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6

Historically, the largest use of relays has been associated with the telephone industry. Before electronic telephone exchanges were introduced in the 1960's, virtually all switching was done using a combination of conventional relays and specialized configurations like crossbar and step-by-step switches. Relays provided a means to connect two parties ...


4

This is exactly why a Reset signal is added to the system. To bring it to an initial known state. There is no way to avoid it when working with actual hardware. As for simulation, you can add an initial block to the design and initialize the state to whatever you want, but it is not synthesizable.


4

A race condition is a timing-related pheonomenon. A standard S-R FF (two cross-coupled NAND or NOR gates) is stable for any stable input. The 'fun' is in the S=1 R=1 input, the memory situation. The state of the FF depends on which state came before the 11, if it was 01 the FF is in Q=1 state, if it was 10 the FF is in the Q=0 state. This is the ...


4

It's not the mystery other comments are saying. Consider the standard CMOS inverter: There are two FETs, one n-channel and one p-channel. Normally one is on and one is off. What happens in the intermediate voltage range? Both of them are partly on. This connects the power rail to the ground rail and a substantial current can flow. This is called ...


4

If the present output depends on previous output, then use the previous output as as one of the variable in truth table and K-map. Take JK flipflop as an example: ------------------- J K Qn Qn+1 ------------------- 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 ------------------ ...


4

Yes, it is usually an IC with 4 of the same gate inside it sharing a common \$V_{CC}\$ and GND. The datasheets sum this up pretty well. Here is a quad NAND gate pinout: Notice that there is one set of power connections and 4 identical gates.


3

You are going to need to re-think how you drive the FB180SA10B FET. The VgsTH spec of 4V max is only going to assure you of a measely 250uA of drain current. One would have to guess that if you plan to use a monster FET looking like this: ...that you are planning to put some serious current through the part. For this you really need to be looking at ...


3

I'll try to cover the parts of your question separately. Multiplexing Is the multiplexer represented in the schematic where it says "buffers"? Nope! The multiplexer is the select digit part of the circuit. As you said, a multiplexer is an electrical switch: if I have \$n\$ "selector" inputs, I can choose from \$2^n\$ outputs. In your case, you have a ...


3

What you understood is correct. Grouping mintems results in sum of product (SOP) form or the AND-OR form as shown in (1) $$Y = a_1a_2a_3 + b_1b_2b_3 + \cdots + \omega_1\omega_2\omega_3\tag1$$ Calculating \$\overline{Y}\$ using De-Morgan's theorem, $$\overline{Y} = (\overline{a_1a_2a_3} )\ (\overline{b_1b_2b_3}) \cdots ...


3

I'm not going to give you a full circuit here, since we don't do that for someone doing homework problems. But hopefully I will give you enough clues to get started. I am going to assume you meant a 4-digit input sequence as I asked in my comment. I also assume you need to design this using discrete logic, and not use a microcontroller, since the latter ...


3

When speaking of logical operations on two numbers, two different kinds of operations might be the case: 1) The most common one - bitwise operation between two numbers, which, as you hopefully know, represented by bits (zeros and ones). Given two numbers, represented with 8bits each, for example, the bitwise logical operator will perform the logical ...


3

What you describe is typical of EEPROM chips. The minimum number of bytes you have to erase at once, the maximum you can write at once, and the minimum you can write at once can all be different. The way I usually deal with this is to have a module that virtualizes reads and writes to the EEPROM. This module presents a procedural interface for reading ...


3

The blocks labeled "FA" are called full adders, and they are fundamental to how binary arithmetic is done at the gate level. The circuit you have presented here is an implementation of a 4-bit adder/subtractor. Combining addition and subtraction in the same operation requires the use of an alternative representation of binary numbers. This is called 2's ...


2

An asynchronous counter means that the states of flip flops dont change simultaneously. This is evident by the fact that all flip flops are clocked by different clocks rather than a single master clock. As far as working is concerned, a timing diagram may help : 1) First flip flop (Q0) changes state on PGT of the clock. 2) Second flip flop changes state ...


2

One can not implement every logical function using only XOR gates. Since XOR is a logical operator which obeys associativity, the function implemented using only XOR gates can always be written as $$f = a_1\oplus a_2 \oplus a_3 \oplus \cdots \oplus a_n $$ where \$a_1, a_2 \ldots\$ are the inputs. So the only possible functions that can be implemented ...


