Consider the Continuous Time System With Transfer Function the Corner Frequencies Are

Active Filter Design Techniques

Bruce Carter , Ron Mancini , in Op Amps for Everyone (Fifth Edition), 2018

16.3.1 First-Order Low-Pass Filter

Figs. 16.12 and 16.13 show a first-order low-pass filter in the inverting and in the noninverting configuration.

Figure 16.12. First-order noninverting low-pass filter.

Figure 16.13. First-order inverting low-pass filter.

The transfer functions of the circuits are given as:

$\text{A}\left(\text{s}\right)=\frac{1+\frac{{\text{R}}_2}{{\text{R}}_3}}{1+{\mathrm{\omega }}_{\text{c}}{\text{R}}_1{\text{C}}_1\text{s}}\text{and}\text{A}\left(\text{s}\right)=\frac{-\frac{{\text{R}}_2}{{\text{R}}_1}}{1+{\mathrm{\omega }}_{\text{c}}{\text{R}}_2{\text{C}}_1\text{s}}$

The negative sign indicates that the inverting amplifier generates a 180   degrees phase shift from the filter input to the output.

The coefficient comparison between the two transfer functions and Eq. (16.3) yields

$\begin{array}{l}{\text{A}}_0=1+\frac{{\text{R}}_2}{{\text{R}}_3}\text{and}{\text{A}}_0=-\frac{{\text{R}}_2}{{\text{R}}_1}\\ {\text{a}}_1={\mathrm{\omega }}_{\text{c}}{\text{R}}_1{\text{C}}_1\text{and}{\text{a}}_1={\mathrm{\omega }}_{\text{c}}{\text{R}}_2{\text{C}}_1\end{array}$

To dimension the circuit, specify the corner frequency (f _C), the DC gain (A₀), and capacitor C₁, and then solve for resistors R₁ and R₂:

$\begin{array}{l}{\text{R}}_1=\frac{{\text{a}}_1}{2\mathrm{\pi }{\text{f}}_{\text{c}}{\text{C}}_1}\text{and}{\text{R}}_2=\frac{{\text{a}}_1}{2\mathrm{\pi }{\text{f}}_{\text{c}}{\text{C}}_1}\\ {\text{R}}_2={\text{R}}_3\left({\text{A}}_0-1\right)\text{and}{\text{R}}_1=-\frac{{\text{R}}_2}{{\text{A}}_0}\end{array}$

The coefficient a₁ is taken from one of the coefficient tables, Tables 16.6 through 16.12 in Section 16.9.

Note, that all filter types are identical in their first order and a₁  =   1. For higher filter orders, however, a₁  ≠   1 because the corner frequency of the first-order stage is different from the corner frequency of the overall filter.

Example 16.1. First-Order Unity-Gain Low-Pass Filter.

For a first-order unity-gain low-pass filter with f_C  =   1   kHz and C₁  =   47   nF, R₁ calculates to:

${\text{R}}_1=\frac{{\text{a}}_1}{2\mathrm{\pi }{\text{f}}_{\text{c}}{\text{C}}_1}=\frac{1}{2\mathrm{\pi }·{10}^3\text{Hz}·47·{10}^{-9}\text{F}}=3.38\text{k}\mathrm{\Omega }$

However, to design the first stage of a third-order unity-gain Bessel low-pass filter, assuming the same values for f_C and C₁, requires a different value for R₁. In this case, obtain a₁ for a third-order Bessel filter from Table 16.6 in Section 16.9 (Bessel coefficients) to calculate R₁:

${\text{R}}_1=\frac{{\text{a}}_1}{2\mathrm{\pi }{\text{f}}_{\text{c}}{\text{C}}_1}=\frac{0.756}{2\mathrm{\pi }·{10}^3\text{Hz}·47·{10}^{-9}\text{F}}=2.56\text{k}\mathrm{\Omega }$

When operating at unity gain, the noninverting amplifier reduces to a voltage follower (Fig. 16.14), thus inherently providing a superior gain accuracy. In the case of the inverting amplifier, the accuracy of the unity gain depends on the tolerance of the two resistors, R₁ and R₂.

Figure 16.14. First-order noninverting low-pass filter with unity gain.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128116487000169

11th International Symposium on Process Systems Engineering

Heinz A. Preisig , in Computer Aided Chemical Engineering, 2012

2 A simple case to illustrate

2.1 The plant and the model

To illustrate the above-sketched problem, we use a nominal case, probably one of the simplest ones: For the process we use a second-order process with two time constants and for the model we use a first-order model. We use standard software, thus no particular sophisticated methods, tricks and no special conditions. The transfer function of the model and the plant are:

(1) $\begin{array}{ccc}\text{plant}\text{:}& g_p\left(s\right):=\frac{1}{\left(T_{1}s+1\right)\left(T_{2}s+1\right)},& \text{with}{T}_1:=1\text{and}{T}_2:=0.1\end{array}$

(2) $\begin{array}{ccc}\text{model}\text{:}& g_m\left(s\right):=\frac{1}{{\tau }_{1}s+1}& \text{with}{\tau }_1\text{to}\text{be}\text{identified}\end{array}$

The Bode plot of the plant has two corner frequencies, one at 1 rad/s and one at 10 rad/s, whilst the model will only have one at the location we are going to identify through our identification experiments.

