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My original signal as

f1=2;
f2=5;

fs=100;
Ts=1/fs;
t=0:Ts:1;

xt=cos(2*pi*f1*t)+cos(2*pi*f2*t);

figure
plot(t,xt)

as shown below figure.

enter image description here

and my sampled signal with fs=12 which is greater than two times of maximum frequency, 2*f2=10;

fs=12;
Ts=1/fs;
tn=0:Ts:1;

xn=cos(2*pi*f1*tn)+cos(2*pi*f2*tn);

enter image description here

and then I want to reconstruct the original signal with sampled signal.

I tried it using the matlab function interp1 as below

xr=interp1(tn,xn,t,'spline');

I compared original signal and reconstructed signal, but they are different as shown below figure.

enter image description here

How to reconstruct original signal?

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Sinc interpolation can exactly reconstruct an above-Nyquist-sampled strictly bandlimited signal from noiseless samples. See the Whittaker-Kotelnikov-Shannon reconstruction or resampling theorem: https://en.wikipedia.org/wiki/Whittaker–Shannon_interpolation_formula and https://ccrma.stanford.edu/~jos/resample/Theory_Ideal_Bandlimited_Interpolation.html

For computation, you can try using a windowed Sinc (or other near-brick-wall linear phase low pass filter) for a more reasonable finite length interpolation kernel. For any finite length of samples, both the band-limit and the low-pass reconstruction filter's stop-band need to be strictly below (NOT equal to) half the sampling rate. Due to finite window length, and finite-precision floating point sampling and arithmetic, you may see some end effects and other reconstruction differences.

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  • $\begingroup$ Thank you so much. $\endgroup$
    – agile
    May 17 at 7:13
  • $\begingroup$ For ideal Sinc interpolation you need an infinite number of samples. A windowed Sinc allows you to use a finite set of samples, but you are going to have end effects and the beginning and end of your data i.e. where you run out of data. $\endgroup$
    – David
    May 17 at 14:18
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To implement a digital-to-analog converter, all you need is an ideal low-pass filter to filter out the periodic frequency response that $\varOmega \geq \varOmega_s/2$.

$$ H(j\varOmega) =\left\{ \begin{aligned} T,\ \ \ \ |\varOmega|<\varOmega_s/2 \\ 0,\ \ \ \ |\varOmega|\geq\varOmega_s/2 \end{aligned} \right. $$

The impulse response of an ideal low-pass filter is a sinc function

$$ h(t) = \frac{\sin(\pi t/T)}{\pi t/T} $$

Convolving the sampled signal with this impulse response gives the analog signal.

$$ \begin{aligned} y_a(t) =& \sum_{n=-\infty}^{\infty} x_a(nt) h(t-nT)\\ =&\sum_{n=-\infty}^{\infty} x(n) \frac{\sin[\pi (t-nT)/T]}{\pi (t-nT)/T} \end{aligned} $$

f1=2;
f2=5;

fs=100;
Ts=1/fs;
t=0:Ts:1;
xt=cos(2*pi*f1*t)+cos(2*pi*f2*t);
figure
plot(t,xt)

fs=12;
Tn=1/fs;
tn=0:Tn:1;
xn=cos(2*pi*f1*tn)+cos(2*pi*f2*tn);
hold on
stem(tn, xn)

y = dac(xn, Tn, t);
plot(t,y)


function y = dac(x, T, t)
m = 0:length(x)-1;
y = zeros(1, length(t));
for ii = 1:length(t)
    h = sinc((t(ii)-m*T)/T);
    y(ii) = sum(x .* h);
end
end
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1
  • $\begingroup$ Thank you so much. $\endgroup$
    – agile
    May 17 at 7:13

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