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(Solved) ECET345 Signals and SystemsLab #4 Page 1 DeVry University ECET345 Signals and Systems Processing Name of Student...

This lab are due by the end of this week. I was wondering if I could get some help with these. I have another class that I'm currently behind in, so I'm trying to finish those up as well. Thank you!!

ECET345 Signals and Systemsâ€”Lab #4 Page 1 DeVry University

ECET345 Signals and Systems Processing

Name of Student ___________________________________________________________

Experimental observation of stable and unstable systems Objective of the lab experiment:

Given two op-amp circuits, one stable and the other unstable, calculate their transfer functions and experimentally

observe their step response. Equipment list: Two LM741 Op-amps

Three 1 Mâ„¦, one 1 kâ„¦, one 2 kâ„¦ resistors

One 0.1ÂµF capicator

One simple on/off switch (if available)

Oscilloscope

Power supply with +15 V and -15 V outputs

Function generator

Thee BNC to alligator Theory

When designing any component for engineering, it is best to have a theoretical model to base the design on. This allows

the designer to come up with a good idea of what to build before any testing occurs. This is no different in electrical

engineering, where designing filters is a crucial aspect of the design process. This lab will cover how to model analog

components in the Laplace domain and the significance of the equations.

First examine each of the analog components; this can be a resistor, capacitor, or inductor: Resistance=Z R=R

Capacitance=Z C = 1

Cs Inductance=Z L =Ls

where s is the Laplace variable, R is the resistance in ohms, C is the capacitance in farads, and L is the inductance in

Henrys. These numbers are known as the impedance of the component, and are usually denoted by the capitol letter Z. ECET345 Signals and Systemsâ€”Lab #4 Page 2 Combining this knowledge with other familiar equations such as the voltage divider, current divider, or any op-amp

circuit allows much more complex circuits to be analyzed, such as the circuit shown below. 0.1ÂµF

1MÎ©

15V

7 1 5 3 Vin 1MÎ© 741 Vout 6 2

4 1MÎ©

-15V 2kÎ© 15V

5 1 7

3 6 741

2 1kÎ© 4 -15V ` This may seem like an insurmountable problem at first. However, by breaking this complex filter into smaller ones, the

answer can be easily calculated. First note that the output of the filter is connected back to the input. This is a telltale

sign of a closed loop filter, which has its own method to calculate the transfer function. This will be discussed later,

because it is the last step in finding the closed loop transfer function (CLTF). Second, note the two op amps in the circuit;

each of these is its own filter with a special transfer function. The top one can be identified as an inverting summing

amplifier while the bottom can be identified as a noninverting amplifier. These have the characteristic equations as

follows. ECET345 Signals and Systemsâ€”Lab #4 Page 3 1

V

Zi 1

1

V ) âˆ’Z f )

(Â¿ Â¿ Â¿ 1+

Z i 2 Â¿2 (

V out (summing)=Â¿

Zf

Zi ( ) V out (nonâˆ’inverting)=V Â¿ 1+ Now fill in the values for each of the impedances to get the equations (ignore V in2 for the time being, it will reappear in

the way the circuit is modeled). V out

âˆ’Z f

(summing)=A

=

V Â¿1

Zi1

summing ( ) First calculate Zf for the inverting summing amplifier. Note that the parallel resistor equation is used to calculate the

parallel impedances. Râˆ—1

1

Z Râˆ—Z C

Cs

R

C

Z f=

=

=

=

Z R +Z C

1

RCs+ 1

1

R+

s+

Cs

RC

1

C

âˆ—1

1

1

s+

âˆ’

RC

RC

A summing =

=

R

1

s+

RC ( )

âˆ’ Enter the given values of resistance and capacitance to get A summing = âˆ’10

10+ s After this, calculating the noninverting input is simple. V out

Z

2k â„¦

( noninverting )= A

= 1+ f =1+

=3

VÂ¿

Zi

1k â„¦

noninverting ( ) Knowing that the circuit is closed loop, finding the CLTF is important in how the circuit will react to different inputs. A

closed loop circuit is modeled as follows: ECET345 Signals and Systemsâ€”Lab #4 Page 4 Vin G(s) + Vout + H(s) In the picture, G(s) is the open loop transfer function and H(s) is the feedback gain. The circle signifies that there is a

summing function in the circuit and the + symbol shows the sign of the values going into it. This has a profound effect on

the output of the system. For such a system, the CLTF is calculated by the following equation: CLTF= G( s)

1âˆ’G ( s )âˆ—H ( s ) Note that the sign of the G(s)*H(s) term is the opposite of what appears on the H(s) terminal on the summing device.

