A real-valued signal 𝑥(𝑡) limited to the frequency band \(\left
A real-valued signal 𝑥(𝑡) limited to the frequency band \(\left| f \right| \le \frac{W}{2}\) is passed through a linear time-invariant system whose frequency response is
\(H\left( f \right) = \left\{ {\begin{array}{*{20}{c}} {{e^{ - j4\pi f,\;\;\;\left| f \right| \le \frac{W}{2}}}}\\ {0,\;\;\;\;\left| f \right| > \frac{W}{2}} \end{array}} \right.\)
The output of the system isA. x(t + 4)
B. x(t - 4)
C. x(t + 2)
D. x(t - 2)
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Concept:
Time-shifting property of Fourier Transform:
Shiting in time domain results in a phase shift in the frequency domain, i.e.
\(If\;x\left( t \right)\mathop \leftrightarrow \limits^{FT} X\left( f \right)\)
\(x\left( {t - {t_0}} \right)\mathop \leftrightarrow \limits^{FT} X\left( f \right){e^{ - j2\pi f{t_0}}}\)
If a signal x(t) is passed through a system having transfer function H(f), then the output of the system is:
\(Y\left( f \right) = X\left( f \right)H\left( f \right)\)
where \(x\left( t \right)\mathop \leftrightarrow \limits^{FT} X\left( f \right)\) and
\(y\left( t \right)\mathop \leftrightarrow \limits^{FT} Y\left( f \right)\)
Calculation:
Since both X(f) and H(f) are band-limited to the same frequency band, we can write:
\(Y\left( f \right) = X\left( f \right).H\left( f \right) = X\left( f \right){e^{ - j4\pi f}} = X\left( f \right).{e^{ - i2\pi f \times 2}}\)
Using time shifting property of the Fourier transform, we can write:
\(X\left( f \right).{e^{ - i2\pi f \times 2}}\xrightarrow{IFT} x\left( {t - 2} \right)\)
