DCSP-14. Jianfeng Feng Department of Computer Science Warwick Univ., UK [email protected] http://www.dcs.warwick.ac.uk/~feng/dsp.html. How to deal with noise? How to transmit signals?. Filter.
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DCSP-14
Jianfeng Feng
Department of Computer Science Warwick Univ., UK
http://www.dcs.warwick.ac.uk/~feng/dsp.html
How to deal with noise?
How to transmit signals?
a filter is a device or process that removes from a signal some unwanted component or feature.
Filtering is a class of signal processing, the defining feature of filters being the complete or partial suppression of some aspect of the signal.
Most often, this means removing some frequencies and not others in order to suppress interfering signals and reduce background noise.
y(n) = a0x(n)+a1x(n-1)+…+aNx(n-N)
+b1y(n-1)+…+bNy(n-N)
Nth order filter
signal
y(n) = a0x(n)+a1x(n-1)+…+aNx(n-N)
+b1y(n-1)+…+bNy(n-N)
Nth order filter
y(n) = [ x(n)+ … x(n-N) ] / N
In this case, when bi=0, it is an MA filter (averaging
of input signals)
In general it is called ARMA filter.
To find the coefficient a’s and b’s for certain purposes:
For example, filter out noise, stop certain band signals, allow certain band signal to pass etc.
w=y+.1*randn(length(y),1);
for i=10:length(y)
z(i)=(w(i)+w(i-1)+w(i-2)+w(i-3)+w(i-4)+w(i-5)+w(i-6)+w(i-7)+w(i-8)+w(i-9))/10;
end
sound(z)
sound(w)
{y(n)} = a0{x(n)}+a1{x(n-1)}+…+aN{x(n-N)}
+b1{y(n-1)}+…+bN{y(n-N)}
Multiplying z-n on both size of the equation above
where z is a complex number and summing over
n
Y(z) = a0X(z)+a1z-1X(z)+…+aNz-NX(z)
+b1z-1Y(z)+…+bNz-NY(z)
Y(z)-( b1 z-1 Y(z) +… + bN z-N Y(z))
= a0 X(z) + a1 z-1 X(z) +…+ aN z-NX(z)
[1-b1z-1-…- bN z-N ] Y(z)
= [ a0+ a1 z-1 +… + aN z-N ] X(z)
Y(z) ={ [ a0+a1z-1+…+aNz-N ] / [1-b1z-1-…-bNz-N] }
X(z)
= H(z) X(z)
H(z) is usually called transfer function:
it characterizes the input output relationship of a filter
When a filter is nonrecursive, its difference equation can be writteny(n) = a0 x(n) + a1 x(n-1) +…+ aN x(n-N)
Such filters are also called finite impulse response filters, for the obvious reason that their IR contain only finitely many nonzero terms.
Correspondingly, the ZT of a nonrecursive filter can be written as
a
z
Sometime we use h=a
Trying to figure out how an FIR filter will behave, is not always so simple.
Trying to figure out how an FIR filter will behave, is note always so simple.
Another way of looking at it is through its frequency domain behaviors.
Trying to figure out how an FIR filter will behave, is note always so simple.
Another way of looking at it is through its frequency domain behaviors.
We can make a start on this by examining the zeros of its transfer function H(z), i.e. those values of z for which
H(z)=0
since H(z) is a polynomial of order N with real coefficients,
Trying to figure out how an FIR filter will behave, is note always so simple.
Another way of looking at it is through its frequency domain behaviors.
We can make a start on this by examining the zeros of its transfer function H(z), i.e. those values of z for which
H(z)=0
since H(z) is a polynomial of order N with real coefficients,
It follows that the equation has N roots, which are either real or occur in complex conjugate pairs.
We can express H(z) in terms of the roots by writing
where in general
zm= |zm| exp ( j arg [zm] )
is the mth root, or zero of the transfer function.
The zeros of a transfer function are usually denoted graphically in the complex z-plane by circles, as shown in the following Fig.
We can express H(z) in terms of the roots by writing
where in general
zm= |zm| exp ( j arg [zm] )
is the mth root, or zero of the transfer function.
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.