Search results

  1. C

    How to get (frequency response) from this [difference equation]?

    I haven't made any progress on this ... anybody help me out, please?
  2. C

    How to get (frequency response) from this [difference equation]?

    I'd appreciate (any) help. Continuing from 1st post: \cdots = \left[ x(k-2K+2) + 2x(k-2K+3) + \cdots + (K-1)x(k-K+1) + \cdots + x(k) \right] Applying Z-transform, we get X(z) \left[ z^{-2K+2} + 2z^{-2K+3} + \cdots + (K-1)z^{-K+1} + \cdots + 1 \right] So, the transfer function H(z) of the whole...
  3. C

    How to get (frequency response) from this [difference equation]?

    Homework Statement Given this difference equation y(k) (of a bandpass FIR filter) ... y(k) = \frac{1}{K^2} \sum_{m = k-K+1}^k \; \sum_{n = m-K+1}^m x(n) - \frac{1}{L^2} \sum_{m = k-L+1}^k \; \sum_{n = m-L+1}^m x(n) ... how does one derive this frequency response H(f)? H(f) = \frac{1}{K^2}...
  4. C

    Z-transform of conjugated sequence ( a straightforward exercise)

    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} =...
  5. C

    Z-transform of conjugated sequence ( a straightforward exercise)

    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.
  6. C

    Z-transform of conjugated sequence ( a straightforward exercise)

    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...
  7. C

    Z-transform of conjugated sequence ( a straightforward exercise)

    What am I doing wrong in trying to show x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)?
  8. C

    Z-transform of conjugated sequence ( a straightforward exercise)

    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...
  9. C

    Audio/Video DIY headphone wiring (nylon-stranded wire)

    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.
  10. C

    Audio/Video DIY headphone wiring (nylon-stranded wire)

    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...
  11. C

    Heating oil: effective or real output/yield (per liter)

    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...
  12. C

    Heating oil: effective or real output/yield (per liter)

    Hope I'm allowed to re-ask (after almost 2 months). :smile:
  13. C

    Heating oil: effective or real output/yield (per liter)

    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...
  14. C

    Given capacitance & energy dissipated, find the charge Q

    I know that \overline{P}_{dissipated}\times t=\overline{P}_{supplied}\times t=3.60\text{ }mJ. How do I find V?
  15. C

    Show that power supplied = power dissipated

    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...
  16. C

    Show that power supplied = power dissipated

    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...
  17. C

    Show that power supplied = power dissipated

    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)...
  18. C

    Show that power supplied = power dissipated

    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]
  19. C

    Show that power supplied = power dissipated

    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{...
  20. C

    Show that power supplied = power dissipated

    But there isn't any current through the 500\text{ }\Omega resistor, is there?
  21. C

    Show that power supplied = power dissipated

    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...
  22. C

    Show that power supplied = power dissipated

    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?
  23. C

    Show that power supplied = power dissipated

    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...
  24. C

    Show that power supplied = power dissipated

    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]...
  25. C

    Find (practical) current source constants, I and R

    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...
  26. C

    Capacitance: given V waveform, find max. power

    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)...
  27. C

    Inductance: given V waveform, find I

    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...
  28. C

    Given I, find V across the elements (resistor, inductor, and capacitor)

    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...
Top