ECET345 Signals and Systems—Lab #2 Page 2

DeVry University – Lab 2

ECET 345 Signals and Systems Processing

Response of RC filters

Objective of the lab experiment:

The objective of this lab is to 1) use Laplace Transform method to find the impulse and step response of an RC circuit, and 2) compare it to the known responses found from solving differential equations.

Theory

Recall that an impulse is a short, high burst of energy, and is the derivative of the step response. These two functions may seem arbitrary in the time domain, but in the Laplace domain they are invaluable for the analysis of circuits.

For example, the following circuit is a generic analog, first-order, low-pass filter.

Procedure:

Step 1 – TF

Calculate the transfer function of the circuit above symbolically using the voltage divider rule. The transfer function (TF) of a circuit is defined as the ratio . For time-invariant systems it contains all of the information about what the system does to any given input signal to transform it into an output signal.

Show all of the solution steps and the final result of your computations. (Hint: Use the fact that Vout/Vin = Zout/Zin and the transformed device impedance values to get an equation for that is a function of s, R and C only, as we did in this week’s lesson. Correct answers will have a numerator that is a constant, containing R1 and C1 only, and a denominator that is first-order in the Laplace parameter s. )

Step 2 – Unit Step Response – Show all solution steps clearly

a) Calculate Vout(s) when a unit step input is applied. (Hint: for Vin(s) recall that the Laplace transform of a unit step input u(t) is equal to . Then,

b) invert Vout(s) back into time domain by using inverse Laplace transform tables with either partial fractions or MATLAB ilaplace(V) command .) When done correctly your time domain result vout(t) should be the familiar RC charging voltage in time given by

vout(t) = vin ( 1 – e^(-t/RC) ), where Vin = 1 Volt in this case of a unit step.

Step 3 – Impulse Response – Show all solution steps clearly

a) Calculate Vout(s) when a unit impulse is applied. (Hint: for Vin(s) recall that the Laplace transform of a unit impulse is just equal to 1 .

b) Find vout(t) by invert Vout(s) back into time domain (simply use inverse Laplace transform tables in this case.)

Step 4

Use MATLAB to plot both the impulse and step responses v(t) that you computed in steps 2) and 3) above. You must label which plot is which; label your axes as well. Let R = 1 and C = 0.5 for this. (Hint: Slide 34 in the Week 2 Video Tutorial pt.2 could be of help in doing this.)

Paste Plots here

Questions

1. How does the value of the time constant of RC circuit change when the value of the capacitor is doubled?

2. Would your results in step 4 have been any different from when R=1 and C=0.5 had you been given the values for R = 1 MΩ and C = 0.5 μF ? Explain.

Prof. Ajeet Singh, Ph.D Richard Butler, EET modified 1-19-2019kj

R1

C1

VoutVin

ECET345 Signals and Systems—Lab #2 Page 2

DeVry University – Lab 2

ECET 345 Signals and Systems Processing

Response of RC filters

Objective of the lab experiment:

The objective of this lab is to 1) use Laplace Transform method to find the impulse and step response of an RC circuit, and 2) compare it to the known responses found from solving differential equations.

Theory

Recall that an impulse is a short, high burst of energy, and is the derivative of the step response. These two functions may seem arbitrary in the time domain, but in the Laplace domain they are invaluable for the analysis of circuits.

For example, the following circuit is a generic analog, first-order, low-pass filter.

Procedure:

Step 1 – TF

Calculate the transfer function of the circuit above symbolically using the voltage divider rule. The transfer function (TF) of a circuit is defined as the ratio . For time-invariant systems it contains all of the information about what the system does to any given input signal to transform it into an output signal.

Show all of the solution steps and the final result of your computations. (Hint: Use the fact that Vout/Vin = Zout/Zin and the transformed device impedance values to get an equation for that is a function of s, R and C only, as we did in this week’s lesson. Correct answers will have a numerator that is a constant, containing R1 and C1 only, and a denominator that is first-order in the Laplace parameter s. )

Step 2 – Unit Step Response – Show all solution steps clearly

a) Calculate Vout(s) when a unit step input is applied. (Hint: for Vin(s) recall that the Laplace transform of a unit step input u(t) is equal to . Then,

b) invert Vout(s) back into time domain by using inverse Laplace transform tables with either partial fractions or MATLAB ilaplace(V) command .) When done correctly your time domain result vout(t) should be the familiar RC charging voltage in time given by

vout(t) = vin ( 1 – e^(-t/RC) ), where Vin = 1 Volt in this case of a unit step.

Step 3 – Impulse Response – Show all solution steps clearly

a) Calculate Vout(s) when a unit impulse is applied. (Hint: for Vin(s) recall that the Laplace transform of a unit impulse is just equal to 1 .

b) Find vout(t) by invert Vout(s) back into time domain (simply use inverse Laplace transform tables in this case.)

Step 4

Use MATLAB to plot both the impulse and step responses v(t) that you computed in steps 2) and 3) above. You must label which plot is which; label your axes as well. Let R = 1 and C = 0.5 for this. (Hint: Slide 34 in the Week 2 Video Tutorial pt.2 could be of help in doing this.)

Paste Plots here

Questions

1. How does the value of the time constant of RC circuit change when the value of the capacitor is doubled?

2. Would your results in step 4 have been any different from when R=1 and C=0.5 had you been given the values for R = 1 MΩ and C = 0.5 μF ? Explain.

