b) in figure (a) below, a system is shown with input signal x(t) and output signal y(t). the input signal…

b) in figure (a) below, a system is shown with input signal x(t) and output signal y(t). the input signal has the fourier transform x(w) shown in figure (b). determine and sketch y(w), the spectrum of y(t). (6 pt.)

b) in figure (a) below, a system is shown with input signal x(t) and output signal y(t). the input signal has the fourier transform x(w) shown in figure (b). determine and sketch y(w), the spectrum of y(t). (6 pt.)

Answer

Explanation:

Step1: Use modulation property

The multiplication of a signal $x(t)$ with $\cos(\omega_0t)$ in time - domain corresponds to a frequency - shift in the frequency - domain. When $x(t)$ is multiplied by $\cos(5\omega t)$, the spectrum $X(\omega)$ is shifted by $\pm5\omega$. Let $x_1(t)=x(t)\cos(5\omega t)$. Then, by the modulation property of the Fourier transform, $X_1(\omega)=\frac{1}{2}[X(\omega - 5\omega)+X(\omega + 5\omega)]$.

Step2: Consider the next multiplication

Next, $g(t)=x_1(t)$ is multiplied by $\cos(3\omega t)$. Let $y(t)=g(t)\cos(3\omega t)$. Using the modulation property again, $Y(\omega)=\frac{1}{2}[G(\omega - 3\omega)+G(\omega + 3\omega)]$. Substituting $G(\omega)=\frac{1}{2}[X(\omega - 5\omega)+X(\omega + 5\omega)]$ into the above formula, we get $Y(\omega)=\frac{1}{4}[X(\omega-(5\omega + 3\omega))+X(\omega+(5\omega + 3\omega))+X(\omega-(5\omega - 3\omega))+X(\omega+(5\omega - 3\omega))]=\frac{1}{4}[X(\omega - 8\omega)+X(\omega + 8\omega)+X(\omega - 2\omega)+X(\omega + 2\omega)]$.

Step3: Analyze the original spectrum of $X(\omega)$

Given that $X(\omega)$ is non - zero in the range $[- 2\omega,2\omega]$.

  • For $X(\omega - 8\omega)$, it is non - zero when $-2\omega\leqslant\omega - 8\omega\leqslant2\omega$, i.e., $6\omega\leqslant\omega\leqslant10\omega$.
  • For $X(\omega + 8\omega)$, it is non - zero when $-2\omega\leqslant\omega + 8\omega\leqslant2\omega$, i.e., $-10\omega\leqslant\omega\leqslant - 6\omega$.
  • For $X(\omega - 2\omega)$, it is non - zero when $-2\omega\leqslant\omega - 2\omega\leqslant2\omega$, i.e., $0\leqslant\omega\leqslant4\omega$.
  • For $X(\omega + 2\omega)$, it is non - zero when $-2\omega\leqslant\omega + 2\omega\leqslant2\omega$, i.e., $-4\omega\leqslant\omega\leqslant0$.

Step4: Sketch the spectrum of $Y(\omega)$

The spectrum $Y(\omega)$ will have non - zero values in the intervals $[-10\omega,-6\omega]\cup[-4\omega,0]\cup[0,4\omega]\cup[6\omega,10\omega]$. The amplitude of $Y(\omega)$ in these non - zero intervals is $\frac{1}{4}$ of the corresponding values of $X(\omega)$ after the appropriate frequency shifts.

Answer:

The spectrum $Y(\omega)$ is non - zero in the intervals $[-10\omega,-6\omega]\cup[-4\omega,0]\cup[0,4\omega]\cup[6\omega,10\omega]$ with amplitude $\frac{1}{4}$ of the corresponding shifted values of $X(\omega)$. Sketch should show rectangular or triangular (depending on the shape of $X(\omega)$ in the non - zero region) functions in these intervals with amplitude $\frac{1}{4}$ of the original $X(\omega)$ values at the shifted frequencies.