RF Hardware Engineer was asked...December 1, 2011

↳

IIP2 & DC Offsets

↳

IN the digital domain, zero crossings and ISI , as well as DC offset can be compromised.Loss of received signal strength with distortion, Overheating and damage to TX drivers Less

↳

Worse is not working.

↳

What he really wanted me to do was draw the schematic of an ADC input circuit. I don't know why he didn't just ask that in the first place. Since it was a broad question, I spent over 5 minutes explaining binary representation, word size (8/16/32bit), endian-ness, and finally when I mentioned ADCs, he latched onto that and tried to get me to dig deeper. So I drew a block diagram of an ADC, focusing mostly on the interface to the microcontroller, and the channel select multiplexer. He grew increasingly frustrated and tried to get me to draw more detailed sub-circuits for each of the blocks in my diagram. Finally, I got to the actual ADC front end, and I drew a classic textbook R2R ladder circuit, and he seemed satisfied with that. By that point we had spent over 10 minutes in an interview where he was already complaining that he had a lot of ground to cover in limited time. I was so frazzled by the question that I did not even think of mentioning a successive approximation ADC as an alternative to the R2R, even though it's a much more common implementation in my experience. Less

Hardware Engineer was asked...July 19, 2017

↳

The pilgrim would enter with 7 flowers and would have worshipped each god with 8 flowers each. Less

↳

explain me how to find

↳

The pilgrim would enter the village with 7 flowers and would have worshipped each god with 8 flowers each. Less

Senior Hardware Engineer was asked...August 1, 2014

↳

@Don you are correct up until your conclusion that "Each bulb will be a little dimmer than in parallel, but 100W still brighter." The 100W bulb is 10x dimmer if you assume resistance does not depend on current. Reality is even worse since R increases with current. Since the 10W bulb has 10x higher resistance than 100W bulb, it will absorb 10x as much power in the voltage divider: 100W bulb = 1A at 100V, or 100 ohms 10W bulb = 0.1A at 100V, or 1000 ohms 100W bulb + 10W bulb in series driven off 110V line is 1100 ohms carrying 0.1A of current. Power delivered to 10W bulb is P = I^2 x R = (0.1A)^2 x (1000 ohm) = 10W. Power delivered to 100W bulb is P = (0.1A)^2 x (100 ohm) = 1W. So the 100W bulb only receives 1W delivered power because the 10W bulb absorbs all the power in the divider. The 10W bulb will be brighter. Less

↳

1. When connected in series: In a series connection, current flowing across each element is same. So when 40W bulb and 60W bulb are connected in series, same current will flow through them. To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation P=(I*I) R since current is same we can say that power dissipation will be higher for the bulb with higher resistance i.e. 40W bulb. Hence 40W bulb will glow brighter in series connection. 2. When connected in parallel: In a parallel connection, voltage across each element is same. So when 40W bulb and 60W bulb are connected in parallel, voltage across them will be same (100 V in the given case). To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation P=(V*V)/R since voltage is same we can say that power dissipation will be higher for the bulb with lower resistance i.e. 60W bulb. Hence 60W bulb will glow brighter in parallel connection. Less

↳

I have attended the same question, 40W & 100W bulb with 110V supply , all are them connected in serial Answer : Who is having more resistance, he will have more power drop also he is the brighter one. Here 100W will have less resistance and 40W bulb has more resistance, So 40W will be glow brighter than 100W bulb R1 = 110 x 110 / 100 = 2.2 ohm R2 = 110 x 110 / 40 = 302.5 ohm Answer is 40W bulb will be brighter than 100W bulb Less

Senior Hardware Verification Engineer was asked...January 30, 2010

↳

Write the number in binary and count the number of ones in that.If the number os ones is only 1 then it the number is indeed a power of 2 Less

↳

For an integer n: If n is less than 1, return false. If the bitwise & of n and n-1 is 0, return true. Otherwise, return false. Less

↳

I think the main idea is to use recursion function, for the integer which is larger than 0, if it is 1 return true, else return function(n-1) Less

Staff Hardware Engineer was asked...May 2, 2013

↳

The answer should be YES: @gustion: your example with 49 (7 pwr2) is correct but with 25 (5 pwr2) is incorrect. 7 in binary is 111 (3 1-bits is half of 6 bits), but 5 in binary is 101 (2 1-bits does not equal half of 6 bits). In general, any binary number with n-bits, half of which are 0's and half are 1's is a square of a binary number with half the number of bits, all being 1-value bits. In addition, the number's magnitude will be n/2-1 1-bits followed by n/2 0-bits followed by the last 1-bit. For example, lets say we have a 16-bit number. The number which will have 8 1-bit and 8 0-bit binary digits and also be a square is: 1111 1110 0000 0001 (7 1-bits followed by 8 0-bits followed by 1-bit) and this number is a square of 1111 1111 Binary number with 600 bits and 300 1-bits will have a magnitude of 299 1-bits followed by 300 0-bits followed by 1-bit, and it will be a square of a 300-bit number with the magnitude of 300 1-bits. Less

↳

Well on a simple note, 9's binary is 1001, a 4 digit binary with two 1's and two 0's, and is a perfect square. The same analogy should also be true for any such number. Less

↳

No

Electrical Hardware Design Engineer was asked...February 14, 2018

↳

A: A capacitor looks like a capacitor, inductor, and resistor in series at high frequency. Less

↳

A short

↳

Its an short circuit at very high frequencies. Look at the impedance for a capacitor and you'll notice that the frequency is inversely related to it. (also note that capacitor impedance is purely reactive). Therefore, as frequency increases the impedance approaches 0, which acts like a short. (an inductor is the opposite) Less

Hardware Engineer was asked...June 25, 2009

↳

a 32 bit number can have unique values from 0 to (2^32)-1. Meaning it can have max value of upto (4 Giga -1). If there are more than 4 Giga locations, then there is repetition which is obvious, even before one starts the question. Usually, questions of repetition can be solved by "XOR"ing all the values. If the value is not zero then some number is repeated. One should solve around that angle. Less

↳

I think the employer was asking about the binary search tree algorithm. This is how I would have answered it. 1. First create a binary tree for all the data(which can be 4 gb only) 2. Every time a new data is to be inserted into the memory, we should traverse the tree to search if the element is present: if present, return "element already stored and so we cannot insert it again" else, insert the element into its appropriate place. Since, its a binary tree we can search an element in (log n) time and with minimum number of passes. Less

↳

Good luck with that. You'd better come up with an answer in <10 minutes.

Hardware Design Engineer was asked...July 15, 2016

↳

Zero..since dc can't produce emf in windings.

↳

Zero. Because, Transformer never allow DC power.

↳

Core Gets Saturated. Lol !

Hardware Engineer was asked...May 2, 2018

↳

Signal reflections, power loss.

↳

Signal reflection, bit error, EMI problem

↳

Crosstalk, signal reflection or defrection on PCB or SMT ; are some effect of an impedance mismatch high speed circuit Less

electrical engineerverification engineernetworking engineerfpga engineerembedded engineerapplication engineercomputer architectcomputer engineerelectronics engineersystem engineerdesign engineervalidation engineeric engineersignal integrity engineerstaff engineeranalog engineerfirmware engineerproduct engineer