Добавил:
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:
Analysis and Application of Analog Electronic Circuits to Biomedical Instrumentation - Northrop.pdf
Скачиваний:
205
Добавлен:
10.08.2013
Размер:
4.41 Mб
Скачать

Digital Interfaces

405

The fast video DACs available in 2003 can convert at approximately 200 MSPS. For example the 10-bit, Analog Devices AD9732 DAC is rated at 200 MSPS and takes 5 ns to settle to one half LSB. The Faraday 8-bit DAC8001 also is rated at 200 MSPS and claims a 4-ns analog settling time.

10.4 Hold Circuits

Just as the sampling process can be interpreted in the frequency domain, so can the process of sample (or track) and hold of analog signals from a DAC. Note that the digital input to a DAC is generally periodic with period Ts. The DAC’s analog output from the nth digital input is generally held constant until the (n + 1)th input updates it. This process generates a stepwise output waveform if {bk} is changing and can be viewed as linear filtering operation. The impulse response of the zero-order hold (ZOH) filter to a unit input word at t = 0, and zero inputs at sampling instants thereafter is:

ho(t) = U(t) − U(t − Ts)

(10.28)

where U(t) is the unit step function, defined as zero for all negative argument, and 1 for all positive argument. ho(t) is thus a pulse of unit height and duration (0, Ts), else zero. The Laplace transform of ho(t) is:

Ho

(s) =

1

1

e−sTs =

1− e−sTs

(10.29)

 

 

s

 

 

s s

 

 

To find the frequency response of the ZOH, let s jω and use the Euler relation for e. The ZOH’s frequency response is easily shown to be:

H

() = T

sin(ωTs 2)

e− jωTs 2

(10.30)

 

o

s

(ωTs 2)

 

Note that the zeros in Ho() occur at ω = n2π/Ts r/s, where n = 1, 2, 3, …, and 2π/Ts is the radian sampling frequency of the system. Thus, the overall process of sampling an analog signal, x(t), and reconverting to (held) analog form by a DAC can be written in the frequency domain (neglecting quantization) as (Northrop, 1990):

sin(ωTs 2)

 

 

X() =

X(jω − jnωs )ejωTs 2

(10.31)

(ωTs 2)

n= 0

 

 

© 2004 by CRC Press LLC