Speech Processing by Computer

 

LAB 2

DIGITAL SYSTEMS

 

This lab session demonstrates some simple linear systems operating upon a digital signal.  A program allows the input of the operating coefficients of a small linear system, displays the frequency response and replays a signal before and after processing.

 

1.         Amplification and Attenuation

 

            (i)         Acquire a speech signal at 20000 samples/second.

            (ii)        Use the 'ltitest' program to scale the input samples x[n] by 2.0 to increase the amplitude by 6dB.  That is, set y[n] = 2 * x[n].

            (iii)       Use the 'ltitest' program to scale the input samples x[n] by 0.5 to decrease the amplitude by 6dB.  That is, set y[n] = 0.5 * x[n].

 

2.         Non-recursive low-pass filtering

 

            (i)         Acquire a speech signal at 20000 samples/second.

            (ii)        Use the 'ltitest' program to output the average of the current sample x[n] with previous sample x[n-1].  That is y[n] = 0.5 * x[n] + 0.5 * x[n-1].  At what frequency is the zero?

            (iii)       Use the 'ltitest' program to output the average of the current sample x[n] with previous two samples x[n-1] and x[n-2].  That is y[n] = 0.33 * x[n] + 0.33 * x[n-1] + 0.33 * x[n-2].  At what frequency is the zero?

            (iv)       Use the 'ltitest' program to output the average of the current sample x[n] with previous three samples x[n-1], x[n-2] and x[n-3].  That is y[n] = 0.25 * x[n] + 0.25 * x[n-1] + 0.25 * x[n-2] + 0.25 * x[n-3].  At what frequencies are the zeros?

 

3.         Non-recursive high-pass filter

 

            (i)         Acquire a speech signal at 20000 samples/second.

            (ii)        Use the 'ltitest' program to output the difference between the current sample x[n] and previous sample x[n-1].  That is y[n] = x[n] - x[n-1].  At what frequency is the zero?

 

4.         Recursive implementation of a simple resonator

 

            The formulae for a simple resonator are as follows:

 

                        Take f = resonator frequency as fraction of sample rate

                        Take b = resonator bandwidth as fraction of sample rate

 

                        Then parameter r  =  (1-b/2)

 

                        and coefficient of y[n-1]  =  2 * r * cos(2*π*f)

                        and coefficient of y[n-2]  =  - r * r

 

            (i)         Acquire a speech signal at 20000 samples/second.

            (ii)        Use the 'ltitest' program to pass the signal through a simple resonator at a natural frequency of 2000Hz and a bandwidth of 500Hz.

            (iii)       Use the 'ltitest' program to pass the signal through a simple resonator at a natural frequency of 1000Hz and a bandwidth of 100Hz.  You may need to scale the input to stop overloading, how might you do this?