Figure 1. Haar wavelet coefficients for a step at t   =   100

2.2 Input signal: step

A step signal is the integral of an impulse, with the impulse exciting all the frequencies equally. The step thus also excites all frequencies. A wavelet analysis of the step clearly shows that the high frequencies are most active at the beginning of the step, which is at time 100, whilst the lower frequencies are becoming more important later. This could also be deducted from a Fourier series representation of a pulse, where the high frequencies compensate each other in the centre of the pulse, whilst adding to form the steep ascent at the beginning and the equally steep decent at the other end.

2.3 Fitting the first-order model

It is apparent that this model, when excited with a step, will respond immediately: the slope at time zero is non-zero, whilst it is zero for our second-order plant. So in terms of the slope, the deviation is relative largest at the beginning and if one looks closer, the difference decreases later for most wellfitted models. The second effect one observes is a shift in the response, a reflection of the phase shift. Also, the extreme phase shift, at high frequencies, is double for the plant compared to the model.

The different frequencies are thus more or less "active" as time progresses when applying a step. In order to get a view on how different frequencies affect the identification, we shall use a package of well-defined frequencies to excite the system, namely a package of 8 sinusoidal functions that are logarithmically spaced in a frequency range.

A number of experiments are performed, whereby the range is changed and the input–output signals are used to fit a first-order model keeping the gain constant to the nominal value. Thus we only adjust the time constant in the identification focusing the dynamics.

The range is changed by setting a lower limit well below the first corner (1 rad/s)and extending the range step-wise at the upper end including more and more of the frequencies covering beyond the second corner (10 rad/s) up to 2.7 rad/s.

The upper limit is set to meet the Shanon condition for sampling, as the simulation of the plant is sampled at 0.001 s. Measuring the frequencies relative to the corner frequencies, the experiments begin with packages that are below the lower corner gradually increase the range including both. Next the lower limit is increased and the set of experiments is again performed until the packages cover a range up to the upper limit. The effect is that we get experiments covering only below and above the two corners and different ranges including both and one or the other.

Figure 3 shows the result. Using only a package of relative low frequencies results a time constant smaller than the dominating one. Increasing the range (see first experiment set) the estimate does change less and less, but increasingly monotonically towards the end (left end). The low estimate is due to not having the first corner frequency (located at 0) included in the range. As one moves the package range covering the two corners, the estimates get relatively constant though always increasing with the inclusion of higher frequencies. Once one is over the limit of including the lower corner Figure 4 . Step responses of different models frequency, which is a 1, the estimated time compared with plant constant is increasing rapidly.

Figure 2. Frequency ranges for the 19 sets of experiments

Figure 3. Experiments with packages of 8 sinusoidal. Each line is the set of experiment with range shown in Figure 2 with the upper limit changing.

Looking at the fit in the time domain, the step responses for an under-estimated, a well estimated and an over-estimated model are shown in Figure 4. The difference in the model is apparently large and most likely significant to any application. The figure also clearly reveals the second-order behaviour of the plant, specifically the zero slope at the beginning.

Figure 4. Step responses of different models compared with plant

The frequency responses look similarly intriguing. Figure 5 shows the same three models and the plant in Bode plots.

Figure 5. Step responses of different models without dead time compared with plant

Apparently the phase is not matched as the model′s high-frequency phase shift is 90 degrees, whilst the plant has 180 degrees. Most likely, at this point control people would suggest a first-order-plus-dead-time model, which when fitted would do a better job on the phase Indeed if we use the same experiments as models without dead time compared with above, this is the case, though only for frequency packages that are below the second time constant. Above things get hairy very quickly and the algorithms fail to provide an estimate. On the lower frequency package side nothing changes, the estimated dead time is zero. Figure 6 shows the situation in the frequency domain. for three models. The three chosen experiments are the very first, thus the package clearly below the dominating time constant (green, top one), the second is the red one (second from the top, down turning in the phase plot) and the third is fitting best (third in the phase plot turning down most quickly). The step responses for the two last estimates look superb, whilst the first is equally bad as shown in Figure 4.

Figure 6. Step responses of different models with dead time compared with plant

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780444595072500664

Frequency response

W. Bolton , in Instrumentation and Control Systems, 2004

11.4 System identification

In Chapter 8, methods were indicated by which models, i.e. differential equations describing the input-output relationship or transfer function, can be devised for systems by considering them to be made up of simple elements. An alternative way of developing a model for a real system is determine its response to some input and then find the model that fits the response; this process of determining a mathematical model is known as system identification.

A particularly useful method of system identification is to use a sinusoidal input and determine the output over a range of frequencies. Bode plots are then plotted with this experimental data. We then find the Bode plot elements that fit the experimental plot by drawing asymptotes on the gain Bode plot and considering their gradients.

1.

If the gradient at low frequencies prior to the first corner frequency is zero then there is no s or 1/s element in the transfer function. The K element in the numerator of the transfer function can be obtained from value of the low frequency gain since the gain in dB = 20 lg K.

2.

If the initial gradient at low frequencies is −20 dB/decade then the transfer function has a 1/s element.

3.

If the gradient becomes more negative at a corner frequency by 20 dB/decade, there is a (1 + s/ω_c) term in the denominator of the transfer function, with ω_c being the corner frequency at which the change occurs. Such terms can occur for more than one corner frequency.

