Question

5. Consider the system in the figure below with X (12) = 0 for 221 > 20007, and the discrete-time system a squarer, i.e. yn = t[n]. What is the largest value of T such that y(t) = I.(t)? x[n] Hiscrete-time y[n] system D/C

Answer 1

(1) Lets define the continuous time frequency and discrete time frequency as

                                $\Omega = \text{continuous time frequency}\hspace{2mm}\& \hspace{2mm}\omega = \text{discrete time frequency}$

We will solve it by assuming that input is a sinusoid at frequency f_in such that

                                                            $x(t) = A\sin(2\pi f_{in}t)$

On sampling with sampling period T, we get the discrete time sequence as

                                                        $x(t) = A\sin[(2\pi f_{in}T)n]$

First condition will come from the Nyquist theorem as

                                                    $T < \frac{1}{2f_{in}} \hspace{10mm}(1)$

Since discrete time system output is squared of its input,

                      $y[n] = x^{2}[n] = \frac{A{2}}{2} - \frac{A{2}}{2}\cos[(4\pi f_{in}T)n]$

For no aliasing in y[n] from x[n],

                                                   $4\pi f_{in}T < \pi$

Or,

                                                  $T < \frac{1}{4f_{in}}\hspace{10mm}(2)$

If there is no aliasing in y[n], then y_c(t) is related to y[n] as

                             $y_{c}(t) = y[n = t/T] = \frac{A{2}}{2} - \frac{A{2}}{2}\cos(4\pi f_{in}t)$

Now, squaring x(t), we get

                    $x^{2}(t) = \left [ A\sin(2\pi f_{in}t) \right ]^{2} = \frac{A^{2}}{2} - \frac{A^{2}}{2}\cos(4\pi f_{in}t)$

So we find that if conditions (1) and (2) are satisfied, then

                                                    

So from (1) and (2), we find that maximum allowed value of T is

                                                                 $T_{max} = \frac{1}{4f_{in}}\hspace{10mm}(3)$

Since here it is given in the problem that x(t) is not a single sinusoid but it is band limited signal, so condition (3) should be satisfied for all frequencies content of x(t). From (2) and (3), it is found that we should select T_max for maximum frequency content of the signal. So,

                                                                   $T_{max} = \frac{1}{4f_{in,max}}$

f_in,max of the given signal = 2000*pi/(2*pi) = 1000Hz. So

                                                                  $T_{max} = \frac{1}{4000} = 250\mu s$