HW4_TOTAL_SOLUTIONS - EECS 314 Winter 2011 Homework set 4...

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Unformatted text preview: EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 1 (30 points) Power transfer from the source to the load resistance This circuit diagram shows a voltage source VS with the source resistance RS to the left of the terminals a, b, and the load resistor RL to the right of the terminals a, b. The source parameters VS and RS are fixed (constant); the load resistance RL can be varied. Your goal is to study and explain what happens in the circuit as RL is varied. Show your work for all parts on additional pages; write the results in the space below. Part 1 (10 points) Derive the algebraic expressions for the current I in the circuit, the voltage V across the load resistor (the same as the voltage across the terminals a and b), and the power PL absorbed by the load resistor in terms of VS , RS , and RL. I = Calculate the maximal values of the current IMAX , voltage VMAX , and power PL,MAX in terms of VS and RS (Hint: The maximal power is absorbed by the load resistance that equals the source resistance). IMAX = VMAX = PL.MAX = V = PL = © 2011 Alexander Ganago Last printed 2/6/2011 11:25:00 AM Page 1 of 6 File: 2011 W 314 HW 04 p1.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2011 Alexander Ganago Last printed 2/6/2011 11:25:00 AM Page 2 of 6 File: 2011 W 314 HW 04 p1.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 1 Part 2 (10 points) Assume VS = 120 V, RS = 100 Ω, and RL varied from 1 Ω to 10,000 Ω = 10 kΩ. Use software, with which you are comfortable (MATLAB, Excel, etc.), to plot 3 curves: 1. Normalized current I/ IMAX 2. Normalized voltage V/VMAX 3. Normalized power P/PMAX as functions of the load resistance RL . Use a logarithmic axis for the load resistance and clearly label which curve shows what on your plot (hand‐written labels are OK). Your plot should show at least 5 points per decade (any interval, over which varies by a factor of 10, such as from 3 to 30 Ω). Attach your plot with your name, discussion section #, and the date of work (all in ink). If you had to write computer code (in MATLAB, etc.) include it along with your plot, with your name and date of work printed as part of the code. If you use EXCEL, attach a one‐page printout of the spreadsheet, with your name and date of work printed on top. % Matlab Script Vs = 120; Rs = 100; Rl = Imax Vmax Pmax 1:.01:10000; = Vs/Rs; = Vs; = Vs^2/(4*Rs); Inorm = zeros(1,length(Rl)); Vnorm = zeros(1,length(Rl)); Pnorm = zeros(1,length(Rl)); for n = 1:length(Rl) Inorm(n) = (Vs/(Rs+Rl(n)))/Imax; Vnorm(n) = (Rl(n)/(Rs+Rl(n)))/Vmax; Pnorm(n) = Vnorm(n)*Inorm(n)/Pmax; end © 2011 Alexander Ganago Last printed 2/6/2011 11:25:00 AM Page 3 of 6 File: 2011 W 314 HW 04 p1.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above subplot(3,1,1) semilogx(Rl,Inorm,'LineWidth',2) grid on; xlabel('Rl (ohms)'); ylabel('I/Imax'); title('I/Imax vs. Rl'); subplot(3,1,2) semilogx(Rl,Vnorm,'LineWidth',2) grid on; xlabel('Rl (ohms)'); ylabel('V/Vmax'); title('V/Vmax vs. Rl'); subplot(3,1,3) semilogx(Rl,Pnorm,'r','LineWidth',2) grid on; xlabel('Rl (ohms)'); ylabel('P/Pmax'); title('P/Pmax vs. Rl'); © 2011 Alexander Ganago Last printed 2/6/2011 11:25:00 AM Page 4 of 6 File: 2011 W 314 HW 04 p1.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above I/Imax vs. Rl 1 I/Im a x 0.5 0 0 10 10 1 10 Rl (ohms) V/Vmax vs. Rl 2 10 3 10 4 0.01 0.008 V /V m a x 0.006 0.004 0.002 0 0 10 x 10 -5 1 2 3 4 10 10 Rl (ohms) P/Pmax vs. Rl 10 10 6 P /P m a x 4 2 0 0 10 10 1 10 Rl (ohms) 2 10 3 10 4 Part 3 (10 points) Use your calculations (Part 1 of this problem) and computer‐generated plot (Part 2 of this problem) to explain why the power P absorbed by the load (or transferred to the load) decreases at low load resistances, and why it decreases at high load resistances. At low voltages: Due to small voltage At high voltages: Due to small current © 2011 Alexander Ganago Last printed 2/6/2011 11:25:00 AM Page 5 of 6 File: 2011 W 314 HW 04 p1.