User talk:Williamborg/Signals

Design of continuous-time linear, time-invariant filters
The student should already know how to characterize a linear, time-invariant system using: & frequency response.
 * convolution,
 * unit impulse response function,
 * transfer functions

This provides the basis to design linear time-invariant systems that can achieve specifications for particular purposes - which is what electrical engineers MUST be able to do.

Several things are key to going this:
 * know what a pole in the complex plane is
 * know what the zeros are
 * understand how these influence the frequency response of a system

Reviewing poles and zeros in the complex plane


A complex number is a number consisting of a real part and an imaginary part. Complex numbers extend the idea of the one-dimensional number line to the two-dimensional complex plane by using the number line for the real part and adding a vertical axis to plot the imaginary part. In this way the complex numbers contain the ordinary real numbers while extending them in order to solve problems that would be impossible with only real numbers.

Complex numbers are used in many scientific fields, including electrical engineering, electromagnetism, most of physics included quantum physics, and applied mathematics.

Argand diagrams are frequently used to plot the positions of the poles and zeroes of a math function in the complex plane.

Poles in the complex plane


Let's review complex poles. In complex analysis (an isolated singularity is one that has no other singularities close to it), a pole of a function is a singularity that behaves like the singularity of $$ \scriptstyle \frac{1}{z^n} $$ at z = 0. This means that any pole of the function f(z) is a point a such that f(z) approaches infinity as z approaches a.


 * The function


 * $$f(z) = \frac{3}{z}$$


 * has a pole of order 1 or simple pole at $$\scriptstyle z\, = \,0$$.


 * The function


 * $$f(z) = \frac{z+2}{(z-5)^2(z+7)^3}$$


 * has a pole of order 2 at $$\scriptstyle z\, = \,5$$ and a pole of order 3 at $$\scriptstyle z\, = \,-7$$.


 * The function


 * $$f(z) = \frac{z-4}{e^z-1}$$


 * has poles of order 1 at $$\scriptstyle z\, = \,2\pi ni\text{ for } n\, = \,\dots,\, -1,\, 0,\, 1,\, \dots.$$ To see that, it helps to write $$\scriptstyle e^z$$ in Taylor series around the origin.


 * The function


 * $$f(z) = z$$


 * has a single pole at infinity of order 1.


 * The cosecant function $$\csc \left(\pi z\right)$$ has an isolated singularity at every integer.

Complex plane zeros
A complex number a is a simple zero of f, or a zero of multiplicity 1 of f, if f can be written as


 * $$f(z)=(z-a)g(z)\,$$

where g is a holomorphic function g such that g(a) is not zero.

Generally, the multiplicity of the zero of f at a is the positive integer n for which there is a holomorphic function g such that


 * $$f(z)=(z-a)^ng(z)\ \mbox{and}\ g(a)\neq 0.\,$$

The multiplicity of a zero a is also known as the order of vanishing of the function at a.

The fundamental theorem of algebra concludes that every nonconstant polynomial with complex coefficients has at least one zero in the complex plane. This is in contrast to the situation with real zeros: some polynomial functions with real coefficients have no real zeros.

Consider the example f(x) = x2 + 1. In this case x = ± i

Consider the example f(x) = x2 - 1. In this case x = ± 1

Characterizing the frequency response
The frequency response is defined in terms of the magnitude response and the phase response.

The magnitude is displayed using a log-log plot, expressed in decibels (dBs), which is given by 20 log10 {|H(jω)|}.

The phase response is usually expressed in either radians or degrees.


 * examples using the decibel

These examples yield dimensionless answers because they are relative ratios expressed in decibels.


 * To calculate the ratio of 1 kW (one kilowatt, or 1000 watts) to 1 W in decibels, use the formula

G_\mathrm{dB} = 10 \log_{10} \bigg(\frac{1000~\mathrm{W}}{1~\mathrm{W}}\bigg) \equiv 30~\mathrm{dB} \, $$


 * To calculate the ratio of $$\sqrt{1000}~\mathrm{V} \approx 31.62~\mathrm{V}$$ to $$1~\mathrm{V}$$ in decibels, use the formula

G_\mathrm{dB} = 20 \log_{10} \bigg(\frac{31.62~\mathrm{V}}{1~\mathrm{V}}\bigg) \equiv 30~\mathrm{dB} \, $$ Notice that $$({31.62\,\mathrm{V}}/{1\,\mathrm{V}})^2 \approx {1\,\mathrm{kW}}/{1\,\mathrm{W}}$$, illustrating the consequence from the definitions above that $$G_\mathrm{dB}$$ has the same value, $$30~\mathrm{dB}$$, regardless of whether it is obtained with the 10-log or 20-log rules; provided that in the specific system being considered power ratios are equal to amplitude ratios squared.


 * To calculate the ratio of 1 mW (one milliwatt) to 10 W in decibels, use the formula

G_\mathrm{dB} = 10 \log_{10} \bigg(\frac{0.001~\mathrm{W}}{10~\mathrm{W}}\bigg) \equiv -40~\mathrm{dB} \, $$


 * To find the power ratio corresponding to a 3 dB change in level, use the formula

G = 10^\frac{3}{10} \times 1\ = 1.99526... \approx 2 \, $$

A change in power ratio by a factor of 10 is a 10 dB change. A change in power ratio by a factor of two is approximately a 3 dB change. More precisely, the factor is 103/10, or 1.9953, about 0.24% different from exactly 2. Similarly, an increase of 3 dB implies an increase in voltage by a factor of approximately $$\scriptstyle\sqrt{2}$$, or about 1.41, an increase of 6 dB corresponds to approximately four times the power and twice the voltage, and so on. In exact terms the power ratio is 106/10, or about 3.9811, a relative error of about 0.5%.



Response of a pure integrator
Consider a pure integrator (effectively an RC circuit). It has a transfer function $$ H(s) = {1 \over s} $$

Its frequency response is $$ H({j \omega}) = {1 \over {j \omega}} $$

The magnitude response, expressed in terms of gain, is:

G = | H(j\omega) | = \left|\frac{V_C(j \omega)}{V_{in}(j \omega)}\right| $$

The magnitude response decreases by a decade (factor of ten) when the frequency increases by a decade.

The magnitude response expressed in decibels decreases by 20 dB every decade.


 * $$ |H(j\omega)| = {1 \over |j\omega|} $$

and the phase angles are:



\phi_R = \angle H(j \omega) = \tan^{-1}\left(\frac{1}{\omega}\right) $$.

Response of a pure differentiator
The magnitude response of a pure differentiator with transfer function


 * $$H(s) = s $$

or equivalently, with frequency response


 * $$H(j\omega) = j\omega$$

Frequency response of a single real pole
Let us consider a transfer function of the form


 * $$H(s) = {1 \over {1 + s\tau}} $$

An arbitrary transfer function of the form ˜H (s) = 1 s + a (7.8) can be written as ˜H (s) = (1/a) 1 + (s/a) (7.9) such that H(s) and ˜H (s) are related through a scaling factor 1/a with τ = 1/a. Consequently, if we know the magnitude response of ˆH (s) = 1/{1 + (s/a)}, we can express the log magnitude response of ˜H(s) as 20 log10 {| ˜H (jω)|} = 20log10 {| ˆH (jω)|} + 20log10 {1/a} (7.10)