Question

TT 1) Consider the following signal x(t) = -5 + 9 sin (61t + 3 cos(12nt) + V V7ej24nt a) (5 points) Write the complex exponential Fourier series for x(t). b) (5 points) Sketch the amplitude and phase spectra of x(t). c) (5 points) Find the total average power of the signal.

Answer 1

Ans)

a) Given

z(t) = -5 + 9sin(6nt ) + 3cos(12mt) + V7j24nt 4.

now using $cos\theta =0.5(e^{j\theta }+e^{-j\theta }),\: \: sin\theta =0.5(e^{j\theta }-e^{-j\theta })$

$x(t)=-5+9*0.5(e^{j(6\pi t-\frac{\pi }{4})}-e^{-j(6\pi t-\frac{\pi }{4})})+3*0.5*(e^{j(12\pi t)}+e^{-j(12\pi t)})+\sqrt{7}e^{j24\pi t}$

$\mathbf{x(t)=1.5e^{-j(12\pi t)}-4.5e^{j\frac{\pi }{4}}e^{-j(6\pi t)}-5+4.5e^{-j\frac{\pi }{4}}e^{j(6\pi t)}+1.5e^{j(12\pi t)}+\sqrt{7}e^{j24\pi t}}$

the above expression is the complex exponential fourier series

i.e $x(t)=\sum_{k=-\infty }^{\infty }c_{k}e^{j2\pi f_{k}t}$

Now by comparision

$\mathbf{c_{-6}=1.5,\: \: c_{-3}=-4.5e^{j\frac{\pi }{4}},\: \: c_{0}=-5,\: \: c_{3}=4.5e^{-j\frac{\pi }{4}},\: \: c_{6}=1.5,\: \: c_{12}=\sqrt{7}}$

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b)

From the above terms i.e

$c_{-6}=1.5,\: \: c_{-3}=-4.5e^{j\frac{\pi }{4}},\: \: c_{0}=-5,\: \: c_{3}=4.5e^{-j\frac{\pi }{4}},\: \: c_{6}=1.5,\: \: c_{12}=\sqrt{7}$

the magnitude and phase spectra of x(t) from above coefficients is

Amplitude Spectra 4.5 1.5 1.5 I 0 - 1 -3 -6 3 6 12 fin Hz -4.5 -5 Phase Spectra pi/4 -3 0 3 fin Hz -pi/4

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c)

The average power is given as

$P=\sum_{k=-\infty }^{\infty }\left | c_{k} \right |^{2}=1.5^{2}+(-4.5)^{2}+(-5)^{2}+4.5^{2}+1.5^{2}+(\sqrt{7}^{2})$

$P=2.25+20.25+25+20.25+2.25+7$

$\mathbf{P=77}$

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