Signals And Systems Problems And Solutions Pdf _verified_ Link

\noindent\textbf13. Use Euler formulas and compare with exponential FS: \(x(t)=\sum a_k e^jk\omega_0 t\) with \(\omega_0=\pi\) (fundamental).

\section*Additional Problems (Brief Solutions) signals and systems problems and solutions pdf

\subsection*Solution Modulation: \(x(t)\cos(\omega_0 t) \leftrightarrow \frac12[X(j(\omega-\omega_0)) + X(j(\omega+\omega_0))]\). \\ Thus \(\textrect(t/T)\cos(\omega_0 t) \leftrightarrow \fracT2\left[\textsinc\left(\frac(\omega-\omega_0)T2\pi\right) + \textsinc\left(\frac(\omega+\omega_0)T2\pi\right)\right]\). \noindent\textbf13

\sectionFourier Transform

\subsection*Solution \(\textsinc(100t)\) has bandwidth 50 Hz (since \(\textsinc(Wt) \leftrightarrow \frac\piW\textrect(\frac\omega2W)\) with max freq \(W/(2\pi)\)? Better: \(\textsinc(Bt)\) has bandwidth \(B/2\) Hz? Actually \(\textsinc(Bt) = \frac\sin(\pi B t)\pi B t\) has Fourier transform \(\frac1B\textrect(\fracfB)\), so bandwidth \(B/2\) Hz? Let's be careful: The standard form: \(\textsinc(2Wt)\) has max freq \(W\) Hz. So \(\textsinc(100t) = \textsinc(2\cdot 50 t)\) → bandwidth 50 Hz. \\ \(\textsinc^2(50t)\) has bandwidth 50 Hz (since square in time = convolution in freq, doubles bandwidth to 100 Hz? Wait: \(\textsinc^2(50t)\) is the Fourier transform of triangle function, bandwidth 50 Hz if 50 is the zero-crossing rate? Let's compute: \(\textsinc(at) \leftrightarrow \frac1a\textrect(f/a)\) in Hz, so \(\textsinc(50t)\) has bandwidth 25 Hz. Squaring doubles bandwidth to 50 Hz. Therefore total signal bandwidth = max(50,50) = 50 Hz. Nyquist rate = 100 samples/second. Actually \(\textsinc(Bt) = \frac\sin(\pi B t)\pi B t\)

\noindent\textbf11. Compute convolution of \(x[n]=u[n]-u[n-3]\) and \(h[n]=u[n]-u[n-2]\). \\ \textitAns: \(y[n]=[1,2,2,1]\) for \(n=0..3\).