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Bode Plots - Recall that there are 5 types of terms in H(jw). The first two types were covered last class.

5 types of terms in H(jw)1) K (a constant)

2) (a zero) or (a pole)

3) jw (a zero) or 1/jw (a pole)

4)

5) Any of the terms raised to a positive integer power.

1

w1 j w

2

2 2 20 0

2 20 0

2 w w 11 j - (a complex zero) or (a complex pole) 2 w ww w 1 j -

w w

1

1w1 j w

2

1

wFor example, 1 j (a double zero) w

Lecture #15 EGR 261 – Signals and SystemsRead: Ch. 14, Sect. 1-5; Ch. 15, Sect. 1-2; and App. D&E

in Electric Circuits, 9th Edition by Nilsson

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1. Constant term in H(jw)If H(jw) = K = K/0 Then LM = 20log(K) and (w) = 0 , so the LM and phase responses are:

LM (dB)

w 0

w 0o

(w)

1

20log(K)

10 100

Lecture #15 EGR 261 – Signals and Systems

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LM

w 0dB = +20dB/dec

w

90o

(w)

= + 6dB/oct 20dB

w1

slope

0o 10w1

45o

w1 10w1 0.1w1

= +45 deg/dec slope

(for 2 decades)

2A) term in H(jw) (a zero) 1

w1 j w

1

1w1 j w

2B) term in H(jw) (a pole)

LM w 0dB

= -20dB/dec

w

-90o

(w)

= - 6dB/oct -20dB

w1 slope 0o

10w1

-45o

w1 10w1 0.1w1

= -45 deg/dec slope

(for 2 decades)

Lecture #15 EGR 261 – Signals and Systems

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Example: (from last class) Sketch the Bode LM and phase plots for: 200(s 50)H(s) s 400

First find H(jw) and put H(jw) into "standard form"200(jw 50) 25(1 + jw/50)H(jw) = (standard form)

jw 400 1 + jw/400Log-magnitude:

The LM will start level since there are no

Solutio

jw ter

:

s a

n

m

nd will begin at LM = 20log(25) = 28 dBThe LM will break up at w = 50 due to the zero.The pole at w = 400 will cancel the effect of the zero causing the graph to level out.Since 50 to 400 is three octaves and the slope is 6 dB/oct, the graph increases by 18 dB from 28 to 46 dB.Phase:The phase plots begins at 0 degrees since there are no jw terms.

The phase begins increasing with a slope of 45 /dec at w = 5 (a decade before the zero).The phase levels out at w = 40 (a decade before the pole) as the decreasing slope due to the pole cancels the effect of the zero. The phase begins decreasing at w = 500 (a decade after the zero) as the zero stops increasing.The phase levels out at w = 4000 (a decade after the pole) as the pole stops decreasing.

Since a zero causes a net increase of 90 in phase and a pole causes a net decrease of 90 in phase, the graph will end at 0 .

LM

w 28 dB

= +20dB/dec

w

90

(w) 46 dB

50

slope o

0 o 400 40 500 5

= - 45 deg/dec slope

4000

Lecture #15 EGR 261 – Signals and Systems

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Example: Sketch the LM and phase plots on the 4-cycle semi-log graph paper shown below for the following transfer function. (Pass out 2 sheets of graph paper.)

50,000) 8,000)(s (s

)000,2 s2,000,000( H(s)

Log-magnitude (LM) plot:

(End of review)

Lecture #15 EGR 261 – Signals and Systems

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Example: (continued)

Phase plot:

50,000) 8,000)(s (s

)000,2 s2,000,000( H(s)

Lecture #15 EGR 261 – Signals and Systems

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3A) jw term in H(jw)If H(jw) = jw = w/90 Then LM = 20log(w) and (w) = 90

Calculate 20log(w) for several values of w to show that the graph is a straight line for all frequency with a slope of 20dB/dec (or 6dB/oct).

LM

w 0 dB

= +20dB/dec

w

90o

(w)

= + 6dB/oct

1

20 dB

40 dB

10 100

slope

0o 1 10 100

The LM and phase for H(jw) = jw are shown below.