4.

If the gradient becomes more positive at a corner frequency by 20 dB/decade, there is a (1 + s/ω_c) term in the numerator of the transfer function, with ω_c being the frequency at which the change occurs. Such terms can occur for more than one corner frequency.

5.

If the gradient at a corner frequency becomes more negative by 40 dB/decade, there is a (s ²/ω_c ² + 2ζs/ω_c + 1) term in the denominator of the transfer function. The damping ratio ζ can be found from considering the behaviour of the system to a unit step input.

6.

If the gradient at a corner frequency becomes more positive by 40 dB/decade, there is a (s ²/ω² _c + 2ζs/ω_c + 1) term in the numerator of the transfer function. The damping ratio ζ can be found from considering the detail of the Bode plot at a corner frequency.

7.

If the low-frequency gradient is not zero, the K term in the numerator of the transfer function can be determined by considering the value of the low-frequency asymptote. At low frequencies, many terms in transfer functions can be neglected and the gain in dB approximates to 20 lg (K/ω²). Thus, at ω = 1 the gain in dB approximates to 20 lg K.

The phase angle curve is used to check the results obtained from the magnitude analysis.

Example

Determine the transfer function of the system giving the Bode magnitude plot shown in Figure 11.18.

Figure 11.18. Example

The initial gradient is 0 and so there is no 1/s or s term in the transfer function. The initial gain is 20 and thus 20 = 20 lg K and so we have K = 10. The gradient changes by −20 dB/decade at a frequency of 10 rad/s. Hence there is a (1 + s/10) term in the denominator. The transfer function is thus 10/(1 + 0.1s).

Example

Determine the transfer function of the system giving the Bode magnitude plot shown in Figure 11.19.

Figure 11.19. Example

There is an initial slope of −20 dB/decade and so a 1/s term. At the corner frequency 1.0 rad/s there is a −20 dB/decade change in gradient and so a 1/(1 + s/1) term. At the corner frequency 10 rad/s there is a further −20 dB/decade change in gradient and so a 1/(1 + s/10) term. At ω = 1 the magnitude is 6 dB and so 6 = 20 lg K and K = 10^6/20 = 2.0. The transfer function is thus 2.0/s(1 + s)(1 + 0.1s).

Example

Determine the transfer function of the system giving the Bode magnitude plot shown in Figure 11.20. This shows both the asymptotes and the departure of the true plot from them in the vicinity of the break point.

Figure 11.20. Example

The gain Bode plot has an initial zero gradient. Since the initial magnitude is 10 dB then 10 = 20 lg K and so K = 10^0.5 = 3.2. The change of −40 dB/decade at 1 rad/s means there is a 1/(s ² + 2ζs + 1) term. The transfer function is thus 3.2/(s ² + 2ζs + 1).

The damping factor ζ can be obtained by considering the departure of the true Bode plot from the asymptotes at the break point. Since it rises by about 4 dB, Table 11.2 indicates that this corresponds to a damping factor of about 0.3. The transfer function is thus 3.2/(s ² + 0.6s +1).

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780750664325500115

Advances in Neural Recording and Stimulation Devices

Jian Xu , ... Zhi Yang , in Engineering in Medicine, 2019

13.2.1.1 Requirement of a high-pass filter

Fig. 13.2A shows a high-pass filter with a sub-Hz corner frequency ( R _f and C _f ) to remove electrode offset. ³⁸ Because low-frequency neural activity extends to sub-Hz, a filter corner frequency (1/2πR _f C _f ) of less than 1   Hz is needed to avoid distortion, which translates to large R _f and C _f . Yet, by operating transistors in cutoff region, R _f achieves near TΩ. A further increase in R _f is difficult and undesired due to leakage currents that will occur on bias circuits, induced nonlinearities to amplification, and tuning efforts needed to stabilize the amplifier. Hence, C _f is accordingly designed in a sub-pF range, causing C _in tens of pF, as reviewed in Harrison et al. ⁹ and Wattanapanitch and Sarpeshkar ¹³ ; a large C _in translates to small amplifier input impedance and large chip area, which become a more eminent problem if the recording density follows a "Moore's law." ⁴⁰

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128130681000130

Integrated Sensors

Edward Ramsden , in Hall-Effect Sensors (Second Edition), 2006

4.6 Bandwidth

The bandwidth, or frequency response, of a linear system can be described by a Bode plot, which is a graph of gain and phase lag vs. frequency. For both traditional and practical reasons, the frequency and gain axis are usually expressed in either logarithmic units (dB for gain) or on a log scale (for frequency). A Bode plot corresponding to what is called a "first-order, low-pass system" is shown in Figure 4-8.

Figure 4-8. Bode plot of first-order linear system.

Some of the key features of the system described in the Bode plot are:

1)

Corner frequency F _c , often called the −3 dB point. At this frequency the value of system gain or sensitivity is only $1/\sqrt{2}\left(0.707\right)$ of its value at DC (zero frequency).

2)

Attenuation rate . Beyond the corner frequency the sensitivity of a first-order system rolls off at the rate of −20 dB per decade of frequency. Another way to look at it is that the response drops a factor of 10 for every 10× increase in frequency.

3)

Phase shift . At the corner frequency, the system will delay a sinusoidal input signal by 45°. As one increases the signal frequency, this phase delay increases asymptotically to 90°.