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Write a brief discussion on whether your plot agrees with your algebraic results. From your plot, determine the ranges of load resistances, within which the power transferred to the load remains: above 90% of the max power above 70% of the max power above 50% of the max power. 52‐192 30‐341 17‐583 In ohms In % of RS 52‐192 30‐341 17‐583 © 2011 Alexander Ganago Last printed 2/6/2011 11:25:00 AM Page 6 of 6 File: 2011 W 314 HW 04 p1.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 2 (30 points) Volt‐Amp characteristics of sources represented as their Thevenin equivalents Part 1 (15 points) Assume that the unknown source (the box on the diagram below) can be represented with its Thevenin equivalent circuit (the voltage source VT in series with the source resistance RT). This table gives values of the current I through the load resistor RL: Load resistance RL Current I 1 Ω 5 Ω 12 A 6 A Calculate the following parameters and write your answers in the table below: open‐circuit voltage VT in volts; short‐circuit current ISC in amps; the source resistance RT in ohms. Also calculate the load resistance connected to the terminals a, b, to which the source transfers the maximal power, and the maximal power that can be dissipated in the load. VT (volts) ISC (amps) RT (ohms) 48 16 3 RL, max power (ohms) 3 PMAX (watts ) 192 Show your work on an additional page. Sketch the volt‐amp characteristic of the source in the format shown here: Clearly indicate the numerical values of the open‐circuit voltage and short‐circuit current, and clearly show the two data points listed in the table above. Your sketch (on an additional page) can be either hand‐written or computer‐generated. © 2011 Alexander Ganago Last printed 2/6/2011 11:26:00 AM Page 1 of 6 File: 2011 W 314 HW 04 p2.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above PART 1 Lets find the Thevenin equivalent circuit first; The box given above can be represented by the following equivalent Thevenin and Norton Circuits Lets Apply KVL around the loop (Red circle); Vth=I1*(Rth+RL1)=12*(Rth+1) Vth=I2*(Rth+RL2)=6*(Rth+5) By solving these equations, Rth=3 ohm Vth= 48 V And Norton Equivalent circuit is; Ino=Vth/Rth = 16 A Maximum Power transfer occurs when Rth=RL= 3 ohm For that Load maximum power dissipated in the load; Pmax= ((Vth)^2)/(4*RL) = (48^2) / (4*3) = 192 W © 2011 Alexander Ganago Last printed 2/6/2011 11:26:00 AM Page 2 of 6 File: 2011 W 314 HW 04 p2.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2011 Alexander Ganago Last printed 2/6/2011 11:26:00 AM Page 3 of 6 File: 2011 W 314 HW 04 p2.doc EECS 314 Winter 2011 Homework set 4 Students name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 2 Part 2 (15 points) As in Part 1, assume that the unknown source can be represented with its Thevenin equivalent circuit. Do not assume that the source is the same as in Part 1. The table below gives values of the power transferred to the load resistor RL: Load resistance RL Power PL 2 Ω 8 Ω 1600 W 1152 W Calculate the following parameters and write your answers in the table below: open‐circuit voltage VT in volts; short‐circuit current ISC in amps; the source resistance RT in ohms. Also calculate the load resistance connected to the terminals a, b, to which the source transfers the maximal power, and the maximal power that can be dissipated in the load. VT (volts) ISC (amps) RT (ohms) 125 51.6 2.4225 RL, max power (ohms) 2.4225 PMAX (watts ) 1612.5 Show your work on an additional page. Sketch the volt‐amp characteristic of the source in the format shown here: Clearly indicate the numerical values of the open‐circuit voltage and short‐circuit current, and clearly show the two data points listed in the table above. Your sketch (on an additional page) can be either hand‐written or computer‐generated. © 2011 Alexander Ganago Last printed 2/6/2011 11:26:00 AM Page 4 of 6 File: 2011 W 314 HW 04 p2.