Engineersphere.com » Linear Systems http://engineersphere.com Thu, 19 Apr 2012 17:52:28 +0000 en hourly 1 http://wordpress.org/?v=3.3.2 The Fourier Integral / Transform Explained http://engineersphere.com/linear-systems/the-fourier-integral-transform-explained.html http://engineersphere.com/linear-systems/the-fourier-integral-transform-explained.html#comments Sat, 02 Apr 2011 03:51:33 +0000 Safa http://engineersphere.com/?p=2824

Magic is real, and it all comes from the Fourier Integral.  But one doesn’t become a wizard without a little reading first – so, the purpose of this article is to explain the Fourier Integral theoretically and mathematically.

Before reading any further, it is important to first understand this: in mathematics, there is a rule that states that any periodic function of time may be “reconstructed” exactly from the summation of an infinite series of harmonic sine-waves.  The generalized theory itself is referred to as a “Fourier Series.”  For use with arbitrary electronic time-domain signals of period T_o, it may be expressed as:

f(t) = a_0 + \displaystyle\sum_{n=1}^{\infty}a_ncos(nw_ot) + b_nsin(nw_ot)

over the range:

t_0 \le t \le t_0 + T_0

where:

a_0 is the magnitude of the 0th harmonic

a_n represents the magnitude of the nth harmonic of cosine wave components

b_n represents the magnitude of the nth harmonic of sine wave components

w_o is the fundamental frequency

t is the variable that represents instances in time

n is the variable that represents the specific harmonic, and is always an integer

This monumental discovery was first announced on December 21, 1807 by historic gentleman Baron Jean-Baptiste-Joseph Fourier.

Joseph Fourier

In order to go from the Fourier Integral to the Fourier Transform, it is necessary to express the previous Fourier Series as a series of ever-lasting exponential functions.  Using an orthogonal basis set of signals described by e^{jnw_ot} of magnitude D_n, we now write the Fourier Series as:

f(t) = \displaystyle\sum_{n=- \infty}^{\infty}D_ne^{jnw_ot}

where j is \sqrt{-1}.

What is the Fourier Integral?

The Fourier Integral, also referred to as Fourier Transform for electronic signals, is a mathematical method of turning any arbitrary function of time into a corresponding function of frequency.  A signal, when transformed into a “function of frequency”, essentially becomes a function that expresses the relative magnitudes of each harmonic of a Fourier Series that would be summed to recreate the original time-domain signal. To see this, observe the following figures:

square-pulse-wave

Figure 1. A Square Wave Pulse, in time

In order to rebuild a square wave with sines and cosines only, it is necessary to determine the magnitudes of each harmonic used in the Fourier Series, or rather, the Fourier Integral (for continuous time-domain signals).  The relative magnitudes of these needed harmonics can be displayed graphically as a function of frequency (widely known as a signal’s frequency spectrum):

sinc-wave

Figure 2. The Fourier Integral, aka Fourier Transform, of a square pulse is a Sinc function. The Sinc function is also known as the Frequency Spectrum of a Square Pulse.

Though the recreation of a signal using an infinite series of sines and cosines is impossible to achieve in the lab, one may get very close.  Close enough that the most advanced lab equipment wouldn’t be able to calculate the error due to tolerance specifications.  This allows engineers to use Fourier Analysis to work with time-domain signals, such as radio signals, television signals, satellite signals and just about any signal you can think of.  By viewing a signal according to what frequency components are contained within it, electrical engineers may concern themselves with magnitude changes in frequency only, and may no longer worry about the signal’s magnitude-changes through time.  Not only is this a very practical concept when working in the lab, it also greatly simplifies the mathematics behind signal conditioning in general.  In fact, the entirety of the Communications industry owes its success to the Fourier Transform for not only antenna design, but a plethora of other applications.

The math behind the Fourier Transform

The derivations that follow have been summarized from Chapter 4 of the textbook “Signal Processing and Linear Systems” by B.P. Lathi, a fine book for students of Communication Systems.

We begin by considering some arbitrary, aperiodic time-domain signal.  An example of this kind of wave would be the output of a microphone after a man speaks a few words into it.  For the actual signal generated by the changes in voltage as the man spoke, we can use Fourier Analysis to describe it as a summation of exponential functions if we instead desire to reconstruct a periodic signal composed of the same voice signal repeating every T_0 seconds.  For an accurate description, it is important that T_0 is long enough such that the repeating arbitrary signals do not overlap.  However, if we let T_0 approach \infty, then this “periodic” signal is simply just the voice signal (or, any general arbitrary function) in time we wanted to describe initially.  Mathematically, we express:

\displaystyle\lim_{T_0\to\infty}f_{T_0}(t) = f(t)

where f(t) is the time-domain function we wish to apply the Fourier Transform on (here, the arbitrary “voice” signal).  For the above equation to be true, f_{T_0} is equal to:

f_{T_0}(t) = \displaystyle\sum_{n=- \infty}^{\infty}D_ne^{jnw_ot}

where:

D_n = \frac{1}{T_0} \int^\frac{T_0}{2}_\frac{-T_0}{2} f_{T_0}(t)e^{-jnw_ot}dt

w_o = \frac{2\pi}{T_o}

It is important to note here that in practice, the shape (aka “envelope”) of a signal’s frequency spectrum is what is of main interest, and the magnitude of the components within the spectrum comes secondary.  This is because amplifiers and other signal-conditioning circuits may be built to alter the magnitude in any way one wishes, and will not affect signal frequencies (so long as the circuits are LTI systems).  Analyzing the envelope of a signal’s Fourier Transform allows one to use intuitive and mathematically-simplified approaches to signal-processing in general, which we shall see later.  For this reason (and also as T_0 approaches \infty) let:

F(w) = \int^{\infty}_{-\infty} f(t)e^{-jw_ot}dt

Notice that F(w) is simply D_n without the constant multiplier \frac{1}{T_0}, such that:

D_n = \frac{1}{T_0}F(nw_o)

which implies that f_{T_0}(t) may be written:

f_{T_0}(t) = \displaystyle\sum_{n=- \infty}^{\infty}\frac{F(nw_o)}{T_0}e^{jnw_ot}