While measurements such as the corner frequency describe a system's response to sinusoidal stimuli, they also provide some insight into how a device behaves in response to other stimuli. One common example is the system's response to an input in the form of an abrupt step, usually called the step response. The time that a first-order system needs to settle to within 37% of its final value (called the system time constant, represented by τ) is given by 1/(2πf _c ). Figure 4-9 shows the step response of a first-order, low-pass system.

Figure 4-9. Step response of first-order, low-pass system.

The amount of time this system needs to settle to within a specified error bound can be expressed as a function of τ, as shown in Table 4-1.

Table 4-1. Error vs. settling time, ideal first-order system.

Settling time in τ units	% Error
1	36.8
2	13.5
3	5.0
4	1.8
5	0.6

A typical linear Hall-effect sensor has a corner frequency in the range of 10 kHz–25 kHz. This corresponds to a τ of about 6 to 16 microseconds.

Although a linear Hall-effect sensor is much more complex than a first-order system, especially if it employs auto-nulling techniques, using a first-order model is a useful approximation for many applications. Manufacturers of linear Hall-effect devices will usually publish "typical" corner frequency values in the data sheets for their devices.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780750679343500053

Op Amp Noise Theory and Applications

Bruce Carter , in Op Amps for Everyone (Third Edition), 2009

12.6 Putting It All Together

This example is provided for analysis only—actual results depend on a number of other factors. Expanding on the techniques of Section 12.2.5, a low noise op amp is needed over an audio frequency range of 20 Hz to 20 kHz, with a gain of 40 dB. The output voltage is 0 dBV (1 V). The schematic is shown in Figure 12.13.

Figure 12.13. Split supply op amp circuit.

It would be nice to use a TLE2027, with a noise figure of $2.5\text{nV}/\sqrt{\text{Hz}}$ . The data sheet, however, reveals that this is a ±15 V part and that noise figure is specified at only ±15 V. Furthermore, the specification for V _OM+ and V _OM– (see Chapter 13) shows that it can swing to only within approximately 2 V of its voltage rails. If they are +5 V and ground, the op amp is close to clipping with a 1 V output signal. This illustrates a common fallacy: The designer chooses an op amp based on one parameter only, without checking others that affect the circuit. An expert analog designer must develop an attention to details or be prepared to spend a lot of time in the lab with false starts and unexpected problems.

So, the only choice is to select a different op amp. The TLC2201 is an excellent choice. It is a low noise op amp optimized for single supply operation.

The first circuit change in this example is to change the TLE2027 to a TLC2201. Visually, the corner frequency, f _NC, appears to be somewhere around 20 Hz (from Section 12.5.2), the lower frequency limit of the band we are interested in. This is good: It means, for all practical purposes, the 1/f noise can be discounted. It has $8\text{nV}/\sqrt{\text{Hz}}$ noise instead of $2.5\text{nV}/\sqrt{\text{Hz}}$ , and from Section 12.2.5,

•

To begin with, calculate the root hertz part: $\sqrt{20,000-20}=\mathrm{141.35.}$

•

Multiply this by the noise spec, 8 × 141.35 = 1.131 μV, which is the equivalent input noise. The output noise equals the input noise multiplied by the gain, which is 100 (40 dB).

The signal to noise ratio can be now be calculated:

(12.22) $\begin{array}{c}1.131\mu \text{V}\times 100=113.1\mu \text{V}\\ \text{Signal to noise}(\text{dB})=20\times \log (1\text{V}÷113.1\mu \text{V})=20\times \log (8842)=78.9\text{dB}\end{array}$

Pretty good, but it is 10 dB less than would have been possible with a TLE2027. If this is not acceptable (let's say for 16 bit accuracy), you are forced to generate a ±15 V supply. Suppose for now that 78.9 dB signal to noise is acceptable, and build the circuit.

When it is assembled, it oscillates. What went wrong?

To begin with, it is important to look for potential sources of external noise. The culprit is a long connection from the half supply voltage reference to the high impedance noninverting input. Added to that is a source impedance that does not effectively swamp external noise sources from entering the noninverting input. There is a big difference between simply providing a correct DC operating point and providing one that has low impedances where they are needed. Most designers know the "fix," which is to decouple the noninverting input, as shown in Figure 12.14.

Figure 12.14. TLC2201 op amp circuit.

Better—it stopped oscillating. Probably a nearby noise source radiating into the noninverting input was providing enough noise to put the circuit into oscillation. The capacitor lowers the input impedance of the noninverting input and stops the oscillation. Much more information on this topic is found in Chapter 23, including layout effects and component selection. For now, it is assumed that all of these have been taken into account.

The circuit is still slightly noisier than the 78.9 dB signal to noise ratio given previously, especially at lower frequencies. This is where the real work of this example begins, that of eliminating component noise.

The circuit in Figure 12.15 has four resistors. Assuming that the capacitor is noiseless (not always a good assumption), that means four noise sources. For now, only the two resistors in the voltage divider that forms the voltage reference are considered. The capacitor, however, has transformed the white noise from the resistors into pink (1/f) noise. From Sections 12.3.2 and 12.2.5, the noise from the resistors and the amplifier itself is

Figure 12.15. Improved TLC2201 op amp circuit.