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above For the given powers at known loads, we can calculate the current passing through these loads as following; 1600=(I1^2)*RL1= (I1^2)*2 1152=(I2^2)*RL2= (I2^2)*8 so I1= 28.28 A and I2 = 12 A , After this point we can follow the exact same procedure given in part 1; Vth=I1*(Rth+RL1)=28.28*(Rth+2) Vth=I2*(Rth+RL2)=12*(Rth+8) Rth= 2.4225 ohm Vth= 125 V Ino= Vth/Rth= 51.6 A © 2011 Alexander Ganago Last printed 2/6/2011 11:26:00 AM Page 5 of 6 File: 2011 W 314 HW 04 p2.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Pmax= Pmax= ((Vth)^2)/(4*RL) = (125^2) / (4*2.4225) = 1612.5 W © 2011 Alexander Ganago Last printed 2/6/2011 11:26:00 AM Page 6 of 6 File: 2011 W 314 HW 04 p2.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 3 (30 points) Practical Perspective: The Load Line Determines the Operating Point of a Non-Ohmic Device Background: If a non-ohmic electronic device such as a semiconductor diode, whose Volt-Amp characteristic is known (for example, as a file of lab data), is connected to a linear circuit that includes sources and resistors, we can apply the concept of the Thevenin equivalent circuit in order to determine the operating voltage and current, as well as the power absorbed by the non-ohmic device. The strategy is straightforward: (1) Replace the linear circuit with its Thevenin equivalent; (2) Obtain its Volt-Amp characteristic and plot it as a straight line, which is also known as the load line; (3) On the same graph, plot the non-linear Volt-Amp characteristic of the electronic device; (4) Determine the crossing point of the two Volt-Amp characteristics; (5) The coordinates of the crossing point are the operating voltage and current. This graph shows lab data for a semiconductor diode obtained with the same equipment and the same VI that you used in DC Lab, along with the Volt-Amp characteristic of the Thevenin equivalent circuit with VT = VS = 4 V and RT = R S = 4 k Ω . The operating point is highlighted with a circle; the arrows show the operating voltage of about 1.62 V and the operating current of about 0.6 mA; their product equals about 0.972 mW and is the power absorbed by the semiconductor diode. © 2011 Alexander Ganago Last printed 2/1/2011 2:30:00 AM Page 1 of 3 File: 2011 W 314 HW 04 p3.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 3 (continued) Use file 2007-07-24 IV diode a4_.xls available on CTools web site (HW 04 folder) and determine the operating voltage in volts, the operating current in mA, as well as the power in mW absorbed by the semiconductor diode connected to the following circuit: Note that for this circuit the Thevenin equivalent resistance equals: RT = RS + RX Assume the following parameters: VS = 5 V; RS = 200 Ω, RP = 1 kΩ. Write your results in the table below. RX = 1 k Ω Operating voltage, V Operating current, mA Power in mW absorbed by the diode 1.73 2.71 4.688 RX = 500 Ω 1.76 4.59 8.078 RT = 200 Ω 1.81 7.93 14.353 RX = 0 1.89 15.49 29.276 Attach the printout of the volt-amp characteristic of the diode. Remember to write on it your name, date of work, and discussion section number (in ink!). Matlab code for Problem 3: Va = xlsread ('2007-07-24 IV diode a4_.xls', 1, 'A2:A201'); Id = xlsread ('2007-07-24 IV diode a4_.xls', 1, 'B2:B201'); Vs = 5; Rs = 200; Rx_arr = [1000, 500, 200, 0]; Iload_arr = zeros(length(Rx_arr)); for Rx_iter = 1:length(Rx_arr) Rx = Rx_arr(Rx_iter); Iload_arr = ((Vs - Va)*1000) / (Rs + Rx); grid on; © 2011 Alexander Ganago Last printed 2/1/2011 2:30:00 AM Page 2 of 3 File: 2011 W 314 HW 04 p3.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above plot (Va, Id, 'r', 'Linewidth', 2); hold on ; plot(Va, Iload_arr); end [x, y] = ginput (1); On your printout show the load lines for all resistance values (either draw them by hand using a ruler or plot them with a computer program, with which you are comfortable), as well as arrows that indicate the operating current and voltage for each resistance value. Show your calculations for RX = 200 Ω. © 2011 Alexander Ganago Last printed 2/1/2011 2:30:00 AM Page 3 of 3 File: 2011 W 314 HW 04 p3.