If I have an RC circuit with transfer function 1/(1+sRC) how do I draw the transfer function using MATLAB?
Num2=[1];
Den2=[R*C 1];
RCcirc=tf(Num2,Den2);
How do I declare the R and the C so that there are no errors?
tf is the wrong tool for plotting the transfer function. Try these instead:
Use linspace to generate a range of values for s. Give R and C reasonable values of your choice.
Read up on arithmetic operations in MATLAB, especially ./
Look at how to use plot and familiarize yourself with the command using some simple examples from the docs.
With these you should be able to plot the transfer function in MATLAB :)
First of all you need to understand what transfer function you want. Without defined values of R and C you won't get any transfer function. Compare it to this, you want to plot a sine wave: x = sin(w*t), I hope you can agree with me that you cannot plot such a function (including axes) unless I specifically say e.g. t is the time, ranging from 0 seconds to 10 seconds and w is a pulsation of 1 rad/s. It's exactly the same with your RC network: without any values, it is impossible for numerical software such as MATLAB to come up with a plot.
If you fill in those values, you can use th tf function to display the transfer function in whatever way you like (e.g. a bode plot).
On the other hand, if you just want the expression 1/(1+s*R*C), take a look at the symbolic toolbox, you can do such things there. But to make a plot, you will still have to fill in the R and C value (and even a value for your Laplace variable in this case).
Related
I am trying trying to graph the polynomial fit of a 2D dataset in Matlab.
This is what I tried:
rawTable = readtable('Test_data.xlsx','Sheet','Sheet1');
x = rawTable.A;
y = rawTable.B;
figure(1)
scatter(x,y)
c = polyfit(x,y,2);
y_fitted = polyval(c,x);
hold on
plot(x,y_fitted,'r','LineWidth',2)
rawTable.A and rawTable.A are randomly generated numbers. (i.e. the x dataset cannot be represented in the following form : x=0:0.1:100)
The result:
second-order polynomial
But the result I expect looks like this (generated in Excel):
enter image description here
How can I graph the second-order polynomial fit in MATLAB?
I sense some confusion regarding what the output of each of those Matlab function mean. So I'll clarify. And I think we need some details as well. So expect some verbosity. A quick answer, however, is available at the end.
c = polyfit(x,y,2) gives the coefficient vectors of the polynomial fit. You can get the fit information such as error estimate following the documentation.
Name this polynomial as P. P in Matlab is actually the function P=#(x)c(1)*x.^2+c(2)*x+c(3).
Suppose you have a single point X, then polyval(c,X) outputs the value of P(X). And if x is a vector, polyval(c,x) is a vector corresponding to [P(x(1)), P(x(2)),...].
Now that does not represent what the fit is. Just as a quick hack to see something visually, you can try plot(sort(x),polyval(c,sort(x)),'r','LineWidth',2), ie. you can first sort your data and try plotting on those x-values.
However, it is only a hack because a) your data set may be so irregularly spaced that the spline doesn't represent function or b) evaluating on the whole of your data set is unnecessary and inefficient.
The robust and 'standard' way to plot a 2D function of known analytical form in Matlab is as follows:
Define some evenly-spaced x-values over the interval you want to plot the function. For example, x=1:0.1:10. For example, x=linspace(0,1,100).
Evaluate the function on these x-values
Put the above two components into plot(). plot() can either plot the function as sampled points, or connect the points with automatic spline, which is the default.
(For step 1, quadrature is ambiguous but specific enough of a term to describe this process if you wish to communicate with a single word.)
So, instead of using the x in your original data set, you should do something like:
t=linspace(min(x),max(x),100);
plot(t,polyval(c,t),'r','LineWidth',2)
i'm new in matlab. I didn't understand how to derive a dirac delta function and then shift it using symbolic toolbox.
syms t
x = dirac(t)
why can't i see the dirac delta function using ezplot(x,[-10,10]) for example?
