EE 5340

April 9, 1997

Instructor: C. Davila

1. Assume the unbonded strain guage wire has a guage factor of 2.1 and each resistance has an equilibrium value of 1 W. Determine the length of wire needed for each resistance if the strain guage is to have a sensitivity of 1 mv/mm. Assume the resistive bridge input voltage is 1 V and each resistance consists of a 5-turn coil of the wire.
   
   

   a 1 mm displacement should produce a 1 mV change in the bridge voltage. The change in each resistor wire length is:
   

   

   or

   for the resistive bridge,

   so

2. Consider the following differential temperature sensor:
   

   R_T1 and R_T2 are thermistors which measure temperatures T₁ and T₂, repspectively. Assume that T₂ = 298 ^oK, T₀ = 273 ^oK, b = 3000 ^oK, and R₀ = 100 W. Plot V_o as a function of T₁ for T₁ = 270, 280, 290, 300, 310, 320 ^oK. Then determine the independent nonlinearity expressed as a percentage of full-scale (320 ^oK), based on the 6 values of V_o. Assume the difference amplifier has high input resistance.
   

   The output of the difference is the voltage across R_T1, which is found by voltage division:

   

   vo_hat = mx + b (least squares line):

   nonlinearity = 0.002471 %

3. Describe what reactions take place and determine the voltage V between the silver chloride electrode and the iron electrode. Use the half-cell potentials in the lecture notes.
   

   

   AgCl electrode:

   Fe electrode:

   

   V = 0.222+0.447 = 0.669 V

4. Three evoked potential waveforms, x₁, x₂, and x₃, are to be averaged using two types of averages. The first is the conventional average discussed in lecture, the second is a weighted average. The signal power for x₁, x₂, and x₃ is 1, 9, and 4, respectively. The weighted average is found by multiplying each waveform by the square root of its signal power (1, 3, and 2, respectively) prior to averaging the three waveforms. Assume the noise in each waveform has zero mean, a variance of 1, and is uncorrelated across trials. Find the signal to noise ratio of the conventional average and the weighted average.

Conventional Average:

average of signal components is:

signal power in average: , assume that

average of noise components:

noise power in average: so SNR = 4/(1/3) = 12

Weighted Average:

average of signal components is:

signal power in average: , assume that

average of noise components:

noise power in average: so SNR = (14/3)²/(14/9) = 14

weighted average gives a higher SNR