The result shows that we can use digital method to

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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
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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:
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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.
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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
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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
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