Questions d'entretiens - Design engineer

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Questions d'entretien pour Design Engineer partagées par les candidats

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On a demandé à un Mechanical Design Engineer...5 novembre 2013

need to join tomorrow are you ready?

13 réponses


Yes sir

Yes I am ready to join

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Given a string find the first non-repeated character.

12 réponses

@Rajiv : Your solution is completely wrong. It will fail for input of "aaa" Reason: on first check, you insert "a". On next check you remove it. On next check you again insert it and return that as your answer, even though it was repeated thrice. Moins

program to find first non repeating character in a string c# - three ways

My python implementation: def firstNonRepeatingCharacter(inputString): hashmap = {} for x in inputString: if (x in hashmap): hashmap[x] = hashmap[x] + 1 else: hashmap[x] = 1 for x in inputString: if (hashmap[x] > 1): continue else: return x return "No nonrepeating character found" Moins

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Dolphin Technology

Basic MOSFET, CMOS questions were asked.

11 réponses

Hey, you should feel glad that you weren't accepted by them. This company is very cheap and doesn't treat its employees well. Moins

I guess...but it is not the right way to interview candidates don't you think?

That's really sad. They should rather go with Skype or any other means to interview. Now,. I have got an on-site call for Digital circuit design and I am giving it a serious thought 🤔 Moins

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There is a refrigerator in a perfectly insulated room. The door is then opened. Is it an open or closed system? After a long time has passed, has the temperature in the room increased or decreased?

8 réponses

It is a closed system as electricity is adding energy into the room. After a long time passes, the temperature increases as there's energy going in, but not leaving. Moins

It's a closed system. Since the efficiency of heat transfer is not 100%, will generate heat and eventually heat up the room. Moins

Since there is energy going into the room, it is open system without mass transfer. The room temperature will increase. Moins

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Create a 8 input AND gate using 3 4:1 muxes

8 réponses

Without an enable bit on at least one of the mux's the maximum inputs would be 7. Moins

I don't see it being possible with three standard 4-1 muxes... Using 4, this question is straight forward... The two selects of each mux are your 8 inputs... tie out put of each mux to the (11) case input to the mux. Moins

We need 3 4:1 MUX and a And gate. Are we allowed to use 'and' gate?

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Show me the back of an iPhone and asked me how it might have been manufactured.

8 réponses

Automated CNC milling

A cast polymere

Still don’t get it

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design a full adder with 2-1 mux

6 réponses

full adder can be got by 2 half adders and one OR gate; one half adder can be got by XOR, AND. Therefore, we need only OR, AND, XOR. All these three gates can be got by using MUX.? Moins

Can be implemented using 8 Muxes.

sum = a xor b xor cin carry = (a xor b) cin + ab You can easiy make XOR, OR AND, NOT using 2:1 mux. So 8 mux ?!? Moins

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If I have a solid rod and hollow rod with the same mass and I let them slide in a ramp. Which one reaches the bottom first and why?

6 réponses

Since they've clearly told sliding. Acceleration along the ramp is independent of the mass or the moment of Inertia. Hence, both reaches down at the same time. Moins

Solid bar, smaller moment of inertia allows it to spin faster

Solid reaches first

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Intel Corporation

CMOS Inverter , how to reduce the drive strength of Minimum size inverter

6 réponses

"more the Ids, the lesser the drive strength is" This is exactly opposite of the actual fact...! Moins

Use body effect, reverse biasing, this will reduce the drain current. another way can be change the gate input, according to the I-V characteristic Moins

the first answer is definitely wrong, to decrease the driving capacity, please size down the ratio! Moins

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design a combinational circuit which counts the number of 1s in a 7-bit input .

6 réponses

@Akash, Adder C should be the 7th bit and the sum of adder a and b right? not carry Moins

It can be done by two ways Clocked Circuit: Use a one bit adder and a register. Output of the register acts as 2nd input to the adder. Half adder can also be used. Combinational Circuit: We can do it in 4 full adders. For the adders A and B, let 6 bits be the inputs. For adders C, use the 7th bit and Carry of adder A and B. Adder C gives sum as bit 0. For adder D, use carry from all three adders. The sum is bit 1 and carry is bit 2. Moins

depending on resource and timing constraints, you can use a cascade of adders, where you repeatdly add each bit starting with bit 7 to each other. this is slow because the critical path is on the Cin -> Cout. to improve, you can go further and use a 7bit decoder/any arrangement of decoder/column muxer + decoder for a lookup, which is essentially an SRAM array design to store this value so that next time you try to do this computation, you can directly access it. Essentially caching computation result. This requires extra circuitry overhead, but means you only have to compute the sums once. Moins

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