We have been going over sampling for some time now. Lets determine some relationships within this system:

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\rput(-0.5,0){$x(t)$}

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\rput(6.5,0){$y(t)$}

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\rput(1,-1.2){$T_s$}

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\rput(5,-1.2){$T_r$}

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\rput(2,0.2){$x_d(n)$}

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How is the transform of $Y_d(\omega)$ related to $X_d$? We can use the fact that convolution in time is multiplication in the frequency domain:

$$
Y_d(\omega) = H_d(\omega)X_d(\omega)
$$ 

After we pass $y_d(n)$ through the ``Kronecker to Dirac'' box, we have $y_g(t) = \sum \limits_{n\in\Z} y_d(n)\delta(t - n T_r)$. This gives us the relationship between $y_g(t)$ and $y_d(n)$, but what is $Y_g(\omega)$ in terms of $Y_d(\Omega)$? 

\begin{align*}
y_g(t) &= \sum \limits_{n\in\Z} y_d(n)\delta(t - n T_r) \\
Y_g(\omega) &= \sum \limits_{n\in\Z} y_d(n) e^{-i\omega n T_r} \\
Y_g(\omega) &= \sum \limits_{n\in\Z} y_d(n) e^{-i\lr{\omega T_r} n} \\
Y_g(\omega) &= Y_d(\omega T_r) = \left. Y_d(\Omega) \right|_{\Omega = \omega T_r}
\end{align*}

The kronecker to dirac conversion requires that you specify a reconstruction period. It may or may not be the same as $T_s$. Remember when the signal goes through the kronecker to dirac box, all that happens is a rescaling of the frequency axis. Going from frequency that is, in this case, radians per second to radians per sample.



Lets follow the process from start to finish, and then derive the various relationships. We started off with a signal $x(t)$, and replicated this signal into a train of impulses every $\omega_s$ with the amplitude scaled:

% image of train
% \imgsize{0.75}{images/sampling/impulse/triangles/zoom/threetris.ps}

This gives us a relation between $X_g(\omega)$ and $X(\omega)$:

\begin{nicebox}
$$
X_g(\omega) = \frac{1}{T_s} X(\omega) \quad \abs{\omega} \lt \frac{\omega_s}{2} 
$$
\end{nicebox}

We then scaled the frequency axis:

% image of scaling
% \imgsize{0.75}{images/sampling/impulse/triangles/zoom/twopi.ps}

This gives us a relation between $X_d(\Omega)$ and $X_g(\omega)$:

\begin{nicebox}
$$
X_d(\Omega) = X_g\lr{\frac{\Omega}{T_s}}
$$

$$
X_g(\omega) = X_d(\omega T_s)
$$
\end{nicebox}

Now we pass this signal through a discrete-time filter, $H_d(\Omega)$:

% image of Y_d(omega)
% \imgsize{0.75}{images/sampling/impulse/triangles/zoom/circles/yd.ps}

This gives us a relation between $Y_d(\Omega)$ and $X_d(\Omega)$:

\begin{nicebox}
$$
Y_d(\Omega) = H_d(\Omega)X_d(\Omega)
$$
\end{nicebox}

We then scale the frequency axis again:

% image of Y_g
% \imgsize{0.75}{images/sampling/impulse/triangles/zoom/circles/yg.ps}

This gives us a relation between $Y_g(\omega)$ and $Y_g(\Omega)$:

\begin{nicebox}
$$
Y_d(\Omega) = Y_g\lr{\frac{\Omega}{T_r}}
$$

$$
Y_g(\omega) = Y_d(\omega T_r)
$$
\end{nicebox}

At last, we pass this through a low pass filter with a gain of $G$ to obtain $Y(\omega)$:

% \imgsize{0.75}{images/sampling/impulse/triangles/zoom/circles/y.ps}

This gives us a relation between $Y(\omega)$ and $Y_g(\omega)$:

\sbx{5cm}{
$$
Y(\omega) = G Y_g(\omega) \quad \abs{\omega} \lt \frac{\omega_r}{2}
$$
}

