University of Pennsyvania
Department of Electrical Engineering
 
USING MATLAB FOR ELECTRICAL CIRCUITS 
1. PLOTTING FUNCTIONS
  2. COMPLEX NUMBERS

Working with complex numbers in MATLAB is easy. MATLAB works with the rectangular representation. To enter a complex number, type at the prompt:

example: z = 5-3j

To find the magnitude and angle of z, use the abs() and angle () function.

The angle function gives the angle in radians. To convert to degrees you can use: Example: V = 

type in MATLAB: V = (5+9j)*(7+j)/(3-2j)

To find the real and imaginary part of a complex number z, type: 3. SOLVING LINEAR EQUATIONS AND MATRICES

Assume you have the following two linear comples equations with unknown I1 and I2:

This can be written in matrix form: A.I = B. To solve this in MATLAB one can use the matrix left division operator:

Or one can also use the following command: I = inv(A)*B

The MATLAB code is as follows:

One uses the abs() operator to find the magnitude of the complex number and the angle() operator to find the angle (in radians). To get the result in degree we have multiplied the angle by 180/pi as shown above.

4. FINDING THE ROOTS of a POLYNOMIAL

To find the roots of a polynomial of the form

Define the polynomial as follows: A = [ am am-1 am-2 ... a1 a0];

The command to for finding the roots is:

roots(A)

As an example consider the following function:

In Matlab:
>> A=[4 12 1];
>> roots(A)

ans =
-2.9142
-0.0858

This works also for complex roots. As an example consider the function:

in Matlab:
>> A=[5 3 2];
>> roots(A)

ans =
-0.3000 + 0.5568i
-0.3000 - 0.5568i

5. FINDING THE POLYNOMIAL WHEN THE ROOS ARE KNOWN

Suppose that you have the following expression F(s) and would like to find the coefficient of the corresponding polynomial:

This is defined in Matlab by a column vector of the roots:

roots = [a1; a2; a3 ];

One finds than the coefficient of the polynomial, using the "poly" command:

poly(roots);

As example consider the function F(s) = s(s+50)(s+250)(s+1000). To find the coefficient of the coresonding polynomials, one first define the column vector of the roots:

roots=[0; -50; -250; -1000 ];

The coefficients are then found from the poly command:

coeff = poly(roots)

which will give:

coeff = 1 1300 312500 12500000 0

corresponding to the polynomial,

6. TRANSFER FUNCTIONS H(s) and BODE PLOT

6a. Using the Bode command when the transfer function is specified as a ratio of two polynomials.

with m < n

in MATLAB, specify D and N:

to plot the transfer function (magnitude and phase), type: Example: H(s) = (0.5s +1)/(1E-4s2 + 0.01s +1)

In Matlab:

One can also specify the frequency axis w (radial frequency):

For the same example one can specify the frequency range as a vector:

w = [0.1: 0.1: 10000];

bode(num,den,w)

grid on

in which w is the frequency. The resulting graph is shown below.

6b. Plotting a transfer function when Poles and Zeros are given:

Often, the transfer function is specified as a function of its poles and zeros, in the form:

H(s)=K(s+s1)(s+s2)(1+as + bs2)/ (s+s3)(s+s4)(1+cs + ds2).

In that case one can find the polynomial of the nominator and denomator first by using the

poly function and the conv function. One first defines a column vector of the roots (-s1, -s2, etc.):

To find the polynomial corresponding to these roots, use If necessary one can multiply the d1 polynomial with the quadratic term using the

conv function. First define the polynomial corresponding to the quadratic term:

The same can be done for the denominator.

