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20082ee102_1_HW_1_solution

20082ee102_1_HW_1_solution - HW1 solution SPRING 2008 EE...

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HW1 solution, SPRING 2008, EE 102 1. (i) ( z - w ) 2 = (1 + i - 3 + 4 i ) 2 = - 21 - 20 i, real:-21,Imag:-20; ( z/w ) = ( - 1 + 7 i ) / 25 , real:-1/25,Imag:7/25; zw = (1 + i )(3 - 4 i ) = 7 - i, real:7,Imag:-1; (ii) Let s=a+bi; Re(i¯ s )=Re(ai+b)=b=Im(s); Im(i¯ s )=Im(ai+b)=a=Re(s); 2. (i) e t ( dy ( t ) dt + y ( t )) = e t ( dx ( t ) dt + 1 x ( t ) - 3 x ( t )) , Z t 0 ( de τ y ( τ ) ) = Z t 0 de τ x ( τ ) - 3 Z t 0 e τ x ( τ ) + k, y ( t ) = x ( t ) - 3 e - t Z t 0 e τ x ( τ ) + e - t k, By initial conditions y(0)=0 and x(0)=0, we obtain k=0. So the answer is y ( t ) = x ( t ) - 3 e - t Z t 0 e τ x ( τ ) dτ. (ii) y ( t ) = ( t - 1) e - t - 3 e - t Z t 0 e τ ( τ - 1) e - τ = e - t ( - 3 2 t 2 + 4 t - 1) . 3. (L or NL?) Because T [ kx ( t )] = Z t -∞ e t + τ kx ( τ ) = k Z t -∞ e t + τ x ( τ ) 1

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= kT [ x ( t )] T [ x 1 ( t ) + x 2 ( t )] = Z t -∞ e t + τ x 1 ( τ ) + Z t -∞ e t + τ x 2 ( τ ) = T [ x 1 ( t )] + T [ x 2 ( t )] It’s linear. (TI or TV?) z ( t ) = T [ x ( t - a )] = Z t -∞ e - t + τ x ( τ - a ) dτ, y ( t - a ) = Z t - a -∞ e - t + a + τ x ( τ ) dτ, Let α = τ - a, by which we do change of variable on z(t). We have z ( t ) = Z t - a -∞ e - t + a + α x ( α ) dα, We obtain z(t)=y(t-a). So it’s TI. 4. (i) y ( t ) = Z -∞ x ( σ ) δ ( t - σ ) - 2 Z -∞ e t - σ x ( σ ) U ( σ - t ) dσ.
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20082ee102_1_HW_1_solution - HW1 solution SPRING 2008 EE...

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