(12.23) $\begin{array}{ll}{E}_{\text{TOTALrms}}=\sqrt{5.73{\mu \text{V}}^2+5.73{\mu \text{V}}^2+113.1{\mu \text{V}}^2}=113.1{\mu \text{V}}_{\text{rms}}\hfill & \hfill \end{array}$

(12.24) $\text{Signal to noise}(\text{dB})=20\times \log (1\text{V}÷113.1\mu \text{V})=20\times \log (8842)=78.9\text{dB}$

So far, so good. The amplifier noise is swamping the resistor noise, which only adds a very slight pinkish component at low frequencies. Remember, however, that this noise voltage is multiplied by 101 through the circuit, but that was previously taken into account for the preceding 78.9 dB signal to noise calculation.

Reducing the value of the resistors to decrease their noise is an option. Changing the voltage divider resistors from 100 kΩ to 1 kΩ while leaving the 0.1 μV capacitor the same changes the corner frequency from 32 Hz to 796 Hz, right in the middle of the audio band.

Note: Resist the temptation to make the capacitor larger to move the pinkish effect below the lower limits of human hearing. The resulting circuit must charge the large capacitor up during power up and down during power down. This may cause unexpected results.

If the noise from the half supply generator is critical, the best possible solution is to use a low noise, low impedance half supply source. Remember, however, that its noise is multiplied by 101 in this application.

The effect of the 100 kΩ resistor on the inverting input is whitish and appears across the entire bandwidth of the circuit. Compared to the amplifier noise, it is still small, just like the noise from the noninverting resistors on the input. The noise contribution of resistors is discounted.

Of much more concern, however, is the 10 MΩ resistor used as the feedback resistor. The noise associated with it appears as a voltage source at the inverting input of the op amp and, therefore, is multiplied by a factor of 100 through the circuit. From Section 12.3.2, the noise of a 10 MΩ resistor is –84.8 dBV, or 57.3 μV. Adding this and the 100 kΩ resistor noise to the amplifier noise,

(12.25) $\begin{array}{ll}{E}_{\text{TOTALrms}}=\sqrt{5.73{\mu \text{V}}^2+113.1{\mu \text{V}}^2}=126.8{\mu \text{V}}_{\text{rms}}=-77.9\text{dBV}\hfill & \hfill \end{array}$

(12.26) $\text{Signal to noise}\left(\text{dB}\right)=20\times \log (1\text{V}÷126.8\mu \text{V})=20\times \log \left(7887\right)=77.9\text{dB}$

The noise contribution from the 10 MΩ resistor subtracts 1 dB from the signal to noise ratio. Changing the 10 MΩ resistor to 100 kΩ and the input resistor from 100 kΩ to 1 kΩ preserves the overall gain of the circuit. The redesigned circuit is shown in Figure 12.15.

For frequencies above 100 Hz, where the 1/f noise from the op amp and the reference resistors is negligible, the total noise of the circuit is

(12.27) $\begin{array}{ll}{E}_{\text{TOTALrms}}=\sqrt{0.57{\mu \text{V}}^2+5.73{\mu \text{V}}^2+113.1{\mu \text{V}}^2}=113.2{\mu \text{V}}_{\text{rms}}=-78.9\text{dBV}\hfill & \hfill \end{array}$

(12.28) $\text{Signal to noise}\left(\text{dB}\right)=20\times \log (1\text{V}÷113.2\mu \text{V})=20\times \log \left(8830\right)=78.9\text{dB}$

Proper selection of resistors, therefore, has yielded a signal to noise ratio close to the theoretical limit for the op amp itself. The power consumption of the circuit, however, has increased slightly, which may be unacceptable in a portable application. Remember, too, that this signal to noise ratio is only at an output level of 0 dBV, an input level of –40 dBV. If the input signal is reduced, the signal to noise ratio is reduced proportionally.

Music, in particular, almost never sustains peak levels. The average amplitude may be down 20 dB to 40 dB from the peak values. This erodes a 79 dB signal to noise ratio to 39 dB in quiet passages. If someone "cranks up the volume" during the quiet passages, the noise becomes audible. This is done automatically with automatic volume controls. The only way a designer can combat this is to increase the voltage levels through the individual stages. If the preceding audio stages connecting to this example, for instance, could be scaled to provide 10 dB more gain, the TLC2201 would be handling an output level of 3.16 V instead of 1 V, which is well within its rail to rail limit of 0 V to 4.7 V. This would increase the signal to noise gain of this circuit to 88.9 dB, almost the same as would have been possible with a TLE2027 operated off of ±15 V! But noise in the preceding stages would also increase. Combatting noise is a difficult problem, and trade-offs are always involved.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9781856175050000120

Frequency response

W. Bolton , in Control Systems, 2002

Example

Determine the transfer function of the system giving the Bode magnitude plot shown in Figure 5.18.

Figure 5.18. Example

There is an initial slope of −20 dB/decade and so a 1/s term. At the corner frequency 1.0 rad/s there is a −20 dB/decade change in gradient and so a 1/(1 + s/1) term. At the corner frequency 10 rad/s there is a further −20 dB/decade change in gradient and so a 1/(1 + s/10) term. At ω = 1 the magnitude is 6 dB and so 6 = 20 lg K and K = 10^6/20 = 2.0. The transfer function is thus 2.0/ s(1 + s)(1 + 0.1s).