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 4 (10 points) Saw-Tooth, or Ramp Waveform and Its Applications The Saw-Tooth waveform is used to move, or sweep material objects whose position depends on the applied voltage. For example, due to piezoelectric effect, voltage causes mechanical displacement, which can be used for making positioning tools with extreme accuracy. In old televisions and oscilloscopes with Cathode Ray Tubes, or CRTs, the applied voltage displaces the beam of electrons (or cathode rays, according to their old name). Behind each television screen (old, not a flat one) sits a CRT, where beams of electrons are produced by electron guns (one for each color) and deflected by the electric fields, which are generated by high-voltage power supplies connected to the pairs of vertical and horizontal plates so that each beam sweeps the entire surface of the screen. To each pair of plates, a high-voltage Saw-Tooth (ramp) signal is applied, which makes the beam sweep the whole the length (or height) of the screen at a constant speed, and ensures nearly instantaneous return of the beam to the original position (during the return motion the beam is blanked, its brightness made “blacker than black”). A simplified sequence of scans is shown on the sketch (the return motion is shown with dashed lines). Note that the beam moves slowly from top to bottom, and moves fast from left to right, because the frequency of the saw-tooth waveform for vertical deflection is low, while the frequency of the saw-tooth waveform for horizontal deflection is high. © 2011 Alexander Ganago Last printed 2/1/2011 2:31:00 AM Page 1 of 2 File: 2011 W 314 HW 04 p4.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 4 (continued) The US standard for television involves 30 frames (called fields) per second; each frame consists of 525 lines (out of which only 483 are visible; others – at the edges of the screen – are blanked). In other words, the frequency in Hz for the vertical sweep equals 30 (the number of frames); the frequency in Hz for the horizontal sweep equals the product of 30 and 525 For the ramp signals used in this TV standard, calculate and write your answers below: Vertical sweep Frequency, in Hz Period, in µsec Rise Time (from 10% level to 90% level), in µsec Fall Time (from 10% level to 90% level), in µsec 30 Hz 1/F = 33,333 µsec 0.8*T = 26,666 µsec Horizontal sweep 30*525 = 15,750 Hz 1/F = 63.49 µsec 0.8*T = 50.792 µsec 0 (Ideal) 0 (Ideal) Show your work below and/or on additional pages. © 2011 Alexander Ganago Last printed 2/1/2011 2:31:00 AM Page 2 of 2 File: 2011 W 314 HW 04 p4.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 5 (50 points) Build a Saw‐Tooth Waveform as a Sum of Its Harmonics Part 1 (10 points) Consider a saw‐tooth waveform at 30 Hz and 5 kVppk = 5,000 Vppk, which can be used for the vertical deflection of electron beams in a TV set. Calculate the peak amplitude in volts of each sinusoid corresponding to the fundamental and the harmonics up to the 5th of this waveform. Write your results in the table below. Component Fundamental 2nd harmonic 3rd harmonic 4th harmonic 5th harmonic Frequency (Hz) 30 60 90 120 150 Vpk (volts) 1591.5 795.774 530.5165 397.887 318.309 Show your work = a sample step‐by‐step calculation for the 5th harmonic. Use additional pages as needed. © 2011 Alexander Ganago Last printed 2/6/2011 11:27:00 AM Page 1 of 6 File: 2011 W 314 HW 04 p5.