Signals and Systems Test 2

Time scaling will not effect the Fourier series coefficient

x (0.5t) → C_k

\(x\left( {t - 0.5} \right) \to {e^{ - j{\omega _0}0.5k}}{C_k}\)

x (2t) → C_k

∴ Fourier series coefficient of the given signal is

x(t) = cos² 8t + sin² 4t

\(= \frac{{1 + \cos 16t}}{2} + \frac{{1 - \cos 8t}}{2}\) 

\({\cos ^2}t = \frac{{1 + \cos 2t}}{2}\) 

\({\sin ^2}t = \frac{{1 - \cos 2t}}{2}\) 

\(x\left( t \right) = 1 + \frac{1}{2}\cos 16t - \frac{1}{2}\cos 8t\) 

Where ω₁ = 16, ω ₂ = 8

\({T_1} = \frac{{2\pi }}{{{{\rm{\omega }}_1}}} = \frac{{2\pi }}{{16}} = \frac{\pi }{8}\) 

\({T_2} = \frac{{2\pi }}{{{{\rm{\omega }}_2}}} = \frac{{2\pi }}{8} = \frac{\pi }{4}\) 

T = π / 4

\({\omega _0} = \frac{{2\pi }}{T} = \frac{{2\pi }}{\pi } \times 4\) 

ω₀ = 8 rad / sec

\(x\left( t \right) = 1 + \frac{1}{2}\cos 16t - \frac{1}{2}\cos 8t\) 

\(= 1 + \frac{1}{2}\cos 2{\omega _0}t - \frac{1}{2}\cos {\omega _0}t\) 

\(= 1 + \frac{1}{2}\left( {\frac{{{e^{j2{{\rm{\omega }}_0}t}} + {e^{ - j2{\omega _0}t}}}}{2}} \right) - \frac{1}{2}\left( {\frac{{{e^{j{\omega _0}t}} + {e^{ - j{\omega _0}t}}}}{2}} \right)\) 

\(x\left( t \right) = 1 + \frac{1}{4}{e^{j2{\omega _0}t}} + \frac{1}{4}{e^{ - j2{{\rm{\omega }}_0}t}} - \frac{1}{4}{e^{j{\omega _0}t}} - \frac{1}{4}{e^{ - j{{\rm{\omega }}_0}t}}\) 

∵ C₀ = 1; C₂ = 1 / 4; C₁ = -1 / 4

C_-2 = 1 / 4; C_-1 = -1 / 4

f(t) = cos (3t) + sin (4t) = a₃ cos (3t) + b₄ sin (4t)

Fourth harmonic is only present in an odd component of the signal i.e. sin (4t)

⇒ b₄ = 1

Third harmonic is only present in even component of the signal i.e. cos (3t)

⇒ a₃ = 1

Component of fourth harmonic is absent in the even component of the signal a₄cos (3t)

⇒ a₄ = 0

x[k] = j δ[k - 1] – j δ [k + 1] + δ [k - 3] + δ [k + 3], ω₀ = 4π

\(x\left( t \right) = j{e^{j\left( 1 \right)4\pi t}}-{e^{j\left( { - 1} \right)4\pi t}} + {e^{j\left( 3 \right)4\pi t}} + {e^{j\left( { - 3} \right)4\pi t}}\)

= -2 sin (4 πt) + 2 cos (12 πt)