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780750654616500052

Bode diagram

Yazdan Bavafa-Toosi , in Introduction to Linear Control Systems, 2019

7.2.8 Manual drawing of the Bode diagram

In practice if we want to produce the diagrams by hand, we specify the asymptotes and then with curved lines around the corner frequencies we modify the diagrams. To this end, we should first "add" the Bode diagrams of the individual components of the system en masse, see Exercise 7.31. However, by sweeping the frequency range from zero to infinity and constructing the diagram step by step as outlined in the subsequent instructive example, we can further simplify the procedure and actually circumvent the "addition." It is advisable to turn your hand to it – it is quite insightful.

Example 7.1

Draw the Bode diagram of the system $L(s)=\frac{1600(s+0.5)}{s(s+2)(s^2+4s+100)}.$

We first rewrite the transfer function as $L(j\omega )=\frac{4(1+j2\omega )}{j\omega (1+j0.5\omega )[1+j0.04\omega +{(j0.1\omega )}^2]}.$ Now the constituting elements of the systems are $l_1=4$ , $l_2=1+j2\omega$ , $l_3={(j\omega )}^{-1}$ , $l_4={(1+0.5j\omega )}^{-1}$ , and $l_5={[1+j0.04\omega +{(j0.1\omega )}^2]}^{-1}$ . The basic approach is to draw the Bode diagram of the five constituting elements of the system and then to add them at different frequencies. However, we circumvent this lengthy procedure and turn it to encapsulation of the Bode diagrams, starting from the zero frequency towards the infinite frequency. The procedure is illustrated below.

First we draw the magnitude diagram. We start from the lowest frequency which is $\omega =0$ and move towards the frequency $\omega =\infty$ and in the process consider the effect of the constituting components of the transfer function as they appear. Note that this results in the modification of the already constructed phase diagram because as we saw the phase diagram of any dynamic element spans approximately two decades. At $\omega =0$ we have the effect of the terms $l_1$ and $l_3$ . In fact the effect of the term $l_1$ is in all frequencies. It suffices to consider it one time in the first step. The effect of the magnitude diagrams of the terms $l_1$ and $l_3$ is given in Fig. 7.11, left panel. Note that the curve has the slope $-20\text{db}/\text{dec}$ in low frequencies and the curve itself or its continuation passes through the point $20\log 4=12\mathrm{db}$ at the frequency $\omega =1$ . (Whether the curve itself passes this point or its continuation does pivot on the smallest corner frequency of the other components.) Now we increase the frequency. The first (i.e., smallest) corner frequency we arrive at is that of the term $l_2$ , being $\omega =0.5$ . At this point the effect of the term $l_2$ starts in the form of a $+20\text{db}/\text{dec}$ slope in the magnitude diagram. Thus the slope of the curve increases from $-20$ to 0. Thus the diagram in the right panel of the same figure is obtained.

Figure 7.11. Construction of the Bode magnitude diagram of Example 7.1 in asymptotic form.

As we further increase the frequency the second corner frequency we reach is that of the term $l_4$ , which is $\omega =2$ . From this point the effect of the term $l_4$ begins in the form of a $-20\text{db}/\text{dec}$ slope in the magnitude diagram. Thus the slope of the curve decreases from 0 to $-20$ . We continue increasing the frequency. The diagram in the upper panel of Fig. 7.12 is obtained. The next corner frequency we come to is $\omega =10$ , which is the corner frequency of the term $l_5$ . The effect of this term is $-40\text{db}/\text{dec}$ on the magnitude diagram and a peak at around this frequency because $\zeta <0.7$ . (In fact $2\zeta /{\omega }_n=0.04$ , thus $\zeta =0.2$ .) Therefore, after this frequency the slope of the diagram will be $-60\text{db}/\text{dec}$ . The whole diagram looks like that in Fig. 7.12.

Figure 7.12. Construction of the Bode magnitude diagram of Example 7.1 in asymptotic form.

Next we draw the phase diagram. Here as well we start from frequency 0 and increase it to infinity and in the process consider the effect of the constituting elements of the transfer function as they appear. At $\omega =0$ we have the effect of $l_3$ . In fact we have its effect in all frequencies. It suffices to consider it one time in the first stage. Thus the phase angle is $-90\mathrm{deg}$ at $\omega =0$ . Now we increase the frequency. The first corner frequency we reach is $\omega =0.5$ , due to $l_2$ . The effect of this term which is a zero is ultimately 90   deg increase in phase. The effect starts from $0.1\omega =0.05$ and ends at $10\omega =5$ . At $\omega =0.5$ its effect is $45\mathrm{deg}$ increase in phase. As we further increase the frequency we arrive at $\omega =2$ due to $l_4$ which is a pole. The effect of this pole starts from $0.1\omega =0.2$ and ends at $10\omega =20$ . At $\omega =2$ its effect is $45\mathrm{deg}$ decrease in phase. This results in the modification of the phase diagram of the previous stage. As we continue to increase the frequency we come to $\omega =10$ due to $l_5$ . The effect of this term is 180   deg decrease in phase. Its effect starts (nearly) at $0.1\omega =1$ and ends (almost) at $10\omega =100$ . At $\omega =10$ its effect is $90\mathrm{deg}$ decrease in phase. Because $\zeta$ is small the drop in phase occurs in the vicinity and above $\omega =10$ . The whole diagram will look like as the one given in Fig. 7.13.