Observation of this fact reveals insight: The shorter the period, T_0, the larger the magnitude of the coefficients.  But, on the other hand, as T_0 \rightarrow \infty, the magnitudes of every frequency component approaches 0 – which is why engineers choose to analyze spectrum envelopes.  So, instead of visualizing absolute frequency magnitudes, instead consider that the frequency spectrum simply expresses the magnitude-density per unit of bandwidth, aka Hz. And since:

T_0 = \frac{2\pi}{w_o}

then:

w_o = \frac{2\pi}{T_0}

and:

\Delta w_o = \frac{2\pi}{\Delta T_0}

so:

f_{T_0}(t) = \displaystyle\sum_{n=- \infty}^{\infty}\frac{F(n\Delta w_o)\Delta w_o}{2\pi}e^{jn\Delta w_ot}

In the limit as T_0 \rightarrow \infty we see:

f(t) = \displaystyle\lim_{T_0\to\infty}f_{T_0}(t) = \frac{1}{2\pi} \int^{\infty}_{-\infty}F(w)e^{jwt}dw

which is referred to as the Fourier Integral.  F(w) is referred to as the Fourier Transform of the original aperiodic function f(t), and we express this concept as:

f(t) \Leftrightarrow F(w)

A fourier transform example

This example is from the same textbook as the previous derivation, and can be found on page 239.

Find the Fourier Transform of: e^{-at}u(t) where a is an arbitrary constant.

To do this, we apply the Fourier Integral to the function e^{-at}u(t) as follows:

F(w) = \int^{\infty}_{-\infty}e^{-at}u(t)e^{-jwt}dt

Because of the u(t) factor, we only integrate from 0 \rightarrow \infty.  We simplify for:

F(w) = \int^{\infty}_{0} e^{-(a+jw)t}dt = \frac{-1}{a+jw}e^{-(a+jw)t}\mid^{\infty}_{0}

Also, we know that |e^{-jwt}| =Re[e^{-jwt}] = 1.  So, for a > 0, as t \rightarrow \infty:

e^{-(a+jw)t} = e^{-at}e^{-jwt} = 0

So:

F(w) = \frac{-1}{a+jw}e^{-(a+jw)t}\mid^{\infty}_{0} = 0 - \frac{-1}{a+jw}e^{-(a+jw)0} = \frac{1}{a+jw}

for: a> 0

Useful Fourier Transform Properties

The relationship between f(t) and F(w) exhibit beautiful symmetry that help one to develop an intuitive approach to signal analysis.  Among all the concepts within electrical engineering, the properties between a time-domain function and its Fourier transform are among the most important to understand.  Observe these following properties that apply for all f(t) \Leftrightarrow F(w):

1.) Fourier Transform: Gives an equation to solve for the time-domain function f(t) from F(w).

f(t) = \frac{1}{2\pi} \int^{\infty}_{-\infty}F(w)e^{jwt}dw

2.) Inverse Fourier Transform: Gives an equation to solve for the frequency-domain function F(w) from f(t).

F(w) = \frac{1}{2\pi} \int^{\infty}_{-\infty}f(t)e^{-jwt}dw

3.) Symmetry Property: For a given pair of a time-domain signal and its Fourier transform, we note that the time-domain envelope is different in shape when compared to the frequency-domain envelope.  However, switching the shape of the two functions with respect to domain (time or frequency), will result in the same envelopes except with different scaling coefficients.  For example, a square pulse through time has a frequency spectrum described by a sinc function, and a sinc function through time results in a frequency spectrum described by a square pulse.

F(t) \Leftrightarrow 2\pi f(-w)

4.) Scaling Property: Time-scaling a time-domain signal (by a constant a) will result in a magnitude-and-frequency-scaling of the signal’s corresponding frequency spectrum.  Also signifies that the longer a signal exists through time, the narrower the bandwidth (collection of frequency components needed to rebuild the signal) of its frequency spectrum.

f(at) \Leftrightarrow \frac{1}{|a|}F(\frac{w}{a})

5.) Time-Shifting Property: By time-shifting, or delaying/advancing, a time-domain signal results in a phase delay in each of the ever-lasting frequency-components needed to rebuild it.  The frequency spectrum is otherwise unchanged – only the phase of each component is shifted.

f(t-t_0) \Leftrightarrow F(w)e^{-jwt_0}

6.) Frequency-Shifting Property: Multiplying a time-domain signal by a sinusoidal signal of some frequency w_0, a method which begets amplitude and frequency modulation (AM/FM), results in the frequency spectrum remains unchanged except for a shift in frequency for each individual frequency component by w_0.

f(t)e^{jw_0t} \Leftrightarrow F(w-w_0)

 

Lastly, these tables (table 1, table 2) can greatly simplify Fourier analysis when used in signal processing.

 

 

 


 

]]> http://engineersphere.com/linear-systems/the-fourier-integral-transform-explained.html/feed 0 The Convolution Integral Explained http://engineersphere.com/math/the-convolution-integral-explained.html http://engineersphere.com/math/the-convolution-integral-explained.html#comments Thu, 10 Mar 2011 06:57:38 +0000 Safa http://engineersphere.com/?p=2179

Introduction to the convolution

Amongst the concepts that cause the most confusion to electrical engineering students, the Convolution Integral stands as a repeat offender.  As such, the point of this article is to explain what a convolution integral is, why engineers need it, and the math behind it.

In essence, the “convolution” of two functions (over the same variable, e.g. f_1(t) and f_2(t)) is an operation that produces a separate third function that describes how the first function “modifies” the second one.  Conversely, the resulting function can be seen as how the second function “modifies” the first function.  Sometimes the result is used to describe how much the first two functions “have in common.”  In all honesty, the concept of the convolution of two functions is quite abstract, but the frequency at which it appears in nature grants its importance to scientists and engineers.  Ultimately the aim here is to identify its use to electrical engineers – so for now do not dwell solely on its mathematical significance.

