I o w a S t a t e U n i v e r s i t y

EE 324: Signals and Systems II Section B Professor: Dr. Ratnesh Kumar

Lab Report 3: TF, poles, zeros By: Brittany Duffy

2/13/14

1

Introduction

During this lab, I extended my practice on testing and system behavior. In order to understand what systems do and how they are expected to behave, I used Laplace representations, determined region of convergence, and many other computations. I predicted system behavior using Laplace analysis. The main goal of this lab was to build a connection between the Laplace transform, the frequency characteristics, and the time-domain features of the response.

Pre-lab

For each of the systems in Figure 1:

1. Use the integration-in-time-domain properties of the Laplace transform to derive the impedance of the capacitors and inductors. Since both capacitors and inductors are causal systems, what are the regions of convergence of their Laplace transforms?

Inductor Capacitor !() = !!" !() ! ! = () ! !() = :

! = !! ! !!! !! = !!" ! ! ! = 1 : > 0

2

2. Derive the transfer function (from input voltage to output voltage, in Laplace representation) of each system. Compute its pole(s) and zero(s), and the region of convergence.

Circuit A:

= ()() = +

Circuit B: = ()() = !! ! + ! + (!! + !! + !!)

ROC Zeros Poles Re{s} > 0 s=0 =

ROC Zeros Poles Re{s} > (!!!!!!!!!!!!!!)!!!!! ! N/A = (!! + !! + !!)! + !

L

R

L R1

R2

R3

3

Circuit C:

= ()() = 112 1212 + 11 + 22 + 12 + 1

3. Substitute s=jw (can you do this? why?). Compute the magnitude of the transfer function and verify analytically your intuition about what each system does to very low ( 0 ) frequency components of the input signal, and to very high frequency components ( ). Can you substitute s=jw in the capacitors impedance? Why? What is the effect of resistors on a passive circuits poles?

Circuit A: = !"!"#!! = !!(!!)!!!! 0 () = 0 () = 1

High-pass filter

Circuit B: = !!!!!(!!) !!!!! !(!!!!!!!!!!!!!!) = !!!!(!!)! !!!!! !!(!!!!!!!!!!!!!!)! 0 () = !!!! + (!!) + (!!)

ROC Zeros Poles

Re{s} > ! !!!!!!!!!!!!!! ! (!!!!!!!!!!!!!!)!!!(!!!!!!!!)!(!!!!!!!!) s=0 = !! + !! + !! (!! + !! + !!)! 4(!!!!)2(!!!!)

R2

R1

C1

C2

4

() = 0 Low-pass filter Circuit C: = !!!!!(!!)! !!!!!!!! !(!!) !!!!!!!!!!!!!! ! ! = !!!!!!!!!!!!!!!!! !!!! !!!!!!!!!!!!!! ! 0 () = 0 () = 0 Band-pass filter

The ROC for a capacitor is Re{s}>0. You cannot substitute s=jw in this case. When s=jw, if we can show that the ROC for the transfer function includes the jw axis, then we can substitute jw for s. To know if we have enough information to find ROC for the transfer function we can do the following: Find poles where the ROC is either to the left or the right. You have to know whether it is right sided or left sided. Physical systems have the causal property. Causal means that the systems are dependent only on past inputs. Transfer function is Y/X, this can be used to find Y. Input of a causal system is in terms of impulse response. Impulse response applies at time zero. If system is causal, we see zero before zero (no input before zero) and after zero, we see the response. The transfer function is nothing but the impulse response on a causal system. On jw axis, if right most pole is less than zero, we can evaluate on jw. The effect resistors have on a passive circuits poles is that when we add a resistor, this normally makes the poles go on the left side of the s-plane.

4. If the input is a unit step function u(t), what are the main features you will expect to see in the output? Justify by formally deriving the time-domain response of each system to a unit step (you will use a partial fraction expansion, and the inverse Laplace transform for each fraction term.) Circuit A: = 1 = 1 + = 1 + 100

Therefore, = !!!!() The output is a right-sided signal. It exponentially decays to zero.

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Circuit B: = 1 = !!! ! + ! + (!! + !! + !!) = 50! + 350 therefore, = !! (1 !!"#!)()

The output signal is zero at t0, and will increase until it reaches a particular value.

Circuit C: = !! = !!!!!! !!!!!!!! !! !!!!!!!!!!!!!! ! ! = !"#!!!!!!!"#!!!"#$$ therefore, = 0.14(!!!.!"! !!"#.!"!)()

The output signal will be zero at the start and end of the system.

Laboratory Assignment

1. Use the Matlab functions tf, pole, zero, residue, impulse, step and bode (for a description of each of the functions, use Matlab Help or the help command) to verify the correctness of you analytical results from the pre-lab.

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Circuit A:

Script Editor Output

0 0.01 0.02 0.03 0.04 0.05 0.060

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Impulse Response

Time (seconds)

Ampli

tude

7

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01Step Response

Time (seconds)

Ampli

tude

-80

-70

-60

-50

-40

Mag

nitud

e (d

B)

100 101 102 103 104-90

-45

0

Phas

e (d

eg)

Bode Diagram

Frequency (rad/s)

8

Circuit B:

Script Editor Output

0 0.2 0.4 0.6 0.8 1 1.20

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18Step Response

Time (seconds)

Ampli

tude

9

0 0.005 0.01 0.015 0.02 0.0250

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16Impulse Response

Time (seconds)

Ampli

tude

-140

-120

-100

-80

-60

-40

-20

Mag

nitud

e (d

B)

101 102 103 104-180

-135

-90

Phas

e (d

eg)

Bode Diagram

Frequency (rad/s)

10

Circuit C:

Script Editor Output

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

0.02

0.04

0.06

0.08

0.1

0.12

0.14Step Response

Time (seconds)

Ampli

tude

11

As you can see, my results are consistent with my pre-lab.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08-10

0

10

20

30

40

50Impulse Response

Time (seconds)

Ampli

tude

-70

-60

-50

-40

-30

-20

-10

Mag

nitud

e (d

B)

10-1 100 101 102 103 104-90

-45

0

45

90

Phas

e (d

eg)

Bode Diagram

Frequency (rad/s)

12

2. Using the transfer function, build the equivalent Simulink diagram (Simulink block Transfer Function).

Circuit A:

Circuit B:

Circuit C:

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You can see that all three systems use the transfer function Simulink block. The difference between each system is the coefficients of the numerator and denominator of the transfer functions.

3. Test your system to different input signals and conditions, and verify that it is consistent with your both your pre-lab analysis and with the otuput of the block diagram you derived in Lab 2.

Circuit A:

Sine Wave with Frequency at 1 rad/sec Sine Wave with Frequency at 10 rad/sec

Impulse Response Impulse Response (zoomed in)

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Step Response Step Response (zoomed in)

Circuit B:

Sine Wave with Frequency at 1 rad/sec Sine Wave with Frequency at 10 rad/sec

Impulse Response Impulse Response (zoomed in)

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Step Response Step Response (zoomed in)

Circuit C:

Sine Wave with Frequency at 1 rad/sec Sine Wave with Frequency at 10 rad/sec

Impulse Response Impulse Response (zoomed in)

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Step Response Step Response (zoomed in)

You will find that MATLAB doesnt start the responses until it gets to 1 second for some of the inputs. If you compare the above simulations with the pre-lab and lab 2, you can conclude that they are consistent.

Conclusion

In this lab, we became more familiarized with how to derive transfer functions, poles, zeros, and bode plots. I am starting to become very comfortable using Simulink to simulate the response of the system to different input signals. I find it very interesting how we can use multiple ways to find the response of a system and get the same results.