Complex Numbers in Electronics

Transcription

1 P5 Computing, Extra Practice After Session 1 Complex Numbers in Electronics You would expect the square root of negative numbers, known as complex numbers, to be only of interest to pure mathematicians. However, complex numbers are very important in Engineering. Particularly in Electronics where they are used to calculate voltages and currents in electric circuits. In this session, you will be using MATLAB to explore why complex numbers are so important in Electronics. If you get lost or stuck, have a look at the program listings at the end of this document. The domestic electrical supply uses an alternating current (AC). AC signals are also used in radio and communications. First, let's plot what an alternating current looks like. Create a new MATLAB script called ACwave1.m. Plot the graph y1 = A cos(2π f t ) Where A = 4, f = 5 and the time t is a vector of 400 points between 0 and 2. Although the results we obtain will work for both sine and cosine waves, it will be easier to understand if we work with cosine waves. The amplitude of the wave A determines the height of the wave. The maximum height is A and the minimum height is -A. The function repeats after a time T = 0.2 seconds. This is called the periodic time. In mathematics, a periodic function is a function that has the following property. f (t ) = f (t+t ) Change the program so that the figure has two subplots, one above the other. Plot y1 in the top subplot, and in the bottom subplot plot :- y2 = A cos(2π f (t+t )) Where T = 0.2 Both graphs should look identical. The first T seconds is one cycle. The frequency f is the number of cycles in one second. If there are f cycles in one second then the periodic time is going to be one divided by f. T = 1 f Instead of setting T = 0.2, change the program so that T is calculated using the above equation. Frequency is measured in Hertz. 1 Hertz = 1Hz = 1 cycle a second. Try changing the amplitude and frequency of the wave. 1

2 Copy ACwave1.m to Acwave2.m. Then in Acwave2.m, return the frequency to 5 cycles per second and the amplitude back to 4. The Greek letter omega is used for the angular frequency. This is the number of radians per second. ω = 2 π f Of course we cannot use omega in a MATLAB program, so we will use w instead of omega. Rewrite the script so that it calculates the angular frequency and uses the following to calculate y1. y1 = A cos(ω t ) As you will find out, using ω instead of f in the calculations is easier. Change y2 to the following y2 = A cos(ω t +ϕ ) Where ϕ = π 2 You can not use ϕ, the greek letter phi, in MATLAB. Instead, call the variable phi. In the top subplot, plot y1 in blue and y2 with a red dash line. Both plots have the same basic shape. They are both called sinusoidal waves. The peaks do not line up. They are said to be out of phase. The variable ϕ is called the phase angle. It is a measure of how much y2 is out of phase with y1. Phase angle can be measured in either degrees or radians. If you plotted a sine wave, it would look identical to y2. The difference between a sine and cosine wave is that they have a different phase angle. They are out of phase by π 2. In the bottom subplot, plot y3 = y1 + y2 Run the program observe what the sum of the two waves looks like. Try changing the phase angle to π 2, π 3, π 4, π and π 3 When the phase angle is π just the very small error. the two waves cancel each other out. So the plot of y3 is In Electronics, out of phase waveforms get added together all the time. We need a way of calculating the amplitude and phase of the sum of the two waves. We cannot just add the amplitude of two waveforms if they are out of phase. We could use trigonometry to find the amplitude and phase, but it is a fairly long and tedious process. There is a much easier method using complex numbers. 2

3 Phasors Phasors are rotating vectors. They rotate in the complex plane. The formula for a phasor is z = Ƶ e j ω t I am using j for 1, The lower case z is a function of time. The upper case Ƶ is a constant that can be a complex number. Electronic engineers often use a Z with a stroke across it in written work to help distinguish Z and 2. It is used here to help see the difference between the upper and lower case z. To plot a phasor, we are going to use a compass plot. You can use compass plots in MATLAB to plot vectors. For example, enter the following to plot a vector to the point x=3, y=4. >> compass(3,4) Compass plots can also be used to plot complex numbers. >> z = 3 + 4j >> compass(z) The compass plot shows the absolute value and the angle of a complex number. Below, the magnitude is 2 and the angle is 30 o. >> z = 2*exp(j*pi/6); >> compass(z) 3

4 Create a new MATLAB script called Phasor1.m and enter the following MATLAB code. % Plot a phasor Z exp(jwt) % The number of points n = 18; %Time in seconds t = linspace(0,1,n); % seconds % The coefficient of the phasor Z = 2; % The frequency of the signal f = 1; % Cycles per second % The angular frequency w = 2*pi*f; %Calculate the phasor z = Z*exp(j*w*t); %Plot the graph compass(z); 4

