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EI quiz 1 spring 2011 solution

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Materials Science For Engineer (ENGR 1600)

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ENGR-

Electronic Instrumentation

Quiz 1

Spring 2011

Name ________________________

Section ___

Question I (20 points) ___________

Question II (20 points) ___________

Question III (20 points) ___________

Question IV (20 points) ___________

Question V (20 points) ___________

Total (100 points) ______________

On all questions: SHOW ALL WORK. BEGIN WITH FORMULAS, THEN

SUBSTITUTE VALUES AND UNITS. No credit will be given for numbers that

appear without justification.

Question I. Resistive circuits (20 points)

V2=12V, R10=R11=R12=6kΩ, R13=R14=R15=R16=R17=1kΩ

V R12 R11 R

R13 R14 R15 R

Find the total resistance of the circuit, seen from the voltage source. (6 pts)

The three resistances in parallel will add up to 2k which then will be in series with the 2k for the last two resistors at the right. Then the two sets of 2k resistors add in parallel to 1k. This then adds to the 3k in series to obtain 4k total.

Find the voltages across R12 and R17. (8 pts)

The total resistance across the terminals of R2 is 1k so the voltage is (1/4)12=3V The voltage across R17 will be half of this or 1

Find the currents through R13 and R11. (6 pts)

The current through R13 is 12/4k or 3mA

Half of this current goes through the three 6k resistors in parallel or a third through each one of the resistors or 0

200Vp-p

Time

-100V 20ms V(R1:1,R2:1)25ms 30ms 35ms 40ms 45ms 50ms 55ms 60ms 65ms 70ms 75ms 80ms 85ms 90ms 95ms 100ms

-50V

50V

100V

Doing an AC sweep, we can see that the four voltages are quite different. The first one shown on the previous page is across the capacitor.

Frequency

0V 1 V(R1:2,R2:2) 3 V(V1:+,R1:2) 10HzV(R1:2) V(R2:2)30Hz 100Hz 300Hz 1 3 10KHz 30KHz 100KHz 300KHz 1

20V

40V

60V

80V

100V

120V

To do this problem, evaluate the impedance of the capacitor and inductor at 60Hz

Z

j C j F j

= = =

1 1

2 60 0 63

1 4210

ω ( π )(. μ )

and Z = j ωL = j 2 π 60 0 01(. ) =j 38. so that the

magnitude of the capacitive impedance is large so can be treated as an open circuit. The impedance of the inductance is smaller than the two resistors in series, so the circuit is mostly resistive. If we neglect the inductive impedance then the p-p voltage across the capacitor will be about 1/6 of the source or about two times 20V. The first plot is the closest one to this (not exactly because the inductance is still relatively large)

What kind of filter response would best represent this circuit? (Please circle one.) (4 pts)

b) Band Pass at very low freq, C is open and L is a short so the load is across the second R so

some signal will pass, but it will be smaller. At high freq. C is a short, so nothing will pass. This

does operate like a low pass to some extent, because some signal will pass attenuated, so 2

points for that answer.

What kind of filter response would best represent this circuit if the inductor were removed, leaving the capacitor and two resistors? (Please circle one.) (4 pts) Hint: Sketch the new circuit diagram.

a) Low Pass The capacitor is open at low freq and a short a high freq so it is a low pass filter.

V FREQ = 60 VAMPL = 120VOFF = 0

R 50

R 10

C 0

What kind of filter response would best represent the original circuit if the capacitor were removed leaving the inductor and two resistors, with the output measured across R2? (Please circle one.) (4 pts) Hint: Sketch the new circuit diagram. There are two possible answers, if an explanation is provided.

a) Low Pass The inductor is open at high frequencies so no current flows through R2 or R and there is no voltage across R2. At low freq the inductor is a short so current flows through the resistors and the signal appears (attenuated some) across R2.

c) High Pass If the output is measured between R1 and R2, as it was in the other parts of this

problem, then the full voltage will appear at high frequencies. Full credit for this answer too.

