Discussion in 'Homework Help' started by James4553, Jun 7, 2008.

1. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
Hello all

I need some help with the following questions.
Please don't just give the answer but rather show me the formula/method to solve the problems.

1. A tachometer displays engine RPMs from zero to 10,000. If the ADC accepts voltage levels from zero to five volts, in order to achieve full-scale representations from the tachometer, what should the output voltage be at 8500RPM?

2. An ADC converter has a binary input of 0010 and an analog output of 20mv. What is the resolution?

3. A given 4-bit digital to analog converter has a reference voltage of 15 volts and a binary inout of 0101. What is the proportionality factor?

4. The _______ of a DAC is the smallest change that can occur in the analog output as a result of a change in the digital input. (Fill in the blank)

I have more unanswered question but I'll leave it at that for now.

Many thanks.

Apr 5, 2008
15,806
2,389
3. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
Thank Bertus but that hasn't helped me.

Can anyone else please give me assistance? I have an exam coming up shortly...

4. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,402
1,227
What is the name of your coursebook?

5. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
We aren't using a coursebook in our classes.

So nobody can show me how to solve these problems?

6. ### roddefig Active Member

Apr 29, 2008
149
0
1. To utilize the full range of the ADC 0 RPM should be equal to 0 volts input and 10,000 RPM should be equal to 5 volts. Therefore, we can just divide 5 volts by 10,000 RPM and determine the number of volts/RPM.

$\frac{5\text{ V}}{10000\text{ RPM}} = 5 \text{ mV/RPM}$

To determine the voltage level for 8500 RPM we simply multiply the previous constant by 8500 RPM.

$8500 \text{ RPM}\,\cdot\, 5 \text{ mV/RPM} = 4.25 \text{ V}$

2. The resolution of an ADC is the smallest voltage that it can detect. This is typically expressed as V/LSB where LSB stands for least significant bit and is equivalent to the one. 0010 in binary is equal to two or two LSBs. To determine V/LSB we can simply divide the analog input voltage by the binary output value.

$\frac{20 \text{ mV}}{2 \text{ LSB}} = 10\text{ mV/LSB}$

Note that the input and output that you specified for this problem implied you were using a DAC, but the math works for either device.

3. I'm not sure what a binary inout or a proportionality factor is so I can't help you there.

4. You should be able to find this in your notes. I'm not sure what term your professor is looking for, but V/LSB is one answer.

Last edited: Jun 8, 2008
7. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
That's great, thank you so much.

Is an ADC conversion both more complicated and more time consuming than a DAC conversion?

And for #3 'binary inout' is a typo. I meant to write 'binary input'.

8. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,402
1,227
Max number for four bits is 15. Your reference is 15 V. Proportionality is 1V/bit.

Get yourself a recorder and record the lectures.

Last edited: Jun 8, 2008
9. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
A lot depends on your ADC and DAC methodology; for example a Flash ADC is very fast however has increased complexity, in terms of numbers of components.

Have you checked our Digital-Analog Conversion section in the e-book?

Dave

10. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Several analog to digital schemes use a DAC, they feed the DAC a number, compare it to the input, and adjust accordingly. Tends to be slow, but reliable.

11. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
That is a successive aproprimation ADC, where the speed is limited by the number of bits. Delta-Sigma is the standard these days as a fast and cost effective implementation (particularly common in DSP applications).

Dave