A convolution of two functions is denoted with the operator “* ”, and is written as:

convolution integral

Convolution of f1(t) and f2(t)

Where \tau is used as a “dummy variable.”  To aid in understanding this equation, observe the following graphic:

Convolution of 2 square pulses

Convolution of two square pulses, resulting in a triangular pulse

Before diving any further into the math, let us first discuss the relevance of this equation to the realm of electrical engineering.

Why is the convolution integral relevant?

Most electrical circuits are designed to be linear, time-invariant (LTI) systems.  Being “linear” implies that the magnitude of a circuit’s output signal is a scaled version of the input signal’s magnitude.  Further, an LTI system that is excited by two independent signal sources will output the sum of the scaled versions of each signal.  This is extended for an infinite number of independent signal sources, and gives rise to the concept of superposition.  Put in another way, if a function x_1(t) causes an LTI system to output y_1(t), then:

a_1 \cdot x_1(t) \to a_1 \cdot y_1(t)

Where a_1 is a multiplicative constant.  In addition to this, superposition allows us to say:

a_1 \cdot x_1(t) + a_2 \cdot x_2(t) + \ldots \to a_1 \cdot y_1(t) + a_2 \cdot y_2(t) + \ldots

Being a “time-invariant” system means it does not matter when the input signal is applied – a specific input signal will always result in the same output signal for a given LTI system.  Put mathematically, time-invariance can be expressed as:

x_1(t) \to y_1(t) \Leftrightarrow x_1(t+\tau) \to y_1(t+\tau)

where \tau can be viewed as a time delay when dealing with signals through time (i.e. “time-domain signals”).  Though not directly, this concept also signifies that an output signal cannot contain frequency components not inherent in the input signal (causality).

The vast majority of circuits are LTI systems, each with a specific impulse response. The “impulse response” of a system is a system’s output when its input is fed with an impulse signal – a signal of infinitesimally short duration.  A real-world “impulse signal” would be something like a lightning bolt – or any form of ESD (electro-static dischage).   Basically, any voltage or current that spikes in magnitude for a relatively short period of time may be viewed as an impulse signal.  The impulse response of a circuit will always be a time-domain signal, and exists because no signal can propagate through a circuit in zero time; each individual electron involved can only move so quickly through each component.  Typically, real-world electronic LTI systems exhibit an impulse response that consists of an initial spike in magnitude, followed by an everlasting and ever-decreasing exponential relationship in signal magnitude.  The following image describes this graphically.

Typical Unit Impulse Response

So, here’s the big deal: the fact that each LTI circuit has a specific impulse response function (here, referred to as h(t)) is very useful in predicting its behavior given a particular input signal (here, referred to as x(t)).  This is because the input signal itself may be viewed as an impulse train – a stream of continuous impulse functions, with infinitesimally short durations of time between each impulse.  This fact, along with superposition, allows one to find the output of an LTI system given an arbitrary input signal by summing the LTI system’s impulse response to each impulse function that make up the input signal. By allowing the time between each “impulse” of the input signal to go to zero, this approach can be used to determine the output time-domain signal of an LTI system for any time-domain input signal.  For example, the following graphic shows the output of an RC circuit when fed with a square pulse:

Convolution of RC network impulse response and square wave input to find the output signal.

What is seen here is the integral of the impulse response and the input square wave as the square wave is stepped through time. In the above convolution equation, it is seen that the operation is done with respect to \tau, a dummy variable.  In reality, we are taking an input signal, flipping it vertically through the origin (not evident with a square wave), and determining what the integral is at each value of \tau, which here is delay through time. Since the output of any LTI system is non-causal (meaning it cannot exist until the signal that excites the output has been applied), we must mathematically step through time to see how each impulse signal of the input affects the LTI system’s impulse response – again, achieved by stepping through \tau – the “time-delay” dummy variable.

A Convolution Example

To see how the convolution integral can be used to predict the output of an LTI circuit, observe the following example:

For an LTI system with an impulse response of h(t) = e^{-2t}u(t) , calculate the output, y(t), given the input of:

f(t) = e^{-t}u(t)

The output of this system is found by solving:

y(t) = h(t)*f(t) = \int_{0}^{\infty} h(\tau) \cdot f(t-\tau) d\tau

We only integrate between 0 and +\infty because, if we define t = 0 as the time that the input signal f(t) is applied, then both h(t) and f(t) have zero magnitude at any time t<0.

From there, we calculate:

y(t) = \int_{0}^{\infty} e^{-2\tau}u(\tau) \cdot e^{-\tau} d\tau= \int_{0}^{t}e^{-\tau} \cdot e^{-2(t-\tau)}d\tau

Next, we can simplify and compute the integral:

y(t) = \int_{0}^{t}e^{-\tau} \cdot e^{-2(t-\tau)}d\tau = e^{-2t} \int_{0}^{t}e^{\tau}d\tau = e^{-2t}(e^t-1) = e^{-t} - e^{-2t}

Since y(t) = 0 for all t < 0, we can write the output y(t) as:

y(t) = (e^{-t}-e^{-2t})u(t)

This result y(t) describes the output function for an LTI system with an impulse response h(t) when fed the input signal f(t).

5 Steps to perform mathematical convolution

Often, one may wish to compute the convolution of two signals that can’t be described with one function of time alone.  For arbitrary signals, such as pulse trains or PCM signals, the convolution at any time t can be computed graphically.  For signals whose individual “sections” can be described mathematically, follow these steps to perform a convolution:

1.) Choose one of the two funtions (h(\tau) or f(\tau)), and leave it fixed in \tau-space.

2.) Flip the other function vertically across the origin, so that it is time-inverted.

3.) Shift the inverted signal through the \tau axis by t_0 seconds.  Choose to shift the signal to the first “section” of the fixed function that is described by the same equation.  The inverted signal (say, f(- \tau) ), now shifted, represents f(t_0 - \tau) , which is basically a “freeze frame” of the output after the input signal has been fed to the LTI system for t_0 seconds.

4.) The integral of the two functions, after shifting the inverted function by t_0 seconds, is the value of the convolution integral (i.e. output signal) at t = t_0.

5.) Repeat this procedure through all “sections” of the function fixed in \tau-space.  By doing this, you can compute the value of the output at any time t!

