This problem deals with another signal operation, namely, scaling operation on a

Question

This problem deals with another signal operation, namely, scaling operation on a continuous time signal. The continuous time signal x(t) is given in problem 1(a) Let x_1(t) be a compressed version or an expanded version of the signal x(t). We can write a scaled version x_ (t) = x(at). x(t) = 2t exp(-1.5t) for -3 lessthanorequalto t lessthanorequalto 3 Use MATLAB to plot x (t) for t =-5 to 5 in steps of 0.1 x_1(t) is a compressed version of x(t) if the scaling factor a >1. The duration of the signal is reduced by a x_1(t) is an expanded version of x(t) if the scaling factor is 0

Explanation / Answer

function [ X ] = expsmooth( X, fs, tau ) % EXPSMOOTH Exponential smoothing. % % Y=EXPSMOOTH(X,FS,TAU) for a given input sequence X sampled % at FS Hertz and for time constant TAU given in milliseconds % returns exponentially smoothed output sequence Y. % % Inputs % X input signal as a column vector, or matrix % of input signals as column vectors. % % FS sampling frequency (Hz). % % TAU time constant (ms), i.e., the time it takes for % the step response of the exponential smoother % to reach 63.21% of its final (asymptotic) value, % i.e., 1-exp(-[TIME=TAU]/TAU)=0.6321. Similarly, % at TIME=5*TAU the output will reach 99.33% % of its final value [1]: 1-exp(-5)=0.9933. The % following inline functions may be useful: % % tau2alpha = @(TAU,T)( exp(-T/TAU) ); % alpha2tau = @(ALPHA,T)( -T/log(ALPHA) ); % ratesteady = @(TIME,TAU)( 1-exp(-TIME/TAU) ); % % where T is the sampling period (T=1/FS) and % ALPHA is a smoothing constant. % % Outputs % Y exponentially smoothed signal(s) as column vectors, % such that: % % Y(n) = ALPHA*Y(n-1) + (1-ALPHA)*X(n) % % where n is discrete-time index. % % Reference % [1] Greg Stanley, 2011, "Exponential filtering", % url: http://tinyurl.com/exponential-filtering % % Example % % define sampling frequency (Hz) % fs = 8E3; % % % define signal duration (s) % duration = 0.1; % % % get signal length (samples) % N = round( fs*duration ); % % % generate signal time vector (s) % time = [0:N-1]/fs; % % % generate input signals (sinusoids in noise) % X = 0.5*randn( N, 2 ) + [ sin(2*pi*50*time+pi/9); sin(2*pi*10*time+pi/3) ].'; % % % time constant (ms) % tau = 5; % % % apply exponential smoothing % Y = expsmooth( X, fs, tau ); % % % plot the results % figure('Position', [10 10 500 350], 'PaperPositionMode', 'auto', 'Visible', 'on', 'color', 'w'); % % subplot(211); % plot( time, X ); % xlabel( 'Time (s)' ); % ylabel( 'Amplitude' ); % title( 'Noisy signals' ); % set( gca, 'box', 'off' ); % axis( [min(time) max(time) min(X(:))-0.1*max(X(:)) 1.1*max(X(:))] ); % % subplot(212); % plot( time, Y ); % xlabel( 'Time (s)' ); % ylabel( 'Amplitude' ); % title( 'Smoothed signals' ); % set( gca, 'box', 'off' ); % axis( [min(time) max(time) min(Y(:))-0.1*max(Y(:)) 1.1*max(Y(:))] ); % % % print results to png file % print( '-dpng', 'expsmooth.png' ); % Author: Kamil Wojcicki, UTD, January 2012. % check for correct number of input arguments if( nargin~=3 ), error( sprintf('Incorrect number of input arguments. Type "help %s" for usage help. ', mfilename) ); end; % T sampling period (ms) T = 1E3/fs; % compute the smoothing constant alpha = exp( -T/tau ); % get number of column vector signals (M) and their lengths (N) [ N, M ] = size( X ); % for each column vector signal for m = 1:M % for each sample for n = 2:N % apply the exponential smoother [1] X(n,m) = alpha*X(n-1,m) + (1-alpha)*X(n,m); end end % EOF

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