ECE301 Fall 2008, Professor C.C. Wang

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I am trying to figure out how to compute the norm of the DT signal $ x[n]= e^{j 2 \pi n} $. According to the solutions, the answer is

$  \left| e^{j 2 \pi n} \right| = 1 $.

I don't get it. Should'nt the answer be a function of n???


Since we are dealing with complex numbers, we can make the following substitution:

$ \left| e^{j 2 \pi n} \right| = \left| \cos{2 \pi n} + j \sin{2 \pi n} \right| $

The magnitude of this expression will always be one since $ \cos^2\theta+\sin^2\theta\equiv 1 $.

Related Question

I see your point. But if what you say is true, then we also have

$ e^{j 2 \pi t} = \left( e^{j 2 \pi} \right)^t= \left( cos{2 \pi } + j sin{2 \pi } \right)^t= 1^t =1, $

which is clearly wrong, because my high school teacher showed us the graph of $ e^{j 2 \pi t} $ and it was oscillating. So what am I doing wrong?

MA181 to the rescue!

The identity $ (e^a)^t=e^{ta} $ is true for real numbers, but it is not always true for complex numbers, even when $ a $ is complex and $ t $ is real. When $ z $ is complex and $ t $ is real, $ z^t $ stands for $ e^{t\log z} $, where $ \log z $ is the complex logarithm of $ z $, which has infinitely many possible values.

By the way, the constant function is a function. The modulus of $ e^{j\theta} $ is equal to one for any value of $ \theta $ because that complex number represents a point on the unit circle.

I saw the plea on our page and thought I'd check it out. While I can't claim to have much understanding of what is going on, it seems like that solution is valid because $ n $ represents either the natural numbers or integrers, not the real numbers. Therefore, it would be like taking the value of the sine function at whole-number products of $ 2\pi $. Even though the function oscillates, the sequence is constant. Hope I haven't made a fool of myself. --Jmason

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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