Respuesta :
Answer:
The three of them are periodic discrete-time signals. Their respective
fundamental periods are:
[tex]T_1=\frac{2 \pi}{0.3m}\\T_2=32\pi\\T_3=\frac{6\pi}{i}[/tex]
Step-by-step explanation:
If a discrete-time signal X is a periodic one, it must exist a non-zero number T such that for any t time given X(t) is equal to X(t+T). The minimum number T that meets this equation is called a period. Let's test each signal.
Signal a:
Is there any T that X1[n]=X1[n+T]?
cos(0.3mn +1/4) = cos(0.3m[T+n] +1/4) =
cos(0.3mn +1/4) = cos(0.3mT +0.3mn] +1/4)
If 0.3mT=2Ï€ both sides of the equation will be balanced. Therefore T exists, this is a periodic discrete-time signal and its period is:
[tex]T=\frac{2 \pi}{0.3m}[/tex]
Signal b:
[tex]X_2[n]=e^{j\frac{n}{16}}=cos(\frac{n}{16})+jsin(\frac{n}{16})[/tex]
This is a sum of complex sinusoidal functions with the same frequency, therefore this is a periodic discrete-time signal and its period is:
[tex]X_2[n]=X_2[n+T][/tex]
[tex]e^{j\frac{n}{16}}=e^{j\frac{n+T}{16}}[/tex]
[tex]e^{j\frac{n}{16}}=e^{j(\frac{n}{16}+\frac{T}{16})}[/tex] if [tex]\frac{T}{16}=2\pi[/tex]
T=32Ï€
Signal c:
This is a sum between sinusoidal functions with the different frequencies and a constant (4cos(2 /7) is a number). The constant don´t affect the fact whether this is a periodic signal or not, so we don´t need to consider it.
For the two sinusoidal functions, their frequencies are different but these are multiples of each other and both of them have a null phase, so the period of the function with the lowest frequency will be the period of the function X3[n]. Therefore this is a periodic discrete-time signal and its period is: Â
[tex]8sin(\frac{i}{3}n)=8sin(\frac{i}{3}(n+T))[/tex] if [tex]\frac{iT}{3}=2\pi[/tex]
[tex]T=\frac{6\pi}{i}[/tex]