Answer Happy

5. Compare the break point frequency from your Bodeplot, expressed in radians/sec, with the inverse of the timeconstant that you measured and calculated in Part 2. Discuss yourresults. Since the method for finding the break point frequencyintroduced here is somewhat vague, you should not be surprised ifthere is some error
Info below
part 2
Corner frequency = 1/T= 1/RC = 1k*0.1u= 10k rad/s
Transfer function= sT/(1+sT)= s/(s+10k)
Magnitude and phase Plot:
Transfer function based on excel by manual plotting:
Yes the corner frequencies are close. For both curves 3dBfrequencies are coming around 10k rad/s
4. The slope observed in starting = -24.054-(-44.037)= 20 db/decwhich true for first order circuits
Magnitude Phase (deg) ܗ ܘ ܬ -15- -25- -35- -45 100 90 80 4 V(PR1)/V(PR2) Ok 70 60 50 40 30- 20 10 -10 0. Ok V(PR1)/V(PR2) 20k 20k A 40k 40k Design1 AC Sweep Frequency (Hz) Frequency (Hz) 60k 60k 80k 80k 100k 100k
Mag (dB) BOSCH 4 3 4 4 5 5 0 -5 50 -10 -15 -20 -25 -30 -35 -40 -45 -50 -44.03657408 500 TF (in dB) -24.05351414 5000 -5.481472749 w (rad/s) 50000 0.108637899 -0.001099941 500000