Prof. Ajeet Singh, Ph.D Richard Butler, EET modified 1-19-2019kj

R1

C1

VoutVin

ECET345 Signals and Systems—Lab #2 Page 2

DeVry University – Lab 2

ECET 345 Signals and Systems Processing

Response of RC filters

Objective of the lab experiment:

The objective of this lab is to 1) use Laplace Transform method to find the impulse and step response of an RC circuit, and 2) compare it to the known responses found from solving differential equations.

Theory

Recall that an impulse is a short, high burst of energy, and is the derivative of the step response. These two functions may seem arbitrary in the time domain, but in the Laplace domain they are invaluable for the analysis of circuits.

For example, the following circuit is a generic analog, first-order, low-pass filter.

Procedure:

Step 1 – TF

Calculate the transfer function of the circuit above symbolically using the voltage divider rule. The transfer function (TF) of a circuit is defined as the ratio . For time-invariant systems it contains all of the information about what the system does to any given input signal to transform it into an output signal.

Show all of the solution steps and the final result of your computations. (Hint: Use the fact that Vout/Vin = Zout/Zin and the transformed device impedance values to get an equation for that is a function of s, R and C only, as we did in this week’s lesson. Correct answers will have a numerator that is a constant, containing R1 and C1 only, and a denominator that is first-order in the Laplace parameter s. )

Step 2 – Unit Step Response – Show all solution steps clearly

a) Calculate Vout(s) when a unit step input is applied. (Hint: for Vin(s) recall that the Laplace transform of a unit step input u(t) is equal to . Then,

b) invert Vout(s) back into time domain by using inverse Laplace transform tables with either partial fractions or MATLAB ilaplace(V) command .) When done correctly your time domain result vout(t) should be the familiar RC charging voltage in time given by

vout(t) = vin ( 1 – e^(-t/RC) ), where Vin = 1 Volt in this case of a unit step.

Step 3 – Impulse Response – Show all solution steps clearly

a) Calculate Vout(s) when a unit impulse is applied. (Hint: for Vin(s) recall that the Laplace transform of a unit impulse is just equal to 1 .

b) Find vout(t) by invert Vout(s) back into time domain (simply use inverse Laplace transform tables in this case.)

Step 4

Use MATLAB to plot both the impulse and step responses v(t) that you computed in steps 2) and 3) above. You must label which plot is which; label your axes as well. Let R = 1 and C = 0.5 for this. (Hint: Slide 34 in the Week 2 Video Tutorial pt.2 could be of help in doing this.)

Paste Plots here

Questions

1. How does the value of the time constant of RC circuit change when the value of the capacitor is doubled?

2. Would your results in step 4 have been any different from when R=1 and C=0.5 had you been given the values for R = 1 MΩ and C = 0.5 μF ? Explain.

Prof. Ajeet Singh, Ph.D Richard Butler, EET modified 1-19-2019kj

R1

C1

VoutVin

DeVry University – Lab 2

ECET 345 Signals and Systems Processing

Response of RC filters

Objective of the lab experiment:

The objective of this lab is to 1) use Laplace Transform method to find the impulse and step response of an RC circuit, and 2) compare it to the known responses found from solving differential equations.

Theory

Recall that an impulse is a short, high burst of energy, and is the derivative of the step response. These two functions may seem arbitrary in the time domain, but in the Laplace domain they are invaluable for the analysis of circuits.

For example, the following circuit is a generic analog, first-order, low-pass filter.

Procedure:

Step 1 – TF

Calculate the transfer function of the circuit above symbolically using the voltage divider rule. The transfer function (TF) of a circuit is defined as the ratio . For time-invariant systems it contains all of the information about what the system does to any given input signal to transform it into an output signal.

Show all of the solution steps and the final result of your computations. (Hint: Use the fact that Vout/Vin = Zout/Zin and the transformed device impedance values to get an equation for that is a function of s, R and C only, as we did in this week’s lesson. Correct answers will have a numerator that is a constant, containing R1 and C1 only, and a denominator that is first-order in the Laplace parameter s. )

Step 2 – Unit Step Response – Show all solution steps clearly

a) Calculate Vout(s) when a unit step input is applied. (Hint: for Vin(s) recall that the Laplace transform of a unit step input u(t) is equal to . Then,

b) invert Vout(s) back into time domain by using inverse Laplace transform tables with either partial fractions or MATLAB ilaplace(V) command .) When done correctly your time domain result vout(t) should be the familiar RC charging voltage in time given by

vout(t) = vin ( 1 – e^(-t/RC) ), where Vin = 1 Volt in this case of a unit step.

Step 3 – Impulse Response – Show all solution steps clearly

a) Calculate Vout(s) when a unit impulse is applied. (Hint: for Vin(s) recall that the Laplace transform of a unit impulse is just equal to 1 .

b) Find vout(t) by invert Vout(s) back into time domain (simply use inverse Laplace transform tables in this case.)

Step 4

Use MATLAB to plot both the impulse and step responses v(t) that you computed in steps 2) and 3) above. You must label which plot is which; label your axes as well. Let R = 1 and C = 0.5 for this. (Hint: Slide 34 in the Week 2 Video Tutorial pt.2 could be of help in doing this.)

Paste Plots here

Questions

1. How does the value of the time constant of RC circuit change when the value of the capacitor is doubled?

2. Would your results in step 4 have been any different from when R=1 and C=0.5 had you been given the values for R = 1 MΩ and C = 0.5 μF ? Explain.

Prof. Ajeet Singh, Ph.D Richard Butler, EET modified 1-19-2019kj

R1

C1

VoutVin