Plan of the Lecture Review: design using Root Locus; dynamic - PowerPoint PPT Presentation

Effect of Lag Compensation on Root Locus s + 1 L ( s ) = ( s + p )( s − 1) Intuition: By choosing p very close to zero, we can make the root locus arbitrarily close to PI root locus (stable for large enough K ). Let’s check: Try p = 0 . 1 Compare to PI root locus: 1.0 1.0 0.5 0.5 - 3 - 2 - 1 1 - 3 - 2 - 1 1 - 0.5 - 0.5 - 1.0 - 1.0 What do we see? Compared to PD vs. lead, there is no qualitative change in the shape of RL, since we are not changing #(poles) or #(zeros).

More Pole Placement As before, we can choose z lag for a fixed p lag (or vice versa) based on desired pole locations.

More Pole Placement As before, we can choose z lag for a fixed p lag (or vice versa) based on desired pole locations. The procedure is exactly the same as the one we used with lead. (In fact, depending on the pole locations, we may end up with either lead or lag.)

More Pole Placement As before, we can choose z lag for a fixed p lag (or vice versa) based on desired pole locations. The procedure is exactly the same as the one we used with lead. (In fact, depending on the pole locations, we may end up with either lead or lag.) Main technique: select parameters to satisfy the phase condition (points on RL must be such that ∠ L ( s ) = 180 ◦ ).

More Pole Placement As before, we can choose z lag for a fixed p lag (or vice versa) based on desired pole locations. The procedure is exactly the same as the one we used with lead. (In fact, depending on the pole locations, we may end up with either lead or lag.) Main technique: select parameters to satisfy the phase condition (points on RL must be such that ∠ L ( s ) = 180 ◦ ). Caveat: may not always be possible!

Pole Placement via RL 1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p

Pole Placement via RL 1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find K and z to place poles at − 2 ± 3 j .

Pole Placement via RL 1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find K and z to place poles at − 2 ± 3 j . Desired characteristic polynomial: ( s + 2) 2 + 9 = s 2 + 4 s + 13 ,

Pole Placement via RL 1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find K and z to place poles at − 2 ± 3 j . Desired characteristic polynomial: 2 ( s + 2) 2 + 9 = s 2 + 4 s + 13 , damping ratio ζ = √ 13 ≈ 0 . 555

Pole Placement via RL 1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find K and z to place poles at − 2 ± 3 j . Desired characteristic polynomial: 2 ( s + 2) 2 + 9 = s 2 + 4 s + 13 , damping ratio ζ = √ 13 ≈ 0 . 555 Im Must have s – given 3 � ψ − ϕ i = 180 ◦ ���� ���� i angle from angles from ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ s to zero s to poles ψ � Re So, we want ψ = 180 ◦ + o x x ϕ i 0 − z − 2 (to be i selected)

Pole Placement via RL Im s – given 3 ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ ψ o x x Re 0 − z − 2 (to be selected)

Pole Placement via RL Im s – given 3 We have ϕ 1 = 135 ◦ , ϕ 2 = 90 ◦ ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ ψ o x x Re 0 − z − 2 (to be selected)

Pole Placement via RL Im s – given 3 We have ϕ 1 = 135 ◦ , ϕ 2 = 90 ◦ ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ ψ o x x Re � We want ψ = 180 ◦ + 0 − z ϕ i − 2 (to be i selected)

Pole Placement via RL Im s – given 3 We have ϕ 1 = 135 ◦ , ϕ 2 = 90 ◦ ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ ψ o x x Re � We want ψ = 180 ◦ + 0 − z ϕ i − 2 (to be i selected) Must have ψ = 180 ◦ + 135 ◦ + 90 ◦

Pole Placement via RL Im s – given 3 We have ϕ 1 = 135 ◦ , ϕ 2 = 90 ◦ ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ ψ o x x Re � We want ψ = 180 ◦ + 0 − z ϕ i − 2 (to be i selected) Must have ψ = 180 ◦ + 135 ◦ + 90 ◦ = 405 ◦