Useful Properties

 

The following is a list of useful properties of the convolution integral that can help in developing an intuitive approach to solving problems:

1.) Commutative Property:

f_1(t)*f_2(t) = f_2(t)*f_1(t)

2.) Distributive Property:

f_1(t)*[f_2(t)+f_3(t)] = f_1(t)*f_2(t) + f_1(t)*f_3(t)

3.) Associative Property:

f_1(t)*[f_2(t)*f_3(t)] = [f_1(t)*f_2(t)]*f_3(t)

4.) Shift Property:

if f_1(t)*f_2(t) = c(t)

then f_1(t-T_1)*f_2(t-T_2) = c(t-T_1-T_2)

5.) Convolution with an Impulse results in the original function:

f(t)* \delta (t) = f(t) where \delta (t) is the unit impulse function

6.) Width Property:

The convolution of a signal of duration T_1 and a signal of duration T_2 will result in a signal of duration T_3 = T_1 + T_2

Convolution Table

Finally, here is a Convolution Table that can greatly reduce the difficulty in solving convolution integrals.

Thank you so much to Safa Khamis @ Kansas State University for taking the time to write this tutorial for Engineersphere and the electrical engineering community.

 

 

]]> http://engineersphere.com/math/the-convolution-integral-explained.html/feed 2 Solving a Linear System Using the Inverse Matrix http://engineersphere.com/math/solving-a-linear-system-using-the-inverse-matrix.html http://engineersphere.com/math/solving-a-linear-system-using-the-inverse-matrix.html#comments Thu, 03 Mar 2011 22:49:47 +0000 Luke http://engineersphere.com/?p=2009

Describing the process of solving a linear system using the adjacent matrix is best done while performing an example. Suppose we have a system A*x = B where A is the coefficient matrix of our system, x is the column vector containing our variables, and B is the solution column vector. We are asked to solve for the column vector x made up of variables x_1, x_2, and x_3.

\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} 12\\-10\\16\end{bmatrix}

Typically, we would divide B by A to solve for x , however there is no method for performing division between matrices. By taking advantage of the inverse matrix property A^{-1}*A = 1 , we can simply the formula to solve for the column vector x . The commutative property does not apply in matrix multiplication so A^{-1}*B \not= B*A^{-1}.  Therefore we have have to be aware of the ‘order’ in which we multiply:

(A^{-1}) * A * x = (A^{-1}) * B       simplifies to      x = A^{-1}*B

Notice that since we multiplied by A^{-1} ‘first’ on the left side of the equation, we also multiply ‘first’ on the right side. Now, multiplying the inverse of matrix A by matrix B will yield a column vector matching our x_1, x_2, and x_3. Below, I have used the equation x = A^{-1}*B and plugged the values for A^{-1} into the equation. The product between A^{-1} and B is shown on the far right. Note: This article assumes you know how to find the inverse of a matrix. This process is described in my article Finding The Inverse of a Matrix.

\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix} \begin{bmatrix} 12\\-10\\16\end{bmatrix} = \begin{bmatrix} 66\\-22\\4\end{bmatrix}

Therefore, x_1=66, x_2=-22, and x_3=4. Simple systems (i.e. this 3×3 system) are much easier to solve with algebra instead of finding the inverse of the coefficient matrix and performing matrix multiplication. This application is more practical for larger systems or while working on Matrix Theory homework.

Please leave comments by signing in and then clicking on the “sticky note” located in the top right corner of this post to show your appreciation to the author!

]]> http://engineersphere.com/math/solving-a-linear-system-using-the-inverse-matrix.html/feed 0 Smith Charts Explained http://engineersphere.com/linear-systems/smith-charts-explained.html http://engineersphere.com/linear-systems/smith-charts-explained.html#comments Mon, 28 Feb 2011 21:35:56 +0000 Safa http://engineersphere.com/?p=2025

Why We Need Smith Charts

In the world of RF (Radio Frequency) electronics, normal “bench-top” circuit components cease to operate the way they were designed to.  This means a normal resistor can become a capacitor, a capacitor can become an inductor, and a normal wire can become a distributed network of inductors and capacitors.  This highly non-ideal behavior occurs because, in reality, no true resistor, capacitor, inductor, or wire exists; rather they are all processed and manufactured to operate within a certain frequency range – at frequencies where the real-world effects are quantitatively insignificant.  However, as one approaches RF frequencies, these real-world effects become much more pronounced in cheap components.   Eventually, the frequency of operation can become so high that the transmission line itself – no longer a simple wire – will exhibit significant signal-loss.  But even with lossless transmission lines, it is important in communications to “match impedances,” i.e. attach an antenna whose impedance matches that of the signal source – this maximizes the transmitting-antenna’s power dissipation (and “reflects” back zero power).  Indeed, being able to calculate and measure the impedances of antennas, transmission lines, etc is very important within RF design, which are almost always complex numbers.  Another reason determining load impedances is important is because of the 1:1 mapping between a value of load impedance and a corresponding value of \Gamma , the reflection coefficient (a ratio of how much a signal is reflected versus how much a signal is radiated for a given load).

How To Read a Smith Chart

One way of simplifying the analytical problems communication engineers typically face is by using a Smith Chart.  Smith Charts provide a graphical representation of the impedance of any load – whether that load be an antenna or simply an open-circuited transmission line, such as a coax cable.  Because these impedances may very well be complex in nature, a Smith Chart is designed such that each point on it represents both the real and imaginary parts of the load’s impedance.   To begin, observe the basic “format” of any Smith Chart:

 

smith chart

Generic Smith Chart

What is seen here is the generic, normalized Smith Chart.  Each point on the chart represents an impedance, and the numbers marked on the chart represent different coefficients needed to multiply by the original load value  (from which the chart was normalized from – If this is confusing, don’t worry about it – the “original load value” will almost always be known).  The chart consists of yellow circles and red arcs.  The yellow circles represent contours of where the Real part of the impedance magnitude is the same, e.g. for any point along the yellow circle marked “2.0,” the real part of the impedance is:

where  is the impedance the chart was normalized from.