As others have noted, the Dirac delta function is not a true function, but a generalized function. The help for dirac indicates this:
dirac(X) is not a function in the strict sense, but rather a
distribution with int(dirac(x-a)*f(x),-inf,inf) = f(a) and
diff(heaviside(x),x) = dirac(x).
Strictly speaking, it's impossible for Matlab to plot the Dirac delta function in the normal way because part of it extends to infinity. However, there are numerous workarounds if you want a visualization. A simple one is to use the stem plot function and the > operator to convert the one Inf value to something finite. This produces a unit impulse function (or Kronecker delta):
t = -10:10;
x = dirac(t) > 0;
stem(t,x)
If t and x already exist as symbolic variables/expressions rather than numeric ones you can use subs:
syms t
x = dirac(t);
t2 = -10:10;
x2 = subs(x,t,t2)>0;
stem(t2, x2)
You can write your own plot routine if you want something that looks different. Using ezplot is not likely to work as it doesn't offer as much control.
First, I've not met ezplot before; I had to read up on it. For things that are functionals like your x, it's handy, but you still have to realize it's exactly giving you what it promises: A plot.
If you had the job of plotting the dirac delta function, how would you go about doing it correctly? You can't. You must find a convention of annotating your plot with the info that there is a single, isolated, infinite point in your plot.
Plotting something with a line plot hence is unsuitable for anything but smooth functions (that's a well-defined term). Dirac Delta definitely isn't amongst the class of functions that are smooth. You would typically use a vertical line or something to denote the point where your functional is not 0.
I'm using octave 3.8.1 which works like matlab.
I have an array of thousands of values I've only included three groupings as an example below:
(amp1=0.2; freq1=3; phase1=1; is an example of one grouping)
t=0;
amp1=0.2; freq1=3; phase1=1; %1st grouping
amp2=1.4; freq2=2; phase2=1.7; %2nd grouping
amp3=0.8; freq3=5; phase3=1.5; %3rd grouping
The Octave / Matlab code below solves for Y so I can plug it back into the equation to check values along with calculating values not located in the array.
clear all
t=0;
Y=0;
a1=[.2,3,1;1.4,2,1.7;.8,5,1.5]
for kk=1:1:length(a1)
Y=Y+a1(kk,1)*cos ((a1(kk,2))*t+a1(kk,3))
kk
end
Y
PS: I'm not trying to solve for Y since it's already solved for I'm trying to solve for Phase
The formulas located below are used to calculate Phase but I'm not sure how to put it into a for loop that will work in an array of n groupings:
How would I write the equation / for loop for finding the phase if I want to find freq=2.5 and amp=.23 and the phase is unknown I've looked online and it may require writing non linear equations which I'm not sure how to convert what I'm trying to do into such an equation.
phase1_test=acos(Y/amp1-amp3*cos(2*freq3*pi*t+phase3)/amp1-amp2*cos(2*freq2*pi*t+phase2)/amp1)-2*freq1*pi*t
phase2_test=acos(Y/amp2-amp3*cos(2*freq3*pi*t+phase3)/amp2-amp1*cos(2*freq1*pi*t+phase1)/amp2)-2*freq2*pi*t
phase3_test=acos(Y/amp3-amp2*cos(2*freq2*pi*t+phase2)/amp3-amp1*cos(2*freq1*pi*t+phase1)/amp3)-2*freq2*pi*t
Image of formula below:
I would like to do a check / calculate phases if given a freq and amp values.
I know I have to do a for loop but how do I convert the phase equation into a for loop so it will work on n groupings in an array and calculate different values not found in the array?
Basically I would be given an array of n groupings and freq=2.5 and amp=.23 and use the formula to calculate phase. Note: freq will not always be in the array hence why I'm trying to calculate the phase using a formula.
Ok, I think I finally understand your question:
you are trying to find a set of phase1, phase2,..., phaseN, such that equations like the ones you describe are satisfied
You know how to find y, and you supply values for freq and amp.
In Matlab, such a problem would be solved using, for example fsolve, but let's look at your problem step by step.
