EE 353

Pennsylvania State University - Penn State Main Campus

EE

### Question Description

I’m working on a Engineering exercise and need support.

Problem 28: (15 points) A signal f(t) is said to be periodic if for some positive constant To f(t) = f(t + To), for all t ≥ 0. The smallest nonzero value of To satisfying the last equality is called the fundamental period of f(t). If a signal f(t) is not periodic, it is called aperiodic. Determine whether or not each of the following continuous-time signals is periodic. If the signal is periodic, determine the fundamental period. 1. (5 points) f(t) = sin2 (10πt + 45◦ ) + cos(4πt + 60◦ ) + sin(28πt − 60◦ ) 2. (5 points) f(t) = sin(6t) + cos(3πt) 3. (5 points) f(t) = ao + X∞ n=1 (an cos (n ωot) + bn sin (n ωot)), where a0, a1, a2, . . ., b1, b2 . . . are real-valued constant coefficients, ωo = 2π/To, and To is some positive realvalued constant. Problem 29: (20 points) 1. (5 points) Consider real-valued periodic functions of the form xn(t) = cos(nωot) where ωo = 2π/To and To is the period. Determine the value of < xn, xm > for two cases, n 6= m and n = m 6= 0. For what condition, if any, are the functions xn(t) and xm(t) orthogonal over the interval [−To/2, To/2]? Hint: Trigonometric identities will be helpful for the first three parts! 2. (5 points) Consider real-valued periodic functions of the form xn(t) = sin(nωot) where ωo = 2π/To and To is the period. Determine the value of < xn, xm > for two cases, n 6= m and n = m 6= 0. For what condition, if any, are the functions xn(t) and xm(t) orthogonal over the interval [−To/2, To/2]? 3. (5 points) Over the interval [−To/2, To/2], show that < sin(nωot), cos(mωot) > = 0 for all n and m, where ωo = 2π/To. 4. (5 points) Consider a set of periodic complex-valued signals xn(t) = e nωot n = 0, ±1, ±2, . . . where ωo = 2π/To and To is the period. Show that over the interval [−To/2, To/2] < xn(t), xm(t) >= 0 m 6= n To m = n. Problem 30: (25 points) When determining the trigonometric Fourier series representation of a signal f(t), the analysis can be simplified by exploiting the symmetry properties of f(t). 1. (6 points) Prove mathematically that if f(t) is an even, real-valued periodic function of time with period To, then a0 = 2 To Z To/2 0 f(t)dt an = 4 To Z To/2 0 f(t) cos(nωot)dt bn = 0 and an is an even function of n. 2. (6 points) Prove mathematically that if f(t) is an odd, real-valued periodic function of time with period To, then a0 = 0 an = 0 bn = 4 To Z To/2 0 f(t) sin(nωot)dt and bn is an odd function of n. 3. (8 points) If a periodic signal f(t) with fundamental period To satisfies f t − To 2 = −f(t), the signal f(t) is said to have a half-wave symmetry. In signals with half-wave symmetry, the two-halves of one period of the signal are of identical shape except that one half is the negative of the other. In this case, show that all the even-numbered harmonic coefficients (a0, a2, a4, . . ., b2, b4, . . .) vanish, and that the odd-numbered harmonic coefficients (a1, a3, . . ., b1, b3, . . .) are given by an = 4 To Z To/2 0 f(t) cos(nωot)dt bn = 4 To Z To/2 0 f(t) sin(nωot)dt, where an is an even function of n while bn is an odd function of n. 4. (5 points) Sketch a signal that exhibits half-wave symmetry. Show mathematically that the function is half-wave symmetric. Note that the symmetry properties of a function may be combined. For example, a function f(t) may exhibit both even symmetry and half wave symmetry simultaneously. In this case, bn would still be zero due to the even symmetry while an would still be calculated using the half wave symmetry formula. Also note that symmetry properties don’t invalidate the standard expressions for the coefficients; they simply provide special cases that can greatly simplify the computation of the coefficients. Problem 31: (36 points) For each of the periodic signals in Figure 1 use the weak Dirichlet condition to verify the existence of their Fourier series and the strong Dirichlet condition to verify the series converges. Determine the trigonometric Fourier series for each of the periodic signals. Place your answers in the form f(t) = ao + X∞ n=1 (an cos (n ωot) + bn sin (n ωot)), Whenever possible, use the results from Problem 30 to simplify the calculations. Use your results to also find the compact trigonometric Fourier series and place your answer in the form f(t) = co + X∞ n=1 (cn cos (n ωot + θn)), f (t) t A -A T0 2 -T0 2 -t1 -t2 t 1 t 2 t 2A+B B T0 -T0 g(t) x(t) T0 -T0 A cos 2πt T0 ( ) 0 0 δT (t) t t T0 -T0 -2T0 2T0 0 0 (1) (1) (1) (1) (1) Figure 1: Periodic signals. Problem 32: (34 points) 1. (10 points) A pulse width modulated (PWM) signal fPWM (t) in Figure 2. The symbol D represents a duty cycle, a number between zero and one. Determine the compact trigonometric Fourier series coefficients (C0, Cn, θn) of the signal f(t). 