Q2 Fourier series of the output voltage of an ideal full-wave Diode Rectifier The following circuit is an ideal full-wave rectifier It is often used as a first stage to
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8 sept 2012 · This is the mathematical theory of Fourier series which uses the periodic waveforms, such as square waves, triangle waves, rectified sinusoids, and so on simple property of the complex exponential signal—the integral of a The mathematical formula for the full-wave rectified sine signal is just the
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Fourier Series: Half-wave Rectifier • Ex A sinusoidal voltage Esinωt, is passed through a half-wave rectifier that clips the negative portion of the wave Find the
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It is also useful to know the values of the cosine, sine, and exponential functions for Determine the Fourier series for the half-wave rectified cosine function
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4 jui 2010 · What is the conditions of Fourier series expansion for a signal? Answer: Exponential Fourier series DEU, Electrical Half-wave Symmetry; If a signal has an half wave symmetry, only Full-wave rectifier DEU, Electrical
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The period of the sinusoid (inside the absolute value symbols) is T1 = 2π/ω1 The period of the rectified sinusoid is one half of this, or T = T1/2 = π/ω1 Therefore,
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Vm / , so the rms value of a half-wave rectified sine wave is Vrms If periodic voltage and current waveforms represented by the Fourier series case), the exponential term in equation 3 1 becomes very small compared to unity and the diode
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Q2 Fourier series of the output voltage of an ideal full-wave Diode Rectifier The following circuit is an ideal full-wave rectifier It is often used as a first stage to
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1 5 cos(1000 )t Figure (1) 2 (a) Find the exponential Fourier series expansion for the time-shifted version of the halfwave rectified sinusoidal signal: y (t) = x2(t
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Problem 10: Show that the exponential form of a Fourier series representation of a function ( ) f x through a half-wave rectifier, which clips the negative portion
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EE-210.SignalsandSystems
Solutions of homework 3
Spring 2010
Exercise Due Date
Week of 8
thFeb.Problems
Q1 (a) Find the fundamental periodT, the fundamental frequencyω0, and the Fourier seriescoefficientsakof the following periodic signalx(t). Expressx(t) as a Fourier series.0.510-0.5-11-1x(t)t......ANSWERThe fundamental period is T=1, hence!0= . First of all, the average over one?LUMS School of Science & Engineering, Lahore, Pakistan.
1 period is 0, soa0= 0. Fork?= 0, a k=1T? 1 0 x(t)e-jk2πtdt 1 0 (1-2t)e-jk2πtdt -1jk2π[(1-2t)e-jk2πt]10-1jkπ? 1 0 e-jk2πtdt1jk2π+1jk2π
1jkπ
-jkπ Note that the coefficients are purely imaginary and form an odd sequence, which is consistent with our real, odd signal. The Fourier series representation ofx(t) is x(t) =+∞? k=-∞a kejk2π=+∞? k=-∞k?=0-jkπejk2πt (b) Find the coefficients of the real form of the Fourier series ofx(t): x(t) =a0+ 2+∞? k=1[Bkcos(kω0t)-Cksin(kω0t)] ANSWERRecall that theCkcoefficients are the imaginary parts of theak"s. Hence C k=-1kπ, Bk= 0, a0= 0 (c) Find the total average power of the second the third harmonic components (taken together) of the signalx(t).ANSWER
P avg= 2|a2|2+ 2|a3|2= 2????-j2π? ???2 + 2????-j3π? ???2 =12π2+29π21318π2
Q2Fourier series of the output voltage of an ideal full-wave Diode Rectifier The following circuit is an ideal full-wave rectifier. It is often used as a first stage to generate a constant voltage from 60Hz sinusoidal line voltage for all kinds of electronic devices The input and output voltages are related through the memoryless, time-invariant, non- linear systemv(t) =|vin(t)|. The input voltage is as given in Q1:vin(t) =x(t) 2 (a) Sketch the rectified voltage signalv(t). Find the fundamental periodTand its fun- damental frequencyω0.ANSWERFundamental period:T= 1, fundamental frequencyω0= 2π00.511.51-0.51-1.5v(t)t......(b) Compute and sketch the Fourier series of coefficients ofv(t)
ANSWERFirst of all, the average over one period is 0.5, soa0=12. Fork?= 0, a k=1T? T2T2v(t)e-jkω0tdt
012(2t+ 1)e-jk2πtdt+?