For simplicity, let me re-write your equations for phase1, phase2, and phase3. For example, your first equation, the one for phase1, would read
amp1*cos(phase1 + 2 freq1 pi t) + amp2*cos(2 freq2 pi t + phase2) + amp3*cos(2 freq3 pi t + phase3) - y = 0
Note that ampX (X is a placeholder for 1, 2, 3) are given, pi is a constant, t is given via Y (I think), freqX are given.
Hence, you are, in fact, dealing with a non-linear vector equation of the form
F(phase) = 0
where F is a multi-dimensional (vector) function taking a multi-dimensional (vector) input variable phase (comprised of phase1, phase2,..., phaseN). And you are looking for the set of phaseX, where all of the components of your vector function F are zero. N.B. F is a shorthand for your functions. Therefore, the first component of F, called f1, for example, is
f1 = amp1*cos(phase1+...) + amp2*cos(phase2+...) + amp3*cos(phase3+...) - y = 0.
Hence, f1 is a one-dimensional function of phase1, phase2, and phase3.
The technical term for what you are trying to do is find a zero of a non-linear vector function, or find a solution of a non-linear vector function. In Matlab, there are different approaches.
For a one-dimensional function, you can use fzero, which is explained at http://www.mathworks.com/help/matlab/ref/fzero.html?refresh=true
For a multi-dimensional (vector) function as yours, I would look into using fsolve, which is part of Matlab's optimization toolbox (which means I don't know how to do this in Octave). The function fsolve is explained at http://www.mathworks.com/help/optim/ug/fsolve.html
If you know an approximate solution for your phases, you may also look into iterative, local methods.
In particular, I would recommend you look into the Newton's Method, which allows you to find a solution to your system of equations F. Wikipedia has a good explanation of Newton's Method at https://en.wikipedia.org/wiki/Newton%27s_method . Newton iterations are very simple to implement and you should find a lot of resources online. You will have to compute the derivative of your function F with respect to each of your variables phaseX, which is very simple to compute since you're only dealing with cos() functions. For starters, have a look at the one-dimensional Newton iteration method in Matlab at http://www.math.colostate.edu/~gerhard/classes/331/lab/newton.html .
Finally, if you want to dig deeper, I found a textbook on this topic from the society for industrial and applied math: https://www.siam.org/books/textbooks/fr16_book.pdf .
As you can see, this is a very large field; Newton's method should be able to help you out, though.
Good luck!
My project require me to use Matlab to create a symbolic equation with square wave inside.
I tried to write it like this but to no avail:
syms t;
a=square(t);
Input arguments must be 'double'.
What can i do to solve this problem? Thanks in advance for the helps offered.
here are a couple of general options using floor and sign functions:
f=#(A,T,x0,x) A*sign(sin((2*pi*(x-x0))/T));
f=#(A,T,x0,x) A*(-1).^(floor(2*(x-x0)/T));
So for example using the floor function:
syms x
sqr=2*floor(x)-floor(2*x)+1;
ezplot(sqr, [-2, 2])
Here is something to get you started. Recall that we can express a square wave as a Fourier Series expansion. I won't bother you with the details, but you can represent any periodic function as a summation of cosines and sines (à la #RTL). Without going into the derivation, this is the closed-form equation for a square wave of frequency f, with a peak-to-peak amplitude of 2 (i.e. it goes from -1 to 1). Recall that the frequency is the amount of cycles per seconds. Therefore, f = 1 means that we repeat our square wave every second.