5 The problem with this plot is that it does not show the true nature of a phasor. A phasor changes with time. It would be nice to see how it changes. Instead of plotting all the vectors at once, we could plot them one at a time, with a half a second pause between each. %Plot the graph compass(z(1)); pause(0.5); compass(z(2)); pause(0.5); compass(z(3)); ETC The above works, but is very tedious to enter. Especially as we are eventually going to use hundreds of points. Instead we are going to use a FOR loop. Replace the compass plot at the bottom of the script with the following. %Plot the graph for k = [1 2 3] compass(z(k)); pause(0.5); end Each line of code between FOR and END is repeated for each number in the vector [1 2 3]. The variable k is first set to 1 and the code in the loop is executed. Then it is executed again with k = 2, then again with k = 3. Now change the vector from [1 2 3] to 1:n so that all 18 vectors are plotted. You will find that the plot is a but jumpy. So increase the number of points n to 360. We also need to reduce the delay to speed things up. Reduce the pause to 0.01 seconds. You should now be able to see how a phasor behaves over time. 5

6 Now we will use MATLAB to plot the real part of the phasor. Above the FOR loop, set a variable x to be the real part of z. Split the figure into two subplots, one above the other. In the top subplot, plot(x,t) so that x is on the x axis and t on the y axis. We want it this way round so that the x is the x axis on both the graph and on the compass plot. This is a static plot for the time being. %The real part of phasor x = real(z); %plot the real part subplot(2,1,1) plot(x,t,'b'); axis square grid on hold on xlabel('value') ylabel('time') In the bottom subplot, plot the animation of the phasor. %Plot the graph for k = 1:n subplot(2,1,2) compass(z(k)); pause(.01); end You should now be able to see that the real part is a cosine wave. It would be nice to be able to see where the cosine wave has got to while the animation is running, so hold the cosine wave on the graph, and plot a red dot over the top of cosine wave in the for loop. for k = 1:n %plot the phasor subplot(2,1,2) compass(z(k)); %plot the real part subplot(2,1,1) plot(x(k),t(k),'r.'); end pause(.01); You should now be able to see the real part of the phasor being plotted as the phasor rotates around the origin. But why is it a cosine wave? 6

7 Euler s formula explains why the real part is a cosine wave. Euler s formula is e j θ = cos(θ )+ j sin(θ ) Which means z = Ƶ e j ω t = Ƶ(cos(ωt )+ j sin(ωt )) So if Ƶ = 2, the real part will be 2 cos(ω t) If you make the constant Ƶ complex, the real part is no longer a cosine wave. Set Ƶ to j and run the program again. This time we get a sine wave. That is because j (cos(ωt )+ j sin(ωt )) = sin(ωt ) j cos(ωt) Also notice that the phasor starts and stops at 270 O = 90 O = π /2 If you convert j into polar form, this is the angle of j. Remember that to describe a sinusoidal wave you need the amplitude and the phase. The amplitude and the phase can be expressed as a single complex number Ƶ. The amplitude of the wave is abs(ƶ) and the phase is angle(ƶ). z = Ƶ e j ω t z = Ae j ϕ e j ω t Where A = abs(ƶ) and ϕ = angle(ƶ) z = Ae j (ω t + ϕ) = A(cos(ωt + ϕ ) + j sin(ωt + ϕ )) real(z) = A cos(ω t + ϕ ) To see how this works, we want to change Ƶ to a number with a particular absolute value and angle. Lets say we want an absolute value of 4 and an angle of π/6. Change the program so that Z = 4*exp(j*pi/6); Notice that the phasor starts at 30 O, that is π/6 radians, the same as the angle of Z. That is because When t = 0 e j ωt = 1 and z = Ƶ The start angle of the phasor also determines the phase of the sinusoidal real part. The length of the arrow in the compass plot is 4 and this is the absolute value of Ƶ. This also determines the amplitude or the real part. The properties of Ƶ describe the features of the phasor. Be careful, sometimes Ƶ is referred to as the phasor. 7

