A/D Converter exercise !HELP!

[ckck20]

Hi, I have the following problems:
1)We have an A/D converter. The analog signals have a maximum change of 2.5V/μs (2.5 V per 10^-6 second). The sampling frequency is 200KHz.The conversion range is from -2.5V to 2.5V and also the converter has 8 bits. Calculate the needed holding time of the S/H (sample and hold) and the conversion time of the A/D.

2)An A/D converter system has a conversion range from -12V to 12V and 10bits.
The system has S/H and an 8input multiplexer. The maximum frequency of the input signal is 4KHz and the sampling frequency is 20KHz per input. Calculate the needed holding time of the S/H and the conversion time of the A/D.

Please, tell me what to do. I have exams on Monday and the book I'm studuying doesn't doesn't say anything on these except that for a bipolar converter
f<=1/(2*pi*Dt*2^n) where f is the frequency of the input signal, n is the number of bits and Dt is the A/D conversion time or the S/H holding time.
Anything is appreciated.
Thanks in advance

 

[VVV]

I think it is about the ability of the A/D to follow the input, which means that the input voltage rate of change is limited. For the A/D to faithfully follow it, the input voltage should not change by more than 1 LSB during the conversion time.

Thus, for the case of the sinusoidal input, with an amplitude Um, the voltage is :
u(t)=Um*sin(ωt).

The rate of change is found by differentiation:
dv/dt=Um*ω*cos(ωt)

Its maximum is for ωt=0:
dv/dt=Um*ω=Um*2*π*f

If the converter has n bits and the input ranges from –Um to +Um, then:
1LSB=2*Um/(2^n)

Therefore, since the input must not change more than 1LSB during Dt, we can write:
dv/dt*Dt <= 1LSB

Um*2*π*f*Dt <= 2*Um/(2^n)

Therefore,
f<= 1/(2*π*Dt*(2^n-1))

I think this is what the book meant.

 

[ckck20]

Thanks for your response. Now it makes sense where that equation came from, but i think (from what I concluded from the net) that the input must not change more than 0.5LSB during Dt,so dv/dt*Dt <= 0.5LSB and this simplifies to the equation in my book.
When the sampling frequency is f then the conversion time of the A/D is 1/f ?
And in the above equation if put all the given data and solve for the Dt, that's the holding time of the S/H ?
Is this correct?
The clock is ticking...