Pole Placement via RL Im s – given 3 We have ϕ 1 = 135 ◦ , ϕ 2 = 90 ◦ ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ ψ o x x Re � We want ψ = 180 ◦ + 0 − z ϕ i − 2 (to be i selected) Must have ψ = 180 ◦ + 135 ◦ + 90 ◦ = 405 ◦ = 45 ◦ mod 360 ◦

Pole Placement via RL Im s – given 3 We have ϕ 1 = 135 ◦ , ϕ 2 = 90 ◦ ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ ψ o x x Re � We want ψ = 180 ◦ + 0 − z ϕ i − 2 (to be i selected) Must have ψ = 180 ◦ + 135 ◦ + 90 ◦ = 405 ◦ = 45 ◦ mod 360 ◦ Thus, we should have z = − 5

Pole Placement via RL Im s – given 3 We have ϕ 1 = 135 ◦ , ϕ 2 = 90 ◦ ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ ψ o x x Re � We want ψ = 180 ◦ + 0 − z ϕ i − 2 (to be i selected) Im Must have s – given 3 ψ = 180 ◦ + 135 ◦ + 90 ◦ = 405 ◦ ϕ 1 = 135 ◦ ϕ 2 = 90 ◦ = 45 ◦ mod 360 ◦ ψ = 45 ◦ o x x Re 0 − z = − 5 − 2 Thus, we should have z = − 5

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j .

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution:

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial:

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial: ( s − 1)( s + p ) + K ( s + z )

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial: ( s − 1)( s + 2) + K ( s + 5)

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial: ( s − 1)( s + 2) + K ( s + 5) s 2 + ( K + 1) s + 5 K − 2

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial: ( s − 1)( s + 2) + K ( s + 5) s 2 + ( K + 1) s + 5 K − 2 ◮ compare against desired characteristic polynomial:

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial: ( s − 1)( s + 2) + K ( s + 5) s 2 + ( K + 1) s + 5 K − 2 ◮ compare against desired characteristic polynomial: s 2 + 4 s + 13

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial: ( s − 1)( s + 2) + K ( s + 5) s 2 + ( K + 1) s + 5 K − 2 ◮ compare against desired characteristic polynomial: s 2 + 4 s + 13 = ⇒ K + 1 = 4 , 5 K − 2 = 13

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial: ( s − 1)( s + 2) + K ( s + 5) s 2 + ( K + 1) s + 5 K − 2 ◮ compare against desired characteristic polynomial: s 2 + 4 s + 13 = ⇒ K + 1 = 4 , 5 K − 2 = 13 so we need K = 3

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial: ( s − 1)( s + 2) + K ( s + 5) s 2 + ( K + 1) s + 5 K − 2 ◮ compare against desired characteristic polynomial: s 2 + 4 s + 13 = ⇒ K + 1 = 4 , 5 K − 2 = 13 so we need K = 3 ◮ compute s.s. tracking error:

1 G c ( s ) = K s + z Let G p ( s ) = s − 1, s + p Problem: given p = 2, find z to place poles at − 2 ± 3 j . Solution: ◮ we already found that we need z = 5 ◮ resulting characteristic polynomial: ( s − 1)( s + 2) + K ( s + 5) s 2 + ( K + 1) s + 5 K − 2 ◮ compare against desired characteristic polynomial: s 2 + 4 s + 13 = ⇒ K + 1 = 4 , 5 K − 2 = 13 so we need K = 3 � � � 1 � � = 1 � � ◮ compute s.s. tracking error: 6 . 5 ≈ 15% � � 1 − Kz � p

Story So Far PD control:

Story So Far PD control: ◮ provides stability, allows to shape transient response specs

Story So Far PD control: ◮ provides stability, allows to shape transient response specs ◮ replace noncausal D-controller Ks with a causal, stable lead controller K s + z s + p , where p > z

Story So Far PD control: ◮ provides stability, allows to shape transient response specs ◮ replace noncausal D-controller Ks with a causal, stable lead controller K s + z s + p , where p > z ◮ this introduces a zero in LHP (at − z ), pulls the root locus into LHP

