Question
Given taple 3,please solve for part(d)-part(f)
hange the mode on the signal generator to sine-wave, with a peak-to-peak amplitude of eak-to-peak amplitude of the voltage ac
ange the mode on the signal generator to sine-wave, with a peak-to-peak amplitude of 2V. Measure the 1500, 2000, and 5000 Hz)
hange the mode on the signal generator to sine-wave, with a peak-to-peak amplitude of eak-to-peak amplitude of the voltage across the capacitor for various frequencies (100, 4 500, 2000, and 5000 Hz), and complete Table 3, (0.4%) Table 3 Frequency (in Hertz) 100 400 800 1000 1500 2000 5000 Vc (peak-to-peak)
ange the mode on the signal generator to sine-wave, with a peak-to-peak amplitude of 2V. Measure the 1500, 2000, and 5000 Hz), and complete Table 3 (04%) amplitude of the voltage across the capacitor for various frequencies (100, 400, 800, 1000, Table 3 Frequency (in Hertz) 100 400 800 1000 1500 2000 5000 vc (peak-to-peak) (d) FromTable 3, sketch the frequency response curve (-,1 versus frequency) of the arout (04%) o-S Plot of vs frequency s 0 707 of the e) From part (a), estimate the -3 dB frequency, c, fo is the frequency at which the gain fc 1045 gain in the passband. (04%) What observation can be made between the time (transient) response and the frequency response? (0 5%) (f)
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Answer #1

(d)

(e) to check for -3dB frequency, the gain should be about 0.707 which on the above graph lies slightly beyond 1000 Hz

we take frequency to be 1045 Hz

(f) The transient response of a sine wave signal generator has the same pattern with change in frequency. The peak to peak voltage time varies with change in frequency for transient response. The characteristic of the curve does not change.

The frequency response graph shows that the gain decreases with increase in frequency. It is a low pass filter.

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