8 Run the program with Z = 1 + j; In the command window, find the angle of Ƶ and convert it into degrees. This is the phase angle of the wave. Its amplitude is the absolute value of Ƶ. Adding Phasors Just like vectors, phasors can be combine mathematically to solve problems. You can add and subtract them. Ƶ 1 e j ω t + Ƶ 2 e j ω t = (Ƶ 1 + Ƶ 2 )e j ω t Let Ƶ 3 = Ƶ 1 + Ƶ 2 Ƶ 1 e j ω t + Ƶ 2 e j ω t = Ƶ 3 e j ωt real(ƶ 1 e j ωt ) + real( Ƶ 2 e j ω t ) = real(ƶ 3 e j ω t ) A 1 cos(ωt +ϕ 1 ) + A 2 cos(ω t +ϕ 2 ) = A 3 cos(ωt+ϕ 3 ) Where A i = Z i and ϕ i = angle(z i ) We can use this to find the amplitude and phase of the sum of two sinusoidal waves of different phase. To see how this works, copy ACwave2.m to ACwave3.m. At the end of the program ACwave3.m add the following. %Use phasors to find the amplitude and phase of the sum Z1 = A; %phasor coefficient for y1 Z2 = A*exp(j*phi); %phasor coefficient for y2 Z3 = Z1 + Z2; %phasor coefficient for the sum % The Amplitude of the sum of the waveforms A3 = abs(z3); % The phase angle of the sum of the waveforms phi3 = angle(z3); % Check that the answer is correct. y4 = A3*cos(w*t+phi3); %In the bottom subplot, plot y4 over y3 subplot(2,1,2) hold on %Plot the wave plot(t,y4,'r.'); You should see that y3 and y4 coincide. 8

9 Try running the program with different phase angles. If you wish, you can change the program so that y1 and y2 have different amplitudes. You could also try subtraction instead of addition. I hope that you now realize that the really easy way to add or subtract sinusoidal waves of different phase, is to just add or subtract the complex numbers that represent their amplitude and phase. The calculations in AC theory mostly use complex numbers. Differentiating Phasors You may be wondering what shifts the phase of a sinusoidal wave in the first place. The answer is differentiation. The electrical components, capacitors and inductors, differentiate electrical signals. When you differentiate a sinusoidal wave, the wave is shifted by 90 O. You can also differentiate phasors If z = Z e j ω t then dz dt = j ω Z e j ω t = j ω z d 2 z dt 2 d n z dt n = ( j ω)2 z = ω 2 z = ( j ω)n z 9

10 As an example, the following differential equation describes the circuit on the right. RC dy dt + y = x R When you know more about resistors and capacitors, the appendix on page 13 explains where this equation comes from. x C i y 0V Assuming that both x and y are sinusoidal, we can use phasors to find y. Let x = X e j ω t The input of the circuit with a known amplitude Let y = Y e j ω t The output. We want to find the amplitude of the output and its phase relative to the input signal. dy dt = j ω Y e j ω t By differentiating the above. We can substitute these equations into differential equation describing the circuit. RC dy dt + y = x RC ( j ω Y e j ω t ) + Y e j ω t = X e j ω t j ω RC Y + Y = X The differential equation is now an algebraic equation. Much easier to solve. Y = X 1 + j ω RC 10

11 Question 1 Suppose that X = 10 Volts f = 1 khz = 1000 Hz R = 16 kω = Ω C = 10 nf = F Write a script called SeriesRC1.m to find the amplitude and phase of y, using X Y = 1 + j ω RC Question 2 Copy SeriesRC1.m to SeriesRC2.m. Change the script to plot the Amplitude and Phase against Frequency. The frequencies should be logarithmically spaced between 1Hz and 1MHz. Hint Look at the documentation for logspace Plot Amplitude against Frequency using a loglog plot. Plot Phase against Frequency using a semilogx plot. 11

12 The Conjugate of a Phasor Although the story as presented so far all works, I am not totally happy with it. Can we really just ignore the imaginary part of a phasor? If you have the same misgivings, there is another way of looking at phasors that makes it much clearer why we can do this. There are many mathematical methods that use complex numbers to produce a totally real solution. That is because the solution consists of a complex number and its conjugate. Let's look at the conjugate of a phasor. Copy phasor1.m to phasor2.m. In phasor2, set zc to the conjugate of z. Change x so that it is the real part of zc instead of z. Change the compass plot, so that it plots zc instead of z. Notice that the conjugate of the phasor rotates clockwise while previously, the phasor itself rotated anticlockwise. The real part is the same as before. Electrical signals cannot be complex, but they can have complex components, provided that each complex component is accompanied by its conjugate. Real sinusoidal signals are the average of a phasor and its conjugate. Change phasor2 so that x = z + zc 2 Change the compass plot so the phasor z, its conjugate zc and their average x are shown on the same plot, as below. %plot the phasor subplot(2,1,2) compass(z(k),'b'); hold on compass(zc(k),'c'); compass(x(k),'r'); hold off The variable x is a completely real sinusoidal wave. Just as before, the variable Ƶ determines the amplitude and the phase of the wave. Try using different values of Ƶ. The differential equations of electronic systems are linear. Which basically means we can find a solution for each component separately and then add them together. An overly complicated way of solving a problem would be to find a solution for the phasor, then find a solution for the conjugate, then add the two solutions together. If you do this, you will find that each of of the solutions is the conjugate of the other. So the imaginary parts cancel out. This should not be a surprise as we know the solution must be real. The real parts are equal. So we only need to find the real part of the phasor solution to obtain a complete solution. 12