120(1-2t)e-jk2πtdt
=...=1-(-1)k(kπ)2 Thus the Fourier series expension of the full-wave rectified voltage is: v(t) = 0.5 ++∞? k=-∞k?=01-(-1)k(kπ)2e-jk2πt (c) What is the total average power of the output voltagev(t) 3 ANSWERThe total average power of the output voltage is P out=1T? T2T2|v(t)|2dt=?
012(2t+ 1)2dt+?
120(1-2t)2dt
= 2 120(1-2t)2dt(by symmetry)
2-6[(1-2t)3]120=13
(d) Expressv(t) as its real Fourier series of the form v(t) =a0+ 2+∞? k=1[Bkcos(kω0t)-Cksin(kω0t)] ANSWERNote that the Fourier series coefficients ofv(t) are real and even. Hecne v(t) =a0++∞? k=1(akejkω0t+a-ke-jkω0t) =a0++∞? k=1a k(ejkω0t+e-jkω0t) =a0+ 2+∞? k=1a kcos(kω0t) and we can identify the coefficientsBk=ak,Ck= 0. Note thatv(t) is an even function, hence it must be an infinite sum of even functions only (cosines and a constant). This is why theCkcoefficients are 0 (the sine function is odd!!!). Q3Fourier Series of the output voltage of an ideal half-wave diode rectifier The following circuit is an ideal half-wave rectifier. The output voltage is given by the memoryless, nonlinear, time-invariant system: v(t) =?vin(t), vin>0 Suppose that the periodic input voltage isvin(t) =Asin(ω1(t) volts,ω1= 2π/T1. 4 (a) Sketch the half-wave rectified voltage signalv(t). Find its fundamental periodTand its fundamental frequencyω0 ANSWERThe fundamental period (T) isT1and the fundamental frequency (ω0) isω1 (b) Compute the Fourier series coefficients ofv(t) and write the voltage as Fourier seriesANSWER
a k=1T? T 0 v(t)e-jkω0tdt=AT? T20sin(ω0t)e-jkω0tdt
A2jT? T20(ejω0t-e-jω0t)e-jkω0tdt
A2jT?πω0
0(ejω0(1-k)t-e-jω0(1+k)tdt
=...=A2π? (-1)k+ 11-k2? Note that we have to make sure that this expression is finite fork=±1 (L"Hopital"s rule) a1=A2π?
?ddk(e-jkπ+ 1)ddk(1-k2)? k=1=A2π?
-jπe-jkπ-2k? k=1=-jA4 a -1=A2π? -jπe-jkπ-2k? k=-1=jA4 Notice that these are imaginary whereas all the other coefficients are real !! Thus the Fourier series expension of the half-wave rectified sinusoid is v(t) =+∞? k=-∞A2π? (-1)k+ 11-k2? e jkω0tQ4Fourier series of a rectangular impulse train
Consider the rectangular impulse train......nx[n]0123456789101-10-5-4-3-2-1(a) Compute the Fourier series coefficients of the signal. Plot its magnitude and phase.
5ANSWERFourier series coefficients are
X[k] =4?
n=0e -jk(2π/10)n1-e-jk(2π/10)51-e-jk(2π/10)
=e-j(4πk/10)sinπk/2sinπk/10 (b) Plot the Fourier Series of the signal for different number of terms considered in the summation. What is the minimum number of terms for which the Fourier seriesresembles the original signal?01234567891z[n]n(c) What will be the F urier series f the signal x[n]z[n]?
ANSWERThe F urier c ecients f the signal y[n]=x[n]z[n] can be c m uted by Y k=? l=<5>X lZk-l The F urier series c ecients f x[n] and z[n] are equal. Using the results fr m art (a) f the r blem, F urier series f y[n] can be calculated as Y k=? l=<5>X lXk-l l=<5>?? 4? n=0e -jl(2π/10)n?? 4? n=0e -j(k-l)(2π/10)n?? l=<5>?? e -j(4π/10)lsin(πl/2)sin(πl/10)?? e =e-j(4πk/10)? 6quotesdbs_dbs14.pdfusesText_20