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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 Systems—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 MΩ, one 1 kΩ, one 2 kΩ resistors
One 0.1µF capicator
One simple on/off switch (if available)
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 Systems—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. 0.1µF
7 1 5 3 Vin 1MΩ 741 Vout 6 2
4 1MΩ
-15V 2kΩ 15V
5 1 7
3 6 741
2 1kΩ 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 Systems—Lab #4 Page 3 1
Zi 1
V ) −Z f )
(¿ ¿ ¿ 1+
Z i 2 ¿2 (
V out (summing)=¿
Zi ( ) V out (non−inverting)=V ¿ 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
−Z f
V ¿1
summing ( ) First calculate Zf for the inverting summing amplifier. Note that the parallel resistor equation is used to calculate the
parallel impedances. R∗1
Z f=
Z R +Z C
RCs+ 1
A summing =
RC ( )
− Enter the given values of resistance and capacitance to get A summing = −10
10+ s After this, calculating the noninverting input is simple. V out
2k Ω
( noninverting )= A
= 1+ f =1+
1k Ω
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 Systems—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)
1−G ( s )∗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 −10
10+ s+30 s +40
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=−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 Systems—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 Systems—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 Systems—Lab #4 Page 7 Step 2
Now reconfigure the circuit so that it looks like the following. The short across the 0.1 μF 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 0.1µF
7 1 5 3 741 Vout 6 2
4 1MΩ
-15V 2kΩ 15V
5 1 7
3 6 741
4 -15V 1kΩ ECET345 Signals and Systems—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
Voltage (V) 6
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 Systems—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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