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$$x(t)=5 e^{-i 20\pi t} +5$$

I need to find the magnitude of this function, $|x(t)|$.

My idea was to convert from the first have from polar to rectangular form.

$$x(t)=5[\cos(-20\pi t)+j \sin(-20\pi t)]+5$$ $$=5\cos(20\pi t)+5-j5 \sin(20 \pi t)$$

I know the answer is, $$|10 \cos(10\pi t)|$$

but, I don't know how to get here from what I have.

What technique/rule do I need to apply here?

cdignam
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  • You did not distribute the $5$ correctly, it should be $-5jsin(20{\pi}t)$. This may help for a start. – imranfat Oct 11 '17 at 01:59
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    Firstly, you have a typo (it should be $-5j\sin(20\pi t)$). Just compute the magnitude from there. $$\sqrt{25(1+\cos(20\pi t))^2+25\sin^2(20\pi t)}$$ Use elementary trigonometric identities to simplify the expression. – Prasun Biswas Oct 11 '17 at 01:59

1 Answers1

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First,$x(t)=5\cos(20\pi t)+5-5i\sin(20\pi t)$, then taking the modulus of $x(t)$. $$|x(t)|=\sqrt{(5\cos(20\pi t)+5)^2+(-5\sin(20\pi t))^2}$$ $$|x(t)|=\sqrt{25\cos^2(20\pi t)+25+50\cos(20\pi t)+25\sin^2(20\pi t)}$$ $$|x(t)|=\sqrt{50(1+\cos(20\pi t)}$$ Using $\cos(\frac{\theta}{2})=\sqrt{\frac{1+\cos 2\theta}{2}}$, then $1+\cos(20\pi t)=2\cos^2(10\pi t)$ So, $$|x(t)|=\sqrt{100\cos^2(10\pi t)}$$ Finally,$$|x(t)|=|10\cos(10\pi t)|$$

Isai Dávila Cuba
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