# The result shows that we can use digital method to

This preview shows pages 24–29. Sign up to view the full content.

The result shows that we can use digital method to realize the desired frequency response. Also, we can use the sampling rate to control the frequency response as well. Example Consider the 11-point running-average filter whose frequency response is 5 ˆ ˆ ) 2 / ˆ sin( 11 ) 2 / 11 ˆ sin( ) ( ϖ ϖ ϖ ϖ j j e e H - = . The discrete-time system is operating at a sampling frequency of 1000Hz and the analog input is ] ) 250 ( 2 sin[ ] ) 25 ( 2 cos[ ) ( t t t x π π + = Find the output. As the sampling rate is above the Nyquist rate (the minimum sampling rate), there is no aliasing in the sampled signal. Hence the frequency response of each sinusoid in the output is respectively given by

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
1000 / ) 5 )( 25 ( 2 1000 / ) 25 ( 2 ) 1000 / ) 25 ( sin( 11 ) 1000 / ) 11 )( 25 ( sin( ) ( π π π π j j e e H - = = 4 / 8811 . 0 π j e - 1000 / ) 5 )( 250 ( 2 1000 / ) 250 ( 2 ) 1000 / ) 250 ( sin( 11 ) 1000 / ) 11 )( 250 ( sin( ) ( π π π π j j e e H - = = 2 / 0909 . 0 π j e - It is noted that the frequency response at 25Hz has larger gain than that at 250Hz (check the gains from the plot of the frequency response). The final output is ] 2 / ) 250 ( 2 sin[ 0909 . 0 ] 4 / ) 25 ( 2 cos[ 8811 . 0 ) ( π π π π - + - = t t t y . )] 200 / 1 )( 250 ( 2 sin[ 0909 . 0 )] 200 / 1 )( 25 ( 2 cos[ 8811 . 0 - + - = t t π π It is interesting to observe that there is a time delay induced by the equivalent continuous-time frequency response. The value of the time delay is the linear phase of the filter given by 2 / ) 1 ( - L T s ϖ . Example: Design a digital signal processing system with the sampling rate of 1 kHz to remove the sinusoid of 25 Hz in the above example. Solution:
The digital signal processing system samples the input signal at the rate of 1 kHz. The discrete-time signal of the input is ] 1000 / 250 2 sin[ ] 1000 / 25 2 cos[ ] [ n n n x × + × = π π ) 5 . 0 sin( ) 05 . 0 cos( n n π π + = The component of ) 05 . 0 cos( n π can be removed by a second order symmetric filter with the input-output equation ] 2 [ ] 1 [ ] [ ] [ 0 1 0 - + - + = n x a n x a n x a n y by setting the frequency response at π ϖ 05 . 0 ˆ = equal to zero. The frequency response is ϖ ϖ ϖ ˆ 2 0 ˆ 1 0 ˆ ) ( j j j e a e a a e H - - + + = )] 05 . 0 cos( ) ˆ cos( 2 [ 1 0 ˆ π ϖ ϖ a a e j + = - We have two unknowns. We can choose a 0 = 1 and set 0 ) ( 05 . 0 = π j e H It turns out ) 05 . 0 cos( 2 1 π - = a . The digital filter is ] 2 [ ] 1 [ ) 05 . 0 cos( 2 ] [ ] [ - + - - = n x n x n x n y π referred to as nulling filter.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
The output of the digital filter to ) 5 . 0 sin( ) 05 . 0 cos( ] [ n n n x π π + = is )) ( 5 . 0 cos( | ) ( | ] [ 5 . 0 5 . 0 π π π j j e H n e H n y + = The frequency response at π ϖ 5 . 0 ˆ = is π π π π π 5 . 0 5 . 0 5 . 0 975 . 1 )] 05 . 0 cos( ) 5 . 0 [cos( 2 ) ( j j j e e e H = - = - Thus the digital output is ) 5 . 0 5 . 0 sin( 975 . 1 ] [ π π + = n n y The output of the processing system is ) 5 . 0 250 2 sin( 975 . 1 ) ( π π + × = t t y
Suggested Problems (1) Determine the DTFT of each of the following sequences (a) ] 5 [ ] [ ] [ - - = n u n u n x (b) 1 | | ]), 8 [ ] [ ( ] [ < - - = α α n u n u n x n (c) 1 | | ]), 8 [ ] [ ( ] [ < - - = α α n u n u n n x n (2) Show that the inverse DTFT of 1 | | , ) 1 ( 1 ) ( ˆ ˆ < - = - α α ϖ ϖ m j j e e X is given by ] [ )! 1 ( ! )! 1 ( ] [ n u m n m n n x n α - - + = . (3) Let ) ( ˆ ϖ j e X denote the DTFT of a length-9 sequence ] [ n x given by 6 2 }, 4 , 2 , 1 , 3 , 4 , 0 , 1 , 3 , 2 { ] [ - - - = n n x Evaluate the following functions of ) ( ˆ ϖ j e X without computing the transform itself: (a) ) ( 0 j e X , (b) ) ( π j e X , (c) - π π ϖ ϖ ˆ ) ( ˆ d e X j , (d) - π π ϖ ϖ ˆ | ) ( | 2 ˆ d e X j (4) A linear time-invariant system has frequency response ) 1 )( 1 )( 1 ( ) ( ˆ ˆ 2 / ˆ 2 / ˆ ϖ ϖ π ϖ π ϖ j j j j j j e e e

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business â€˜17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. Itâ€™s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania â€˜17, Course Hero Intern

• The ability to access any universityâ€™s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLAâ€™s materials to help me move forward and get everything together on time.

Jill Tulane University â€˜16, Course Hero Intern