13 Appendix Finding the voltage across a capacitor in the circuit on the right. R From Ohms law. i R = v R R x C i y i = x y R 0V From the definition of capacitance. q = C v C q = C y i = dq dt = C dy dt Current is the rate of change of the charge. C dy dt = x y R RC dy dt + y = x 13

14 ACwave1.m Program Listings % Plot what an AC signal looks like % The number of points n = 400; %Time in seconds t = linspace(0,2,n); % seconds % The Amplitude of the signal A = 4; % The frequency of the signal f = 5; % Cycles per second %Calculate the AC wave y1 = A*cos(2*pi*f*t); %The period of the wave T = 1/f; %Calculate the AC wave shifted by one period y2 = A*cos(2*pi*f*(t+T)); %In the top subplot, plot y1 subplot(2,1,1) %Plot the wave plot(t,y1); %Add a grid grid on %Label the x axis xlabel('time in seconds') %In the bottom subplot, plot y2 subplot(2,1,2) %Plot the wave plot(t,y2); %Add a grid grid on %Label the x axis xlabel('time in seconds') 14

15 ACwave2.m % Plot what an AC signal looks like % The number of points n = 400; %Time in seconds t = linspace(0,2,n); % seconds % The Amplitude of the signal A = 4; % The frequency of the signal f = 5; % Cycles per second % phase angle phi = -pi/3; % The angular frequency w = 2*pi*f; %Calculate the AC wave y1 = A*cos(w*t); %Calculate the AC wave shifted phi y2 = A*cos(w*t + phi); %Add the two waves together y3 = y1 + y2; %In the top subplot, plot y1 and y2 subplot(2,1,1) %Plot the wave y1 in blue plot(t,y1,'b'); hold on %Plot the wave y2 as a red dash line plot(t,y2,'r--'); %Add a grid grid on %Label the x axis xlabel('time in seconds') pto 15

16 ACwave2.m cont %In the bottom subplot, plot y3 subplot(2,1,2) %Plot the wave plot(t,y3); %Add a grid grid on %Label the x axis xlabel('time in seconds') Phasor1.m version1 % Plot a phasor Z exp(jwt) % The number of points n = 18; %Time in seconds t = linspace(0,1,n); % seconds % The coefficient of the phasor Z = 2; % The frequency of the signal f = 1; % Cycles per second % The angular frequency w = 2*pi*f; %Calculate the phasor z = Z*exp(j*w*t); %Plot the graph compass(z); 16

17 Phasor1.m version 2, animated. % Plot a phasor Z exp(jwt) % The number of points n = 360; %Time in seconds t = linspace(0,1,n); % seconds % The coefficient of the phasor Z = 2; % The frequency of the signal f = 1; % Cycles per second % The angular frequency w = 2*pi*f; %Calculate the phasor z = Z*exp(j*w*t); %The real part of phasor x = real(z); %Plot the graph for k = 1:n compass(z(k)); pause(.01); end 17

18 Phasor1.m version 3, animated with real part. % Plot a phasor Z exp(jwt) % The number of points n = 360; %Time in seconds t = linspace(0,1,n); % seconds % The coefficient of the phasor Z = 1 + j; % The frequency of the signal f = 1; % Cycles per second % The angular frequency w = 2*pi*f; %Calculate the phasor z = Z*exp(j*w*t); %The real part of phasor x = real(z); %plot the real part subplot(2,1,1) plot(x,t,'b'); axis square grid on hold on xlabel('value') ylabel('time') for k = 1:n %plot the phasor subplot(2,1,2) compass(z(k)); %plot the real part subplot(2,1,1) plot(x(k),t(k),'r.'); end pause(.01); 18

