A continuous-time linear systemS with input x(t) and output y(t) yields the following input-output pairs:
x(t) = ej21 ~ y(t) = ej31,
x(t) = e-j 21 ~ y(t) = e- j 31.
(a) If x1 (t) = cos(2t), determine the corresponding output y1 (t) for systemS.
(b) If x2(t) = cos(2(t – 1/2)), determine the corresponding output y2(t) for system
The correct answer and explanation is :
Problem Overview
In this problem, we are dealing with a continuous-time linear system ( S ) that has two given input-output pairs and we need to determine the corresponding outputs for two different inputs: ( x_1(t) = \cos(2t) ) and ( x_2(t) = \cos(2(t – 1/2)) ).
The input-output relationships are given by:
- ( x(t) = e^{j2t} ) yields ( y(t) = e^{j3t} ),
- ( x(t) = e^{-j2t} ) yields ( y(t) = e^{-j3t} ).
System Behavior
The system ( S ) appears to act on complex exponentials of the form ( e^{j\omega t} ). Based on the given pairs:
- For ( x(t) = e^{j2t} ), the output is ( y(t) = e^{j3t} ),
- For ( x(t) = e^{-j2t} ), the output is ( y(t) = e^{-j3t} ).
This suggests that the system has a frequency shift property. Specifically, the system seems to map an input frequency ( \omega ) to an output frequency ( \omega + 1 ). This implies that the system acts as a frequency shifter that adds 1 to the frequency of the input signal.
Part (a): ( x_1(t) = \cos(2t) )
The input ( x_1(t) = \cos(2t) ) can be expressed in terms of complex exponentials using Euler’s formula:
[
\cos(2t) = \frac{1}{2} \left( e^{j2t} + e^{-j2t} \right)
]
From the system’s behavior:
- For ( x(t) = e^{j2t} ), the output is ( y(t) = e^{j3t} ),
- For ( x(t) = e^{-j2t} ), the output is ( y(t) = e^{-j3t} ).
Therefore, applying the system’s frequency shift to both parts of the cosine function:
[
y_1(t) = \frac{1}{2} \left( e^{j3t} + e^{-j3t} \right) = \cos(3t)
]
Thus, the corresponding output for ( x_1(t) = \cos(2t) ) is:
[
y_1(t) = \cos(3t)
]
Part (b): ( x_2(t) = \cos(2(t – 1/2)) )
Now, for ( x_2(t) = \cos(2(t – 1/2)) ), we first simplify the expression:
[
x_2(t) = \cos(2t – 1) = \frac{1}{2} \left( e^{j(2t – 1)} + e^{-j(2t – 1)} \right)