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Will be interfacing with the NI 6356 next week, but ultimately plan to use an NI PCI-6123.

Oh looks like the datasheet is more specific. 100Gohms in parallel with 10pF. Though I'm still not sure what to do with this information. I'm thinking the DAQ input impedance should be really high while the preamp output impedance should be really low since the DAQ is sensing voltage. Similarly...

Yes I know it won't be simple but I need to start somewhere and I thought impedance requirements was a suitable starting point, but I'm not sure how the input and output impedances should relate. Everywhere I search says something different or isn't clear. For now I would just like to know how...

Oh yes, sorry forgot to mention but I'll have the hydrophone go to a preamplifier first, then the DAQ. So would the preamplifier need a high input impedance and a low output impedance? I plan to design/build one

It's a Reson TC4013 hydrophone, and it's datasheet: http://www.teledyne-reson.com/download/hydrophone_data_sheets__/TC4013.pdf gives me some information about its impedance at different frequencies. It's much lower than the ADC input of the DAQ (>100 Gohms). So would I need to make these...

Hello, I'm confused about the concept of impedance matching in regards with hydrophones and DAQs. If the input impedance to an ADC channel on a DAQ is very high, would I need whatever's being fed into it (hydrophone) to also need a high impedance? Or would it need a low impedance? I've tried...

It looks like I'd actually need to feed Instruction[10:6] and Instruction[15:0] into another Mux and have it select Instruction[10:6] when I want to shift, which will then be sent through the sign extender and into the ALU when ALUSrc is set to 1. Right? Anyone?

Homework Statement Here's the datapath: So this seems like a pretty common question but I can't seem to find any answers on how to extend the datapath to implement SLL and SRL. The Attempt at a Solution This is how I would think to do it but I'm not entirely sure: It would need another...

Yeah we're assuming ideal components. The non-idealities come into play for the lab portion of this class.

Ohh okay I think I'm starting to get it now. Ideally, a FWR with cap filter would output a pure DC voltage. That's why (depending on the capacitor value) I can use the DC power formulas (P = IV, P = V²/R), even though there's a small voltage ripple. But with a FWR without a cap filter, I need to...

Right, I understand that. Okay I'm sorry but I'm still having trouble understanding. I know what rms means but it's confusing me in this instance. In a full-wave rectifier without cap filter, the output voltage would be the fully rectified input sinusoid, and the rms would be double the hwr...

Using the equation it'd be 6 Ω. But is it different for half-wave rectifiers then? I'm looking at this example in my book and it's using RMS to calculate the power (part b): But the output of a HWR is DC too isn't it?

Oh okay so then with my turns ratio, ΔVo = 12.12*0.03 = 0.3636 V Hm I'm a bit confused, how else would I calculate the power absorbed by the resistor here? Everything else we've done so far has it as either P = V2rms/R or P = I2rmsR. Or because it's DC, should it just be 12.122/R without...

Homework Statement I'm having trouble understanding part C of this question. Homework Equations ΔVo = Vout/2fRC The Attempt at a Solution So to get ~12V at the output, the turns ratio should be 14 because 120√2/12 = 14.14. Using a turns ratio of 14, the output voltage will be 12.12 V. To...

Yep my mistake, 0.025 mm. Thanks for your help; much appreciated.

Oh that's not quite right. I need to take 180 - θ4 - upper angle of height triangle from before. Then I can say 0.023451mm*tan(θ) = f [mm], and I'm getting 0.0025 mm as the focal length.

Okay, and then I can use snell's law again to determine, θ4, the refracted exit ray's angle. Then I can say 0.023451mm*tan(θ4) = f [mm]

An isosceles triangle. So the angle of the incident ray angle when it exits the sphere is the same as the refracted angle when it enters the sphere, θ2 = θ3 = 22.8854°. Trying to find the height it leaves the sphere now, I cut the triangle in two and using the refracted angle and radius...

Okay I've determined that sinθ1 = 0.7. Using Snell's law, the refracted ray inside the sphere's θ2 = sin-1(0.7/1.8) = 22.8854°. But then how do I get the angle of the incident ray when it exits the sphere? I'm just not seeing the geometry.

Snell's Law: n1sinθ1 = n2sin2 But how would I find θ1?

Homework Statement Homework Equations I know that yin = 0.7mm, y2 = 0mm, the index of refraction for the sphere is 1.8 and the indices of refraction for the air surrounding it is 1. The Attempt at a Solution Not sure how to even begin with the given information. I was thinking I could...

bump

Okay so like they said the initial flux is in the -x direction into the page and is maximum at t = 0. It varies as a cosine initially, so it is decreasing. Here's where my confusion is: in the -x direction, cos(0) = -1, maximum cos(pi/4) = -1/√2 Is it not getting more positive (i.e, closer to...

Homework Statement It's Problem 6.6 on this page: http://web.ece.ucdavis.edu/~lxgliu/eec130a/2012winter/hw9s.pdf It has the solution and everything. I don't understand part (b) though. Homework Equations The Attempt at a Solution [/B] For part (b), I thought the direction of the induced...

Okay this right here makes sense. Thanks!

Bump. That's still not where it should be though. Right now I have: (1/jπk)[(1/2)exp(jkπ) + (1/2)exp(-jkπ) - 1] but I'm not sure how to simplify it further.

Okay so distributing that (1/2) from the term outside the brackets, I have: cos(kπ) - (1/2)cos(k2π) - (1/2) Then using the identity cos(2x) = 1 - 2sin²(2x), I have: cos(kπ) - 1, since the sine term becomes 0 Using Euler's equation, I then have: (1/2)exp(jkπ) + (1/2)exp(-jkπ) - 1

This is where I thought it'd be helpful to expand it with Euler's equation. The term in the brackets becomes: 2coskπ - j2sinkπ - cosk2π + jsink2π - 1 If k can be any integer value, all the sines should become 0 right? so I'm left with: 2coskπ - cosk2π - 1 in the brackets. But like you said...

