You mean those "R" and "I" indices (as in X_R and X_I)? Say x[n]=\{..., 1-j2, 5+j, ...\}. Then, x_R[n]=\{..., 1, 5, ...\}, x_I[n]=\{..., -2, 1, ...\}, and so x[n] = x_R[n] + jx_I[n] ... also, x^*[n]=\{..., 1+j2, 5-j, ...\}.
So, Z\{x[n]\} = \sum_{n=-\infty}^\infty (x_R[n] + jx_I[n]) z^{-n} =...
That's like cat chasing its tail. :) And it also doesn't help shedding light on my 1st, erroneous attempt. Hope you'll find something out; I'll most surely post as well if I find the error.
rude man, thank you for the nudge in the right direction:
Z\{x^*[n]\} = \sum_{n=-\infty}^\infty x^*[n]z^{-n} = \sum_{n=-\infty}^\infty \left(x[n](z^*)^{-n}\right)^* = \left(\sum_{n=-\infty}^\infty x[n](z^*)^{-n}\right)^* = \left(X(z^*)\right)^* = X^*(z^*)
Is this correct? If so, what did I do...
Z-transform of a conjugated sequence ("a straightforward" exercise)
Homework Statement
The conjugation property is expressed as x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)
This property follows in a straightforward maner from the definition of the z-transform, the details of which are left...
Yes, that is a mess. :) I'll clean it up and solder it.
Thank you for your reassurance that I wasn't making some blunder regarding the casing behaviour.
I've connected the 3 wires as shown in the picture: GROUND, GREEN, and RED (the latter 2 having their enamel removed).
[PLAIN]http://img821.imageshack.us/img821/4133/nylonstranded.jpg [Broken]
But I'm getting only low-volume (un-amplified) sound. Now, if I connect either GREEN or RED wire to...
The 120,400 are in btu units, right? So, from this number (presuming 86% efficiency, etc.) it follows that the NET yield is 9.28 kWh/L, correct?
In short, the net yield (using an average boiler) certainly comes closer to 8 kWh/L (than to 6.5 kWh/L)?
The latter number of 6.5 kWh/L was put...
Heating oil: effective or "real" output/yield (per liter)
http://en.wikipedia.org/wiki/Heating_oil" [Broken] says that
which is about 11 kWh/L.
I was given an unbiased (but not necessarily correct) number that the effective output is about 8 kWh/L ...
... and a potentially biased number...