Story So Far PD control: ◮ provides stability, allows to shape transient response specs ◮ replace noncausal D-controller Ks with a causal, stable lead controller K s + z s + p , where p > z ◮ this introduces a zero in LHP (at − z ), pulls the root locus into LHP ◮ shape of RL differs depending on how large p is

Story So Far PD control: ◮ provides stability, allows to shape transient response specs ◮ replace noncausal D-controller Ks with a causal, stable lead controller K s + z s + p , where p > z ◮ this introduces a zero in LHP (at − z ), pulls the root locus into LHP ◮ shape of RL differs depending on how large p is PI control:

Story So Far PD control: ◮ provides stability, allows to shape transient response specs ◮ replace noncausal D-controller Ks with a causal, stable lead controller K s + z s + p , where p > z ◮ this introduces a zero in LHP (at − z ), pulls the root locus into LHP ◮ shape of RL differs depending on how large p is PI control: ◮ provides stability and perfect steady-state tracking of constant references

Story So Far PD control: ◮ provides stability, allows to shape transient response specs ◮ replace noncausal D-controller Ks with a causal, stable lead controller K s + z s + p , where p > z ◮ this introduces a zero in LHP (at − z ), pulls the root locus into LHP ◮ shape of RL differs depending on how large p is PI control: ◮ provides stability and perfect steady-state tracking of constant references ◮ replace unstable I-controller K/s with a stable lag controller K s + z s + p , where p < z

Story So Far PD control: ◮ provides stability, allows to shape transient response specs ◮ replace noncausal D-controller Ks with a causal, stable lead controller K s + z s + p , where p > z ◮ this introduces a zero in LHP (at − z ), pulls the root locus into LHP ◮ shape of RL differs depending on how large p is PI control: ◮ provides stability and perfect steady-state tracking of constant references ◮ replace unstable I-controller K/s with a stable lag controller K s + z s + p , where p < z ◮ this does not change the shape of RL compared to PI

What About PID Control? Obvious solution — combine lead and lag compensation. We will develop this further in homework and later in the course using frequency-response design methods — which are the subject of several lectures, starting with today’s.

The Frequency-Response Design Method Recall the frequency-response formula: sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω )

The Frequency-Response Design Method Recall the frequency-response formula: sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Derivation:

The Frequency-Response Design Method Recall the frequency-response formula: sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Derivation: 1. u ( t ) = e st �− → y ( t ) = G ( s ) e st

The Frequency-Response Design Method Recall the frequency-response formula: sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Derivation: 1. u ( t ) = e st �− → y ( t ) = G ( s ) e st 2. Euler’s formula: sin( ωt ) = e jωt − e − jωt 2 j

The Frequency-Response Design Method Recall the frequency-response formula: sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Derivation: 1. u ( t ) = e st �− → y ( t ) = G ( s ) e st 2. Euler’s formula: sin( ωt ) = e jωt − e − jωt 2 j 3. By linearity, → G ( jω ) e jωt − G ( − jω ) e − jωt sin( ωt ) �− 2 j

The Frequency-Response Design Method Recall the frequency-response formula: sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Derivation: 1. u ( t ) = e st �− → y ( t ) = G ( s ) e st 2. Euler’s formula: sin( ωt ) = e jωt − e − jωt 2 j 3. By linearity, → G ( jω ) e jωt − G ( − jω ) e − jωt G ( jω ) = M ( ω ) e jφ ( ω ) sin( ωt ) �− 2 j

The Frequency-Response Design Method Recall the frequency-response formula: sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Derivation: 1. u ( t ) = e st �− → y ( t ) = G ( s ) e st 2. Euler’s formula: sin( ωt ) = e jωt − e − jωt 2 j 3. By linearity, → G ( jω ) e jωt − G ( − jω ) e − jωt G ( jω ) = M ( ω ) e jφ ( ω ) sin( ωt ) �− 2 j = M ( ω ) e j ( ωt + φ ( ω )) − M ( ω ) e − j ( ωt + φ ( ω )) 2 j