2. (10 points) One use of PWM is to generate variable DC voltages. While the PWM signal is not DC, you should be able to see from your results in part 1 that it has a DC component that depends on the duty cycle, D. If the voltage is applied to a system that is low pass in nature, it will respond to the DC component of the signal while being largely unaffected by the other frequency components. Assume the PWM signal is applied to the low-pass system represented by the frequency response function H(ω) = 1 ωτ + 1 , A T0 -T0 DT0 f PWM(t) Figure 2: A pulse width modulated signal. where the time constant τ . Show that the sinusoidal steady-state output y(t) can be expressed as y(t) = Co + X∞ n=1 Cn p (ωo n τ ) 2 + 1 cos n ωo t + θn − tan−1 (ωo n τ ) . 3. (14 points) Use MATLAB to generate and plot the signal y(t) for three values of duty cycle, 0.25, 0.58 and 0.8 using the following parameters and guidelines: • Set A = 12 V, To = 8 ms, and τ = 16.5 ms. • Use a time vector that spans the interval [−2To, 3To] with 5,001 uniformly spaced points. • Approximate the output of the filter network as y(t) ≈ Co + X 120 n=1 Cn p (ωo n τ ) 2 + 1 cos n ωo t + θn − tan−1 (ωo n τ ) . • Plot all three results for y(t) as a function of time in ms. – For D = 0.2, plot y(t) using a solid red line. – For D = 0.5, plot y(t) using a dashed blue line. – For D = 0.8, plot y(t) using a dotted black line. • Use the MATLAB function legend to distinguish the plots. • Appropriately label the x and y axes and title the graph; no credit is given for MATLAB plots whose axes are unlabeled! • Use the MATLAB command gtext to place your name and section name within the figure. To receive credit: • Turn in the figure along with a copy of your m-file. • Include your name and section number at the top of m-file using the comment symbol %. Problem 33: (20 points) Consider a periodic signal f(t), with fundamental period To, that has the exponential Fourier series representation f(t) = X∞ n=−∞ Dne ωont , where ωo = 2π/To and Dn = 1 To Z To f(t)e −ωontdt. 1. (2 points) When f(t) is a real-valued, show that D−n = D∗ n. This is known as the complex conjugate symmetry property or the Hermitian property of real signals. 2. (2 points) Show that when f(t) is an even function of time that Dn is an even function of n. 3. (2 points) Show that when f(t) is an odd function of time that Dn is an odd function of n. 4. (2 points) Using the results from parts 1 and 2, show that if f(t) is real-valued and an even function of time, then the coefficients Dn are real-valued and an even function of n. 5. (2 points) Using the results from parts 1 and 3, show that if f(t) is real-valued and an odd function of time, then the coefficients Dn are imaginary and an odd function of n. 6. (4 points) Show that the signal f(−t) has the Fourier series coefficients D−n. This is known as the time reversal property. 7. (6 points) Consider three periodic signals whose Fourier series representations are f1(t) = 2 π X∞ n=−∞ 1 9 + 3n2 e 6πnt f2(t) = X 7 n=1 cos(3n)e 3nt 2 f3(t) = X∞ n=−∞ sin(4πn)e 6nt Using the properties of the exponential Fourier series, determine if the signals are (a) (3 points) real or complex valued, (b) (3 points) an even or odd function of time, or neither. Problem 34: (20 points) 1. (8 points) Suppose that f(t) is a periodic signal with exponential Fourier series coefficients Dn. Show mathematically that the power Pf of f(t) is Pf = X∞ n=−∞ |Dn| 2 , This is Parseval’s theorem for the exponential Fourier series. 