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DCSP-17 Jianfeng Feng Department of Computer Science Warwick Univ., UK

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Frequency Response of an MA filter We can formally represent the frequency response of the filter by substituting z = exp( j ) and obtain z N H( )=A( ) =K[(exp (j ) – 1 )…(exp (j ) – N )] Consider an example

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Assume that H( )= K[(exp (-j ) – 1 )(exp (-j ) – 2 )] with 1 = exp (-j ), we then have Y( )= H( ) X( ) and Y( )=0 whenever = Therefore, any signal with a frequency of will be stopped

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h=0.01; for i=1:314 x(i)=i*h; f(i)=(exp(-j*x(i))-exp(-j*pi/4))*(exp(-j*x(i))- exp(j*pi/4)); end plot(x,abs(f))

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Moving 1 along the circle, we are able to stop a signal with a given frequency The original difference expression can be recovered by H( z ) = k [(z -1 – 1 )(z -1 – 2 )] = k [z -2 – ( ) z -1 + ( 1 2 )] y(n) = k [x(n-2) – ( ) x(n-1) + ( 1 2 )x(n)] = Matlab/work/simple_filger_design_ma.m

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Recursive Filters Of the many filter transfer function which are not FIR, the most commonly use in DSP are the recursive filters, so called because their current output depends not only on the last N inputs but also on the last N outputs. y(n) = a(0)x(n)+ a(1) x(n-1) +…+ a(N) x(n-N) + b(1) y(n-1) +…+ b(N)y(n-N)

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Transfer function

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Block diagram X(n) a(N) D D a(N-1) + a(0) y(n) + b(N) b(N-1) a(1) b(1) + D +

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Poles and zeros We know that the roots n are the zeros of the transfer function. The roots of the equation B(z)=0 are called the poles of the transfer function. They have greater significance for the behaviour of H(z): it is singular at the points z= n. Poles are drawn on the z-plane as crosses, as shown in the next Fig.

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ON this figure, the unit circle has been shown on the z- plane. In order for a recursive filter to be bounded input bounded output stable, all of the poles of its transfer function must lie inside the unit circle. A filter is BIBO stable if any bounded input sequence gives rise to a bounded output sequence. Now if the pole of the transfer lie insider the unit circle, then they represent geometric series with a coefficient whose magnitude | m |<1, i.e. a sequence {1, m, ( m ) 2, … } which is convergent.

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Consequently, the filter output, which is a sum of such sequences weighted by the appropriate input terms, is bounded if the input is. If, on the other hand, | m |>1, the geometric series diverges and the filter output will grow without bound as n is large enough. If | m |=1, the filter is said to be conditionally stable: some input sequence will lead to bounded output sequence and some will not. Since FR filters have no poles, they are always BIBO stable: zeros have no effect on stability.

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Poles, Zeros and Frequency Response Now suppose we have the ZT transfer function of a filter We can formally represent the frequency response of the filter by substituting z = exp( j ) and obtain H( )= A( ) / B( )

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Poles, Zeros and Frequency Response Obviously, H( ) depends on the locations of the poles and zeros of the transfer function, a fact which ca n be made more explicit by factoring the numerator and denominator polynomials to write

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Each factor in the numerator or denominator is a complex function of frequency, which has a graphical interpretation in the rms of the location of the corresponding root in relation to the unit circle Thus we can make a reasonable guess about the filter frequency response imply by looking at the pole-zero diagram.

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Filter Types There are four main classes of filter in widespread use: lowpass, highpass, bandpass bandstop filters. The name are self-explanatory, but he extent to which the ideal frequency responses can be achieved in practice is limited. This four types are shown in the next figure

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