102_1_hw4_sol - 1 EE102 Spring 2009-10 Systems and Signals...

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1 EE102 Spring 2009-10 Lee Systems and Signals Homework #4 Solutions Due: Tuesday, May 4, 2010 at 5 PM. 1. A signal f ( t ) is periodic with period T 0 . You compute the Fourier series assuming the period is T 0 = 2 T 0 and hence ω 0 = ω 0 / 2 . Find the resulting values of the Fourier series coefficients, D n,T 0 , D n,T 0 = 1 T 0 t 0 + T 0 t 0 f ( t ) e - jn ω 0 t dt Express your answer in terms of D n,T 0 , the Fourier coefficients when you assume the cor- rect fundamental period of T 0 . Solution If we write an expression for the Fourier coefficients over a period of T 0 we get D n,T 0 = 1 T 0 t 0 + T 0 t 0 f ( t ) e - jn ω 0 t dt = 1 2 T 0 t 0 +2 T 0 t 0 f ( t ) e - jn ω 0 2 t dt = 1 2 T 0 t 0 +2 T 0 t 0 f ( t ) e - jn ω 0 2 t + f ( t - T 0 ) e - jn ω 0 2 t dt = 1 2 T 0 t 0 + T 0 t 0 f ( t ) e - jn ω 0 2 t + f ( t ) e - jn ω 0 2 ( t - T 0 ) dt = 1 2 T 0 t 0 + T 0 t 0 f ( t ) e - jn ω 0 2 t + e - jn ω 0 2 t e - jn ω 0 2 T 0 dt = 1 2 T 0 t 0 + T 0 t 0 f ( t ) e - jn ω 0 2 t + e - jn ω 0 2 t e - jn 2 π 2 T 0 T 0 dt = 1 2 T 0 t 0 + T 0 t 0 f ( t ) e - jn ω 0 2 t + e - jn ω 0 2 t e - jn π dt The term e - jn π is equal to ± 1 , so the two exponentials either add if n is even, or cancel if n is odd. If n is even, D n,T 0 = 2 2 T 0 t 0 + T 0 t 0 f ( t )( e - jn ω 0 2 t dt = 1 T 0 t 0 + T 0 t 0 f ( t )( e - j n 2 ω 0 t dt = D n 2 ,T 0 . If n is odd, D n , T 0 = 0 .
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2 2. Suppose that f ( t ) is a periodic signal with period T 0 , and that f ( t ) has a Fourier series. If τ is a real number, show that f ( t - τ ) can be expressed as a Fourier series identical to that for f ( t ) except for the multiplication by a complex constant, which you must find. Solution: f ( t - τ ) = n = -∞ D n, τ e jan ω 0 t where D n, τ = t 0 + T 0 t 0 f ( t - τ ) e - j ω 0 nt dt Let t = t - τ , D n, τ = t 0 + τ + T 0 t 0 + τ f ( t ) e - j ω 0 n ( t + τ ) dt = e - jn ω 0 τ t 0 + τ + T 0 t 0 + τ f ( t ) e - j ω 0 nt dt = e - jn ω 0 τ D n where we have used the fact that f ( t ) is periodic in the last step. We get the same Fourier series no matter where we choose the one period to integrate over. The Fourier series is then f ( t - τ ) = n = -∞ D n e - jn ω 0 τ e jan ω 0 t 3. Switching amplifiers are a very efficient way to generate a time-varying output voltage from a fixed supply voltage. They are particularly useful in high-power applications. The basic idea is that an output voltage a is generated by rapidly switching between zero and the supply voltage A . The output is then lowpass filtered to remove the harmonics generated by the switching operation. For our purposes we can consider the lowpass filter as an integrator over many switching cycles, so the output voltage is the average value of the switching waveform.. Varying the switching rate varies the output voltage. In this problem we will only consider the case where the desired output voltage is constant. We can analyze this system with the Fourier series. If the output pulses are spaced by T , the waveform the amplifier generates immediately before the lowpass filter is T 2 T - 2 T - T 0 t ! T A The duty cycle of the switching amplifier is α , and the width of the pulses is α T . When α = 1 , the amplifier is constantly on and produces its maximum output A .
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3 (a) Reducing the duty cycle reduces the output voltage. After the lowpass filter, only the zero-frequency spectral component D 0 remains, and this will be the output voltage.
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