2. (12 points) If f(t) is real-valued, Parseval’s theorem can be expressed as Pf = D2 o + 2X∞ n=1 |Dn| 2 . (a) (3 points) Find the power of the PWM signal shown in figure 2. Hint: for this part don’t use Parseval’s theorem. (b) (9 points) Utilize your CTFS from your solution for Problem 32 and convert it to the exponential Fourier series. Use Parseval’s theorem for a real-valued signal to determine the power in the harmonics 0, 1, and 2 of the PWM signal (that is, the DC component and the first two harmonics). Plot the fraction of the total power contained in each of these three harmonics as a function of duty cycle for values of D ranging from 0 to 1 using 101 evenly spaced points. To receive credit: • Include your M-file with your solutions. Include your name and section number at the top of M-file using the comment symbol %. • Appropriately label the x and y axes; no credit is given for MATLAB plots whose axes are unlabeled! • Include a legend to distinguish the three curves. Use the default Matlab colors for the curves. Make the line styles solid for the DC term, dashed for the first harmonic and dotted for the second harmonic. • Use the title command to appropriately label the figure. • Use the MATLAB command gtext to place your name and section name within the figure. • Include your figure with your solution. Problem 35: (30 points) 1. (10 points) Consider the periodic signal f(t) = −2 + 3 cos(3t) − 2 √ 3 sin(3t) cos(9t) − 4 cos 9t + 2π 3 . (a) (2 points) Determine the fundamental frequency ωo. (b) (5 points) Determine the complex exponential Fourier series coefficients. (c) (3 points) Sketch the Fourier magnitude spectrum |Dn| versus ω and the Fourier phase spectrum 6 Dn versus ω. 2. (20 points) A full wave rectifier is often used to change residential AC power into DC power. For an ideal full wave rectifier, the output waveform would be given by: f(t) = |170 cos(120πt)|. (a) (10 points) Determine the complex exponential Fourier series representation of the full-wave rectified waveform. (b) (10 points) Show that the power of the rectified signal is Pf = 14450. Use Parseval’s theorem to determine what fraction of the total signal power is provided in the DC component. Problem 36: (20 points) This problem considers the response of a LTI system to a periodic input represented by a complex exponential Fourier series. 1. (8 points) A periodic signal f(t) with period To and complex exponential Fourier series coefficients Df n is passed through a LTI system with frequency response function H(ω). Show that the sinusoidal steady-state response of the system is given by y(t) = X∞ n=−∞ D f nH(nωo)e nωot . 2. (12 points) Apply the result from part 1 to determine the response of a lowpass filter. (a) (4 points) Determine the fundamental frequency and non-zero complex exponential Fourier series coefficients of the periodic signal f(t) = −2 − 5 sin(2πt) + 10 cos(6πt + 2π 3 ) and sketch the Fourier magnitude spectrum |Df n| versus ω and the Fourier phase spectrum 6 Df n versus ω. (b) (2 points) Use Parseval’s theorem for the exponential Fourier series to find the power of the signal f(t). (c) (4 points) The signal f(t) from part (a) is passed through an ideal low-pass filter whose frequency response function is H(ω) = 2e −ω/4 |ω| ≤ 4π 0 |ω| > 4π to produce an output signal y(t). Find an expression for the resulting sinusoidal steady-state response y(t) of the system and express your result as the sum of real-valued sinusoidal term(s) and if necessary, a DC offset. (d) (2 points) Use Parseval’s theorem for the exponential Fourier series to find the power of the signal y(t). Problem 37: (20 points) Consider a linear time-invariant system represented by the ODE y¨(t) + 2 ˙y(t) + 10y(t) = 20f(t). 1. (4 points) Determine the zero-state unit-step response of the system. 2. (2 points) Using your result in part 1, determine the impulse response function representation, h(t), of the system. 3. (2 points) You have seen that the frequency response function of a system is determined from its ODE representation as H(ω) = Y˜ F˜ = P(ω) Q(ω) . Determine H(ω) for the system considered in this problem. 4. (6 points) Determine the Fourier transform of the impulse response function determined in part 2, and express your answer in the standard form H(ω) = bm(ω) m + bm−1(ω) m−1 + · · · + b1(ω) + b0 (ω) n + an−1(ω) n−1 + · · · + a1(ω) + a0 , and verify that it matches the result obtained in part 3.

