Chapter 6 The frequency response

In terms of the Fourier transform we have

\[ \hat {x}(\omega )=(i\omega I-A)^{-1}B\hat {u}(\omega ),\qquad \hat {y}(\omega )=F(\omega )\hat {u}(\omega ). \]

6.1 Examples

6.1.1 First order systems

The frequency response of the first order system

\[ T\dot {y}(t)+y(t)=Ku(t), \]

is

\[ F(\omega )=\frac {K}{Ti\omega +1}. \]

We plot this frequency response in Figures 6.1a (the Nyquist diagram) and 6.1b (the Bode diagram); the shape does not depend on the parameters \(K\) and \(T\) (through scaling as for the step response). In Figures 6.2a and 6.2b we illustrate how the Bode magnitude diagram depend on the steady state gain \(K\) and the time constant \(T\).

6.1.2 Second order systems (low pass)

The frequency response of the second order system

\[ T^2\ddot {y}(t)+2\zeta T\dot {y}(t)+y(t)=Ku(t), \]

is

\[ F(\omega )=\frac {K}{-T^2\omega ^2+2\zeta Ti\omega +1}=\frac {K\omega _0^2}{-\omega ^2+2\zeta \omega _0i\omega +\omega _0^2}. \]

The Nyquist diagrams for relevant values of \(\zeta \) are given in Figure 6.3a. The bode diagrams for relevant values of \(\zeta \) are given in Figure 6.3b. When \(\zeta \in [0,\frac {1}{\sqrt {2}})\phantom {]}\), \(|F(\omega )|\) has a maximum at \(\omega _{\rm max}:=\omega _0\sqrt {1-2\zeta ^2}\). Instead, when \(\zeta \geq \frac {1}{\sqrt {2}}\), \(|F(\omega )|\) is monotonically decreasing (for \(\omega >0\)). So whereas \(\zeta =1\) is the value for which a qualitative difference in the step response occurs, it is for \(\zeta =\frac {1}{\sqrt {2}}\) that a qualitative difference in the frequency response occurs.

6.1.3 Second order systems (band pass)

The frequency response of the second order system

\[ T^2\ddot {q}(t)+2\zeta T\dot {q}(t)+q(t)=Ku(t),\qquad y(t)=\dot {q}(t), \]

is

\[ F(\omega )=\frac {Ki\omega }{-T^2\omega ^2+2\zeta Ti\omega +1}=\frac {K\omega _0^2i\omega }{-\omega ^2+2\zeta \omega _0i\omega +\omega _0^2}. \]

The Nyquist diagram is in Figure 6.4a and the Bode diagram in Figure 6.4b.

6.2 Case study: microphones

A microphone operates as follows. Sound is a pressure wave in air. This causes a diaphragm (which is essentially a mass-spring-damper system) to vibrate. This vibration is then converted into an electrical signal. This is done either through an inductor based conversion (this is done in what are called dynamic microphones) or through a capacitor based conversion (this is done in what are called condenser microphones). Behind the diaphragm there is a small cavity (called the air gap) which is connected by tubes to a larger cavity (called the back cavity).

In Figure 6.5 we give the Bode magnitude diagrams of several microphones as provided by their manufacturers. The hearing range of humans is 20 Hz to 20,000 Hz (which is why this is the range on the horizontal axis). Typically for a microphone we want the Bode magnitude diagram to be approximately constant over this frequency range: this means that all frequencies are equally represented in the recording made through the microphone as they are in the live performance.

To obtain a model of a microphone, it seems easiest to think in terms of energy. There are three sources of potential energy in a capacitor-based microphone:

• • The potential energy in the diaphragm (i.e. the spring) is \(\frac {1}{2k}F^2\) where \(k\) is the spring constant and \(F\) is the force across the spring.

• • The potential energy due to pressure in the cavity is \(\frac {C_c}{2}p^2\) where \(C_c\) is a constant (related mainly to the volume of the cavity) and \(p\) is the pressure in the cavity.

• • The potential energy in the capacitor is \(\frac {C}{2}V^2\) where \(C\) is the capacitance and \(V\) is the voltage across the capacitor.

Additionally there are two sources of kinetic energy:

• • The kinetic energy in the diaphragam (i.e. the mass) is \(\frac {m}{2}v^2\) where \(m\) is the mass and \(v\) is the velocity.

• • The kinetic energy of the particles in the cavity is \(\frac {L_c}{2}q^2\) where \(L_c\) is a constant and \(q\) is the volumetric flow rate.