Similarly, the red arcs represent contours where the Imaginary part of the impedance magnitude is the same, e.g. for any point along the red arc marked “0.5,” the imaginary part of the impedance is:

This is very handy for displaying complex impedance values.  For instance, notice the following Smith Chart:

 

Smith Chart for Z0 = 50 Ohms

What is the impedance for the load represented on the Smith Chart by the blue dot?  This is easy to determine, because:

  1. The blue dot is along the yellow circle marked “2.0,” so the real part of the impedance must be:
  2. The blue dot is also along the red arc marked “0.5,” so the imaginary part of the impedance must be:

So, from this, we have determined that the impedance of the “load” represented by the blue dot is:

By using this method, it is simple to find the impedance represented by any point on the Smith Chart!

Useful Smith Chart Relations

As mentioned before, a Smith Chart is really just a 1:1 mapping between a value of load impedance and a value of \Gamma, the reflection coefficient of a load.  The reflection coefficient is defined as:

\Gamma =\frac{V_{reflected}}{V_{incident}} = \frac{V_{refl}}{V_{inc}}

The reflection coefficient is a very important metric.  For antennas, a reflection coefficient expresses how much signal voltage is used in exciting the antenna and how much signal voltage is reflected back to the source.  For an ideal antenna, \Gamma would be zero – corresponding to Z = Z_0 \Omega , which is the origin of the Smith Chart!  In fact, simple relations exist between Z and \Gamma :

and

Useful Notes

Simple observations allow for a more “intuitive” approach when using Smith Charts.  Note the following:

  • The straight red line in the center is an “arc” representing all points where the imaginary part of the impedance is zero
  • The furthest-left point on the straight red line represents where the impedance is zero (or, a short circuit), and the furthest-right point on the straight red line represents where the impedance is \infty (or, an open circuit)
  • The very top point of the Smith Chart is where the impedance is +j
    • For this reason, the top half of the Smith Chart represents inductive loads
  • The very bottom point of the Smith Chart is where the impedance is -j
    • Similarly, the bottom half of the Smith Chart represents capacitive loads

____

Safa Khamis

Questions?  Comments?  safa@ksu.edu – Or please leave a comment below for the author!

 

 

]]> http://engineersphere.com/linear-systems/smith-charts-explained.html/feed 0 Frequency Response for MOSFET/BJT http://engineersphere.com/basic-electrical-concepts/frequency-response-for-mosfetbjt.html http://engineersphere.com/basic-electrical-concepts/frequency-response-for-mosfetbjt.html#comments Sun, 30 May 2010 01:48:25 +0000 Riley http://engineersphere.com/?p=926

The frequency response of a BJT or MOSFET can be found using nearly the exact same process, with the only variations being caused by a single resistor and simple naming conventions that differ between the two devices.

Before we start let’s think a little bit about what we’re doing:
Our goal is going to be to find the pole(s) of the circuit.
Okay? What is a pole and why do I care where it is?
A pole is a frequency at which the gain of the device rolls off. (remember that when it rolls off , it will be at the -3dB frequency with a slope of -20dB/decade)

We care because if the gain of a device rolls off at a certain frequency, then we won’t be able to amplify a signal above that frequency very well because the gain will be decreasing by 20dB/decade.

The procedure is nearly identical whether we are using a BJT of a MOSFET, but we will work each of them side by side just in case there might be any confusion, and we’ll follow these steps as we go through.  (we will also use some values that came from the output file when running a simulation of this circuit in Cadence (or PSPICE) )mosfet-amplifier

bjt-amplifier
1. Take a look at one of the circuits and see what you notice, how about the MOSFET.  This step is just to help us with our knowledge understanding of the circuit.
- At a glance it just looks just like another MOSFET right? Sure is, but let’s take a look at a few things just for kicks. Notice that it is using a bypass capacitor at the source so we don’t have to worry about R_s (at when working with high frequency).  Since the capacitor C_s bypasses R_s to ground, you should notice that this is a common-source amplifier.  You could notice the Values for R_1 and R_2 and start to think about what the Gate voltage is and how that may affect the circuit.
2. We are talking about frequency response so that means we are probably going to want to draw the small signal equivalent circuit.
Remember that the capacitors C_1 and C_2 will act like short circuits at high frequencies so we will ignore them, but we will have to account for some of the capacitance internal to the device.

Both devices have internal capacitances that are very similar.  As you can see from the small signal models for a MOSFET (above) and BJT (below), the only significant difference is that the BJT has an additional resistance Rpi between the Base and Emitter.

Most of the analysis we will do is based on the small signal model. Note that small signal models are not typically used in PSPICE so this picture may look a bit odd, especially the controlled source but for our purpose it is good to have a visual reference. To start we will point out what everything is. Cgs is an internal capacitance betwemosfet-small-signal-model

en the gate and source. The

values for Cgs was similar to one the a PSPICE simulation may give.  CM1 and CM2 are Miller capacitances which we will find values for laterbjt-small-signal-model.  ro is a Norton equivalent resistance that makes the model more ideal.  And just pretend that the G2 looks more like a voltage controlled current source and that their gains are gm*Vgs and gm*Vpi. For the BJT CM1 and CM2 are both Miller capacitances, Cpi is similar to Cgs and Rpi the additional component used for BJTs but not MOSFETs. The other part should look familiar from the other figures.

ON TO THE ANALYSIS!!!

We will find the device gain, overall gain, equivalent input and output capacitances, and the input and output poles. The process for both is essentially the same.

Device Gain: This is the gain from the control source to the output so we are looking for Vout/Vgs (or Vout/Vpi for a BJT). We will ignore CM2 for this process. Notice the resistances ro, RD, and RL are in parallel. Vout should be given by that equivalent resistance times the current though it which is gm*Vgs from the control source. So the equation for device gain is,

V_{out} / V_{gs} = gm*(r_o//R_D//R_L)   (MOSFET)

V_{out} / V_{\pi} = gm*(r_o//R_C//R_L)   (BJT)

Overall Gain: This will be the gain from the source (Vs) to the output (Vout). We already know what Vout/Vgs is so if we find Vgs/Vs, we can multiply them to get Vout/Vs = (Vout/Vgs) * (Vgs/Vs).  Vgs/Vs is a simple voltage divider. Hopefully you can see this from the small signal model (remember that we are ignoring the capacitors for now but they will play a part later).  The equations we will get for Vgs/Vs and the overall gain are.