Example:

H(s)=72x(s+2)/s(s+50)(s+250)(s+1000)(s2 + 2.4s + 144)

First find the coefficent of the polynomial corresonding to s(s+50)(s+250)(s+1000) of the denominator:

Then, one multiply this polynomial with the remaining quadratic term (s2 + 2.4s + 144) in the denominator (d2 defines the quadratic term):

d2=[1 2.4 144];

In which "den" corresponds to the polynomial of the denominator. Finally, one can plot the Bode Diagram:

 

An altnerative command to plot the magnitude and phase of a transfer function is:

freqs(num, den)

6c. Using the plot command when the transfer function is not specifed as a ratio of polynomials

The previous transfer function

H(s) = 72x(s+2)/s(s+50)(s+250)(s+1000)(s2 + 2.4s + 144)

in which s=jw, can be plotted using the plot function in Matlab.We need to define the range of the indpendent variable w before plotting the fucntion H(jw).

>> w=[0.1: 0.1: 10^5];
>> H=72*(2+j*w)./(j*w.*(50+j*w).*(250+j*w).*(2000+j*w).*(144+2.4*j*w+(j*w).^2));subplot(2,1,1);
>> subplot(2,1,1);
>> semilogx(w,20*log10(abs(H)));
>> grid on
>> ylabel('|H(j\omega)|');
>> title('Bode Plot: Magnitude in dB');
>> subplot(2,1,2);
>> semilogx(w,unwrap(angle(H))*180/pi);
>> grid on;
>> xlabel('\omega(rad/s)');
>> ylabel('\angleH(j\omega)(\circ)');
>> title('Bode plot: Phase in degrees');

Notice:

The results is shown below. As can be seen it is the same graph as the one obtained from the Bode command.

 

7. OUTPUT of a SYSTEM DEFINED BY A TRANSFER FUNCTION

7.a Response to any input x(t)

Assume that the system is described by a transfer function H(s)=NUM(s)/DEN(s) where NUM and DEN contain the polynomial coefficients in descending powers of s.

One can then plot the output of the system for a given input signal x(t). As an example, lets consider a simple 2nd order low pass filter with a cutoff frequency of 100 rad/s:

in which NUM = [0 1] and DEN = [10^-4 0.02 1].

Lets apply two sinusoidial input signals x1 and x2 of frequency 50 Hz:and 500 Hz, respectively. The the corresonding output sigals are called y1 and y2, respectively. Since the system filters out higher frequencies we expect that the output y2 is considerably smaller than y1. We can plot the outputs of y1 and y2 using the lsim comment:

LSIM(NUM,DEN,U,T) command.

in which U is the input and T is the time. The matlab code for the filter and the input and output signals is as follows.

>> t=0:0.0001:0.1;
>> x1=10*sin(2*pi*50*t);
>> x2=10*sin(2*pi*500*t);
>> y2=lsim(NUM,DEN,x2,t);
>> y1=lsim(NUM,DEN,x1,t);
>> subplot(2,1,1);
>> plot(t,x1);
>> hold on
>> plot(t,x2);
>> subplot(2,1,2);
>> plot(t,y1);
>> hold on
>> plot(t,y2) ;
>> text(0.015, 2.5, 'y1(t)');
>> text(0.01, 0.35, 'y2(t)');
>> title('Filtering action of a low pass filter');
>> ylabel('y1 and y2');
>> xlabel('time (s)');
>> subplot(2,1,1);
>> ylabel('x1 and x2') ;

The graphs of the inputs x1 and x2 (top graph) and the outputs y1 and y2 (bottom graph) illustrates the effect of the filtering action.

7.b Step Response

Again, assume that the system is described by a transfer function H(s)=NUM(s)/DEN(s) where NUM and DEN contain the polynomial coefficients in descending powers of s. The step response y(t) of the system with the above transfer function is obtained by the command step.

Step(NUM,DEN

Again An example, lets plot the step response to the 2nd order low-pass filter used above,

in which NUM = [0 1] and DEN = [10^-4 0.02 1].

>> num=[0 1];
>> den=[10^-4 0.02 1];
>> step(num,den)

 

Go to ESE215, ESE216

Created by J. Van der Spiegel <jan@ee.upenn.edu>; Feb. 11, 1997;
Updated by J. Van der Spiegel, March 17, 2009.