Question

) Dula) allow)
6) What is the Nyquist frequency for the signal
X(t) = 3 cos 50 nt + 10 sin 300 nt -cos 100 nt?
a) 50 Hz
b) 100 Hz C 200 Hz
d) 300 Hz
​

Answer 1

Answer:

The Nyquist frequency for the single X is 100Hz C 200 Hz

Answer 2

Answer:

Therefore,  the Nyquist frequency for the signal  is 300 Hz.

Step-by-step explanation:

Given that:

x(t) = 3 cos 50$\pi$t + 10 sin 300$\pi$t - cos 100$\pi$t ….(1)

Let us assume that there are three frequencies present in the signal.

This signal may be represented as:

x(t) = $A_{1}$ cos ω₁ t + $A_{2}$ cos ω₂ t + $A_{3}$ cos ω₃ t …(2)

where $A_{1}$, $A_{2}$ and  $A_{3}$  are amplitudes and ω₁, ω₂ and ω₃ are angular frequencies.

Comparing (1) and (2), we get:

ω₁ = 50$\pi$ radian/sec

ω₂ = 300$\pi$ radian/sec

 ω₃ = 100$\pi$ radian/sec

Therefore, $F_{1}$ = ω₁/2$\pi$ = $\frac{50\pi }{2\pi }$ = 25 Hz

                  $F_{2}$ = ω₂/2$\pi$ = $\frac{300\pi }{2\pi }$ = 150 Hz

                  $F_{3}$ = ω₃/2$\pi$ = $\frac{100\pi }{2\pi }$ = 50 Hz

The highest frequency component of the given message signal will be

                 $F_{max}$ = Max { $F_{1}$ ,  $F_{2}$ , $F_{3}$}

                          = Max { 25, 150, 50}

                          = 150 Hz.

Nyquist rate = $2 * F_{max}$

Therefore, $F_{S}$ = 2 x 150 = 300 Hz.

Hence, the Nyquist frequency for the signal  is 300 Hz.

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