While studying frequency transforms ,I get confused with the terms like $X(j \omega) ,X(\ e^{j \omega })$ and $ X(\omega)$ ,where $ \omega = 2 \pi f $. So what is the difference between them ?
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1These two answers should answer your question: answer 1, answer 2 – Matt L. Jun 19 '16 at 08:44
1 Answers
they should have different subscripts.
bilateral Laplace Transform:
$$ \mathscr{L} \left\{ x(t) \right\} \triangleq X_\mathcal{L}(s) \triangleq \int\limits_{-\infty}^{+\infty} x(t) e^{-st} \ dt $$
continuous Fourier Transform:
$$ \mathscr{F} \left\{ x(t) \right\} \triangleq X_\mathcal{F}(\omega) \triangleq \int\limits_{-\infty}^{+\infty} x(t) e^{-j\omega t} \ dt $$
it's pretty easy to see that
$$ X_\mathcal{F}(\omega) = X_\mathcal{L}(j \omega) $$
likewise in discrete-time, the bilateral $\mathcal{Z}$ Transform:
$$ \mathcal{Z} \left\{ x[n] \right\} \triangleq X_\mathcal{Z}(z) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] z^{-n} $$
and the Discrete-Time Fourier Transform (DTFT):
$$ \mathcal{DTFT} \left\{ x[n] \right\} \triangleq X_{\small{DTFT}}(\omega) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $$
it's also easy to see that
$$ X_{\small{DTFT}}(\omega) = X_\mathcal{Z}(e^{j\omega}) $$
now often the context is understood and everyone knows that $X(e^{j\omega})$ is the Z transform evaluated at $z=e^{j \omega}$, which is the DTFT and we don't bother with the clunky subscript notation. we might leave the space for subscripts for another function, like if we have a vector of signals as with a state-variable system representation in control theory.
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Or like distinguishing between periodicities:
$$ X_{2\pi}(\omega) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $$
$$ X_{1/T}(f) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] e^{-j2\pi f nT} $$
– Bob K Sep 11 '20 at 01:42