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I have a signal $x(t) = \sin(2\pi t)$ with $t \in [0,1]$. The signal $x_s(t)$ is the sampled signal with a sampling period of $T_s = 0.05$. Now I want to compute the reconstructed $x_r(t)$ signal using a zero-order-hold.

The ZOH has the following impulse response

$$h_{\mathrm{ZOH}}(t,T_s)\,= \frac{1}{T_s} \mathrm{rect} \left(\frac{t}{T_s}-\frac{1}{2} \right)$$

This I implemented in Matlab:

zohImpl = @(t,Ts) 1/Ts*rectangularPulse(t/Ts - 1/2);

Now I simply did a convolution between the impulse response $h_{\mathrm{ZOH}}$ and the sampled signal $x_s(t)$

xZoh = conv(zohImpl(t,Ts),xSampled);

However I do not get the result expected, the reconstructed signal. I get a sinus with a higher amplitude and in the convolution, xZoh gets signal points padded which are 0.

What am I doing wrong here? I would namely expect the same results as which I obtain using...

k = 0; for i = 1:length(t) if k*Ts <= t(i) && t(i) < (k + 1)*Ts k = k + 1; end xZoh(i) = xSampled(k); end

Full code:

clear all; close all; clc; % Impulse ZOH zohImpl = @(t,Ts) 1/Ts*rectangularPulse(t/Ts - 1/2); % Impulse FOH fohImpl = @(t,Ts) 1/Ts*triangularPulse(t/Ts); % Impulse FOH delayed fohDelayImp = @(t,Ts) 1/Ts*triangularPulse((t - Ts)/Ts); % Impulse FOH predictive fohPreImp = @(t,Ts) 1/Ts*(rectangularPulse(t/Ts - 1/2) - ... rectangularPulse(t/Ts - 3/2) + ... triangularPulse(t/Ts - 1)); % Time domain tStart = 0; tStep = 1e-4; tEnd = 1; t = tStart:tStep:tEnd; % Original signal x = @(t)(sin(2*pi*t)); % Sampling period Ts = 5e-2; % Samples nSamples = tEnd/Ts; samples = 0:nSamples; % Sampled signal xSampled = x(samples*Ts); % Convolution with impulse response xZoh1 = conv(zohImpl(t,Ts),xSampled); % Plot figure(4); hold all; plot(t,x(t)); stem(samples*Ts,xSampled); % plot(t,xZoh1); does not work k = 0; for i = 1:length(t) if k*Ts <= t(i) && t(i) < (k + 1)*Ts k = k + 1; end xZoh2(i) = xSampled(k); end plot(t,xZoh2);

Created signal using convolution (plot(xZoh1)): enter image description here

Created signal using for loop (plot(xZoh2)): enter image description here

latest code

clear all; close all; clc; % Time domain tStart = 0; tStep = 1e-4; tEnd = 1; t = tStart:tStep:tEnd; beta = 0.2; % Original signal x = @(t)(sin(2*pi*t)); % Sampling period Ts = 5e-2; % Impulse ZOH zohImpl = @(t,Ts) 1/Ts*rectangularPulse(t/Ts - 1/2); % Impulse FOH fohImpl = @(t,Ts) 1/Ts*triangularPulse(t/Ts); % Impulse FOH delayed fohDelayImp = @(t,Ts) 1/Ts*triangularPulse((t - Ts)/Ts); % Impulse FOH predictive fohPreImp = @(t,Ts) 1/Ts*(rectangularPulse(t/Ts - 1/2) - ... rectangularPulse(t/Ts - 3/2) + ... triangularPulse(t/Ts - 1)); % Impulse FOH fractional fohFractImp = @(t,Ts,beta) 1/Ts*(rectangularPulse(t/Ts - 1/2) - ... beta*rectangularPulse(t/Ts - 3/2) + ... beta* triangularPulse(t/Ts - 1)); tt = -1:1e-4:1; figure(1); subplot(5,1,1); plot(tt,zohImpl(tt,Ts)); grid on; box on; axis tight; subplot(5,1,2); plot(tt,fohImpl(tt,Ts)); grid on; box on; axis tight; subplot(5,1,3); plot(tt,fohDelayImp(tt,Ts)); grid on; box on; axis tight; subplot(5,1,4); plot(tt,fohPreImp(tt,Ts)); grid on; box on; axis tight; subplot(5,1,5); plot(tt,fohFractImp(tt,Ts,beta)); grid on; box on; axis tight; % Samples nSamples = tEnd/Ts; samples = 0:nSamples; % Sampled signal xSampled = zeros(1,length(t)); xSampled(1:Ts/tStep:end) = x(samples*Ts); % Convolution with impulse response xZoh = conv(zohImpl(t,Ts),xSampled); xFoh = conv(fohImpl(t-t(Ts/tStep),Ts),xSampled); xFohDelayImp = conv(fohDelayImp(t,Ts),xSampled); xFohPreImp = conv(fohPreImp(t,Ts),xSampled); xFohFractImp = conv(fohFractImp(t,Ts,beta),xSampled); xZoh = xZoh(1:length(t)) / nSamples; xFoh = xFoh(1:length(t)) / nSamples; xFohDelayImp = xFohDelayImp(1:length(t)) / nSamples; xFohPreImp = xFohPreImp(1:length(t)) / nSamples; xFohFractImp = xFohFractImp(1:length(t)) / nSamples; % Plot figure(2); hold all; plot(t,x(t)); % stem(t,xSampled); plot(t,xZoh); plot(t-t(Ts/tStep),xFoh); plot(t,xFohDelayImp); plot(t,xFohPreImp); plot(t,xFohFractImp);
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  • $\begingroup$ can you add your plots? That would save us needing to run your code to understand what's happening. $\endgroup$ Commented Apr 16, 2017 at 17:57
  • $\begingroup$ @MarcusMüller added. $\endgroup$ Commented Apr 16, 2017 at 18:23
  • $\begingroup$ sorta tl;dr (i'm too lazy to read code), but you plot looks correct for a ZOH. what's wrong?? $\endgroup$ Commented Apr 16, 2017 at 18:45
  • $\begingroup$ @robertbristow-johnson I wanted to create the output using convolution not by the for-loop implementation. $\endgroup$ Commented Apr 16, 2017 at 18:53
  • $\begingroup$ and you convolved with a rectangular pulse of the same width as your step? $\endgroup$ Commented Apr 16, 2017 at 18:55