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above SOLUTION: The saw tooth wave of peak amplitude A can be mathematically expressed as The peak to peak voltage is 5000 V so Peak voltage Vpk=A=2500, fundamental freq is 30 Hz Amplitude of Fundamental Freq: (2/pi)*Vpk=1591.5 Amplitude of Sec. Harmonic: (2/pi)*Vpk/2=795.774 Amplitude of third Harmonic: (2/pi)*Vpk/3=530.6165 Amplitude of fourth Harmonic: (2/pi)*Vpk/4=397.887 Amplitude of fifth Harmonic: (2/pi)*Vpk/5=318.3099 © 2011 Alexander Ganago Last printed 2/6/2011 11:27:00 AM Page 2 of 6 File: 2011 W 314 HW 04 p5.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Part 2 (20 points) Make a computer‐generated plot of the sinusoidal components that sum up to this waveform. Your plot should show 3 curves: (1) The fundamental only (2) The sum of the fundamental and harmonics up to the 3rd (3) The sum of the fundamental and harmonics up to the 5th. Use equation 6 of the file Saw­tooth harmonics. On the plot show 2 to 5 periods of the waveforms. Use distinct line styles to show different curves and clearly explain what is what. Make sure that the peak‐to‐peak amplitude of the waveforms on your plot is reasonably close to 5,000 Vppk, and that the period is close to 33 msec. SOLUTION: Sinusoidal components of 30Hz, 5000Vppk Alternative Waveform 4000 Fundamental Freq. Fundamental to 3rd Harmonic Fundamental o 5th Harmonic 3000 2000 Amplitude (V) 1000 0 -1000 -2000 -3000 0 0.02 0.04 0.06 0.08 0.1 Time (s) 0.12 0.14 0.16 0.18 © 2011 Alexander Ganago Last printed 2/6/2011 11:27:00 AM Page 3 of 6 File: 2011 W 314 HW 04 p5.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above The matlab codes for the graph given above; clear all; close all; f0=30; % the fundamental frequency of the SawTooth wave A=5000/2; % Amplitude or the peak (Vpk) of the SawTooth wave t=[0:1/(100*f0):5/f0]; % time resolution of simulation Wave=(2*A/pi)*sin(2*pi*f0*t); % fundamental wave function plot(t,Wave); hold % fundamental wave + 2nd harmonic+ 3rd harmonic Wave_to_3rd_harm=(2*A/pi)*sin(2*pi*f0*t)(2*A/(pi*2))*sin(2*pi*2*f0*t)+(2*A/(pi*3))*sin(2*pi*3*f0*t); plot(t,Wave_to_3rd_harm,'r'); % fundamental wave + 2nd harmonic+ 3rd harmonic + 4th harmonic + 5th harmonic Wave_to_5th_harm=(2*A/pi)*sin(2*pi*f0*t)(2*A/(pi*2))*sin(2*pi*2*f0*t)+(2*A/(pi*3))*sin(2*pi*3*f0*t)(2*A/(pi*4))*sin(2*pi*4*f0*t)+(2*A/(pi*5))*sin(2*pi*5*f0*t); plot(t,Wave_to_5th_harm,'g'); xlabel('Time (s)'); ylabel('Amplitude (V)'); Title('Sinusoidal components of 30Hz, 5000Vppk Alternative Waveform'); legend('Fundamental Freq.','Fundamental to 3rd Harmonic','Fundamental o 5th Harmonic'); axis([0 .18 -3000 4000]); Problem 5 Part 3 (20 points) Appreciate the importance of the phase shifts in the summation of harmonic components. The same sinusoidal components that comprise the saw‐tooth wave above can also sum up to an alternative waveform if you neglect the phase shifts of even‐numbered harmonics: Make a computer‐generated plot of the alternative waveform according to equation 7 of the file Saw­tooth harmonics. Your plot should show 3 curves: (1) The fundamental only (2) The sum of the fundamental and harmonics up to the 3rd (3) The sum of the fundamental and harmonics up to the 5th. © 2011 Alexander Ganago Last printed 2/6/2011 11:27:00 AM Page 4 of 6 File: 2011 W 314 HW 04 p5.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above SOLUTION: Sinusoidal components of 30Hz, 5000Vppk Alternative Waveform 4000 Fundamental Freq. Fundamental to 3rd Harmonic Fundamental to 5th Harmonic 3000 2000 Amplitude (V) 1000 0 -1000 -2000 -3000 0 0.02 0.04 0.06 0.08 0.1 Time (s) 0.12 0.14 0.16 0.18 The codes are as; clear all; close all; f0=30; % the fundamental frequency of the SawTooth wave A=5000/2; % Amplitude or the peak (Vpk) of the SawTooth wave t=[0:1/(100*f0):5/f0]; % time resolution of simulation Wave=(2*A/pi)*sin(2*pi*f0*t); % fundamental wave function plot(t,Wave); hold © 2011 Alexander Ganago Last printed 2/6/2011 11:27:00 AM Page 5 of 6 File: 2011 W 314 HW 04 p5.