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Transfer Function

In its simplest form for continuous-time input signal $x(t)$ $x(t)$ and output $y(t)$ $y(t)$ , the transfer function is the linear mapping of the Laplace transform of the input, $X(s)$ $X(s)$ , to the output $Y(s)$ $Y(s)$ :

$$Y(s) = G(s)X(s)$$ $$Y(s) = G(s)X(s)$$

or

$$G(s) = \frac{Y(s)}{X(s)},$$ $$G(s) = \frac{Y(s)}{X(s)},$$

where $G(s)$ $G(s)$ is the transfer function of the LTI system. In discrete-time systems, the function is similarly written as:

$$G(z) = \frac{Y(z)}{X(z)}.$$ $$G(z) = \frac{Y(z)}{X(z)}.$$

The transfer function is derived using the Laplace transform. It was the primary tool used in classical control engineering. However, it has proven to be unwieldy for the analysis of multiple-input multiple-output (MIMO) systems, and has been largely supplanted by state space representations for such systems. In spite of this, a transfer matrix can be always obtained for any linear system, in order to analyze its dynamics and other properties: each element of a transfer matrix is a transfer function relating a particular input variable to an output variable.

Bode Plot

A Bode plot, named after Hendrik Wade Bode, is usually a combination of a Bode magnitude plot and Bode phase plot:

A Bode magnitude plot is a graph of log magnitude against log frequency often used in signal processing to show the transfer function or frequency response of a linear, time-invariant system.


Figure 1: The Bode plot for a first-order (one-pole) lowpass filter

It makes multiplication of magnitudes a simple matter of adding distances on the graph, since 

    
$$\log(a \cdot b) = \log(a) + \log(b)$$
$$\log(a \cdot b) = \log(a) + \log(b)$$

The Bode plot describes the output response of a frequency-dependent system for a normalised input. The magnitude axis of the Bode plot is often converted directly to decibels.

A Bode phase plot is a graph of phase against log frequency, usually used in conjunction with the magnitude plot, to evaluate how much a frequency will be phase-shifted. For example a signal described by: $A\sin(\omega t)$ $A\sin(\omega t)$ may be attenuated but also phase-shifted. If the system attenuates it by a factor $x$ $x$ and phase shifts it by $-\phi$ $-\phi$ the signal out of the system will be $(A/x) \sin(\omega t -\phi)$ $(A/x) \sin(\omega t -\phi)$ . The phase shift $\phi$ $\phi$ is generally a function of frequency.

The magnitude and phase Bode plots can seldom be changed independently of each other — changing the amplitude response of the system will most likely change the phase characteristics and vice versa. For minimum-phase systems the phase and amplitude characteristics can be obtained from each other with the use of the Hilbert transform.

If the transfer function is a rational function, then the Bode plot can be approximated with straight lines. These asymptotic approximations are called straight line Bode plots or uncorrected Bode plots and are useful because they can be drawn by hand following a few simple rules. Simple plots can even be predicted without drawing them.

The approximation can be taken further by correcting the value at each cutoff frequency. The plot is then called a corrected Bode plot.

'Rules for hand-made Bode plot'

The main idea about Bode plots is that one can think of the log of a function in the form: 

    
$$f(x) = A \prod (x + c_n)^{a_n}$$
$$f(x) = A \prod (x + c_n)^{a_n}$$

as a sum of the logs of its poles and zeros: 

    
$$\log(f(x)) = \log(A) + \sum a_n \log(x + c_n)$$
$$\log(f(x)) = \log(A) + \sum a_n \log(x + c_n)$$

This idea is used explicitly in the method for drawing phase diagrams. The method for drawing amplitude plots implicitly uses this idea, but since the log of the amplitude of each pole or zero always starts at zero and only has one asymptote change (the straight lines), the method can be simplified.

Straight-line amplitude plot

Amplitude decibels is usually done using the 20log10(X) version. Given a transfer function in the form 

    
$$H(s) = A \prod \frac{(s + x_n)^{a_n}}{(s + y_n)^{b_n}}$$
$$H(s) = A \prod \frac{(s + x_n)^{a_n}}{(s + y_n)^{b_n}}$$

where $s = j\omega$ $s = j\omega$ , $x_n$ $x_n$ and $y_n$ $y_n$ are constants, and $H$ $H$ is the transfer function:

  • at every value of $s$ $s$ where $\omega = x_n$ $\omega = x_n$ (a zero), increase the slope of the line by $20 \cdot a_n$ $20 \cdot a_n$ dB per decade.
  • at every value of s where $\omega = y_n$ $\omega = y_n$ (a pole), decrease the slope of the line by $20 \cdot b_n$ $20 \cdot b_n$ dB per decade.
  • The initial value of the graph depends on the boundaries. The initial point is found by putting the initial angular frequency $\omega$ $\omega$ into the function and finding $\vert H(j\omega) \vert$ $\vert H(j\omega) \vert$ .
  • The initial slope of the function at the initial value depends on the number and order of zeros and poles that are at values below the initial value, and are found using the first two rules.

