Answer to Question #200835 in Electrical Engineering for Bahawal tahir

Question

a. A 20 V dc voltage is measured by analog and digital multi-meters. The analog instrument is on 

its 25 V range, and its specified accuracy is ± 2%. The digital meter has a 3½ digital display 

and an accuracy of ± (0.6+1). Determine measurement accuracy in each case.

b. A digital frequency meter has a time base from 1 MHz clock generator frequency divided by 

decade counters. Determine the measured frequency when a 1.512 kHz sine wave is applied 

and the time base uses (a) six decade counters and (b) four decade counters.

Answer 1

Part a

Analog instrument:

Voltage error $= ±2 \% *25 V = ±0.5 V$

error $=± \frac{0.5 V}{20 V}*100=±2.5\%$

Digital instrument for 20 V displayed on a 3 ½ digit display

1 Digit = 0.1 V

Voltage error = ± (0.6% of reading + 1 Digit)

= ± (1.2 V + 0.1 V)

= ± 0.22 V

error $=± \frac{0.22 V}{20 V}*100=±1.1\%$

Part b

(a) Using six decade counters:

Time base frequency (f1) = $1MHz/ (10*10*10*10*10*10) = 1Hz$

Time base period (T1) = $1/f1 = 1/1Hz = 1 second$

Input frequency period (Ti) = $1/fi = 1/1.512kHz$

Cycles counted = $T1/Ti = T1*fi = 1 second * 1.512kHz = 1512 cycles$

measured frequency on the display = 1.512 kHz.

(b) Using four decade counters:

Time base frequency (f2) = $1MHz/ (10*10*10*10) = 100Hz$

Time base period (T2) = $1/f2 = 1/100Hz = 10ms$

Input frequency period (Ti) $= 1/fi = 1/1.512kHz$

Cycles counted =$T2/Ti = T2*fi = 10ms * 1.512kHz = 15 cycles <lowest integer>$

Measured frequency on the display = 001.5kHz.