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...
Then the circuit would look like this, right:
[PLAIN]http://img852.imageshack.us/img852/7263/dsc01026b.jpg [Broken]
And KVL would be like this: 5000V-(500+3)\Omega\times I+3V_R+1.75V=0
Now, how did you find the I_{3\Omega}? There are 2 unknowns in the KVL equation, you can't just take V_R from...
Yes, the I_{500\Omega} is entering the resistor from the top ... so the V_{500\Omega} (from the post #11) has a minus before it:
I_{500\Omega}=\frac{V_{500\Omega}}{500\text{ }\Omega}=\frac{-\left[(3V_R+1.75)-V_R\right]}{500\text{ }\Omega}
From this follows that V_R=30.375\text{ }V and...
OK, so the voltage across the current source is the same as that over the 500\text{ }\Omega resistor, right?
Then, P_{10A}=V_{500\Omega}\times 10A=\left(3V_R+1.75-V_R\right)\times 10A=57.518\text{ }V\times 10\text{ }A=575.18\text{ }W, correct?
As I've already (hopefully correctly)...
Please help me out. :frown: Are the steps {1,2,3}) in previous post OK?
Here is the picture, so that we are on the same "page".
[PLAIN]http://img171.imageshack.us/img171/2010/dsc01005kn.jpg [Broken]
If I should have any hope of ever solving this, I should go step-by-step. :smile: (Please point out at which step the first mistake occurs.)
1) OK, so the current through the 500\text{ }\Omega resistor is:
I_{500\Omega}=\frac{V_{500\Omega}}{500\text{...
The KCL for the upper principal node:
I_{10A}-I_{3\Omega}-I_{500\Omega}=0
I_{3\Omega}=\frac{V_R}{3\text{ }\Omega}}
I_{500\Omega}=\frac{V_{500\Omega}}{500\text{ }\Omega}}
Now, here is (my) difficulty: what is V_{500\Omega} ? :confused:
I'd say there is no voltage drop over the...
You're correct. In my second try (last post) I didn't use this incorrect assumption and have relied only on KCL ... but, I still don't get the "correct" solution? Can you help me out, please?
The factor of 500/503 is due to the current divider ... guess I didn't do this step correctly? Or is the current source ideal (so that it represents zero resistance; same for 2 voltage sources) and so no current flows through 500\text{ }\Omega resistor (and the whole of 10 A then flow through...
Show that "power supplied" = "power dissipated"
Homework Statement
For the circuit of Fig. 3-42, show that the power supplied by the sources is equal to the power dissipated in the resistors.
Ans. P_T=940.3\text{ }W
[PLAIN]http://img269.imageshack.us/img269/2010/dsc01005kn.jpg [Broken]...
Homework Statement
A practical current source has a current of 22 A. Loading the source with 50 \Omega results in terminal voltage of 390.3 V. Obtain the source constants, I and R.
Homework Equations
V=I\times R
The Attempt at a Solution
In my clumsy picture below, shouldn't I_{load} be...
Yes, that is easier. Now I got the correct solution of P_{max}=1.5\text{ }A\times 50\text{ }V=75\text{ }W ... but, I've got two more questions:
1) As the derivative \frac{du(t)}{dt} at t=\{2,4,6,8\} is undefined: are the end-points at those t "empty" or "filled" (see picture)...
So the last part can be isolated? So that the initial current at t=4 is 10 A ... and t=4 can be thought of as t=0? Then t=5 is equivalent to t=1:
i(t=5)=\frac{1}{3}(-30)(1)+10=0\text{ }A
I also found it easier to reverse i(t) into v(t) and look at derivatives of current...
Homework Statement
The current after t=0 in a single circuit element is as shown in Fig. 2-37. Find the voltage across the element at t=6.5\mu s, if the element is (a) 10 k\Omega, (b) 15 mH, (c) 0.3 nF with Q(0)=0.
Ans. (a) 25 V; (b) -75 V; (c) 81.3 V...