We have the following relation between the various variables

\[ F=Ap,\quad Av=q,\qquad F=\alpha V, \]

where \(A\) is the area of the diaphragm and \(\alpha \) is a constant. It follows that the total energy is \(1/2\) times

\[ \left (\frac {1}{k}+\frac {C_c}{A^2}+\frac {C}{\alpha ^2}\right )F^2 +(m+A^2L_c)v^2. \]

From the above it seems reasonable (and this can indeed be justified) that the overall system is of mass-spring-damper type where the “mass” is \(m+A^2L_c\) (i.e. there is a contribution of the mechanical mass and of the mass of the particles) and the “spring constant” is \(k+\frac {A^2}{C_c}+\frac {\alpha ^2}{C}\) (i.e. there are contributions from the mechanical spring, from the pressure in the cavity and from the capacitance of the capacitor). The “damping constant” equals \(d+A^2R_c\) where \(d\) is the damping constant for the diaphragm and \(R_c\) is due to energy loss by the air particles in the cavity. The input to the system is pressure caused by sound, which equivalently is a force and the output is the voltage across the capacitor which is proportional to the force across the spring which is proportional to the displacement of the diaphragm. Therefore we have a mass-spring-damper system with force input and displacement output. This we saw before is a second order (low pass) system and the transfer function is a constant times

\[ \frac {1}{(m+A^2L_c)s^2+(d+A^2R_c)s+\left (k+\frac {A^2}{C_c}+\frac {\alpha ^2}{C}\right )}. \]

As we see in Figure 6.3b, the frequency response of such a system is indeed (relatively) flat for low frequencies. To obtain a good microphone, this flat part should extend to 20,000 Hz.

For an inductor-based microphone the energy stored in the inductor instead is \(LI^2\) where \(L\) is the inductance and \(I\) is the current through the inductor. We have \(\alpha v=I\) so that the total energy now is

\[ \left (\frac {1}{k}+\frac {C_c}{A^2}\right )F^2 +\left (m+A^2L_c+\frac {L}{\alpha ^2}\right )v^2. \]

The output is the current through the inductor (or actually: the voltage across a resistor parallel to the inductor which by Ohm’s law is a constant times the current) which is proportional to the velocity. So now we effectively have a mass-spring-damper system with force input and velocity output. This we saw before is a second order (band pass) system and the transfer function is a constant times

\[ \frac {s}{\left (m+A^2L_c+\frac {L}{\alpha ^2}\right )s^2+(d+A^2R_c)s+\left (k+\frac {A^2}{C_c}\right )}. \]

As we see in Figure 6.4b, the frequency response of such a system is indeed (relatively) flat for an intermediate range of frequencies. To obtain a good microphone, this flat part should extend from 20 Hz to 20,000 Hz.

6.3 Filters

In Section 6.2 we saw that we want to study (second order) systems which have a frequency response which is “flat” for certain frequencies. The general name for such systems is “filters”.

6.3.1 Low-pass filters

The second order (low-pass) system with \(\zeta =\frac {1}{\sqrt {2}}\) and therefore transfer function

\[ G(s)=\frac {K\omega _0^2}{s^2+\sqrt {2}\omega _0s+\omega _0^2}, \]

is such that the magnitude of the frequency response is \(\frac {1}{\sqrt {2}}\) times the maximum magnitude (reached at frequency zero) at \(\omega _0\). In this context \(\omega _0\) is then called the cut-off frequency. The idea is that below that frequency the magnitude of the frequency response is relatively flat and that above it this is no longer the case.

In the microphone application (for the capacitor-based case since that results in a second order (low-pass) system), this cut-off frequency would be chosen around the upper limit of the hearing range, i.e. 20,000 Hz.

6.3.2 Band-pass filters

When we want to pick out frequencies in a given range \([\omega _L,\omega _H]\), then we can use a band-pass filter

\[ G(s)=\frac {(\omega _H-\omega _L)s}{s^2+(\omega _H-\omega _L)s+\omega _L\omega _H}. \]

The quantity \(\omega _H-\omega _L\) is called the bandwidth. The frequencies \(\omega _L\) and \(\omega _H\) are such that the magnitude of the frequency response is \(\frac {1}{\sqrt {2}}\) times the maximum magnitude at these frequencies. We note that this transfer function is of the form

\[ \frac {2\zeta \omega _0s}{s^2+2\zeta \omega _0s+\omega _0^2}, \]

where

\[ \omega _0=\sqrt {\omega _L\omega _H},\qquad \zeta =\frac {\omega _H-\omega _L}{2\sqrt {\omega _L\omega _H}}. \]

See Figure 6.6 for the Bode diagrams for various bandwidths.

In the microphone application (for the inductor-based case since that results in a second order (band-pass) system), we would choose \(\omega _L\) around 20 Hz and \(\omega _H\) around 20,000 Hz so that the pass band \([\omega _L,\omega _H]\) corresponds to the human hearing range. This results in \(\omega _0\approx 632\) Hz and \(\zeta \approx 15.8\) (i.e. very overdamped).

