C(u) (d)To compute c(t), we multiply b(t) with q(t). BGw)is the sum of two ideal frequency shifted unity-gain filters. Filters in the frequency domain become sinc functions in the time domain. In addition, a frequency shift of wo corresponds to multiplying by e-Jwot in the time domain hence b(t) rt + -/art sin 2t e- art sin2丌t sin 2It cos 4t Therefore c(t)=b(t)q(t)=cos 4t sin- 2rt Problem 3 O&W 4.44. In addition to parts(a) and(b), answer the following (c) Find the differential equation relating the input and output of this system Solution We are given the following equation that relates the output y(t) of a causal LTI system to the input a (t dy(t) dt-+10v(t) r(Tz(t-r)dr-a(t) where z(t)=e-t(t)+36(1) We want to find the frequency response H(u)=Y(w)/X(w)of this system. To do this we take the transform of each term in Equation 5. Because of linearity, to find the Fourier transform of the entire equation, we take the Fourier transform of each term in the equation. For the first term, du(), we use the differentiation property to find that x{0}乙,jYGa)
C(jω) ω −8π −4π 0 4π 8π 1 (d) To compute c(t), we multiply b(t) with q(t). B(jω) is the sum of two ideal frequencyshifted unity-gain filters. Filters in the frequency domain become sinc functions in the time domain. In addition, a frequency shift of ωo corresponds to multiplying by e −jωot in the time domain. Hence, b(t) = 1 2 e −j4πt sin 2πt πt + 1 2 e j4πt sin 2πt πt = cos 4πt sin 2πt πt . Therefore c(t) = b(t)q(t) = cos 4πt sin2 2πt π 2 t 2 . Problem 3 O&W 4.44. In addition to parts (a) and (b), answer the following . (c) Find the differential equation relating the input and output of this system. Solution: (a) We are given the following equation that relates the output y(t) of a causal LTI system to the input x(t). dy(t) dt + 10y(t) = Z ∞ −∞ x(τ )z(t − τ )dτ − x(t) (5) where z(t) = e −tu(t) + 3δ(t). We want to find the frequency response H(jω) = Y (jω)/X(jω) of this system. To do this we take the transform of each term in Equation 5. Because of linearity, to find the Fourier transform of the entire equation, we take the Fourier transform of each term in the equation. For the first term, dy(t) dt , we use the differentiation property to find that F dy(t) dt F←→ jωY (jω). 6
For the second term, we have 10r(w). For the next term, the integral, we recognize this integral as being the convolution of x(t) and a(t). Thus, the convolution property tells us that Fc(t)*2(t))+ X(w)z(ju) Finally, .(t)becomes X(w). The new equation in the frequency domain is juY(ju)+10Y(ju)=X(j)Z(ju)-X(ju) Algebraic manipulations are used to separate out Y(u)on the left side of the equation nd separate out X(u) on the right side of the equation. We then find H(j)Y(0j)_2(u)-1 10+ju We insert the function Z(w). By linearity, F((t)= Fle-tu(t)+F38(t).The Fourier transform for each of the terms can be found ino w's Table 4.2 4+3j 1+u This can be plugged into Equation( 6)to give Y 3+2 H(0u)=x(0)=0+11+0) (b) The impulse response can be found by doing a partial fraction expansion of H(u)as found in Equation(7) HGw) 3+2 17/9 1/9 (10+ju)(1+j) +Jw 10+ The inverse Fourier transform of the last two terms can be determined from table 4.2 of o& w to give a(t) (c)To find the differential equation relating the input to the output we go back to Equa- tion 7 and rewrite it as 3+2 HGw) Xou) (ju)2+llu+ After cross-multiplying we get
For the second term, we have 10Y (jω). For the next term, the integral, we recognize this integral as being the convolution of x(t) and z(t). Thus, the convolution property tells us that F{x(t) ∗ z(t)} F←→ X(jω)Z(jω). Finally, x(t) becomes X(jω). The new equation in the frequency domain is jωY (jω) + 10Y (jω) = X(jω)Z(jω) − X(jω). Algebraic manipulations are used to separate out Y (jω) on the left side of the equation and separate out X(jω) on the right side of the equation. We then find H(jω) = Y (jω) X(jω) = Z(jω) − 1 10 + jω (6) We insert the function Z(jω). By linearity, F{z(t)} = F{e −tu(t)} + F{3δ(t)}. The Fourier transform for each of the terms can be found in O & W’s Table 4.2. Z(jω) = 1 1 + jω + 3. = 4 + 3jω 1 + jω This can be plugged into Equation( 6) to give H(jω) = Y (jω) X(jω) = 3 + 2jω (10 + jω)(1 + jω) (7) (b) The impulse response can be found by doing a partial fraction expansion of H(jω) as found in Equation( 7). H(jω) = 3 + 2jω (10 + jω)(1 + jω) = A 10 + jω + B 1 + jω = 17/9 10 + jω + 1/9 1 + jω . The inverse Fourier transform of the last two terms can be determined from Table 4.2 of O & W to give: h(t) = 1 9 e −t + 17 9 e −10t u(t). (c) To find the differential equation relating the input to the output we go back to Equation 7 and rewrite it as H(jω) = Y (jω) X(jω) = 3 + 2jω (jω) 2 + 11jω + 10 . After cross-multiplying we get 7