# Convert to polar Form

1. Sep 20, 2010

### amiv4

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

H(F) = 5/(1+j2piF/10)

Rewrite in polar form, that is, in terms of magnitude and phase.

2. Relevant equations

3. The attempt at a solution

phase is the 2piF/10 but I'm not sure how I account for it being on the bottom of the fraction
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 20, 2010

### xcvxcvvc

$$\frac{ A \angle \theta}{B \angle \phi} = \frac{A}{B} \angle \theta - \phi$$

also "2piF/10" is not the phase, not even for the complex number in the denominator alone.

to find an angle, you think of it in terms of real and imaginary parts forming a right triangle on the unit circle. The angle is this:
$$\arctan\frac{Im(A \angle \theta)}{Re(A\angle \theta)}$$

Last edited: Sep 20, 2010
3. Sep 20, 2010

### amiv4

So I should convert the top and bottom to polar separately and then divide them?

4. Sep 20, 2010

### xcvxcvvc

we must be careful when you ask questions with so many undefined pronouns. If by them, you mean the magnitudes and if you meant "should i ... to find the resultant magnitude", then the answer is yes.

if you are talking about angles, however, the answer is no. you must subtract the denominator's angle from the numerator's angle to find the resultant angle.

Last edited: Sep 20, 2010
5. Sep 20, 2010

### amiv4

k this is what i got

5$$\angle$$0$$/$$1.18$$\angle$$32.14

but idk where the F goes

6. Sep 20, 2010

### xcvxcvvc

the f remains symbolically.

for example, the magnitude of the bottom vector is:
$$\sqrt{1^2 + (\frac{2piF}{10})^2}$$

7. Sep 20, 2010

### amiv4

and then I would do the arctan of those two things to get the angle, but I dont see how they are gonna simplify. Cuz I have to use this answer. Is there a way I could get the answer into a form similar to this H(F)=|5F| exp(-jpiF/10)

8. Sep 20, 2010

### xcvxcvvc

it doesn't need to simplify. it will be messy but correct. as for putting it in the form you've specified:

$$A \angle \theta = Ae^{j\theta}$$
and your theta for the overall transfer function will be:
$$\theta = -\arctan \frac{\pi F}{5}$$

the negative comes from 0 - stuff