Figure 7.13. Bode phase diagram of Example 7.1.

Now in the magnitude diagram we should modify the asymptotes to curves, either above them or below them as appropriate. The answer is shown using MATLAB® in Fig. 7.14.

Figure 7.14. Bode diagram of Example 7.1 produced by MATLAB®.

Remark 7.1

As we discussed in Remark 6.2 of Chapter 6, there is the problem of inconsistent definition of angles in MATLAB® 2015a release. As we stated, to get consistent answers we must agree on a convention re phase definition. For instance, the points $1\angle -90$ and $1\angle 270$ clearly refer to the same point, but we should agree to use one of them not both! This is discussed in the next example.

Example 7.2

In some examples MATLAB® 2015a defines the angles CCW (thus using $\angle 270$ and not $\angle -90$ ), like for the system $L_1(s)=(s^3+1)/s^3$ , see Fig. 7.15, left panel, and sometimes it defines both CCW and CW (clockwise) angles, like for the system $L_2(s)=(s^2+1)/{(s-1)}^3$ , see Fig. 7.15, right panel. This is explained below.

Figure 7.15. Inconsistent definition of angles in MATLAB® 2015a release.

Fig. 7.15 , left panel: For the system $L_1(s)=(s^3+1)/s^3$ all angles are defined CCW. Define $\alpha =\angle (s-e^{j\pi /3})$ , $\beta =\angle (s-e^{-j\pi /3})$ , $\gamma =\angle (s+1)$ , and $\theta =\angle (s-0)$ . Thus the phase angle is given by $\alpha +\beta +\gamma -3\times \theta$ . At $\omega =0^{+}$ : $\alpha =240$ , $\beta =120$ , $\gamma =0$ , and $\theta =90$ . As frequency increases from $\omega =0^{+}$ to $\omega =+\infty$ , $\alpha$ and $\beta$ decrease to $\alpha =90$ and $\beta =90$ , $\gamma$ increases to $\gamma =90$ , and $\theta$ remains constant. Therefore, the phase angle decreases from $90$ to $0$ .

Fig. 7.15 , right panel: For the system $L_2(s)=(s^2+1)/{(s-1)}^3$ some angles are defined CW and some CCW. Define $\alpha =\angle (s+j)$ , $\beta =\angle (s-j)$ , and $\theta =\angle (s-1)$ . The phase angle is hence given by $\alpha +\beta -3\times \theta$ . At $\omega =0^{+}$ : $\alpha =90$ , $\beta =-90$ , $\theta =\angle (s-1)=180$ . As frequency increases from $\omega =0^{+}$ to $\omega =+\infty$ , $\alpha$ remains constant, $\beta$ increases to $\beta =90$ , and $\theta$ decreases to $\theta =90$ . As a result the phase angle increases from $-540\mathrm{deg}$ to $-90\mathrm{deg}$ . However, as you recall this system was considered in detail in Example 6.11 of Chapter 6, Nyquist Plot. There we saw that the phase varies between −180 and 270   deg.)

Inconsistent definition of angles is certainly not a good practice, at least a hotbed of some snafu especially for students. Our take is that the concord should be to always define angle values CCW, which is the worldwide consensus in mathematics as well. In the preceding Example 7.2 it is not a serious issue in some ways, nevertheless from time to time the answer is miserably wrong; try, e.g., the system $L(s)={(s^2+1)}^2/{(s+1)}^4$ . We know that there should be an increase of $2\times 180=360\mathrm{deg}$ at $\omega =1$ as a jump, however due to inconsistent definition of angles the MATLAB-produced diagram shows no jump of phase at this frequency. (There are two angles $\angle (s-j)$ . One of them is defined CCW and the other CW!) MATLAB® also produces a wrong NKMH chart for this system, as we shall see in Exercise 8.39 of Chapter 8.

We close this section by the sequel instructive discussion.

Discussion: It is essential to correctly interpret a positive or negative slope due to an individual component in a system: A positive slope (due to a zero) does not necessarily mean that the magnitude diagram of the system goes to infinity, unless we are aware that we implicitly assume the increasing direction of frequency as the context of our argument "and" the system is improper. (For instance, in the decreasing direction of frequency a plot with a positive slope due to zero at origin goes to zero, i.e., minus infinity in db scale.) The counterpart is true for a system with negative slope due to a pole: A negative slope (due to a pole) does not necessarily mean that the magnitude diagram of the system goes to zero (minus infinity in db scale), unless we are aware that we implicitly assume the increasing direction of frequency as the context of our argument "and" the system is strictly proper. (For instance, in the decreasing direction of frequency the magnitude diagram goes to infinity if the system has an integrator).

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128127483000070

The Systems Approach to Control and Instrumentation

William B. Ribbens , in Understanding Automotive Electronics (Seventh Edition), 2013

Filter-Design Techniques

Filter design begins with a low-pass prototype function normalized frequency (Ω   = ω/ω_c ) where ω_c is the passband corner frequency. The other three filter types are derived by a linear transformation involving complex frequency s as explained below.

