0
$\begingroup$

This is something I've always wondered about in DSP class, but just accept as a fact because I never really fully understand why this is the case:

Given CTFT:

$$X_s(j\Omega) = 6000 \pi \sum \limits_{k=-\infty}^{\infty} \Bigg[ \delta(\Omega - 1200\pi k - 4000 \pi) ~+~ \delta(\Omega - 1200\pi k + 4000 \pi)\Bigg]$$

Then converting to DTFT using formula:

$$X(e^{j\omega})=\frac{1}{T_s} X_s\bigg(\frac{\omega}{T_s}\bigg)$$

But, Why is answer this:

$$X(e^{j\omega}) = 6000 \pi \sum \limits_{k=-\infty}^{\infty} \Bigg[ \delta(\omega + T_s(-1200\pi k - 4000 \pi)) ~+~ \delta(\omega +T_s(- 1200\pi k + 4000 \pi))\Bigg]$$

instead of this:

$$X(e^{j\omega}) = 6000 \pi \sum \limits_{k=-\infty}^{\infty} \Bigg[ \delta(\frac{\omega}{T_s} - 1200\pi k - 4000 \pi) ~+~ \delta(\frac{\omega}{T_s} - 1200\pi k + 4000 \pi)\Bigg]$$


seems like this formula: $$X(e^{j\omega}) = \frac{1}{T_s} X_s\bigg( \frac{\omega}{T_s}\bigg)$$ should really be: $$X(e^{j\omega}) = \frac{1}{T_s} X_s\bigg( \omega = \Omega T_s\bigg)$$

$\endgroup$
2
  • $\begingroup$ Even in answer as per you, shouldn't $T_s$ be in numerator? Then it would be of the form $\delta(\omega-\omega_0 k-\omega_1)$. Also, I am assuming $1/T_s$ is $6000\pi$. $\endgroup$
    – jithin
    Apr 23 '20 at 18:12
  • $\begingroup$ I think it’s really two steps. First, make substitution $\Omega= \omega/T_s$. Next, you realize that output horizontal scale for delta functions are still in radians/sec so you multiply input function to delta functions by Ts to convert to output horizontal scale of radians. It’s a two step process... both equations above are correct bit the horizontal units are different... the later equation is the finishing step to make scale in radians to math $\omega$ input. $\endgroup$
    – Pico99
    Apr 23 '20 at 21:48
1
$\begingroup$

It is due to the property of unit impulse function

\begin{equation} \delta(\alpha t) = \frac{ 1}{|\alpha|} \delta(t) \end{equation}

(Usually $\omega $ is used for angular frequency and $\Omega $ is used for angle. Interpret my answer accordingly. )

suppose \begin{equation} X_a(\omega) = \delta(\omega -x_0 ) \end{equation} and if we substitute '$\omega$' with $\frac{\Omega}{T_s}$ then \begin{equation} X_a(\frac{\Omega}{T_s}) = \delta(\frac{\Omega}{T_s} -x_0 ) = \delta(\frac{\Omega - x_0 T_s}{T_s} ) = \delta(\frac{ 1}{T_s} (\Omega - x_0 T_s) ) = {T_s} *\delta(\Omega - x_0 T_s) \end{equation} where '$\alpha$' is $\frac{ 1}{T_s}$ and $$ \frac{ 1}{|\alpha|} = T_s $$

and when you do DTFT calculation on

\begin{equation} X(\Omega) = \frac{ 1}{T_s} X_a(\frac{\Omega}{T_s}) =\frac{ 1}{T_s} * ( T_s * \delta(\Omega - x_0 T_s) ) = \delta(\Omega - x_0 T_s) \end{equation}

in the final answer $T_s$ and $\frac{ 1}{T_s}$ cancels

and you will get the expression based on the independent variable '$\Omega - x_0 T_s$'

So there is nothing wrong in your actual answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.