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There are a few issues:

  • The type of impulse we need here is not a discrete delta but a Dirac delta.

  • The type of convolution we need is a continuous not a discrete one.

  • If you want to implement this on a computer, the rectangular pulse should also be sampled with a sampling rate higher than $1/T_s$ (e.g. 1/tStep).

An easy way to address the above issues is to use kron function and sample the rectangular pulse with an appropriate rate. Simply replace the line

xZoh1 = conv(zohImpl(t,Ts),xSampled);

with the following

xZoh1 = kron(xSampled, ones(1,Ts/tStep)); % Clip it to adjust the length xZoh1 = xZoh1(1:length(t));

enter image description here

Note that the scaling by $T_s$ in the formula you provided is due to time-scaling the Dirac delta function and is dropped here.

If you want to use conv, then the code should be changed into something like 

clear all; close all; clc; % Time domain tStart = 0; tStep = 1e-4; tEnd = 1; t = tStart:tStep:tEnd; % Original signal x = @(t)(sin(2*pi*t)); % Sampling period Ts = 5e-2; % Impulse ZOH zohImpl = ones(1,Ts/tStep); % Samples nSamples = tEnd/Ts; samples = 0:nSamples; % Sampled signal xSampled = zeros(1,length(t)); xSampled(1:Ts/tStep:end) = x(samples*Ts); % Convolution with impulse response xZoh1 = conv(zohImpl,xSampled); xZoh1 = xZoh1(1:length(t)); % Plot figure(4); hold all; plot(t,x(t)); stem(t,xSampled); plot(t,xZoh1);
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  • $\begingroup$ I use a 'higher' sampling rate, see tStep and Ts, Ts is the sample rate for the sampling the sinusoid and tStep is used to create the "original" sinusoid. I also use it on the convolution for the impulse response. Also I would prefer to use conv because I also wanted to zohImpl, fohImpl, et cetera. However, am I to understand that this is not possible? It is not clear to me why not. Currently, I have the feeling that I am using rectangularPulse incorrect, see my comment towards @robert bristow-johnson $\endgroup$ Commented Apr 16, 2017 at 22:52
  • $\begingroup$ I see what you do. You can also use conv (maybe it is a little more tricky). In such case, You should stuff enough zeros in between the samples (i.e. in xSampled) and also sample the rectangular pulse with rate 1/tStep. Just make sure you understand the points in my answer. $\endgroup$ Commented Apr 16, 2017 at 23:23
  • $\begingroup$ I added a code based on conv. You can use something similar for FOH as well. $\endgroup$ Commented Apr 16, 2017 at 23:42
  • $\begingroup$ Still not what I wanted, because I want to use the zohImpl. You are now just shifting your first solution more or less. However, based on your solution I altered my code. It is now working for all methods, accept for the basic FOH. This is because it is acasual. So therefor I shifted the time signal in both the xFoh = conv(fohImpl(t-t(Ts/tStep),Ts),xSampled); and the plot(t-t(Ts/tStep),xFoh);. I am not sure however if this is actually also correct because it is actually simply equal to the xFohDelayImp but shifted manually. $\endgroup$ Commented Apr 17, 2017 at 8:50
  • $\begingroup$ The FOH impulse is [linspace(0,1,Ts/tStep-tStep) linspace(1,0,Ts/tStep)];. If you want to implement the non-causal FOH, you should compensate a duration of one $T_s$ at the beginning of xSampled. Then it becomes just like the others. $\endgroup$ Commented Apr 17, 2017 at 10:01

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