6.4 Case study: color

Different colors of light are sinusoids of different frequencies (see Figure 6.7). Each object acts as a light filter: it filters out all frequencies in the incident light except for the frequency corresponding to the color we perceive the object as having. Normally the input is white light which contains all frequencies. Therefore illumination with white light allows us to see objects of all different colors. Lasers produce light of a particular color (i.e. of a particular frequency). If we therefore use a laser to illuminate an object (i.e. our input is \(\sin (\omega t)\) for a specific frequency \(\omega \) determined by the laser color) then we can only see objects of the corresponding color (objects of a different color appear black).

6.5 Case study: control of a tape drive

In Figure 6.8 we plot the frequency response from the disturbance \(v_e\) to each of the state variables \(v_1\), \(v_2\) and \(T\). Note that we already calculated the corresponding transfer functions in Section 5.2. We also indicated the approximate location of the peak of each of the frequency responses, namely

\begin{equation} \label {eq:tapedrivefreq} \sqrt {\frac {k}{M_1}+\frac {k}{M_2}}, \end{equation}

which is the natural frequency of an undamped mass-spring system with spring constant \(k\) and mass

\[ \frac {1}{\frac {1}{M_1}+\frac {1}{M_2}}, \]

(you might recognize this way of “adding” the masses as the same as adding parallel resistances).

From the Bode magnitude diagram we can see at which frequencies a periodic disturbance of the form \(v_e(t)=\sin (\omega t)\) is most detrimental, namely for frequencies around the peak, so close to the frequency (6.2).

6.6 Case study: a suspension system

We consider the fixed structure suspension system. In Figure 6.9 we give the Bode magnitude diagram from the road velocity to each of the three performance outputs. Here we have chosen values for the various physical parameters as in the introduction. On the horizontal axis the frequency is given in terms of Hertz rather than radians per second (see Remark 6.4). An input of \(\sin (\omega t)\) corresponds to a sinusoidal road velocity, and therefore a sinusoidal road profile, namely \(\frac {-1}{\omega }\cos (\omega t)\) as a function of time \(t\). Noting that \(t=x/V\) where \(V\) is the horizontal velocity of the car (i.e. the speed we are travelling at), the corresponding road profile as a function of position \(x\) is \(\frac {-1}{\omega }\cos \left (\frac {\omega }{V}x\right )\). This can be thought of as representing an infinite sequence of equally spaced speedbumps. If we write \(\omega _0=\frac {\omega }{V}\), then the road profile is \(\frac {-1}{\omega _0V}\cos (\omega _0 x)\) and therefore \(\omega _0\) is related to the distancing of the speed bumps and \(\omega =\omega _0 V\) depends on both this spacing and on the speed of the car. So to find the relevant value of \(\omega \) in Figure 6.9 we have to consider both the distancing of the speed bumps and the speed of the car.

In Figure 6.9 we see that the Bode diagram has two “peaks” which correspond roughly to complex conjugate pairs of eigenvalues of the matrix \(A\) in the input-state-output system (or equivalently: poles of the transfer function). There is a clear peak around 2 Hz which in the automotive industry is called vehicle heave. This is because it is essentially due to the sprung mass and spring: we note that \(\frac {1}{2\pi }\sqrt {\frac {k_s}{m_s}}\approx 2.5\) Hz, which is the natural frequency of the subsystem consisting of only the sprung mass and spring. There is also a notable peak around 10 Hz which in the automotive industry is called wheel hop (or tyre hop). We note that \(\frac {1}{2\pi }\sqrt {\frac {k_{us}+k_s}{m_{us}}}\approx 12\) Hz, which is the natural frequency of the subsystem consisting of only the unsprung mass and both springs.

Assume a given spacing of speedbumps and consider what happens if we increase the speed \(V\) of the car (so that \(\omega =\omega _0 V\) increases). As we increase the speed up to where vehicle heave occurs, the amplification factor \(|F(\omega )|\) increases. This will make driving over these speedbumps less comfortable. Once we pass the point where vehicle heave occurs, the amplification factor \(|F(\omega )|\) decreases again. However, for handling, it starts to increase again around the frequency where wheel hop occurs (and after that it decreases again).

From the transfer functions computed in Section 5.3, we see that all three frequency responses are zero in zero, which we indeed see in Figure 6.9. The handling transfer function behaves like \(s^{-1}\) near infinity, the suspension stroke transfer function like \(s^{-3}\) and the comfort transfer function like \(s^{-2}\). In Figure 6.9 (since we have a loglog scale), we can see these powers \(-1\), \(-3\) and \(-2\) respectively in the slope for large frequencies.

6.7 Problems

Consider the (scaled) frequency response

\[ F(\omega )=\frac {1}{-\omega ^2+2\zeta i\omega +1}, \]

of a second order (band-pass) system. Show that when \(\zeta \in [0,\frac {1}{\sqrt {2}})\phantom {]}\), \(|F(\omega )|\) has a maximum at \(\omega _{\rm max}:=\sqrt {1-2\zeta ^2}\) and that when \(\zeta \geq \frac {1}{\sqrt {2}}\), \(|F(\omega )|\) is monotonically decreasing for \(\omega >0\).