The Frequency-Response Design Method Recall the frequency-response formula: sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Derivation: 1. u ( t ) = e st �− → y ( t ) = G ( s ) e st 2. Euler’s formula: sin( ωt ) = e jωt − e − jωt 2 j 3. By linearity, → G ( jω ) e jωt − G ( − jω ) e − jωt G ( jω ) = M ( ω ) e jφ ( ω ) sin( ωt ) �− 2 j = M ( ω ) e j ( ωt + φ ( ω )) − M ( ω ) e − j ( ωt + φ ( ω )) 2 j = M ( ω ) sin( ωt + φ ( ω ))

The Frequency-Response Design Method sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω )

The Frequency-Response Design Method sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Let’s apply this formula to our prototype 2nd-order system: ω 2 n G ( s ) = s 2 + 2 ζω n s + ω 2 n

The Frequency-Response Design Method sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Let’s apply this formula to our prototype 2nd-order system: ω 2 n G ( s ) = s 2 + 2 ζω n s + ω 2 n � � ω 2 � � n M ( ω ) = | G ( jω ) | = � � − ω 2 + 2 jζω n ω + ω 2 � � n

The Frequency-Response Design Method sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Let’s apply this formula to our prototype 2nd-order system: ω 2 n G ( s ) = s 2 + 2 ζω n s + ω 2 n � � ω 2 � � n M ( ω ) = | G ( jω ) | = � � − ω 2 + 2 jζω n ω + ω 2 � � n � � � 1 � � � = � ω � 2 + 2 ζ ω � � 1 − ω n j � � ω n

The Frequency-Response Design Method sin( ω t ) M sin( ω t + φ ) G ( s ) where M = M ( ω ) = | G ( jω ) | and φ = φ ( ω ) = ∠ G ( jω ) Let’s apply this formula to our prototype 2nd-order system: ω 2 n G ( s ) = s 2 + 2 ζω n s + ω 2 n � � ω 2 � � n M ( ω ) = | G ( jω ) | = � � − ω 2 + 2 jζω n ω + ω 2 � � n � � � 1 � � � = � ω � 2 + 2 ζ ω � � 1 − ω n j � � ω n 1 = �� � 2 � 2 � ω + 4 ζ 2 � ω � 2 1 − ω n ω n

The Frequency-Response Design Method For our prototype 2nd-order system: ω 2 n G ( s ) = s 2 + 2 ζω n s + ω 2 n 1 1 M ( ω ) = = �� � � ω � 2 + � ω � 4 � 2 � 2 1 + (4 ζ 2 − 2) � ω + 4 ζ 2 � ω � 2 1 − ω n ω n ω n ω n M H w L 1.0 z= 1 ê 2 0.8 z= 1 ê 2 0.6 z= 1 0.4 0.2 w 0.5 1.0 1.5 2.0 2.5 3.0 w n

Frequency Response Parameters Here is a typical frequency response magnitude plot: M H Ω L M r 1.0 0.8 √ 1 / 2 0.6 0.4 0.2 ω r ω BW 3.0 Ω 0.5 1.0 1.5 2.0 2.5 ω r – resonant frequency M r – resonant peak ω BW – bandwidth

Frequency Response Parameters M H Ω L M r 1.0 0.8 √ 1 / 2 0.6 0.4 0.2 ω r ω BW 3.0 Ω 0.5 1.0 1.5 2.0 2.5 We can get the following formulas using calculus:  � 1 − 2 ζ 2 ω r = ω n  1 1 1 (valid for ζ < √ 2; for ζ ≥ √ 2, ω r = 0) M r = 1 − ζ 2 − 1 �  2 ζ � � (1 − 2 ζ 2 ) 2 + 1 (1 − 2 ζ 2 ) + ω BW = ω n � �� � √ =1 for ζ =1 / 2 — so, if we know ω r , M r , ω BW , we can determine ω n , ζ and hence the time-domain specs ( t r , M p , t s )

Frequency Response & Time-Domain Specs All information about time response is also encoded in frequency response!! M H Ω L M r 1.0 0.8 √ 1 / 2 0.6 0.4 0.2 ω r ω BW 3.0 Ω 0.5 1.0 1.5 2.0 2.5 small M r ← → better damping large ω BW ← → large ω n ← → smaller t r

