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Output F needs to be active for all odd numbers except 7. So... LSB to mux 20 input, mux outputs '0' for even numbers and '1' for odd numbers. AND gate detects 7, output to mux 21 input forces it to output '3' instead of '1'.


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I generally take a top-down design approach, and I start by drawing a block diagram that shows the interfaces among the top-level blocks. I then draw additional diagrams that represent the implementations of the top-level blocks in terms of lower-level blocks. This hierarchy of block diagrams translates pretty much directly to the hierarchy of the HDL ...


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A few things about your basic circuit before I address your specific questions. i. Every IC needs a decoupling capacitor. For this circuit, that applies to the linear regulator and the microcontroller. For the microcontroller, as with most ICs, a 0.1uF capacitor between VCC and GND is typical. The capacitor should be as close as possible to the physical ...


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They're arranged this way so that moving a single unit in either axis changes the value of ideally only a single variable. This makes it easier to read and interpret the map.


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That depends on where you place your variables in the gray code running along the edges. If you write out the graycode you will see that every square contains the combined graycode from the horizontal and vertical edge (with x_5x_4x_3x_2x_1 in the the right order of course) interpreted as a binary number.


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I agree that illustration is confusing. The top half of the page is intended to describe the TLB. It sounds like you understand TLB stuff pretty well. The entire bottom half of the page is intended to describe the data cache. (The label "cache" on the left is intended to apply to the entire bottom half of the page. How could it be redrawn to make it more ...


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I do not believe the perforated metal cage is there to act as a Faraday Cage. It is there for safety reasons to prevent contact with the high voltages within. The hole sizes are required to be smaller than the standard finger defined in safety ratings. A perforated rather than solid part is used for ventilation and cooling.


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While you can't trigger a relay directly from the audio jack, you can trigger a transistor, which in turn triggers a relay, to do what you want. An external power supply would be needed for the relay. DSLR Remote is an Android app that is designed to work with DSLR remote release cable ports. One option is IR diodes, but transistors on the audio jack ...


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Its doable, but I suspect that you'd get annoyed by the delays involved, not to mention mis-activations. Delay wise, the phone would have to link to your wi-fi when you got within range. That could be when you're halfway up the block, or maybe not until to get close to the door. If you have to wait there even 5 seconds, thats not so good right? I assume if ...


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So you want to be able to display the ALU result both before and after it is stored in memory, and be able to load that result into the ALU's a or b input? Connect all the data lines together to form a bus. Obviously this bus can only carry one 8 bit value at a time, so each device connected to it needs an input register (8 bit latch) to write (STORE) ...


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That's simple — you just need to AC-couple the switch, like this: simulate this circuit – Schematic created using CircuitLab R4 and C4 are new components; their time constant is set to 10 ms, a small fraction of your timer period. Regardless of how long the switch is held, the TRIGGER input of the 555 will only see a short low-going ...


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In the last few pictures, it looks like you've configured the "sub flag" as an output, not an input. Since there's nothing driving that net, it's resulting in an error.


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This is not possible, as you view it. Conservation of energy IN: 10V*10uA == 100uW OUT: 50*1A = 50W. You will need a power amplifier to take the small-signal AC to increase the power capability


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Additional specifications as clarified by "tman" Signal is not expected between 75deg and 90 deg and also between 0deg and -90deg elevation, and rejection in these sectors is not explicitly required. An omni-directional antenna is an acceptable solution. The antenna is to be used indoors in a fixed installation -- inside a home at about ceiling height. ...


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There is a nice white paper from SCHAFFNER explaining the 0.1/100 Ohm curves. Note that attenuation is measured as a voltage ratio (not a power ratio), so when working at different input/output impedances, you can actually have a voltage gain.


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Passive circuits cannot produce a power gain but they can produce a voltage gain. A resonant circuit can produce a voltage gain equal to its Q. If the impedance levels are considered, there is no power gain.


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There is no problem with your circuit. although I would suggest that you set pull-down resistors on the outputs. that's because the decoders usually set their outputs to high-impedance (high-Z) when they're not enabled. so the output may remain the same on the output node (because of node capacitance) and the wrong value may be read by the device that is ...


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I added the same object in eagle now too and it works fine here. Maybe the 'random' pins are also connected ground for example. If the shift register is closer to a component that is also connected to ground you will see a yellow line (from the unrouted layer) to that component. To check if this is the case you can use the show function on the board ...


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You need to download a simulator. Sigasi is only for design entry (typing and inspecting code). Look here for a list of free simulators: http://www.sigasi.com/faq/which-free-vhdl-simulator-can-i-use


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Is there a special conversion happening?! Your image for slide 25 corresponds to figure 9.10 found on Page 506 and slide 27 corresponds to table 9.3 found on Page 507 in the book "Digital Arithmetic" by Miloŝ D. Ercegovac and Tomás Lang, 2004, ISBN: 1-55860-798-6. If you look in the book the text below figure 9.10 (b) on Page 506, Example 9.2: The ...



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