2

You're not using Next_State so I'm not clear why you bother to have it at all. The (recommended) approach would be to do a blocking assignment to Next_State in your combinational always block, and a non-blocking asignment from Next_State to State in your synchronous always block. You're over-allocating the State and Next_State variables; they only need 5 ...


2

You skipped the reset operation in your test-bench, Reset should go high and some time later go low. If you are targeting for FPGA, you could also put initial State = Idle; in your design. Note that initial only works for simulation and FPGA synthesis, not IC synthesis. Other issues: Input_Data and Next_Round_Data are missing form the sensitivity list that ...


2

Yes, it can be either. It will work with unsigned 3-bit numbers. It will also work with 2s complement signed numbers. You'll need to do a bit of study, but the thing to remember with 2s complement is that a 0 sign bit means positive, and a 1 means negative.


2

When performing a NOT on a single boolean value, then you're correct in pointing out that it's an inversion of the value. A = 1 A' = 0 When A is represented by more than one bit, then all the bits in A will all be NOT'd independently. A = 00110101 A' = 11001010 The representation of binary in hex allows us to shorten how these numbers are ...


2

I agree with you, and disagree with @Kynit: the multiplexer is the part marked "buffers" on your schematic. But more on that in a moment: let's walk through the circuit backwards, from the display to the oscillator. I'm going not going to provide any schematics, and may be a little vague at times - as you're learning, I imagine it's best not to give too ...


2

OP says: You may use any MSI modules as well as any other gates Therefore you want to save yourself a lot of work and use a 4-bit MSI (medium scale integration) IC comparator such as the CD4585,or the CD4063. The pinout below is for the latter. The inputs connect to A0-A3 and B0-B3. Ignore the three lines (xxx)IN, they are used for cascading. The ...


2

1100 in four bits is -4 extending this to 6 bits, do sign extension to the left, adding 1's because the left bit above is 1, and get 111100 this is still -4 in two's complement 0101 is 5 in decimal extending it to the left, adding 0's, because the left bit is 0, and get 000101 adding these two together: 111100 000101 ====== 1000001 the carry bit ...


2

Erm, that's just \$A \bar B\$.


2

An XOR gate and a pair of inverters will do this if you don't need precise control over the pulse width. simulate this circuit – Schematic created using CircuitLab How it works: An XOR gate output is high only when its inputs are at different states (i.e. 10 or 01). The two inverters add a small amount of delay to the signal seen on the bottom ...


1

Nice project. Starting at the tail end, with the oscillator, you're proposing using a 555 in its standard astable mode, so it'll free-run with an output frequency of: $$f = \frac{1.46}{(R1+2R2)\ C} $$ and a duty cycle of $$D = \frac{R2}{R1+2Rb}$$ For your application, duty cycle doesn't matter much, so make R1 equal to about 1000 ohms and work out R2 ...


1

The 555 provides only the oscillator, to produce the digit drive signals you also need a counter. One example is an MC14017. Use Q0, Q1, and Q2 as the inputs to your three digit drivers, then have the Q3 signal feedback to the Reset pin. So the output counts as 0,1,2,0,1....... See the data sheet for other connection info: ...


1

Race around condition in digital circuits occur when the final state of the output depends on how the inputs arrive. Digital circuits have inherent delays. So it is possible that one of the inputs arrive a little earlier or later than others ie the inputs which were meant to be present at the same time actually arrive at different times due to different ...


1

The reason you are getting incorrect expression is because your K-map is incorrect. Look carefully, 0110 (6) and 0111(7) should be zero, not one. If you construct the K-map properly, you will arrive at the same expression ie Q = W + Y' which using Demorgans rule becomes Q = (W'.Y)' This corresponds to (01)' . But since your outputs are active low, ...


1

Depending on the transfer ratio of the optocouplers inside the stepper driver you could attach VCC to 3.3V and drive the LEDs directly from your beaglebones ports. They can source/sink ~4mA. I believe you will be even more limited by Rled. If you're not willing to give a try or if you have long cables then you can use a single buffer/line driver(such as ...


1

A three-input gate which outputs true when exactly one of its inputs is true may be used as a two-input NOR gate by tying one of its inputs high; it may also be used as an "A and not B" input by tying one input to the desired "A" and two to the desired "B", or as a two-input exclusive-or gate by tying one input low. Such two-input gates may be combined to ...



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