• So a jw (zero) term in H(jw) adds an upward slope of +20dB/dec (or +6dB/oct) to the LM plot.

• And a jw (zero) term in H(jw) adds a constant 90 to the phase plot.

Lecture #15 EGR 261 – Signals and Systems

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3B) 1/(jw) term in H(jw)If H(jw) = 1/(jw) = (1/w)/-90 Then LM = 20log(1/w) = -20log(w) and (w) = -90

A few calculations could easily show that the graph of 20log(1/w)is a straight line for all frequency with a slope of -20dB/dec (or -6dB/oct).

The LM and phase for H(jw) = 1/(jw) are shown below.

• So a 1/(jw) (pole) term in H(jw) adds a downward slope of -20dB/dec (or -6dB/oct) to the LM plot.

• And a 1/(jw) (pole) term in H(jw) adds a constant -90 to the phase plot.

LM

w 0 dB

= -20dB/dec

w

-90o

(w)

= - 6dB/oct -20 dB

-40 dB

10 100

slope

0o 10 100

Lecture #15 EGR 261 – Signals and Systems

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200(s)H(s) s 5000

Example: Sketch the LM and phase plots for the following transfer function.

Lecture #15 EGR 261 – Signals and Systems

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500(s 500)H(s) (s)(s 8000)

Example: Sketch the LM and phase plots for the following transfer function.

Lecture #15 EGR 261 – Signals and Systems

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Calculation of exact points to check Bode PlotsEvaluating H(jw) at a particular value of w is helpful to check Bode Plots. An example is shown below.

Example:

5000(s + 200)Suppose H(s) = (s + 4000)

w250 1 + j200so H(jw) =

w1 + j4000

Evaluating this function at w = 4000 rad/s yields:4000250 1 + j 250 1 + j200H(jw) = =

40001 + j4000

20 = 3539.95 42.1

1 + j

so (4000) = 42.1LM(4000) = 20log(3539.95) = 70.98 dB

Lecture #15 EGR 261 – Signals and Systems

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Example:Evaluate the H(jw) on the last page at w = 500 rad/s and w = 8000 rad/s. Compare the values with the Bode plots. Do they appear to be correct?

Lecture #15 EGR 261 – Signals and Systems

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5. Poles and zeros raised to an integer power in H(s)In the last class it was demonstrated that terms in H(jw) are additive. Therefore, a double terms (such as a pole or zero that is squared) simply acts like two terms, a triple term acts like three terms, etc.

Illustration: Show that (1 + jw/w1)N results in the following responses:

LM plot:Has a 0dB contribution before its break frequencyWill increase at a rate of 20NdB/dec after the breakThere will be an error of 3NdB at the break between the Bode straight-line approximation and the exact LM

Phase plot:Has a 0 degree contribution until 1 decade before its break frequencyWill increase at a rate of 45Ndeg/dec for two decades (from 0.1w1 to 10w1).

The total final phase contribution will be 90N degrees.

Lecture #15 EGR 261 – Signals and Systems

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So (1 + jw/w1)N results in the following responses:

LM

w 0 dB

= +20NdB/dec

w

90No

w

= + 6NdB/oct

20N dB

w1

slope

0o 10w1

45No

w1 10w1 0.1w1

= +45N deg/dec slope

(for 2 decades)

Lecture #15 EGR 261 – Signals and Systems

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Example: Sketch the LM plot for the following transfer function.12 2

2 2

2x10 s H(s) (s 800) (s 2,000)

Lecture #15 EGR 261 – Signals and Systems

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Generating LM and phase plots using Excel and MathCAD: Refer to the handout entitled “Frequency Response” which includes detailed examples of creating LM and phase plots using Excel and MathCAD.

Generating LM and phase plots using ORCAD: Refer to the handout entitled “PSPICE Example: Frequency Response (Log-Magnitude and Phase)” which includes a detailed example of creating LM and phase plots using ORCAD.

Lecture #15 EGR 261 – Signals and Systems