V_{gs} / V_s = \frac{ (R_1//R_2)}{(R_1//R_2) + R_s}  (MOSFET)

Overall Gain: V_{out} / V_s = \frac{ (R_1//R_2)}{(R_1//R_2) + R_s} * gm(r_o//R_D//R_L)  (MOSFET)

V_{gs} / V_s = \frac{ (R_1//R_2//r_\pi)}{(R_1//R_2//r_\pi) + R_s}  (BJT)

Overall Gain: V_{out} / V_s = \frac{ (R_1//R_2//r_\pi)}{(R_1//R_2//r_\pi) + R_s} * gm(r_o//R_C//R_L)  (BJT)

Now we will find the input and output poles.  For this we will need to look at the capacitances and use a formula to find the Miller capacitances, CM1 and CM2.  Any explanation for the miller capacitance will have to wait for another post or check out your Electronics Book, Wikipedia, Google, etc. but we will need to use a couple of special equations.  Overall we will need to find the input resistance and input capacitance for the input pole and the output resistance and output capacitance for the output pole.

Each pole will be at a frequency w=1/RC where the R and C are the equivalent R and C at that point, so to find the input pole, we will need to find the input resistance and the input capacitance.  These are found by looking into the input (the left side of the small signal model).  The voltage source will  act like a short so we see Rs in parallel with R1//R2 for the MOSFET (the BJT will have Rpi in parallel also).  The input capacitance will be Cgs in parallel with CM1 (the BJT will be the same).  The output resistance and capacitance are found the same way only looking in from the output (the right side of the small signal model).

\omega_{IN} = \frac{1}{R_{IN}C_{IN}}  \omega_{OUT} = \frac{1}{R_{OUT}C_{OUT}}     (MOSFET or BJT)

So the input pole will be: (MOSFET)

R_{IN} = R_S//R_1//R_2  =  950                                     R_{OUT} = r_o//R_D//R_L =

C_{IN} = C_{gs} + C_{M1}  =                                               C_{OUT} = C_{M2} =

\omega_{IN} =                                                                          \omega_{OUT} =

(BJT)

and the output pole will be: (MOSFET)

(BJT)

R_{IN} = R_S//R_1//R_2//r_\pi =                                  R_{OUT} = r_o//R_D//R_L =

C_{IN} = C_{BE} + C_{M1} =                                                 C_{OUT} = C_{M2} =

\omega_{IN} = \frac{}{}                                       \omega_{OUT} =\frac{}{}

To Do:

finish input & ouput R, input C, Pole (& calculate answers)

]]> http://engineersphere.com/basic-electrical-concepts/frequency-response-for-mosfetbjt.html/feed 0 Time Shifting and Scaling of Functions http://engineersphere.com/math/time-shifting-and-scaling-of-functions.html http://engineersphere.com/math/time-shifting-and-scaling-of-functions.html#comments Wed, 07 Apr 2010 21:19:17 +0000 Luke http://engineersphere.com/?p=1317

We’ll begin with a square function, f(t), that has a an amplitude of 1, a start time of 2 seconds and an end time of 4 seconds.

square-wave

Next, a time shift is demonstrated. Here our function is changed from f(t) to f(t-2). Notice that subtracting 2 from t in the function results in a positive shift of the graph.

positive-shifted-square-wave

In the next two graphs t will be scaled. Scaling t is not quite as intuitive as we may have expected. When we multiply t by 2, corresponding points of the function now occur at 1/2 the time they previously had. When we divide t by 2, each corresponding time on the graph occurs at a t that is now multiplied by 2. Notice that each of these factors directly affects the duration of the signal.

time-shifted-square-wave
time-shifted-square-wave2

Scaling the amplitude has more intuitive results. If we multiply f(t) by 2, the amplitude of 1 is changed to 2. Multiplying f(t) by 1/2 results in an amplitude of 1/2.

taller-square-wave
truncated-square-wave

Finally, multiplying t by -1 mirrors our function over the y-axis. Each time now occurs at its negative.

mirrored-square-wave


Example:
Here we will attempt to convert f(t) into 2*f(.5t+3). The graph of f(t) is shown below.

triangle-wave

The easiest way to handle this type of problem without error is to manipulate the function one step at a time. First, I have converted f(t) into 2*f(t). Only the peaks are changed here (by a factor of 2).

triangle-wave2

Next, I convert 2*f(t) into 2*f( \frac{1}{2} t). Notice how the \frac{1}{2} actually expands our graph duration by a factor of 2 (from a 6 sec duration to a 12 sec duration).

shifted-triangle-wave

Finally, we move from 2*f( \frac{1}{2} t) to 2*f( \frac{1}{2} t + 3). As shown in the discussion above, this is a time shift. Time shifts can be a little confusing because adding results in a negative shift of our graph. Try to think of it as our signal occurring 3 seconds earlier than before, reading from left to right on the graph. The easiest way to do this part is shift each x-intercept by 3 seconds (to the left, of course).

shifted-triangle-wave2

]]> http://engineersphere.com/math/time-shifting-and-scaling-of-functions.html/feed 3 Solving a System Equation http://engineersphere.com/math/differential-equations/solving-a-system-equation.html http://engineersphere.com/math/differential-equations/solving-a-system-equation.html#comments Wed, 30 Dec 2009 00:02:11 +0000 Jeff http://engineersphere.com/?p=1117

Why do we need to solve system equations?