Frequency-Response Design Method: Main Idea + K G ( s ) Y R −

Frequency-Response Design Method: Main Idea + K G ( s ) Y R − Two-step procedure:

Frequency-Response Design Method: Main Idea + K G ( s ) Y R − Two-step procedure: 1. Plot the frequency response of the open-loop transfer function KG ( s ) [or, more generally, D ( s ) G ( s )], at s = j ω

Frequency-Response Design Method: Main Idea + K G ( s ) Y R − Two-step procedure: 1. Plot the frequency response of the open-loop transfer function KG ( s ) [or, more generally, D ( s ) G ( s )], at s = j ω 2. See how to relate this open-loop frequency response to closed-loop behavior.

Frequency-Response Design Method: Main Idea + K G ( s ) Y R − Two-step procedure: 1. Plot the frequency response of the open-loop transfer function KG ( s ) [or, more generally, D ( s ) G ( s )], at s = j ω 2. See how to relate this open-loop frequency response to closed-loop behavior. We will work with two types of plots for KG ( jω ):

Frequency-Response Design Method: Main Idea + K G ( s ) Y R − Two-step procedure: 1. Plot the frequency response of the open-loop transfer function KG ( s ) [or, more generally, D ( s ) G ( s )], at s = j ω 2. See how to relate this open-loop frequency response to closed-loop behavior. We will work with two types of plots for KG ( jω ): 1. Bode plots: magnitude | KG ( jω ) | and phase ∠ KG ( jω ) vs. frequency ω (could have seen it earlier, in ECE 342)

Frequency-Response Design Method: Main Idea + K G ( s ) Y R − Two-step procedure: 1. Plot the frequency response of the open-loop transfer function KG ( s ) [or, more generally, D ( s ) G ( s )], at s = j ω 2. See how to relate this open-loop frequency response to closed-loop behavior. We will work with two types of plots for KG ( jω ): 1. Bode plots: magnitude | KG ( jω ) | and phase ∠ KG ( jω ) vs. frequency ω (could have seen it earlier, in ECE 342) � � � � 2. Nyquist plots: Im KG ( jω ) vs. Re K ( jω ) [Cartesian plot in s -plane] as ω ranges from −∞ to + ∞

Note on the Scale Horizontal ( ω ) axis: we will use logarithmic scale (base 10) in order to display a wide range of frequencies. Note: we will still mark the values of ω , not log 10 ω , on the axis, but the scale will be logarithmic: ω ... ... 0.01 0.1 1 10 100 1000 Equal intervals on log scale correspond to decades in frequency.

Note on the Scale Vertical axis on magnitude plots: we will also use logarithmic scale, just like the frequency axis.

Note on the Scale Vertical axis on magnitude plots: we will also use logarithmic scale, just like the frequency axis. Reason: | ( M 1 e jφ 1 )( M 2 e jφ 2 ) | = M 1 · M 2 log( M 1 M 2 ) = log M 1 + log M 2

Note on the Scale Vertical axis on magnitude plots: we will also use logarithmic scale, just like the frequency axis. Reason: | ( M 1 e jφ 1 )( M 2 e jφ 2 ) | = M 1 · M 2 log( M 1 M 2 ) = log M 1 + log M 2 — this means that we can simply add the graphs of log M 1 ( ω ) � � and log M 2 ( ω ) to obtain the graph of log M 1 ( ω ) M 2 ( ω ) , and graphical addition is easy.

Note on the Scale Vertical axis on magnitude plots: we will also use logarithmic scale, just like the frequency axis. Reason: | ( M 1 e jφ 1 )( M 2 e jφ 2 ) | = M 1 · M 2 log( M 1 M 2 ) = log M 1 + log M 2 — this means that we can simply add the graphs of log M 1 ( ω ) � � and log M 2 ( ω ) to obtain the graph of log M 1 ( ω ) M 2 ( ω ) , and graphical addition is easy. Decibel scale: ( M ) dB = 20 log 10 M (one decade = 20 dB)