Starting with the relationship between $Y(\omega)$ and $Y_g(\omega)$, lets work backwards and determine a relationship between $Y(\omega)$ and $X(\omega)$. For simplicity, lets restrict the domain of $\omega$. Consider $\abs{\omega} \lt \frac{\omega_s}{2}$.

\begin{align*}
Y(\omega) &= G Y_g(\omega) \\
Y(\omega) &= G Y_g(\omega) = \left. GY_d(\Omega) \right|_{\Omega = \omega T_r} \\
Y(\omega) &= G Y_g(\omega) =  GY_d(\omega T_r) \\
\end{align*}

We can then use what we know about $Y_d(\omega)$ to expand this further. We know that $Y_d(\Omega) = H_d(\Omega)X_d(\Omega)$, and $Y(\omega) = GY_d(\omega T_r)$. Combining these equations gives us

\begin{align*}
Y(\omega) &= G Y_d(\omega T_r) \\
Y(\omega) &= G H_d(\omega T_r)X_d(\omega T_r) \\
\end{align*}

Recall that $X_d(\Omega) = X_g\lr{\frac{\Omega}{T_s}}$, and that $X_g(\omega)$ is just an amplitude-scaled replica of $X(\omega)$: $X_g(\omega) = \frac{1}{T_s} X(\omega)$ for $\abs{\omega} \lt \omega_s$. Combining these relations:

\begin{align*}
Y(\omega) &= G H_d(\omega T_r)X_d(\omega T_r) \\
Y(\omega) &= G H_d(\omega T_r)X_g\lr{\frac{\omega T_r}{T_s}} \\
Y(\omega) &= G H_d(\omega T_r)\frac{1}{T_s}X\lr{\frac{\omega T_r}{T_s}} \\
\end{align*}

So we finally can arrive at an equation relating $X(\omega)$ and $Y(\omega)$ from start to finish:

\begin{nicebox}
$$
Y(\omega) = \frac{G}{T_s} H_d(\omega T_r)X\lr{\frac{T_r}{T_s} \omega} \\
$$
\end{nicebox}

Lets look at the different results of various cases. Consider the case where there is no aliasing, and the reconstruction period is equal to the sampling period and the gain on the final low pass filter. In other words, the signal is bandlimited and the sampling is fast enough, and in addition, $T_r = T_s = G = T$.

\begin{align*}
Y(\omega) &= \frac{G}{T_s} H_d(\omega T_r)X\lr{\frac{T_r}{T_s} \omega} \quad \abs{\omega} \lt \frac{\omega_s}{2}\\
Y(\omega) &= H_d(\omega T)X\lr{\omega} \quad \abs{\omega} \lt \frac{\omega_s}{2}\\
\end{align*}

Basically we find here that $H(\omega) = H_d(\omega T)$. Lets look at another more interesting example. Consider the case where $H_d(\Omega) = 1$, $T_s = G \neq T_r$.

\begin{align*}
Y(\omega) &= \frac{G}{T_s} H_d(\omega T_r)X\lr{\frac{T_r}{T_s} \omega} \quad \abs{\omega} \lt \frac{\omega_s}{2}\\
Y(\omega) &= X\lr{\frac{T_r}{T_s} \omega} \\
\end{align*}

What is $y(t)$? We can look at some fourier transform pairs to find a relation:

\begin{align*}
x(t) &\ftp X(\omega) \\
x(\alpha t) &\ftp \frac{1}{\abs{\alpha}} X\lr{\frac{\omega}{\alpha}} \\
\abs{\alpha} x(\alpha t) &\ftp X\lr{\frac{\omega}{\alpha}} \\
\end{align*}

Now we can solve for the time-domain representation:

\begin{align*}
Y(\omega) &= X\lr{\frac{T_r}{T_s} \omega} \\
Y(\omega) &= X\lr{\frac{\omega}{T_s/T_r}} \\
y(t) &= \abs{\frac{T_s}{T_r}}x\lr{\frac{T_s}{T_r}t} \\
\end{align*}

\begin{example}
What if $T_s = 2T_r$? We get $y(t) = 2 x(2t)$. The signal will be twice as loud and twice as fast. If $G=T_s/2$, then $y(t) = x(2t)$.
\end{example}


Relations in Sampling