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## Final Answer

Attached.

Solution:

1.

𝑓(𝑡) =

1 1

− cos(20𝜋𝑡 + 90𝑜 ) + cos(4𝜋𝑡 + 60𝑜 ) + sin(28𝜋 − 60𝑜 )

2 2

𝑓(𝑡) =

1 1

𝜋

𝜋

𝜋

− cos (20𝜋𝑡 + ) + cos (4𝜋𝑡 + ) + sin (28𝜋 − )

2 2

2

3

3

LCM for the time constants we have

𝑇0 = 𝐿𝐶𝑀 (

2𝜋 2𝜋 2𝜋

1 1 1

7 35 5

35 1

,

,

) = 𝐿𝐶𝑀 ( , , ) = ( , , ) =

=

20𝜋 4𝜋 28𝜋

10 2 14

70 70 70

70 2

𝑡ℎ𝑒 𝑠𝑖𝑔𝑛𝑎𝑙 𝑖𝑠 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐

2.

𝑓(𝑡) = sin(6𝑡) + cos (3𝜋𝑡)

2𝜋 2𝜋

𝜋 2

2𝜋

, ) = 𝐿𝐶𝑀 ( , ) =

6 3𝜋

3 3

3

𝑡ℎ𝑒 𝑠𝑖𝑔𝑛𝑎𝑙 𝑖𝑠 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐

The signal is periodic, since it is Fourier series,

2𝜋

𝑇0 =

𝑤𝑜

𝑇0 = 𝐿𝐶𝑀 (

3.

Solution

1.

𝑥𝑛 = cos(𝑛𝑤𝑜 𝑡) 𝑎𝑛𝑑 𝑥𝑚 = cos(𝑚𝑤𝑜 𝑡)

𝑇𝑜

2

∫

𝑇

− 0

2

cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡

Since cosine is even

𝑇𝑜

2

∫

𝑇

− 0

2

𝑇𝑜

2

cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = 2 ∫ cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡

0

For 𝑛 ≠ 𝑚

𝑇𝑜

2

𝑇𝑜

2

2 ∫ cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = ∫ cos((𝑚 + 𝑛)𝑤𝑜 𝑡) + cos((𝑚 − 𝑛)𝑤𝑜 𝑡) 𝑑𝑡

0

0

(𝑚 + 𝑛) 𝑤𝑜 𝑇0

(𝑚 − 𝑛) 𝑤𝑜 𝑇0

) sin (

)

2

2

+

(𝑚 + 𝑛)𝑤𝑜

(𝑚 − 𝑛)𝑤𝑜

sin (

=

For 𝑚 = 𝑛 ≠ 0

𝑇𝑜

2

𝑇𝑜

2

2 ∫ cos(𝑛𝑤𝑜 𝑡) cos(𝑛𝑤𝑜 𝑡) 𝑑𝑡 = 2 ∫

0

cos 2(𝑛𝑤𝑜 𝑡)

𝑇𝑜

2

𝑑𝑡 = ∫ (1 + cos (2𝑛𝑤0 𝑡)𝑑𝑡

0

=

0

𝑇0

1

𝑛𝑤𝑜 𝑇0

+

sin (

)

2 𝑛𝑤0

2

2.

𝑥𝑛 = sin(𝑛𝑤𝑜 𝑡) 𝑎𝑛𝑑 𝑥𝑚 = sin(𝑚𝑤𝑜 𝑡)

𝑇𝑜

2

∫

𝑇

− 0

2

cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡

Since sine is odd

𝑇𝑜

2

∫

𝑇

− 0

2

𝑇𝑜

2

sin(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = 2 ∫ sin(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡

0

For 𝑛 ≠ 𝑚

𝑇𝑜

2

2∫

𝑇

− 0

2

𝑇𝑜

2

sin(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = ∫ cos((𝑚 − 𝑛)𝑤𝑜 𝑡) − cos((𝑚 + 𝑛)𝑤𝑜 𝑡) 𝑑𝑡

0

(𝑚 − 𝑛) 𝑤𝑜 𝑇0

(𝑚 + 𝑛) 𝑤𝑜 𝑇0

) sin (

)

2

2

−

(𝑚 − 𝑛)𝑤𝑜

(𝑚 + 𝑛)𝑤𝑜

sin (

=

For 𝑚 = 𝑛 ≠ 0

𝑇𝑜

2

2∫

𝑇

− 0

2

𝑇𝑜

2

sin(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = 2 ∫

sin2 (𝑛𝑤𝑜 𝑡)

𝑑𝑡 = ∫ (1 − sis (2𝑛𝑤0 𝑡)𝑑𝑡

0

=

𝑇𝑜

2

0

𝑇0

1

𝑛𝑤𝑜 𝑇0

−

sin (

)

2 𝑛𝑤0

2

3.

𝑥𝑛 = cos(𝑛𝑤𝑜 𝑡) 𝑎𝑛𝑑 𝑥𝑚 = sin(𝑚𝑤𝑜 𝑡)

𝑇𝑜

2

∫

𝑇

− 0

2

cos(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡

𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑜𝑑𝑑 𝑎𝑛𝑑 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑔𝑖𝑣𝑒𝑠 𝑜𝑑𝑑 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙

𝑇𝑜

2

∫

𝑇

− 0

2

cos(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = 0

Thus, the two function are orthogonal on the interval −

𝑇0

2

𝑡𝑜

𝑇𝑜

2

4.

𝑥𝑛 (𝑡) = 𝑒 𝑗𝑛𝑤0 𝑡 𝑎𝑛𝑑 𝑥𝑚 (𝑡) = 𝑒 𝑗𝑚𝑤𝑜 ...