Example consider an fir system with the impulse

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Example Consider an FIR system with the impulse response coefficients
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{ k b }={1, - 1, 1}. The frequency response is given by ϖ ϖ ϖ ˆ 2 ˆ ˆ 1 ) ( j j j e e e H - - + - = otherwise for e e j j 3 / | ˆ | ) ˆ cos 2 1 ( ) 1 ˆ cos 2 ( ) ˆ ( ˆ π ϖ ϖ ϖ π ϖ ϖ < - - = + - - Find the response to ) 21 20 cos( 3 ) 2 3 cos( 3 4 ] [ n n n x π π π + - + = The input has frequencies at 0, 3 / π , and 21 / 20 π . The frequency responses at the input frequencies are 1 1 2 ) ( 0 = - = j e H ; 0 ) 1 1 ( ) ( 3 / 3 / = - = - π π j j e e H 21 / ) 21 / 41 ( 21 20 ) ( 98 . 2 ) 9777 . 1 1 ( )) cos( 2 1 ( ) ( 21 20 21 20 π π π π π π j j j j e e e e H = + = - = - + - + + - + = )] ( 2 3 cos[ 3 | ) ( | 4 ) ( ] [ 3 3 0 π π π π j j j e H n e H e H n y ) ( 21 20 cos( 3 | ) ( | 21 20 21 20 π π π j j e H n e H + ) 21 21 20 cos( 3 98 . 2 ] 3 2 3 cos[ 3 0 4 1 π π π π π + × + - - × + × = n n ) 21 21 20 cos( 94 . 8 4 π π + + = n
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Design a Simple FIR Filter A simple digital filter can be designed to give the desired gains to the specified sinusoidal signals. The following example is used to illustrate the design of a digital filter for the given frequency specification: The input signal consists of a sum of two sinusoidal signals of normalized radian frequencies π ϖ 1 . 0 ˆ 1 = and π ϖ = 2 ˆ , respectively. We need to design a highpass filter that will pass the high frequency component of the input but block the low frequency part. To solve this problem, we need at least a second order digital filter with the difference equation as ] 2 [ ] 1 [ ] [ ] [ 2 1 0 - + - + = n x a n x a n x a n y We set the frequency response as: 0 | ) ( | 1 . 0 ˆ ˆ = = π ϖ ϖ j e H and 1 | ) ( | ˆ ˆ = = π ϖ ϖ j e H , We choose a symmetric filter; i.e. 2 0 a a = because the design is simple and the magnitude response is much easily satisfied. The frequency response of the filter is ] ) ( [ ) ( 1 ˆ ˆ 0 ˆ ˆ 2 0 ˆ 1 0 ˆ a e e a e e a e a a e H j j j j j j + + = + + = - - - - ϖ ϖ ϖ ϖ ϖ ϖ ] ) ˆ cos( 2 [ 1 0 ˆ a a e j + = - ϖ ϖ The two constraints give the following two equations. 0 902 . 1 0 ) 1 . 0 cos( 2 1 0 1 0 = + = + a a a a π
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1 2 1 ) cos( 2 1 0 1 0 = + - = + a a a a π The two equations give 256 . 0 0 - = a , 4874 . 0 1 = a The filter is given by ] 2 [ 256 . 0 ] 1 [ 4874 . 0 ] [ 256 . 0 ] [ - - - + - = n x n x n x n y -3 -2 -1 0 1 2 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 frequency magnitude response
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Digital processing of continuous-time signals After sampling the continuous-time signal, generally, the sequence will be processed by a digital filter as shown in the following figure Assume the input is a complex sinusoid ) ( ) ( φ ϖ + = t j Ae t x where the input frequency is less than half of the sampling frequency (or π ϖ < | ˆ | ). It means there is no aliasing in the sampling. The digital system has the frequency response ) ( ˆ ϖ j e H . The output of the digital system is given by ) ˆ ( ˆ ) ( ] [ φ ϖ ϖ + = n j j Ae e H n y Since there is no aliasing in the frequency, the continuous-time sinusoidal signal can be reconstructed by the DAC converter by setting s T ϖ ϖ = ˆ and s nT t = we have ) ( ) ( ) ( φ ϖ ϖ + = t j T j Ae e H t y s . Ideal ADC LTI System Ideal DAC ) ( t x s T ] [ n x ] [ n y s T ) ( t y
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The output has exactly the original signal under the desired processing as long as there is no aliasing.
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