The Fundamental Period Of The Signalx(t)=exp(j(7/3)n)is B. Find Whether The Signal Is Periodic Or Notx(t)=2cos(10t+1)sin(4t1)c.
Understanding the fundamental period of signals is essential in signal processing, communications, and electronics. In this comprehensive guide, we analyze two distinct signals: the discrete-time complex exponential signal \( x(n) = e^{j \frac{7}{3} n} \) and the continuous-time real-valued signal \( x(t) = 2 \cos(10 t + 1) \sin(4 t) \). Our goal is to determine the fundamental period of the first signal and assess whether the second signal is periodic, providing detailed explanations, mathematical derivations, and practical insights.
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Analyzing the Fundamental Period of the Discrete-Time Signal \( x(n) = e^{j \frac{7}{3} n} \)
Understanding Discrete-Time Complex Exponentials
The discrete-time complex exponential signals are of fundamental importance in signal processing because they form the basis for Fourier analysis and spectral representation. The general form:\[
x(n) = e^{j \omega n}
\]
where:
- \( \omega \) is the angular frequency (in radians per sample),
- \( n \) is the discrete-time index.
Such signals are inherently periodic if \( \omega \) is a rational multiple of \( 2\pi \). The key to finding the fundamental period is understanding when the exponential repeats itself.
Determining the Fundamental Period \( N_0 \)
For the signal:\[
x(n) = e^{j \frac{7}{3} n}
\]
the angular frequency:
\[
\omega = \frac{7}{3}
\]
is in radians per sample.
The periodicity condition for a discrete-time complex exponential is:
\[
x(n + N_0) = x(n) \quad \text{for all } n
\]
which implies:
\[
e^{j \omega (n + N_0)} = e^{j \omega n}
\]
or:
\[
e^{j \omega N_0} = 1
\]
Since \( e^{j \theta} = 1 \) when \( \theta = 2 \pi k \), where \( k \) is an integer, we have:
\[
\omega N_0 = 2 \pi k
\]
Plugging in \( \omega = \frac{7}{3} \):
\[
\frac{7}{3} N_0 = 2 \pi k
\]
Rearranged as:
\[
N_0 = \frac{2 \pi k \times 3}{7}
\]
To find the fundamental period \( N0 \), we seek the smallest positive integer \( N0 \) such that this equation holds for some integer \( k \).
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Finding the Minimal \( N_0 \)
Rearranged:
\[
N_0 = \frac{6 \pi k}{7}
\]
Since we need \( N_0 \) to be an integer, and \( \pi \) is irrational, the above expression suggests that the exponential will be periodic if and only if \( \omega \) is a rational multiple of \( 2 \pi \).
Alternatively, consider the frequency ratio:
\[
\frac{\omega}{2 \pi} = \frac{7/3}{2 \pi} = \frac{7}{3 \times 2 \pi}
\]
which is irrational because \( \pi \) is irrational. Therefore, the exponential \( e^{j \frac{7}{3} n} \) is not periodic in the discrete-time domain because there is no finite \( N_0 \) satisfying the periodicity condition.
Conclusion:
The discrete-time complex exponential \( x(n) = e^{j \frac{7}{3} n} \) is not periodic because its frequency is an irrational multiple of \( 2\pi \).
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Analyzing the Periodicity of \( x(t) = 2 \cos(10 t + 1) \sin(4 t) \)
Understanding Continuous-Time Signals and Periodicity
A continuous-time signal \( x(t) \) is periodic if there exists a positive real number \( T \) such that:\[
x(t + T) = x(t) \quad \text{for all } t
\]
The fundamental period \( T_0 \) is the smallest such positive \( T \).
In the case of products of sinusoidal functions, the periodicity depends on the individual frequencies involved.
Expressing the Signal Using Trigonometric Identities
The given signal:\[
x(t) = 2 \cos(10 t + 1) \sin(4 t)
\]
can be simplified using the product-to-sum identities:
\[
2 \cos A \sin B = \sin(A + B) - \sin(A - B)
\]
Applying this, we obtain:
\[
x(t) = \sin(10 t + 1 + 4 t) - \sin(10 t + 1 - 4 t)
\]
which simplifies to:
\[
x(t) = \sin(14 t + 1) - \sin(6 t + 1)
\]
Now, the signal is expressed as the difference of two sinusoidal functions:
\[
x(t) = \sin(14 t + 1) - \sin(6 t + 1)
\]
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Determining Periods of Individual Components
The two sinusoidal components are:- \( \sin(14 t + 1) \)
- \( \sin(6 t + 1) \)
\[
T_1 = \frac{2 \pi}{14} = \frac{\pi}{7}
\]
\[
T_2 = \frac{2 \pi}{6} = \frac{\pi}{3}
\]
Since phase shifts (the additional "+ 1") do not affect periodicity, these are the fundamental periods of the individual sinusoids.
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Finding the Fundamental Period of \( x(t) \)
The overall periodicity of \( x(t) \) depends on whether these two sinusoidal components are commensurate—that is, whether their periods are rational multiples of each other.To check:
\[
\frac{T1}{T2} = \frac{\pi / 7}{\pi / 3} = \frac{3}{7}
\]
which is rational. Therefore, the two components are commensurate, and the combined signal \( x(t) \) is periodic with a fundamental period equal to the least common multiple (LCM) of \( T1 \) and \( T2 \).
The least common multiple of \( T1 \) and \( T2 \):
\[
T0 = \text{LCM}(T1, T_2)
\]
Since:
\[
T_1 = \frac{\pi}{7}
\]
\[
T_2 = \frac{\pi}{3}
\]
the fundamental period \( T_0 \) is:
\[
T_0 = \text{LCM} \left( \frac{\pi}{7}, \frac{\pi}{3} \right ) = \pi \times \text{LCM} \left( \frac{1}{7}, \frac{1}{3} \right)
\]
Expressed as:
\[
T_0 = \pi \times \frac{\text{LCM}(1/7, 1/3)}{1}
\]
But more straightforwardly, the period of the sum of two sinusoids with periods \( T1 \) and \( T2 \) is:
\[
T0 = \text{LCM}(T1, T2) = \text{smallest positive } T \text{ such that } T = m T1 = n T_2, \text{ for integers } m, n
\]
Calculating:
\[
m T1 = n T2
\]
\[
m \times \frac{\pi}{7} = n \times \frac{\pi}{3}
\]
Dividing both sides by \( \pi \):
\[
\frac{m}{7} = \frac{n}{3}
\]
Cross-multiplied:
\[
3 m = 7 n
\]
Since \( m, n \) are integers, the minimal solution occurs at:
\[
m = 7, \quad n = 3
\]
This gives:
\[
T0 = m T1 = 7 \times \frac{\pi}{7} = \pi
\]
or equivalently:
\[
T0 = n T2 = 3 \times \frac{\pi}{3} = \pi
\]
Conclusion:
The fundamental period of \( x(t) \) is:
\[
\boxed{
T_0 = \pi
}
\]
The signal \( x(t) \) is periodic with a fundamental period of \( \