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above % fundamental wave + 2nd harmonic+ 3rd harmonic Wave_to_3rd_harm=(2*A/pi)*sin(2*pi*f0*t)+(2*A/(pi*2))*sin(2*pi*2*f0*t)+ (2*A/(pi*3))*sin(2*pi*3*f0*t); plot(t,Wave_to_3rd_harm,'r'); % fundamental wave + 2nd harmonic+ 3rd harmonic + 4th harmonic + 5th harmonic Wave_to_5th_harm=(2*A/pi)*sin(2*pi*f0*t)+(2*A/(pi*2))*sin(2*pi*2*f0*t)+ (2*A/(pi*3))*sin(2*pi*3*f0*t)+(2*A/(pi*4))*sin(2*pi*4*f0*t)+(2*A/(pi*5) )*sin(2*pi*5*f0*t); plot(t,Wave_to_5th_harm,'g'); xlabel('Time (s)'); ylabel('Amplitude (V)'); Title('Sinusoidal components of 30Hz, 5000Vppk Alternative Waveform'); legend('Fundamental Freq.','Fundamental to 3rd Harmonic','Fundamental to 5th Harmonic'); axis([0 .18 -3000 4000]); Notice that the shapes of waveforms on your plots in Parts 2 and 3 of this problem are distinct (mirror images of each other). If you use MATLAB, attach the printouts of your computer code and plot for each part of this problem. If you use EXCEL or another software package that does not generate the code, write by hand the formulas for each of the sums (with numerical values for the amplitudes and the frequencies). On each page with your work, including printouts of the plots, include the following information: Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking this information clearly written. © 2011 Alexander Ganago Last printed 2/6/2011 11:27:00 AM Page 6 of 6 File: 2011 W 314 HW 04 p5.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 6 (30 points) Equivalent Resistance Determine the equivalent resistance between terminals C and D in each circuit. In Circuits 2, 3, and 4, wires (thick curved lines) have zero resistance. Hint: Redrawing the circuit may be among the key tools for successful solution. Part 1 (20 points) For each circuit, write the equivalent resistance RCD in the simplified algebraic form, such as given in the following example RAB = [R1 || (R5+R4+R3)]+R2 for Circuit 1. Write your answers below. Show your work below and/or on additional pages. Circuit 1 RCD = R4//(R5+R1+R3) SOLUTION: R2 is open, it has no connection to anywhere in its other side, so no current passes through it Rcd= R4//(R5+R1+R3) © 2010 Alexander Ganago Last printed 2/6/2011 11:28:00 AM Page 1 of 5 File: 2011 W 314 HW 04 p6.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Circuit 2 NEW_NODE RCD =( R5+R3+ (R1//R2) ) // (R4) SOLUTION: Here point A and point B are same points, so one can simplify the circuit as R R1 D R R4 R R5 AB R R2 C R R3 NEW_NODE So Rcd= ( R5+R3+ (R1//R2) ) // (R4) © 2010 Alexander Ganago Last printed 2/6/2011 11:28:00 AM Page 2 of 5 File: 2011 W 314 HW 04 p6.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 6 Part 1 Continued Circuit 3 NEW NODE RCD = SOLUTION: Here is R5 shorted out by the wire, so point A and point D are the same nodes and R2 is open This circuit can be simplified as D R R4 R R1 NEW_NODE R R3 C Rcd=R4//(R1+R3) © 2010 Alexander Ganago Last printed 2/6/2011 11:28:00 AM Page 3 of 5 File: 2011 W 314 HW 04 p6.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Circuit 4 NEW_NODE RCD = R4//(R1+(R2//R3)) SOLUTION: Here R5 is shorted by making A and D the same point. B and C is same point and the circuit can be drawn as following; D R R4 R R1 R R2 NEW_NODE R R3 CB Rcd= R4//(R1+(R2//R3)) © 2010 Alexander Ganago Last printed 2/6/2011 11:28:00 AM Page 4 of 5 File: 2011 W 314 HW 04 p6.doc EECS 314 Winter 2011 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Part 2 (10 points) Assume the resistance values: R1= 10 Ω, R2 = 20 Ω, R3 = 30 Ω, R4 = 40 Ω, R5 = 50 Ω, and calculate RCD in circuit 4. Write your answer: RCD = _______________ Ω. Show your work. SOLUTION: Rcd= R4//(R1+(R2//R3)) =R4// (R1+ (R2*R3/(R2+R3))) Rcd= { R4* (R1+ (R2*R3/(R2+R3))) }/ { R4+ (R1+ (R2*R3/(R2+R3))) } Rcd= 14.1935 © 2010 Alexander Ganago Last printed 2/6/2011 11:28:00 AM Page 5 of 5 File: 2011 W 314 HW 04 p6.doc ...
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This note was uploaded on 03/20/2011 for the course EECS 314 taught by Professor Ganago during the Winter '07 term at University of Michigan.

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