PART B

1. Prove that the transfer function of the system shown in the figure above is

[

]

(

)

Proof:

Applying KCL to the non – inverting input (V1) ;

…

Applying KCL to the inverting input (V2) and using the Laplace transform for

capacitive impedance;

…

Simplifying Equation 1;

[

]

Simplifying Equation 2;

[

]

Simplifying equation 2 further;

[

]

(

)

Op-amp amplifies the difference between the inverting input and the non-inverting by

its open loop gain A, hence;

( )

Substituting the values obtained for V1 and V2;

( ) {(

)

(

)}

Simplifying;

(

)

(

)

Taking Vin and Vo common on both sides;

[ (

) ( )

( ) (

)] [

(

)]

[

(

)] = [

( ) (

)]

[

(

)]

( ) [(

) ]

[

(

)]

(

)

Hence proved.

2. Determine the expression for transfer functions assuming A = (GB/s). What is the

GB of TL082?

(

) (

)

The gain bandwidth product of the TL082 is 3MHz.

3. Simulation Results:

3.1. Include the connection diagram of the simulated system in your report.

3.2. Obtain the frequency response (magnitude and phase) plots and include them in

your report.

3.3. Calculate the GB from the plots

Maximum gain = 80.19dB

Gain at 3Db attenuation = 80.19 – 3 = 77.19dB = 7.36*103

Frequencies at 77.19dB:

1. f1 = 15.19kHz

2. f2 = 15.39kHz

Hence bandwidth = 0.2kHz

Therefore, gain bandwidth product = 1.44MHz

4. Hardware Implementation

4.1. Vary the input frequency and obtain the output voltage and phase difference

between input and output for different values of input frequency. Tabulate the

readings.

Input = 5mVp-p

Frequency (Hz)

Output Voltage (Vp-p)

Gain(A) Gain in dB Phase Difference in degrees

1k 50m 10 20.00 72

2k 100m 20 26.02 72

3k 150m 30 29.54 72

4k 200m 40 32.04 72

5k 260m 52 34.32 72

6k 340m 68 36.65 72

7k 420m 84 38.49 72

8k 520m 104 40.34 70

9k 650m 130 42.27 70

10k 830m 166 44.40 70

11k 1.1 220 46.84 68

12k 1.5 300 49.54 68

13k 2.2 440 52.87 68

14k 4 800 58.06 68

15k 19 3800 71.59 64

15.5k 20 4000 72.04 62

16k 7.8 1560 63.86 -64

20k 1.3 260 48.29 -64

30k 0.5 100 40.00 -64

40k 320m 64 36.12 -68

50k 240m 48 33.62 -70

60k 200m 40 32.04 -72

70k 160m 32 30.10 -72

80k 140m 28 28.94 -72

90k 120m 24 27.60 -72

100k 110m 22 26.85 -72

200k 50m 10 20.00 -72

300k 40m 8 18.06 -72

400k 30m 6 15.56 -72

500k 20m 4 12.04 -72

600k 18m 3.6 11.13 -72

700k 16m 3.2 10.10 -72

800k 14m 2.8 8.94 -72

900k 12m 2.4 7.60 -72

1M 10m 2 6.02 -72

4.2. Plot the frequency response (magnitude and phase plot) using the tabulated

readings and include them in the report.

4.3. Calculate the GB from the plots.

The gain bandwidth obtained from the magnitude plot is 2.5MHz.

5. Compare the theoretical value, simulation value and hardware implementation

value of GB.

Theoretical value = 3MHz (obtained from the datasheet)

Simulation value = 1.44MHz

Hardware implementation value = 2.5MHz

6. Apply square wave of 1V and 100Hz to the system you built using ASLKv2010 Starter

Kit. Study the transient response on the oscilloscope and explain the nature of the

plot you obtained. Capture the hardware set-up, including the CRO plot in the form

of picture and include it in the report.

The transfer function of the system is:

(

)

This is of the form:

( ) ( )

( )

The characteristic equation of the system is given by D(s) = 0, which is

(

)

(

) (

)

Substituting values for C, R1 and GB from the simulated graphs

(( )( )

( )) (

( )

( )

( ) ( ))

Solving the equation, we get the roots of the equation as

s = -727.27+ ( )

s = -727.27- ( )

We know from control system theory that the roots of the characteristic equation are

the poles of the transfer function.

The positions of the poles in the s- plane determine the stability of the system. Hence

we can deduce the following:

The imaginary part of the poles that exist on both sides the imaginary axis as

complex conjugates add a sinusoidal component to the transfer characteristic.

The negative real part adds a decaying exponential component to the transient

characteristic.

Hence the transient characteristic displays an exponentially decaying sinusoidal.

7. Conclusions

7.1. What challenges did you face in designing the circuit?

The circuit schematic design was provided, and therefore we did not have to design the

circuit.

7.2. What challenges did you face during simulation?

Problems:

We faced problems in using PSPICE to simulate the circuit because the simulation file

for the TL082 was unavailable in the standard libraries.

Even upon downloading the required files from the Texas Instruments website, we

were still unable to obtain the required results.

Solution: We simulated the circuit on TINA.

Using TINA for this was a breeze as it was very simple and intuitive. All the standard library

files for the TL082 were already installed and it was a simple matter of dragging and

dropping components and generating the transient with the simple click of a button.

7.3. What challenges did you face during practical realization?

The problems that we faced during practical realization of the given circuit were two-fold.

i. Lack of male header pins

Problem: There were no male to female header wires included in the kit. This proved to be a problem as we could not make direct connections from the IC pins to the bread board part of the kit. As a result, we could not make optimum use of the bread board for placing the resistors as we had intended to. Solution: We made minimal use of the bread board and instead, we directly connected the resistor and capacitor leads to the header pins.

Advantage: We were able to connect the resistors and capacitors. Also, we made minimal use of single strand wires and other components which were external to the kit. Disadvantages: The disadvantage of the above solution was that we could not make the best use of the bread board. Also, this resulted in excessive number of wires on the kit making it slightly more difficult to debug any errors in connections.

ii. Non-standard power supply

Problem: The power supply of +10V/-10V for the ASLKv2010 Starter Kit were non-standard in comparison to the standard +5V/-5V and +12V/-12V power sources that were available in our labs. Solution: We used two separate variable power sources to supply the required +10V/-10V to the kit. Disadvantage: Due to the connection of two separate variable power sources, there is a possibility of an offset occurring in the output. Care had to be taken to make sure that both the variable voltage sources were supplying the exact same voltage.

7.4. Summarize any new learning based on your experience.

We gained a lot of knowledge in the practical application of control system theory

specifically to electronic circuit design.

We learned the difference between unity gain bandwidth of an open loop op-amp

and the gain bandwidth of a closed loop op amp.

On the software side, we learnt how to use the Tina - TI software in order to get the

transfer characteristics of a given circuit.

Additional Note:

We have made a small video presentation indicating the problems we faced during the

hardware implementation of the experiment. It is available at the following link:

http://youtu.be/SbjZt7k7Ig4