# Homework Help: How do I find the magnitude of this function?

1. Oct 4, 2011

### interxavier

1. The problem statement, all variables and given/known data
I'm asked to find the magnitude of the following complex function:

$$R(j \omega) = 1 + \exp{(-j \omega)} + \exp{(-j2 \omega)} + \exp{(-j3 \omega)} + \exp{(-j4 \omega)}$$

2. Relevant equations
None

3. The attempt at a solution

I did the following but got stuck immediately as I'm not sure on how to proceed:

$$|R(j \omega)| = |1 + \exp{(-j \omega)} + \exp{(-j2 \omega)} + \exp{(-j3 \omega)} + \exp{(-j4 \omega)}|$$

2. Oct 4, 2011

### Staff: Mentor

I'm not sure what you are doing. Maybe finding the peak value of a signal and its harmonics?

I think the idea would be to use the identity from maths: ejW=cosW + j sinW

3. Oct 4, 2011

### interxavier

I'm trying to find the magnitude. If I use euler's identity, then I would get multiple values of sin's and cosine's:
$$R(j \omega) = 1 + \cos{(\omega)} - j\sin{(\omega)} + \cos{(2\omega)} -j\sin{(2\omega)} ...$$

How do you find the magnitude of that?

4. Oct 4, 2011

### Hootenanny

Staff Emeritus
Initially, I would stick with exponential notation and recall that $|z|^2=zz^*$. Once you have computed $|z|^2=zz^*$ and simplified, you should find that you are left with a rather nice expression, which you should be able to easily convert into a real, non-negative, trigonometric expression.

5. Oct 4, 2011

### interxavier

Does that mean I take the conjugates like this:

$$R^{*} = 1 - \exp{(-j \omega)} - \exp{(-j2 \omega)} - \exp{(-j3 \omega)} - \exp{(-j4 \omega)}$$

or

$$R^{*} = 1 + \exp{(j \omega)} + \exp{(j2 \omega)} + \exp{(j3 \omega)} + \exp{(j4 \omega)}$$

6. Oct 4, 2011

### Hootenanny

Staff Emeritus
No. Like this:
Remember that if $z = Re^{i\theta}$, then $z^* = (e^{i\theta})^*R^*$, but since R is real: $z^* = R(e^{i\theta})^* = Re^{-i\theta}$ (check this using Euler's formula if you are unsure).