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Given the signal: $$x(t) = −4\cos(2\pi 4t) + 3\cos \left(\pi t − \frac{\pi}{3}\right)$$

By sampling theorem $Fs = 2 Maxf = 8 Hz$

The question ask what is the output of system y[n] when the signal x[n] is the input and the frequency response is: $H(e^{jω}) = 2 + 2e^{j2ω}$

I try to do this, but i don't undestand how to separate magnitude and phase from this result:enter image description here

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2 Answers 2

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Euler's formula is you friend :) $$e^{j\theta} = \cos(\theta) + j\sin(\theta)$$ After applying it, you will have $$H(e^{j\omega}) = \Re\{H(e^{j\omega})\} + j\Im\{H(e^{j\omega})\}$$ and then it's easy to find magnitude $$|H(e^{j\omega})|=\sqrt{\Re^2\{H(e^{j\omega})\} + \Im^2\{H(e^{j\omega})\}}$$ and phase $$\angle H(e^{j\omega}) = \tan^{-1}\frac{\Im\{H(e^{j\omega})\}}{\Re\{H(e^{j\omega})\}} $$

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  • $\begingroup$ i propose a solution, it's correct? $\endgroup$
    – Owner
    Jan 16, 2020 at 13:12
  • $\begingroup$ Partly. It is correct that $H(e^{j\omega})\Big|_{\omega = \pi} = 4$, and thus its magnitue is $4$ and its phase is zero. However, you have to try again for $H(e^{j\omega})\Big|_{\omega=\pi/8}$ (magnitude part only). $\endgroup$
    – GKH
    Jan 16, 2020 at 14:07
  • $\begingroup$ I just try for $\omega = \pi/8$ in the second line. Why only for magnitude? $\endgroup$
    – Owner
    Jan 16, 2020 at 14:25
  • $\begingroup$ I mean, your phase is OK ($\pi/8$) but your magnitude is wrong. $\endgroup$
    – GKH
    Jan 16, 2020 at 14:31
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    $\begingroup$ Ahh ok, i made a mistake. The magnitude is $2sqrt(2+sqrt(2))$, right? But it's correct to compute y[n] in the way of my solution? $\endgroup$
    – Owner
    Jan 16, 2020 at 15:24
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with $2+2e^{j2\pi}$ = $2+2\cos(2\pi) + 2j\sin(2\pi) = 4 $ so phase is equal to $0$?

and with $2+2e^{j\pi/4}$ = $2 + 2\cos(\pi/4) + 2j\sin(\pi/4)$ = $2 +\sqrt{2}+ j\sqrt2$ here magnitude is $2\sqrt{2}$ and phase is $\pi/8$

finally $y[n] = -4 * 4\cos(\pi n) + 2\sqrt{2} *3\cos (\pi n/8 - \pi/3 + \pi/8)$

It's correct?

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