%TPOSTS  Determine transient characteristics of a response.
% [tp, os, ts] = tposts(t, y)
% Determines the peak time, overshoot, and
% settling time for a response vector.
% Inputs:
%  t: time vector
%  y: response vector
%   t and y must be of the same dimensions. If the response has not reached
%   steady state, there will be an error.  In addition, there must be at
%   least ~100 samples in the response vector.  Note that if there is no
%   overshoot in the response, the returned values for tp and os are
%   essentially meaningless.  Also note that this function only works when
%   the steady state value is a constant (e.g. a step response).
% Outputs:
%  tp: Peak time = the time at which the peak value in y occurs
%  os: Overshoot = percent difference between the peak and steady state values
%  ts: settling time = time at which steady state is reached (i.e. the
%                      response stays within 2% of the steady state value)
%
% Edited by Luke Reisner.  The author of the original version is unknown.

function [tp, os, ts] = tposts(t, y)

% Check for valid input vectors
[mt,nt] = size(t);
if (mt < nt) % If t is a row vector
    m = mt;
    mt = nt; % Swap indexes for the row vector
    nt = m;
end

[my,ny] = size(y);
if (my < ny) % If y is a row vector
    m = my;
    my = ny; % Swap indexes for the row vector
    ny = m;
end

if (mt ~= my)
    error('Input vectors are not of same size.');
end
if (nt ~= ny)
    error('Input vectors are not of same size.');
end

if (nt ~= 1) % If time variable is a matrix, not a vector (N by 1 or 1 by N)
    error('t must be a vector, not a matrix.');
end

m5pc = ceil(mt/20);  % Determine the number of samples / 20
%if (m5pc < 10) % If not at least ~200 samples
if (m5pc < 5) % If not at least ~100 samples
    error('Not enough data points.');
end

if (y(my) < 0)
    y = -y;  % Invert the signal if its last sample (steady state) is negative
end

for i=1:m5pc
    if (abs(y(mt-i)-y(mt)) / y(mt) >= 0.001)
        % If output y is not very steady by the end of the signal
        error('Steady state may not have reached.');
    end
end

% Find the peak time and overshoot
yss = y(mt);  % Steady state = last sample in the y vector (assumes steady state is dc)
[yp,kp] = max(y);  % Peak value = max value in y
os = (yp-yss) / yss;  % Overshoot (os) = percent difference between the peak and steady state values
tp = t(kp);  % Peak time (tp) = the time at which the peak value in y occurs

% Find the settling time via the following procedure:
%  Working backwards from the end of the response vector, find the first
%  sample whose value is more than 2% different from the steady state value.
%  Then select the subsequent sample, which will be <= 2% different from the
%  steady state value.  This sample marks the beginning of the steady state.
k = mt;
while (abs((y(k)-yss) / yss) <= .02)
    k = k-1;
end
k = k+1; % to bring back y(k) within 2% of the steady state
ys = y(k); % ys = first sample in the final group of samples
           % that are within +/- 2% of the steady state value.
           % i.e. it is the first sample that can be considered steady
           % enough to be steady state
ts = t(k); % settling time (ts) = time at which the first sample of
           % the steady state portion of the signal occurs