V FREQ = 60VAMPL = 120

VOFF = 0

R 50

R 10 L110mH

B. We want to determine what type of filter circuit B is

What are the simplified transfer function, the magnitude, and the phase of circuit B at low frequencies? (3 points)

HBLO(jω) = 1 1

1 =

|HBLO| = 1

∠H BLO = 0

What are the simplified transfer function, the magnitude, and the phase of circuit B at high frequencies? (3 points)

HBHI(jω) = 2 LC

1

− ω

|HBHI| = 0

∠HBHI = π or - π

What type of filter is circuit B? (1 point)

Low Pass Filter

C. We want to know what the output of circuit A will look like for the input shown below

Time

-1 0s V(R1:1) 20us 40us 60us 80us 100us

Write an equation in the form Vin(t) = Ain sin(ωt + φin) which describes the input signal shown. (2 points)

Vin(t) = 800mV sin(40K π t - π/2)

If C=0μF and R=10K, what are the magnitude and phase of the transfer function of circuit A? (2 points)

|HA| = 0.

0. 079

12. 57 1

1 ( 12. 57 ) 1

1 ( 40 )( 10 )( 0. 01 ) 1

1 = 2 + 2 =

+

=

+

=

+

=

H

j RC j k k j

H

ω π μ

∠HA = -1 radians

∠(1) - ∠(1+j(12) = 0 - tan-1(12) = 0 - 1 = - 1.

What are the amplitude and phase of the output of circuit A, when the input signal from part C-1 is applied to the circuit? (2 points)

Aout = (0)(800m) = 63

φout = -1 – 1 = -3 radians

Find the value for RL that allows the circuit in 2) to see the same load on the primary (source) side of the transformer as in 1). (4 pts)

The load has to go down by a factor of 4 to 10Ω

Knowing that a real transformer’s behavior deviates from that of an ideal, what would be an appropriate minimum value for the inductance on the primary of the transformer in 1), given the source’s frequency of 60Hz? (4 pts)

a) 0

b) 0

c) 30mH

d) 300mH

Frequency

0V 1 V(RI:1) V(RI:2)3 V(RL:1)10Hz 30Hz 100Hz 300Hz 1 3 10KHz 30KHz 100KHz 300KHz 1

20V

40V

This question is basically a gift because the largest inductance will be the best choice to assure that the inductive impedance dominates.

Question V – Instrumentation, PSpice and components (20 points)

Function Gen 2

Function Gen 1

Channel 2

Channel 1

Shown above is the pinout diagram for the Mobile Studio. Shown also at the right are the 10

input/output connections we have used so far in the course.

Which three connections are used to input signals to channel 1 of the oscilloscope? Circle

and label your answer. (1pt)

Which three connections are used to input signals to channel 2 of the oscilloscope? Circle

and label your answer. (1pt)

Function Generator 1 is modified so that it outputs the triangle wave shown below as it is

observed on oscilloscope channel 1. What are the frequency, the period, the peak-to-peak amplitude and the offset voltage for this wave? (8pts)

3 cycles in 10x50us so the frequency is 3/0=6kHz, the period is 0.5ms/3=1

p-p amplitude is 4x50mV or 200mV or 0, offset is 2x50mV or 100mV or 0

Function Generator 1 is modified so that it outputs the square wave shown below as it is

observed on oscilloscope channel 1, saved as a comma separated file, then displayed in Excel. The frequency of this wave is 3kHz. The horizontal scale shown in the Excel plot is just the number of points so it does not really tell us anything about the wave. Label the plot with the correct time scale, assuming that t = 0sec occurs at the left end of the plot. (2pts)

For a frequency of 3kHz, the period is 0 or 3 cycles is 1ms

‐0.

14895 14218923628333037742447151856561265970675380084789494198810351082112911761223127013171364141114581505

0ms 0 1ms

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