Often during a course you will need to be able to solve a system equation for its roots.  These roots can be complex, distinct, or repeated.  These problems usually arise when working with linear systems or differential equations.  A system equation is formatted as follows:

System Equation: Q(D)y_{0}(t) = 0

How to solve a system equation

For example purposes, I will solve a system equation with complex roots.  A system equation with complex roots as a function of \lambda will appear in the following format (if it does not, you need to manipulate your equation to be in the form):

Q(\lambda) = (\lambda - \alpha - j\beta)(\lambda - \alpha + j\beta)

Roots: \lambda = \alpha \pm j\beta

So we have y_{0}(t) = C_{1}e^{(\alpha + j\beta)t}+C_{2}e^{(\alpha - j\beta)t}

which also equals y_{0}(t) = Ce^{\alpha t}cos(\beta t + \theta)

so your first step is to look at your equation and determine your roots, then write out your y_{0}(t) equation with constants.

Example \frac{d^{2}v}{dt^{2}} + 4\frac{dv}{dt} + 4v(t) = 0 with initial conditions V(0) = 3v and V^{1}(0) = -4v

Q(\lambda) = \lambda^{2} + 4\lambda + 4 = 0

(\lambda + 2)(\lambda + 2) = 0

\lambda_{1} = \lambda_{2} = -2

so now we can write our y_{0}(t) equation as follows:

y_{0}(t) = C_{1}e^{-2t}+C_{2}te^{-2t}

In order to solve for C_{1} and C_{2} we need to use our initial conditions.  To evaluate the first derivative initial condition, we must first take the derivative of our y_{0}(t) = C_{1}e^{-2t}+C_{2}e^{-2t} that we just found.

y_{0}^{1}(t) = -2C_{1}e^{-2t} - 2C_{2}*t*e^{-2t} + C_{2}e^{-2t}

evaluating this equation with t = 0 and the response equal to -4v, we get this: -4 = -2C_{1} + C_{2}

evaluating our y_{0}(t) equation with t = 0 and the response equal to 3v, we calculate C_{1} = 3

Using these two equations, we calculate our constants:

C_{1} = 3 and C_{2} = 2

Fill these into our y_{0}(t) equation to determine the final result.

y_{0}(t) = 3e^{-2t}+2te^{-2t}

Now you know how to solve this common differential equations and linear systems problem, determine characteristic roots and modes, and write system equations. :)

]]> http://engineersphere.com/math/differential-equations/solving-a-system-equation.html/feed 0 Zero Input Response http://engineersphere.com/linear-systems/zero-input-response/zero-input-response.html http://engineersphere.com/linear-systems/zero-input-response/zero-input-response.html#comments Fri, 25 Sep 2009 22:37:15 +0000 Jeff http://engineersphere.com/?p=918

Characterizing the zero-input response

The total response of a given system can be expressed as the sum of two components: the zero-input component and the zero-state component:

Total response = zero-input response + zero-state response.

Our system: Q(D)y_{1}(t) = P(D)f_{1}(t)

Where y_{1}(t) represents our input and f_{1}(t) represents our output.

The zero-input response, which we will be solving for here, is the system response when the input f(t) = 0 so that is the result of the internal system conditions.  It is independent of the external input f(t).  Therefore, in order to solve a differential equation that represents the zero-input response component of a system, we will need to have the initial conditions of the system, or solve for them.  Keep this in mind while working your problem.

A zero-input response example problem

Okay, let’s go ahead and solve a problem shall we?

zero-input-responseOur first step is to find a differential equation that will represent the system for t \ge 0 .  This is because we only care about the behavior after t = 0 for a zero-input response problem.  We can notice that when our switch opens, this is a natural response.  In other words, we do not switch into a circuit that has a current or voltage source in it, just a capacitor, inductor and two resistors.  In order to solve our ZIR problem, we also need initial conditions do we not?  We will worry about those after we get our differential equation.

The problem tells us that we need to find V_{c}(t) , so lets look at this circuit after the switch opens and decide what kind of equation we need to write.  Can you see that once the switch opens we have a capacitor, an inductor, and two resistors in series?  If not, you need to study up on parallel and series circuits.  After we combine these two resistors, we have three elements in series, a series RLC circuit.

The sum of voltages around this loop should equal zero by KVL

V_{c} + V_{L} + V_{R1} + V_{R2} = 0

We should know the following relationships to help us with our substitutions:

V_{L} = L\frac{di}{dt} ;   V_{c}(t) = \frac{1}{C}\int_{t_{0}}^{t}i(t)dt + v(0) ;   I_{c} = C\frac{dV}{dt} ;  I_{L}(t) = \frac{1}{L}\int_{0}^{t}V(t)dt + i(0)

Because we are solving for V_{c} , we will leave this term alone and replace as many of the others as we can.  Note that all four of the elements are in series and therefore share the same current (i).

V_{c} + L\frac{di}{dt} + (R_{1} + R_{2})*i

We need this equation in terms of constants and V_{c} so we need to perform substitutions on all terms that involve the current (i).  From the relationship I_{c} = C\frac{dV}{dt} we note that \frac{di}{dt} = C\frac{dV^{2}}{dt^{2}} and can perform the substitutions that we need:

V_{c}(t) + LC\frac{dV^{2}}{dt^{2}} + (R_{1} + R_{2}) C \frac{dV_{c}}{dt} = 0

We divide all terms by the amount in front of our highest order term which in this case is LC.

\frac{dV^{2}}{dt^{2}} + \frac{(R_{1} + R_{2})}{L}\frac{dV_{c}}{dt} + \frac{1}{LC}V_{c}(t) = 0

and our differential equation becomes: D^{2} + 3D + 5

By using the quadratic equation, we can form our characteristic equation (\lambda + \frac{3}{2} + j\frac{\sqrt{11}}{2})(\lambda + \frac{3}{2} - j\frac{\sqrt{11}}{2}) = 0

Roots: -\frac{3}{2} \pm j\frac{\sqrt{11}}{2}

Now we have some work to do, and a little bit of thinking.  Using the roots that we just derived, we can now write a general equation for this system, remembering that our roots are complex:

V_{c}(t) = C_{1} e^{\frac{3}{2} t} cos(\frac{\sqrt{11}}{2}t) + C_{2} e^{\frac{3}{2} t} sin(\frac{\sqrt{11}}{2}t)

Evaluating this at t = 0

V_{c}(0) = C_{1} We know that V_{c}(0) is 2V because after time has elapsed prior to the switch opening the capacitor has become an open and assumes all available potential across it.  When the switch opens, voltage can’t change instantaneously across a capacitor so it initially remains at the same potential as it was prior to the switch opening.

