-(1.25)x^-1}{1-(0.8)x^-1})(\frac{1-(1.25)x}{1-(0.8)x})$$

$$|H(e^{j\omega})|^2=(\frac{1-(1.25)x^-1-(1.25)x + 1.25^2}{1-(0.8)x^-1-(0.8)x+0.8^2})(\frac{x}{x})$$

$$|H(e^{j\omega})|^2=\frac{x-(1.25)-(1.25)x^2 + 1.25^2x}{x-(0.8)-(0.8)x^2+0.8^2x}$$

$$|H(e^{j\omega})|^2=\frac{-1.25}{-0.80}\frac{x^2 + 2.05x - 1}{x^2 + 2.05x -1}$$

$$|H(e^{j\omega})|^2=\frac{-1.25}{-0.80} = 1.5625$$

$$|H(e^{j\omega})|= G = 1.25 = Magnatude$$

I got the Phase by computing the two angles, one on top with arctan (Im/Re), and the one on bottom, then phase total = top - bottom.

6. Jan 31, 2013

### Simon Bridge

See - you didn't need me :)
Hmmm ... I notice that 1.25 is the magnitude of one of the phasors in the combination which is a little startling. The second to last line says that the magnitude is the square-root of the ratio of the magnitudes of the phasors. You could explore to see if this is a general result or just a judicious choice of amplitudes.

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Some latex notes (just sayin'):
you can use \left ( <some stuff> \right ) to fit brackets around the bigger stuff so
\left ( \frac{q}{p-1} \right ) gives you $$\left ( \frac{q}{p-1} \right )$$

the complex conjugate is usually better represented by a star notation as in $z^*$ or $z^\star$ - the bar gets tricky to typeset after a bit and you risk confusing $z_{ave}=\bar{z}$ ...

7. Feb 1, 2013

### JeeebeZ

I've never used latex before I just quoted yours and edited it as needed but I'll keep it in mind for the future. Thx

8. Feb 1, 2013

### Simon Bridge

You did well at that too :) Most people just go "meh".
Them I don't usually bother to give pointers ;)