How do I find the magnitude of this function?

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To find the magnitude of the complex function R(jω) = 1 + exp(-jω) + exp(-j2ω) + exp(-j3ω) + exp(-j4ω), it's suggested to use Euler's identity to express the exponential terms in terms of sine and cosine. The magnitude can be calculated using the formula |z|^2 = z * z*, where z* is the complex conjugate. The correct form for the conjugate is R* = 1 + exp(jω) + exp(j2ω) + exp(j3ω) + exp(j4ω). After computing |z|^2 and simplifying, a real, non-negative trigonometric expression should emerge, allowing for the determination of the magnitude.
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Homework Statement


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)}

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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)}|
 
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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
 
NascentOxygen said:
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

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?
 
interxavier said:
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?
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.
 
Hootenanny said:
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.

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)}
 
interxavier said:
Does that mean I take the conjugates like this:

R^{*} = 1 - \exp{(-j \omega)} - \exp{(-j2 \omega)} - \exp{(-j3 \omega)} - \exp{(-j4 \omega)}
No. Like this:
interxavier said:
R^{*} = 1 + \exp{(j \omega)} + \exp{(j2 \omega)} + \exp{(j3 \omega)} + \exp{(j4 \omega)}
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).
 
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