We should also note that the current i_{c}(0) is -1 Amp.  Why is that?  The initial voltage across the inductor (which is a short prior to switch opening) is 2 V, leaving all of this potential for the 2 ohm resistor (remember that the cap is an open and no current flows in that branch).  Therefore there is a 1 Amp current flowing in the clockwise direction when the switch opens.  CURRENT CAN CHANGE INSTANTANEOUSLY IN A CAPACITOR.  Therefore, immediately after the switch opens, the 1 Amp current flows through the capacitor in the direction opposite it’s polarity, resulting in -1 Amp.

We will need to evaluate the derivative of this function in order to get a second equation (need 2 equations to solve for 2 unknowns C_{1} and C_{2} ).  We now know that C_{1} = 2 V .

\frac{dV_{c}(0)}{dt} = \frac{3}{2}C_{1}e^{\frac{3}{2} t}cos(\frac{\sqrt{11}}{2}t) + \frac{\sqrt{11}}{2}C_{2}e^{\frac{3}{2} t}cos(\frac{\sqrt{11}}{2}t)

(I have removed the SINE terms because at 0 they evaluate to 0, in this case.)

\frac{dV_{c}}{dt} = \frac{I_{c}(0)}{C} = -10

Therefore,

-10 = 3 + \frac{\sqrt{11}}{2} C_{2}

C_{2} = \frac{-26}{\sqrt{11}}

Now that we have both of our constants, we can insert them into our equation which represents the zero-input response of our system where y_{0}(t) = V_{c}(t) .

V_{c}(t) = 2 e^{\frac{3}{2} t} cos(\frac{\sqrt{11}}{2}t) + \frac{-26}{\sqrt{11}} e^{\frac{3}{2} t} sin(\frac{\sqrt{11}}{2}t) t\ge 0

Conclusion

This example was a little complex for teaching a new concept, but I believe it covers a wide variety of important concepts and steps that you will likely see.  If you can do this whole problem, not only do you understand how to evaluate the ZIR of a system, but you also understand characterisic roots and modes for complex roots, series RLC circuit analysis, initial conditions, basic inductor and capacitor physics concepts, as well as frequency and time domain analysis for such circuits.  I hope this helped!

]]> http://engineersphere.com/linear-systems/zero-input-response/zero-input-response.html/feed 0 Laplace Transforms http://engineersphere.com/math/laplace-transforms.html http://engineersphere.com/math/laplace-transforms.html#comments Sun, 06 Sep 2009 18:06:20 +0000 Luke http://engineersphere.com/?p=701

What is the Laplace Transform method?

The Laplace Transform is a method that simplifies integral and differential equations into algebraic equations. This practice is commonly used to solve for a function out of a differential equation, which otherwise may have been unsolvable or very difficult. The following integrals can be used to transform between F(s) \Leftrightarrow f(t) where \mathcal{L} denotes Laplace and \mathcal{L}^{-1} denotes Inverse Laplace:

\mathcal{L} [f(t)] = F(s) = \int\limits_0^\infty {f(t)*e ^ {-st}d} t

\mathcal{L}^{-1}[F(s)] = f(t) = \frac{1}{2 \pi j} \int\limits_{c - j \omega}^{c + j \omega} {F(s)*e ^ {-st}d}s

Table of Laplace Transforms

Since these integrals can be tedious and certain functions tend to reoccur, a table of Laplace Transforms has been linked:

Laplace Transforms Table

A Laplace Transform example

This table can be a little complex to use at first so an example is provided below to get you started. In this problem we implement the Laplace Transform and Inverse Laplace Transform to solve for y(t) .

\frac{d^2y}{dt^2} + 7 \frac{dy}{dt} + 12y = 10 \quad y(0)=3, y'(0)

The first step is to take the Laplace Transform of both sides of the equation. Use element 1 of our table for the right side and element 18 for the left side. Note that the initial conditions are necessary to take the Laplace Transform of the left side.

s^2 Y(s) - s y(0) - y'(0) + 7 (s Y(s) - y(0)) + 12 Y(s) = \frac{10}{s}

Inputting our initial conditions:

s^2 Y(s) - 3s - 0 + 7 (s Y(s) - 3) + 12 Y(s) = \frac{10}{s}

Assuming you are an engineering student and can do a little alegebra, our next step is to find the terms that have Y(s) in common and factor it out. Our goal is to find Y(s) = F(s) because, after all, \mathcal{L}^{-1}[Y(s)] = y(t) :

(s^2 +7s + 12)Y(s) - 3s -21 = \frac{10}{s}

After solving for Y(s) and factoring the denominator:

Y(s) = \frac{3s^2 + 21s + 10}{s(s+4)(s+3)}

Taking the Inverse Laplace Transform

Now we arrive at the trickier part of this procedure. We must take the Inverse Laplace of Y(s) to find y(t) . If our function Y(s) does not match anything in the table, such as this case, factoring is a good place to start. This problem can easily be factored using the expand function on your TI-89. Just go to catalog \rightarrow expand and enter your function in parenthesis. Using this function:

Y(s) = \frac{\frac{5}{6}}{s} - \frac{\frac{13}{2}}{s+4} + \frac{\frac{26}{3}}{s+3}

Noting \mathcal{L} [f_1(t) + f_2(t)] = \mathcal{L} [f_1(t)] + \mathcal{L} [f_2(t)] and \mathcal{L} [k*f(t)] = k*\mathcal{L} [f(t)] , we can take the Laplace Transform of each term independently and also manipulate the constant terms if necessary or just pull them out. Using the 2nd property in our Laplace Transform table:

y(t) = \frac{5}{6} - \frac{13}{2}e^{-4t} + \frac{26}{3}e^{-3t}

To check your work you can plug y(t), \frac{dy}{dt}, \frac{d^2y}{dt^2} into the original, differential equation and at t=0 we find that 10=10.

]]> http://engineersphere.com/math/laplace-transforms.html/feed 3