Basically, what you have to do is code up the first line of the equation... but how in the world would you do that? Welcome to the world of the Symbolic Math Toolbox. What we will need to do before hand is declare what our frequency is. Let's assume f = 1 for now. With the Symbolic Math Toolbox, you can define what are considered as mathematics variables within MATLAB. After, MATLAB has a whole suite of tools that you can use to evaluate functions that rely on these variables. A good example would be if you want to use this to define a closed-form solution of a function f(x). You can then use diff to differentiate and see what the derivative is. Try it yourself:
syms x;
f = x^4;
df = diff(f);
syms denotes that you are declaring anything coming after the statement to be a mathematical variable. In this case, x is just that. df should now give you 4x^3. Cool eh? In any case, let's get back to our problem at hand. We see that there are in fact two variables in the periodic square function that need to be defined: t and k. Once we do this, we need to create our function that is inside the summation first. We can do this by:
syms t k;
f = 1; %//Define frequency here
funcSum = (sin(2*pi*(2*k - 1)*f*t) / (2*k - 1));
That settles that problem... now how do we encapsulate this into an infinite sum!? The sum command in MATLAB assumes that we have a finite array to sum over. If you want to symbolically sum over a function, we must use the symsum function. We usually call it like this:
funcOut = symsum(func, v, start, finish);
func is the function we wish to sum over. v is the summation variable that we wish to use to index in the sum. In our case, that's k. start is the beginning of the sum, which is 1 in our case, and finish is where we wish to finish up our summation. In our case, that's infinity, and so MATLAB has a special keyword called Inf to denote that. Therefore:
xsquare = (4/pi) * symsum(funcSum, k, 1, Inf);
xquare now contains your representation of a square wave defined in terms of the Symbolic Math Toolbox. Now, if you want to plot your square wave and see if we have this right. We can do the following. Let's go between -3 <= t <= 3. As such, you would do something like this:
tVector = -3 : 0.01 : 3; %// Choose a step size of 0.01
yout = subs(xsquare, t, tVector);
You will notice though that there will be some values that are NaN. The reason why is because right at a multiple of the period (T = 1, 2, 3, ...), the behaviour is undefined as the derivative right at these points is undefined. As such, we can fill this in using either 1 or -1. Let's just choose 1 for now. Also, because the Fourier Series is generally a complex-valued function, and the square-wave is purely real, the output of this function will actually give you a complex-valued vector. As such, simply chop off the complex parts to get the real parts only:
yout = real(double(yout)); %// To cast back to double.
yout(isnan(yout)) = 1;
plot(tVector, yout);
You'll get something like:
You could also do this the ezplot way by doing: ezplot(xsquare). However, you'll see that at the points where the wave repeats itself, we get NaN values and so there is a disconnect between the high peak and low peak.
Note:
Natan's solution is much more elegant. I was still writing this post by the time he put something up. Either way, I wanted to give a more signal processing perspective to how to do this. Go Fourier!
A Fourier series for the square wave of unit amplitude is:
alpha + 2/Pi*sum(sin( n * Pi*alpha)/n*cos(n*theta),n=1..infinity)
Here is a handy trick:
cos(n*theta) = Re( exp( I * n * theta))
and
1/n*exp(I*n*theta) = I*anti-derivative(exp(I*n*theta),theta)
Put it all together: pull the anti-derivative ( or integral ) operator out of the sum, and you get a geometric series. Then integrate and finally take the real part.
Result:
squarewave=
alpha+ 1/Pi*Re(I*ln((1-exp(I*(theta+Pi*alpha)))/(1-exp(I*(theta-Pi*alpha)))))
I tried it in MAPLE and it works great! (probably not very practical though)
I'm trying to plot the following:
syms f_ih n
bi = (1-f_ih)/2
where f_ih is constrained by: 0 ≤ f_ih ≤ (n-1)/n. I'd like to show bi as a surface plot with independent variables n and f_ih, but ezsurf doesn't allow for variable bounds. How can I do this?
The command I'd love to run would simply be ezsurf(bi,[0,(n-1)/n]), but it's not that simple.
Thanks!
There's a good reason for requiring the domain to be numeric as opposed to symbolic: the function has to be evaluated in order to obtain actual numbers that can be plotted. It makes no sense to plot something that is purely symbolic like your equations (in Matlab or on paper) unless you specify n. Some functions may be scale invariant (not quite the right term mathematically), meaning they look the same when evaluated/plotted on different domains, but Matlab has no way of knowing. Choose a value of n that results in a plot that looks how want. Then, if you like, you can remove the numbers from the axes afterwards and label them as if it were for arbitrary n even though it's not.