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

  • $\begingroup$ 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

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.


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