Välkomna till TSRT15 Reglerteknik Föreläsning 5 Summary of lecture 4 Frequency response Bode plot
Summary of last lecture 2 Given a pole polynomial with a varying parameter P(s)+KQ(s)=0 We draw the location of the poles with respect to K in a root-locus An approximate root-locus can be sketched using some simple drawing rules
Summary of last lecture 3 Solution time is roughly 3/(distance to the origon for the pole closest to the origin) for systems with real poles and oscillating systems with a reasonable relative damping ξ (between 0.5 and 1) A relative damping ξ of 0.7 gives an overshoot of 5%, which typically is what we want Im 45º Re Zeros are fun (will be added to the lecture notes)
Frequency response 4
Frequency response 5
Frequency response 6 Speaker test: A test signal (sinusoidal voltage) is sent to the speaker A microphone measures the sound and registers the amplification from voltage to sound level. Typical behavior: The measurement signal (the sound) has the same frequency (it would sound terrible otherwise) but the amplification depends on the frequency
Frequency response 7 A similar experiment can be performed on any system with an input Car slalom: We perform a test with a sinusoidal movement of the steering whell, and measure the lateral position of the car (distance from center line) The car dynamics (from steering wheel angle u(t) in radians to lateral position y(t) in meter) can approximately be modeled with the following linear system
Frequency response 8 Input: Output: : Sinusoidal with an amplitude of roughly 3 meters
Frequency response 9 Input: Output: : Sinusoidal with an amplitude of roughly 80 centimeters
Frequency response 10 Input: Output: : Sinusoidal with an amplitude of roughly 6 meters
Frequency response 11 Experimental hypothesis: A sinusoidal input leads to a sinusoidal output (asymptotically after transients from initial conditions have faded away)
Frequency response 12 Linear systems are described by linear differential equations whose solutions are constructed from the homogenuous part (which depends on the initial conditions) and the particular part (which depends on the input signal) Now assume that a sinusoidal input has been used since t=-. The homogeneous part will then have disappeared at t=0 and we can use the convolution theorems for Laplace transforms
Frequency response 13 A sinusoidal input with frequency ω passed through a linear system system G(s) is amplified by a factor G(iω) and the phase changes arg(g(iω)) radians
Frequency response 14 We use these formulas for our car
Frequency response 15 Phase change
Frequency response 16 The amplification in the car from steering wheel angle to lateral position
Bode plot 17 Amplitude-gain plots are typically hard to interpret in standard linear-linear scale, and log-log scale is used instead. Additionally, the gain is multiplied with 20 to obtain a decibel scale Hence, we plot 20 log G(iω) with a logarithmically growing frequency This is called a Bode plot. We will now learn how to sketch these
Bode plot 18 Assume that the system is given in the following factorized form (i.e. n poles, p integrators and m zeros easily seen) Amplitude and phase:
Bode plot 19 Sketch method: Start with very small frequencies This function is linear in log-log scale, with a slope of p (i.e., it falls with p*20db per decade, if we convert to db) When we increase the frequency, the other terms will increase, and we must finally take them into account too When ω approaches a zero or a pole, the corresponding term must be added, since terms of the type starts to become significant When such a terms is added, the amplitude curve changes direction, and either increases the slope with one unit (when passing a pole, it bends downwards) or decreases the slope with one unit (when passing a zero, it bends upwards)
Bode plot 20 For small frequencies When we increase the frequency, the other terms will start to become significant, just as in the amplitude curve. A stable zero gives a asymptotic phase addition of while a stable pole gives a negative phase loss of The phase plot is typically harder to sketch manually.
Bode plot 21
Bode plot 22
Bode plot 23 Complex roots are harder to draw manually They give rise to resonance frequencies where input signals have an extra large amplification It occurs at ω 0, and its size depends on the relative damping ξ
Periodic signals 24 Why are we interested in frequency responses? In practice, we will probably use other, more general, signals The reason is that all perioduic signals with period T can be written as a sum of sine and cosine signals with frequencies nω 0 where ω 0 =(2π/T) and n=0,1,2, This is called a Fourier expansion
Periodiska signaler 25 Exemple: Given the following linear system The input signal is a square-wave signal with frequency ω 0. The Fourier expansion is given by The output is (asymptotically) If the relative damping ξ is low, we will have a significant resonance frequency at ω 0 leading to a large value for G(jω 0 ). Hence, the first term will dominate the output
Summary 26 Summary of todays lecture A system can be analysed and evaluated based on how it responds to sinusoidal inputs All linear systems driven by a sinusoidal input give a sinusoidal output with the same frequency, but another amplitude and phase A plot showing this amplitude gain and phase change is called a Bode plot Bode plots are drawn in log-log scale since this highlights some of the features better A pole bends the amplitude curve in the Bode plot downwards while a zero bends the curve upwards
Summary 27 Important words Frequency response: Output from a system when the input is sinusoidal. Bode plot: Plot showing the amplitude gain and phase change of a sinusoidal input for a linear system. Most often drawn in log-log and loglinear scale Resonance frequency: A frequency for which input signals are amplified significantly more than at other frequencies Resonance peak: The peak in the bode plot at a resonance frequency