To handle irreducible 2nd order polynomials, $ax^2 + bx + c$ $ax^2 + bx + c$ can, in many cases, be approximated as $(\sqrt{a}x + \sqrt{c})^2$ $(\sqrt{a}x + \sqrt{c})^2$ .

Note that zeros and poles happen when $\omega$ $\omega$ is equal to a certain $x_n$ $x_n$ or $y_n$ $y_n$ . This is because the function in question is the magnitude of $H(j\omega)$ $H(j\omega)$ , and since it is a complex function, $\vert H(j\omega)\vert = \sqrt{H \cdot H^* }$ $\vert H(j\omega)\vert = \sqrt{H \cdot H^* }$ . Thus at any place where there is a zero or pole involving the term $(s + xn)$ $(s + xn)$ , the magnitude of that term is $\sqrt(x_n + j\omega) \cdot (x_n - j\omega)= \sqrt{x_n^2+\omega^2}$ $\sqrt(x_n + j\omega) \cdot (x_n - j\omega)= \sqrt{x_n^2+\omega^2}$ .

Corrected amplitude plot

To correct a straight-line amplitude plot:

  • at every zero, put a point $3 \cdot a_n\ \mathrm{dB}$ $3 \cdot a_n\ \mathrm{dB}$ above the line,
  • at every pole, put a point $3 \cdot b_n\ \mathrm{dB}$ $3 \cdot b_n\ \mathrm{dB}$ below the line,
  • draw a smooth line through those points using the straight lines as asymptotes (lines which the curve approaches).

Note that this correction method does not incorporate how to handle complex values of $x_n$ $x_n$ or $y_n$ $y_n$ . In the case of an irreducible polynomial, the best way to correct the plot is to actually calculate the magnitude of the transfer funcition at the pole or zero corresponding to the irreducible polynomial, and put that dot over or under the line at that pole or zero.

Straight-line phase plot

Given a transfer function in the same form as above: 

    
$$H(s) = A \prod \frac{(s + x_n)^{a_n}}{(s + y_n)^{b_n}}$$
$$H(s) = A \prod \frac{(s + x_n)^{a_n}}{(s + y_n)^{b_n}}$$

the idea is to draw separate plots for each pole and zero, then add them up. The actual phase curve is given by - $\mathbf{arctan}\bigg(\frac{\mathbf{im}[H(s)]}{\mathbf{re}[H(s)]}\bigg)$ $\mathbf{arctan}\bigg(\frac{\mathbf{im}[H(s)]}{\mathbf{re}[H(s)]}\bigg)$

To draw the phase plot, for each pole and zero:

  • if A is positive, start line (with zero slope) at 0 degrees,
  • if A is negative, start line (with zero slope) at 180 degrees,
  • at every $\omega = x_n$ $\omega = x_n$ (a zero), slope the line up at $45 \cdot a_n$ $45 \cdot a_n$ degrees per decade, beginning one decade before $\omega = x_n$ $\omega = x_n$ (that is, start at $\frac{x_n}{10}$ $\frac{x_n}{10}$ ,
  • at every $\omega = y_n$ $\omega = y_n$ (a pole) slope the line down at 45 \cdot b_n degrees per decade, beginning one decade before $\omega = y_n$ $\omega = y_n$ (that is, start at $\frac{y_n}{10})$ $\frac{y_n}{10})$ ,
  • flatten the slope again when the phase has changed by $90 \cdot a_n$ $90 \cdot a_n$ degrees (for a zero) or $90 \cdot b_n$ $90 \cdot b_n$ degrees (for a pole),
  • After plotting one line for each pole or zero, add the lines together to obtain the final phase plot; that is, the final phase plot is the super-position of each earlier phase plot.

Frequency Response

Frequency response is the measure of any system's response at the output to a signal of varying frequency (but constant amplitude) at its input. It is usually referred to in connection with electronic amplifiers, loudspeakers and similar systems. The frequency response is typically characterized by the magnitude of the system's response, measured in dB, and the phase, measured in radians, versus frequency. The frequency response of a system can be measured by:

  • applying an impulse to the system and measuring its response (see impulse response)
  • sweeping a constant-amplitude pure tone through the bandwidth of interest and measuring the output level and phase shift relative to the input
  • applying a signal with a wide frequency spectrum (e.g., maximum length sequence, white noise, or pink noise), and calculating the impulse response by deconvolution of this input signal and the output signal of the system.

Once a frequency response has been measured (e.g., as an impulse response), providing the system is linear and time-invariant, its characteristic can be approximated with arbitrary accuracy by a digital filter. Similarly, if a system is demonstrated to have a poor frequency response, a digital or analog filter can be applied to the signals prior to their reproduction to compensate for these deficiencies.

Frequency response curves are often used to indicate the accuracy of amplifiers and speakers for reproducing audio. As an example, a high fidelity amplifier may be said to have a frequency response of 20 Hz - 20,000 Hz ±1 dB. This means that the system amplifies all frequencies within that range within the limits quoted. 'Good frequency response' therefore does not guarantee a specific fidelity, but only indicates that a piece of equipment meets the basic frequency response requirements.

Page last modified on January 31, 2007, at 04:15 PM