Filters are designed as well as classified by the function F(Ω) from which they are derived. For example, the so-called Butterworth filter is derived from the function

(83) $F(\Omega )=\frac{1}{1+{\Omega }^{2n}}$

where n  is the   order of the filter. Butterworth filters are characterized by maximally flat passbands. This function is taken to be the squared magnitude of the sinusoidal frequency response of the desired filter:

(84) $F(\Omega )={\left|H(j\Omega )\right|}^2$

(85) $=H(S)H(-S){|}_{S=j\frac{\omega }{{\omega }_c}}$

where S is a normalized complex frequency. To find the transfer function (H(s)), the substitution Ω²  →   −S ² is made and the resulting function factored to find the roots of the numerator and denominator. For the example Butterworth filter, these roots lie along the unit circle in the normalized complex S-plane. These roots are equally spaced and except for odd n where a single root exists at S  =   −1, they occur in complex conjugate pairs. The roots of the denominator (i.e., poles at S  = P_m ) in the left half S-plane are determined and the transfer function is given by

(86) $H(S)=\frac{1}{\prod _{m=1}^n(S-P_m)}$

For example, the third-order (i.e., n  =   3) Butterworth low-pass filter prototype is

(87) $H(S)=\frac{1}{(S+1)(S^2+S+1)}$

The transfer function is then un-normalized by replacing S with s/ω_c .

The magnitude and phase of a third-order Butterworth filter are given in Figure 1.23 plot for H(s).

Figure 1.23. Magnitude and phase frequency response of third-order Butterworth filter.

Note that 20   log |H(j1)|² which is equivalent to:

$H(j\Omega ){|}_{\Omega =1}=\frac{1}{\sqrt{2}}$

The normalized frequency Ω   =   1 is the corner frequency of this filter in normalized frequency. The steepness of the drop from passband to stopband increases with increasing filter order n. The slope in the transition from the passband is –20 n dB/decade.

The other major filter types are based upon polynomials in normalized frequency. The two commonest of these are Chebyshev or Cauer. The Chebyshev filter prototypes are derived from the following polynomials:

(88) $F({\Omega }^2)={\in }^2{C}_n^2(\Omega )$

where

$C_n(\Omega )=\cos (n{\cos }^{-1}(\Omega ))\left|\mathit{\omega }\le 1\right|$

$=\cosh (n{\cosh }^{-1}(\Omega ))\left|\mathit{\omega }>1\right|$

and $\in$ is a parameter determined by the passband "ripple." Cauer filter prototypes are derived from Jacobi elliptic function of Ω and are beyond the scope of the present text. However, any standard filter-design reference will supply design parameter tables. Also, filter design is readily accomplished using MATLAB or other design software.

The design of a high-pass filter normalized prototype by replacing S with 1/S. Similar linear transformations of S are available for bandpass or bandstop filters, but the transformation is dependent upon the relative values for ω ₁ and ω ₂. Fortunately, modern software such as MATLAB has the capability of calculating the transfer functions of any of the filters discussed.

It is clear from the above discussion of filter-design techniques that, in principle, it is possible to design a filter for measurement of a given quantity that optimizes the signal/noise. Such an optimum design is based on a priori information about the spectrum of the measured quantity. It is normally not possible (nor is it necessary) to know the exact spectrum of this quantity. However, it is often possible to be able to determine upper and lower bounds of this "signal spectrum." The signal-processing filter passband can be selected to enclose these spectral band limits. Noise suppression occurs in the associated stopbands for the optimal filter.

In automotive electronic instrumentation, the sensor often measures a mechanical variable. The dynamic model for the associated mechanical system is often known with great accuracy, thereby allowing the "signal" spectral bounds to be closely estimated. In such cases, the signal-processing filter optimization can readily proceed.

This chapter has reviewed some basic principles of continuous time system theory that are applicable through the remainder of the book. Specific applications of this theory are found in nearly all automotive electronic systems. However, as explained earlier, modern automotive electronics are digital and are modeled and analyzed using discrete time methods. The next chapter reviews basic principles of such discrete time system modeling/analysis/design.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780080970974000011

ICs, Oscillators, and Filters

Keith Brindley , in Starting Electronics (Fourth Edition), 2011

Answers at the end of the book.

1.

A signal of frequency 1 kHz is applied to a low-pass filter with a corner frequency of 10  kHz. What happens?:

a.

The output signal is one-tenth of the input signal

b.

The output signal is larger than the input signal

c.

There is no output signal

d.

The output signal is identical to the input signal

e.

All of these.

2.

A high-pass filter consisting of a 10   kΩ resistor and an unknown capacitor has a corner frequency of 100   Hz. What is the value of the capacitor?:

a.

1   nF

b.

10   nF

c.

100   nF

d.

1000   nF

e.

1   μF.

3.

A low-pass filter consisting of a 1   μF capacitor and an unknown resistor has a corner frequency of 100   Hz. What is the value of the resistor?:

a.

10   kΩ

b.

1   kΩ

c.

100   kΩ

d.

All of these

e.

It makes no difference

f.

d and e

g.

None of these.

4.

In a circuit similar to that in Figure 5.2, resistor R1 is 10   kΩ, resistor R2 is 100   kΩ, and capacitor C1 is 10   nF. The output frequency of the astable multi-vibrator is:

a.

A square wave

b.

About 680   Hz

c.

Too fast to see the LED flashing

d.

a and b

e.

c and d

f.

None of these.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780080969923000056

bergdidese.blogspot.com

Source: https://www.sciencedirect.com/topics/engineering/corner-frequency

Share this post