The Impulse Signal A signal I(k), which is defined as zero everywhere except at a single point, where its value is equal to 1 I = [0 0 1 0 0 0]; stem(i)
Filter Impulse through Feedforward Filter 0 0.25.5.25 0 b= [.25.5.25]
The Impulse Response The response of a filter to an impulse signal is referred to as the impulse response of that filter. In the case of a feedforward filter, the impulse response will be the sequence of filter coefficients themselves! This gives us a new interpretation of the filter coefficient vector b it is, in fact, the time signal produced by the filter in response to an impulse.
Given the definition of a feedforward filter: Y(k) = b(1)*x(k) + b(2)*x(k-1) +... + b(m)*x(k-m+1) why should we expect this impulse response? Implications: The number of non-zero points in the response will be equal to the number of coefficients. The length of the impulse response will always be finite. Filters of this type (feedforward) are also known as Finite Impulse Response filters, or FIR filters. If we are handed a feedforward filter with an unknown transfer function (a black box), we can characterize it precisely by seeing how it behaves to the impulse response.
Feedforward (FIR) Filters... a review Three salient properties of a filter of this kind: there are no feedback pathways output is a finite sequence if the input is a finite sequence the transfer function H(z) can be represented as polynomial in negative powers of z. Advantages: simplicity: designing and building one is easy phase: it is possible to keep phase linear, which is desirable for some (mainly recording) applications
Feedback Filters (IIR) Feedback: the value of the output sample value at a given point, Y(k), is dependent on previous values of the output. Three important properties of a filter of this kind: output is an infinite sequence even if the input is a finite sequence there are feedback pathways from the output to input the transfer function H(z) can be represented as a ratio of two polynomials in negative powers of z
X(k) * Y(k) D b(1) * (FIR) D b(2) * b(3) X(k) Y(k) D * -a(1) * D (IIR) -a(2)
A Simple Feedback (IIR) Filter Y(k) = X(k) + 0.5*Y(k-1) Impulse Impulse Response
Feedback filter is a dynamical system I Equation (or set of equations) that quantitatively describes the change in something over time. I Two parts: 1. State of the system (quantity whose change is being described) I amount of money in the bank I number of individuals in a population I volume of water in a bathtub I activation level of a neuron I the concentration of Sodium in a solution I the speed of tra conthe101 I the temperature in a room 2. Rule for how the state changes, depending on the current state. Current state predicts the state at the next instant in time. I System is constant, even while state is changing.
Example Dynamical System I State: Amount of water in bathtub (x) I Rule for change: Change in x = 1 2 x Time H20 1 100 2 50 3 25 4 12.5 Water in bathtub 100 90 80 70 60 50 40 30 20 10 5 6.25 0 0 5 10 15 Time x n+1 = x n + dx dx = kx x n+1 = x n + kx n x n+1 =(1+k)x n if k=-.5 x n+1 =.5x n x n =.5x n 1 y n =.5y n 1
Example Dynamical System I State: Amount of water in bathtub (x) I Rule for change: Change in x = 1 2 x Time H20 1 100 2 50 3 25 4 12.5 Water in bathtub 100 90 80 70 60 50 40 30 20 10 5 6.25 0 0 5 10 15 Time x n+1 = x n + dx dx = kx x n+1 = x n + kx n if k=-.5 x n+1 =.5x n if k=-1.5 x n+1 =.5x n x n+1 =(1+k)x n x n =.5x n 1 y n =.5y n 1 y n =.5y n 1
Simple Feedback (IIR) Filters Y(k) = X(k) - 0.5*Y(k-1) * Note that because of the negative sign, the system oscillates as it moves toward its asymptote
Transfer Functions for Simple (IIR) Filters Take the simple feedback filter: (3) Y = X -.5z-1Y Now rearrange terms: (4) X = Y+.5z-1Y X = (1 +.5z-1)Y The transfer function H(z) is the ratio of the the output to the input: (5) H(z) = Y = 1. X 1+ 0.5z -1
Amplitude Response for Simple (IIR) Filters Y(k) = X(k) - 0.5*Y(k-1) H(z) = Y = 1 X 1+ 0.5z -1 Impulse Response Transfer Function
Amplitude Response for Simple (IIR) Filters Y(k) = X(k) + 0.5*Y(k-1) H(z) = Y = 1. X 1-0.5z -1 Impulse Response Transfer Function
Transfer Functions for Simple (IIR) Filters Filter Equation Y = X -.5z -1 Y Transfer Function: H(z) = Y = 1. X 1+ 0.5z -1 Thus, just like moving average filters, the transfer function of this IIR filter involves a polynomial of terms in negative powers of z. In this case, however, the polynomial is the denominator of a ratio. The numerator equals 1 in this case, because there are no terms that delay and weight the input sequence (i.e. feedforward terms). If there were such terms there would be another polynomial in z in the numerator, derived exactly as we have before for moving average filters.
