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I am studying Digital Signal Processing course for my Computer Science and Engineering major.

I have to calculate magnitude and phase of following $$x(n) = e^{(-0.1+0.3j)n}$$ where $$-10\le n \le 10$$

Here the "e" is Euler's constant. I can calculate the magnitude for an n, by putting a value of n from range and calculating value of imaginary and real part. But how do i calculate the phase? My concept is not clear on this. Kindly explain in detail and as easily as possible.

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For a complex number expressed in the polar form, $x = r e^{j\phi}$ , the magnitude is $$|x| = |r e^{j\phi}| = |r| |e^{j\phi}| = |r|$$ where $r$ is a real number and the phase is $\phi$ in radians.

When the complex number is expressed in rectangular form $$x = a + j b =r \cos(\phi) + j r \sin(\phi)$$ then the angle is expressed as $\phi = \arctan(b/a)$

For a sequence of complex numbers $$x(n) = e^{(-0.1+0.3j)n}$$ for each n, one can find its magnitude and phase once writing it in the polar or rectangular form where the polar form is $x[n] = ( e^{-0.1 n}) (e^{j 0.3 n})$.

Then its magnitude is $|x[n]| = |e^{-0.1 n}| |e^{j 0.3 n}| = e^{-0.1 n}$ and its phase is $\phi = 0.3 n$ radians for each n.

Note that principal value of the phase is the angle beetween $-\pi$ and $\pi$, any value other than this range can be equivalently mapped into it.

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  • $\begingroup$ So, if $$x(n) = e^{an} e^{bjn}$$ the phase is bn? Also how do i calculate angle? $\endgroup$ – Miruza Aug 15 '17 at 17:32
  • $\begingroup$ let me add it to the answer... $\endgroup$ – Fat32 Aug 15 '17 at 18:22

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