Explore BrainMass
Share

# Bode Plot for a Transfer Function

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Q9A 4: 1 step down transformer uses a 60 Vac source on the primary windings. There is 120 mA flowing through the load resistance on the secondary windings. That is the resistive load?

a. 500 Ohms
b. 2 K-Ohms
c. 125 Ohms

Q10: Sketch a Bode plot for the following transfer function (see attached).

Q11: The outputs of two "NAND" gates are connected to the inputs of an "EXCLUSIVE OR" gate. Each of the "NAND" gates has one input at logic high level, and the other input at a logic low level. What is the output of the "EXCLUSIVE OR" gate>

a. Logic high
b. Logic low
c. None of the above

Q12: A fully loaded 440 V, three phase motor draws 12.4 A at 86% power factor (pf). Its speed is 1730 rpm and its torque output to a mechanical load is 29.1 lb-ft x 1.356 = 39.5 Nm. What is: 1. The output horsepower? 2. The efficiency?

a. 1. Horsepower = 7.15 hp; 2. Efficiency = 84%
b. 1. Horsepower = 9.58 hp; 2. Efficiency = 88%
c. 1. Horsepower = 8.35 hp; 2. Efficiency = 86%

Q13 Consider the circuit shown in the following figure: Vs is a sinusoidal input source waveform with zero (0) DC value.

a) Which diodes are ON when Vz > 0?
b) Which diodes are ON when Vs < 0?
c) What is the average value of the load voltage, VL.

a. a) D2 and D4, b) D1, D3 and D5, c) 0.25 * Vs
b. a) D1, D3, and D5, b) D2 and D4, c) 0.5*Vs
c. none of the above

https://brainmass.com/physics/dc/bode-plot-for-a-transfer-function-116933

#### Solution Preview

Q9. Turn Ratio = 4:1
Primary Voltage = 60 V
Hence, Secondary Voltage = 60/4 = 15 V
Current thro' Secondary = 120 mA = 0.12 A
Hence resistive load = 15/0.12 = 125 Ohms ==> Ans (c)
--------------------------------------------------------------------
Q10) Corner frequencies correspond to the poles of the Transfer function; i.e. where the denominator goes to zero.

Hence corner frequency F1 = 10,000 = 10^4 Hz
corner frequency F2 = 1,000,000 = 10^6 Hz

From the given Magnitude ...

#### Solution Summary

The solution creates a bode plot for a transfer function.

\$2.19