19 ACwave3.m % Plot what an AC signal looks like % The number of points n = 400; %Time in seconds t = linspace(0,2,n); % seconds % The Amplitude of the signal A = 4; % The frequency of the signal f = 5; % Cycles per second % phase angle phi = -pi/3; % The angular frequency w = 2*pi*f; %Calculate the AC wave y1 = A*cos(w*t); %Calculate the AC wave shifted phi y2 = A*cos(w*t + phi); %Add the two waves together y3 = y1 + y2; %In the top subplot, plot y1 and y2 subplot(2,1,1) %Plot the wave y1 in blue plot(t,y1,'b'); hold on %Plot the wave y2 as a red dash line plot(t,y2,'r--'); pto 19

20 ACwave3.m cont %Add a grid grid on %Label the x axis xlabel('time in seconds') %In the bottom subplot, plot y3 subplot(2,1,2) %Plot the wave plot(t,y3); %Add a grid grid on %Label the x axis xlabel('time in seconds') %Use phasors to find the amplitude and phase of the sum Z1 = A; %phasor coefficient for y1 Z2 = A*exp(j*phi); %phasor coefficient for y2 Z3 = Z1 + Z2; %phasor coefficient for the sum % The Amplitude of the sum of the waveforms A3 = abs(z3); % The phase angle of the sum of the waveforms phi3 = angle(z3); % Check that the answer is correct. y4 = A3*cos(w*t+phi3); %In the bottom subplot, plot y4 over y3 subplot(2,1,2) hold on %Plot the wave plot(t,y4,'r.'); 20

21 SeriesRC1.m % Series RC circuit calculations X = 10; f = 1000; R = 16e3; C = 10e-9; w = 2*pi*f; % The voltage across the R and C % Hz The frequency of the wave % Ohms The value of the resistor % Farads The value of the capacitor % rads/s The angular frequency % Voltage across the capacitor Y = X./(1+j*w*R*C); A = abs(y); rads = angle(y); deg = rads*180/pi; % The amplitude of y % The phase in radians % The phase in degrees fprintf('the amplitude is %4.3f Volts \n',a); fprintf('the phase is %4.3f degrees\n',deg); 21

22 SeriesRC2.m % Series RC circuit plots X = 10; R = 16e3; C = 10e-9; % The voltage across the R and C % Ohms The value of the resistor % Farads The value of the capacitor n = 1000; f = logspace(1,6,n); % The number of points to plot % Hz The frequency of the wave w = 2*pi*f; % rads/s The angular frequency % Voltage across the capacitor Y = X./(1+j*w*R*C); A = abs(y); rads = angle(y); deg = rads*180/pi; % The amplitude of y % The phase in radians % The phase in degrees %Plot the amplitude against frequency subplot(2,1,1) loglog(f,a); xlabel('frequency') ylabel('amplitude') grid on %plot the phase against frequency subplot(2,1,2) semilogx(f,deg); xlabel('frequency') ylabel('phase degrees') grid on 22

23 Phasor2.m version 1, Plot the conjugate of the phasor % Plot a phasor Z exp(jwt) % The number of points n = 360; %Time in seconds t = linspace(0,1,n); % seconds % The coefficient of the phasor Z = 1 + j; % The frequency of the signal f = 1; % Cycles per second % The angular frequency w = 2*pi*f; %Calculate the phasor z = Z*exp(j*w*t); %The conjugate of z zc = conj(z); %The real part of phasor x = real(zc); %plot the real part subplot(2,1,1) plot(x,t,'b'); axis square grid on hold on xlabel('value') ylabel('time') for k = 1:n %plot the phasor subplot(2,1,2) compass(zc(k)); %plot the real part subplot(2,1,1) plot(x(k),t(k),'r.'); end pause(.01); 23

24 Phasor2.m version 2, Plot the phasor, conjugate and average % Plot a phasor Z exp(jwt) % The number of points n = 360; %Time in seconds t = linspace(0,1,n); % seconds % The coefficient of the phasor Z = 1+j; % The frequency of the signal f = 1; % Cycles per second % The angular frequency w = 2*pi*f; %Calculate the phasor z = Z*exp(j*w*t); %The conjugate of z zc = conj(z); %The real part of phasor x = (z+zc)/2; %plot the real part subplot(2,1,1) plot(x,t,'b'); axis square grid on hold on xlabel('value') ylabel('time') pto 24

25 Phasor2.m version 2, cont for k = 1:n %plot the phasor subplot(2,1,2) compass(z(k),'b'); hold on compass(zc(k),'c'); compass(x(k),'r'); hold off %plot the real part subplot(2,1,1) plot(x(k),t(k),'r.'); end pause(.01); 25