Summation over k, the harmonic components

x(t) = a summation from negative infinity to positive infinity of ak*exp(jkwot) sorry about my ignorance on this subject, it's really difficult for me to grasp this stuff just from reading my textbook

The k's are the harmonics, and they can be from 0 to 8 because of the function's range?

Sorry I'm not sure what you mean by that, but I'm looking at a table of properties for coefficients and it says if x(t) is real and odd, then ak should be jIm(ak). Using that with Euler's equation, I get: (1/kπ)[cos(kπ)-(1/2)cos(k2π)-(1/2)], which doesn't seem anywhere near right.

True, my mistake. Take out the t's and that's what I have. I did the integration right I think (plugged in for t) but forgot to remove the t also: (1/-jk2π)[2*exp(-jkπ)-exp(-jk2π)-1] Still can't seem to get it in the form they have, which is necessary for the next part of the problem.

Homework Statement http://puu.sh/gGhdb.jpg [Broken] Solution:[/B] http://puu.sh/gGh3E.jpg [Broken] Homework Equations The Attempt at a Solution How did they get that solution for the Fourier coefficient? When I evaluate the integral I can only seem to get it to...

Think I got it now. The current through the 2nd diode is 0.62 mA, and the cathode voltage of the 2nd diode is -1.2 - 0.7 = -1.9 V. (Solved for the anode voltage above).

Hm okay if I move the 0.7 V source below that 5k resistor, would I need to "step up" by 0.7 V from the -5 V source? So: (Vn-5)/10k + (Vn-(-5+0.7))/5k = 0 (Vn-5)/10k + (Vn+4.3)/5k = 0 Vn = -1.2 V

Doing node voltage analysis at the anode: (Vn-5)/10k + (Vn-0.7)/5k = 0 I get the voltage at the node, Vn = 2.133 V. This isn't less than 0.7 V though, did I do the analysis incorrectly?

I was under the impression that it'd be something like this: Even though the diode's off I thought the anode's voltage would still be 0.7 V because of the other diode's anode, and its cathode would be 0 V because of the ground. If this isn't correct what would it be?

"With diode 1 (left) off and diode 2 (right) on, the voltage drop across diode 1 is still 0.7 V because of the shared node with diode 2, but the current would be 0 A which isn't consistent with the model." I mean with this configuration the anode voltage of diode 1 is 0.7 V but the cathode...

Yeah it looks like v2 in (b) can't be 0 with both the diodes on because then the current I would be negative, which isn't consistent with the Constant Voltage Drop Model. I'm having trouble figuring out what diode configuration to use though. With diode 1 (left) off and diode 2 (right) on, the...

Thanks, that's exactly what I wasn't sure about. So if the voltage at the node connecting the diodes is 0.7V, I would need to "step down" 0.7V to get to the V terminal right? So the voltage at V is 0 V? Also, aside from the wrong V value, is the method I used to solve for the currents correct?

Homework Statement Homework Equations Whenever assuming a diode is "on", replace that diode with a 0.7 voltage source. Whenever assuming a diode is "off", replace that diode with an open circuit. The Attempt at a Solution The problem is pretty straightforward but the format of the circuit...

Ahh okay, thanks. I think I get the general gist of it now, I just need to be careful with the connections.

If I make R3/R2 = 4 that would cause problems for my V2 term wouldn't it? If R3/R2 = 4: For V2, -[R3(R6+R2)/(R6R2)] * V2 becomes -[4(R6+R2)/R6]V2 To make this -4V2, R6 = 1 R2 = 0 which can't be Also sorry but could you show me what's wrong with my solution? Did I mess up on the math somewhere?

Okay here's my new solution: I connected a resistor to V2 V1 only is still the same relation: Vo = s(R3/R2)R1*C*V1 V2 only is now: Vo = -[R3(R6+R2)/(R6R2)] * V2 V3 only is now: (1 + [R3(R6+R2)/(R6R2)]) * [(R4R5)/(R4+R5)] * V3 If I make R3/R2 = 2, then R1*C must equal 1 to satisfy the...

Wait now that I'm looking at it again maybe I didn't calculate V2 only correctly in the original solution. For this circuit Shouldn't V2 only be: Vo = -[4R2 / (4R2||R2) ]*V2 = -[4R2 / 4R2R2/5R2]*V2 = -[4R2*5R2/4R2*R2]*V2 = -(20/4)V2 = -5V2 If so I'll be back later to work on my...

I made it similar to the first solution's circuit: where the Vo for V2 here is -(4R2/4R2) = -V2 the resistor connected to V2 didn't matter here so I thought I didn't need it in my solution. Was that incorrect to assume?

Hm okay since I didn't come up with this particular circuit I'm going to try and start from scratch since I'm probably going to have to do this on the coming exam. Using two op-amps as a general template: Start with V1 on the first op-amp, and because there's an "s" there's probably a...

Hey guys, thanks for your replies. I forgot to mention in my post that that solution isn't actually mine. It's my professors and I kind of just assumed it was right without checking it. It's problem 3 on this past exam: http://www.ece.tamu.edu/~spalermo/ecen325/exam1_spring2014.pdf If I use...

Homework Statement Homework Equations The Attempt at a Solution So I know how to derive a transfer function from a given circuit, but not how to design a circuit from a transfer function like the one above. It seems like there's a huge amount of possible solutions and I don't know where to...