All that is left is to plug in the values we calculated previously. In this case, the summing amplifier is G(s) and the

noninverting amplifier is H(s). Because the gain of Vin1 and Vin2 are the same for the summing amplifier, they are just

added together instead of weighted in addition. The equation that comes out is âˆ’10

10+s

âˆ’10

âˆ’10

CLTF=

=

=

10+ s+30 s +40

âˆ’10

1âˆ’

âˆ—3

10+ s ( ) That is the transfer function for the entire circuit; important information such as stability and response can be calculated

with this information. Stability can be calculated by simply finding the poles and zeroes of the function. Poles occur when

the function approaches infinity and zeroes occur when the function equals zero. In other words, if the variable s can

make the numerator equal zero, then a zero is present, and if s can make the denominator equal zero, then a pole is

present. Note that s can take on imaginary values so any equation with s in the denominator will have a pole. For this

equation the pole can be located at the location below. poleat s=âˆ’40

Stability only focuses on the poles, because if the real part of s is negative, the system is said to be stable. For systems

with more than one pole, all of the poles must have negative real parts in order for the system to be considered stable. ECET345 Signals and Systemsâ€”Lab #4 Page 5 Procedure:

Step 1

Build the system shown in the theory section and observe its step response on the oscilloscope. Set up the function

generator for the following values and apply this to V in.

Vpp 5V

Frequency 1 Hz

Signal Type Square wave

The observed response should look like the following: Stable Circuit Step Response

3 2 1 Voltage (V) 0 0 0.2 0.4 0.6 0.8 1 1.2 Input Voltage

Output Voltage -1 -2 -3

Time (s) The system is indeed stable, just as predicted in the theory section. Note that the output voltage approaches a constant

value and does not saturate the op amps. Paste a photo of the oscilloscope when it shows the response of the circuit which

should look similar to the graph above. ECET345 Signals and Systemsâ€”Lab #4 Page 6 Now answer the following question.

1. What is the ratio of output to input after the system has reached steady state, or equilibrium (i.e., when the

output shows a flat response)? What is the number telling us about the circuit? (Hint: look at the transfer

function.) ECET345 Signals and Systemsâ€”Lab #4 Page 7 Step 2

Now reconfigure the circuit so that it looks like the following. The short across the 0.1 Î¼F cap is created using a toggle

switch. Until instructed otherwise, keep the toggle switch closed. If you do not have a switch, connecting a wire and

removing it when the switch needs to be open will work. Key = A 0.1ÂµF

1MÎ©

15V

7 1 5 3 741 Vout 6 2

4 1MÎ©

-15V 2kÎ© 15V

5 1 7

3 6 741

2

4 -15V 1kÎ© ECET345 Signals and Systemsâ€”Lab #4 Page 8 Measure the output on an oscilloscope; it will be a constant zero. However when the switch is opened, the output may

look as below (if the reading is hard to capture, reset the system by closing the switch, and then open the switch to

observe the response). Unstable Circuit Response with no input applied

14

12

10

8

Voltage (V) 6

4

2

0 0 0.2 0.4 0.6 0.8 1 1.2 -2

Time (s) Note: The observed output may approach a negative number, depending on the op-amp used. This is indicative of an unstable system. Note that the output voltage approaches a high value even though there is no

input applied. This is due to a small offset voltage which is always present at the input of the op amp. Also note that each

op amp has different offset characteristics; therefore, the observed output may approach a negative maximum instead of

the positive maximum output shown above. Paste a photo of the oscilloscope when it shows the response of the circuit which

should look similar to the graph above. ECET345 Signals and Systemsâ€”Lab #4

Now answer the following question.

Prove that the system is unstable by deriving the transfer function and finding the poles and zeroes. Page 9

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