General Expression for Feedback (IIR) Filters Recall that the general expression that we derived for a moving average filter was as in (7): (7) Y(k) = b(1)*x(k) + b(2)*x(k-1) +... + b(m)*x(k-m+1) For a general recursive filter, including terms that can depend on both past input and past output, the expression is in (8): (8) Y(k) = b(1)*x(k) + b(2)*x(k-1) +... + b(m)*x(k-m+1) -a(1)*y(k-1) - a(2)*y(k-2) -... - a(l)*y(k-l) By convention, the terms that depend on previous output are subtracted, rather than added (if you want the contribution of the term to be positive you can, of course, chose a negative value for a).
Filtering in Matlab The filter function in MATLAB can be used to filter an input signal through a feedforward filter, a feedback filter, or both. Y = filter(b,a,x) % Y filtered signal % b vector of feedforward coefficients, numerator of transfer function % a - vector of feedback coefficients, denominator of transfer function % X input signal To filter with feedforward only, set a=1 filter([0.5 0.5], 1, X) To filter with feedback only, set b=1 filter(1, [1 0.5], X) * Note that the first element of the vector of a coefficients must be one. This is the coefficient of the undelayed output on the output. If you think about it, it has to be 1. So a(1) in equation (8) will be the second element of the a vector, a(2) the third element, etc.
General Expression for H(z) of a Feedback (IIR) Filter Using simple algebra like we employed in (3)-(5) above we can derive the following expression for the transfer function of a recursive filter: (9) H(z) = Y = b(1) + b(2)z -1 +... + b(m)z -(M-1). X 1+ a(1)z -1 +... + a(l)z -L Note that the extra 1 that is in the a vector shows up in the denominator here.
Poles H(z) = Y = b(1) + b(2)z -1 +... + b(m)z -(M-1). X 1+ a(1)z -1 +... + a(l)z -L The denominator polynomial will have roots, just as the numerator does. The roots of the denominator are values of z at which H(z) gets very large (possibly infinite). These roots are the poles of the filter.
function [mag phase] = transfer(b,a) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Calculate the Transfer Function %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % first find the number of coefficients (filter order) m = length(b); n = length(a); % fill a vector of test frequencies to plot nfreqs = 100; w = linspace(0, pi, nfreqs); % Compute a vector of z from w; % This will result in a vector of z's z = exp(j*w); % Create a vector Hnum that will eventually contain % the transfer function. % This first one will be for the numerator Hnum = zeros(1,nfreqs); % Compute Hnum by summing the terms of the polynomial in z for i=1:m Hnum = Hnum + b(i).* z.^ (-(i-1)); end
% Create a vector Hnum that will eventually contain % the transfer function. % This first one will be for the numerator Hden = zeros(1,nfreqs); % Compute Hnum by summing the terms of the polynomial in z for i=1:n Hden = Hden + a(i).* z.^ (-(i-1)); end H = Hnum./Hden; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Plot the results %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %Magnitude (M) and Phase (P) outputs mag = abs(h); phase = angle(h); %Plot the Magnitude response subplot(3,1,1); plot(w/pi,mag); grid on; xlabel('frequency (pi radians)'); ylabel('magnitude');
%Plot the Phase response subplot(3,1,2); plot(w/pi,phase*180/pi); grid on; xlabel('frequency (pi radians)'); ylabel('phase (degrees)'); %Plot the zeros subplot(3,1,3); z = roots(b); scatter(real(z),imag(z),'ro'); hold on; t = 0:pi/32:2*pi; plot(cos(t),sin(t),'b'); axis('square'); axis([-1 1-1 1]); grid on; %Plot the poles p = roots(a); scatter(real(p),imag(p),'r+'); hold on; t = 0:pi/32:2*pi; plot(cos(t),sin(t),'b'); hold off return