MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 PROBLEM SET 6 SOLUTIONS Issued: October 21. 2003 Due: October 29 2003 Exercise for home study O&W6.49 Solution: (a)To determine the time constants of the differential equation P649-1 in O&w,we perform a Fourier transform of the equation. By linearity, the Fourier transform of the quation is the Fourier transform of each of the individual equation terms. This gives The time constants are the zeros of the denominator(1=-1 and c2=-10) (b)Equation 1 can be rewritten as 9 To make this a parallel interconnection of two first order systems, we do a partial fraction expansion on Equation 2 to find u+1 By linearity, the inverse Fourier transform, h(t), is a parallel interconnection of two first order systems. h(t) is the sum of the inverse Fourier transform of each term in Equation 3. Each term can be quickly determined using Table 4.2 of O&W. Thus h(t)=eu(t)-eu(t (c)The dominant time constant in a system with multiple decaying exponentials is the time constant that takes the longest to die out. For this problem, the dominant time constant is C1=-1. We can approximate this to Equation 1 as shown: HGu) (ju)2+110ju)+10
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 6 Solutions Issued: October 21, 2003 Due: October 29, 2003 Exercise for home study: O&W 6.49 Solution: (a) To determine the time constants of the differential equation P6.49-1 in O&W, we perform a Fourier transform of the equation. By linearity, the Fourier transform of the equation is the Fourier transform of each of the individual equation terms. This gives Y (j�) X(j�) = 9 (j�)2 + 11(j�) + 10 = H(j�). (1) The time constants are the zeros of the denominator (c1 = −1 and c2 = −10). (b) Equation 1 can be rewritten as 9 H(j�) = . (2) (j� + 1)(j� + 10) To make this a parallel interconnection of two first order systems, we do a partial fraction expansion on Equation 2 to find 1 H(j�) = + −1 . (3) j� + 1 j� + 10 By linearity, the inverse Fourier transform, h(t), is a parallel interconnection of two first order systems. h(t) is the sum of the inverse Fourier transform of each term in Equation 3. Each term can be quickly determined using Table 4.2 of O&W. Thus, u(t) − e−10t h(t) = e−t u(t). (4) (c) The dominant time constant in a system with multiple decaying exponentials is the time constant that takes the longest to die out. For this problem, the dominant time constant is c1 = −1. We can approximate this to Equation 1 as shown: 9 1 H(j�) = . (5) (j�)2 + 11(j�) + 10 � j� + 1 1
This approximation has a maximum difference of 10% at w=0 and the difference (d)We want to approximate the faster component of Equation 4 as an impulse with a height equal to its final value. We will then check to see how this approximation affects the overall h(t)of the entire second order system The faster component, h(t)=-eotu(t). Let h,(t)denote the approximation. For h,(t)=s(oo)8(t), we need to determine sf(oo). The step response is defined as s(t)=_o h(t)dt Thus SfIO) h, (t)dt= h(t)=-oo(t) and the approximation to the overall system is h(t)=h(t)+hs(t)=-i(t)+e-u(t Equation 6 enables us to determine the frequency response. We determine B(元u)=-01+-1 0.1ju+0.9 jw+l Jw+1 We then do algebraic manipulations and an inverse Fourier transform to Equation 7 to get the differential equation of the approximation. This gives +y()=0.9(t)-0 The original frequency response, H(u) and the approximate frequency response H jw) are shown below
� This approximation has a maximum difference of 10% at � = 0 and the difference decreases as � increases. (d) We want to approximate the faster component of Equation 4 as an impulse with a height equal to its final value. We will then check to see how this approximation affects the overall h(t) of the entire second order system. The faster component, hf (t)= −e−10t u(t). Let hˆf (t) denote the approximation. For hˆf (t) = sf (∗)�(t), we need to determine sf (∗). The step response is defined as ⎩ t s(t) = h(t)dt. −� Thus, � � � 1 −e−10t sf (∗) = hf (t)dt = dt = − . −� 0 10 hˆf (t) = − 1 �(t) and the approximation to the overall system is 10 hˆ(t) = hˆf (t) + hs(t) = − 1 �(t) + e−t u(t). (6) 10 Equation 6 enables us to determine the frequency response. We determine: 1 Hˆ (j�) = −0.1 + = −0.1j� + 0.9 . (7) j� + 1 j� + 1 We then do algebraic manipulations and an inverse Fourier transform to Equation 7 to get the differential equation of the approximation. This gives, dy dx + y(t) = 0.9x(t) − 0.1 . dt dt The original frequency response, H(j�) and the approximate frequency responseHˆ (j�) are shown below: 2
Exercise: Problem 6.49-IH(joo) Exercise: Problem 6.49- This is the approximate to H(jo) requency(o) requency(oo) The two are shown on the same plot below. The plot of H(w) is solid and the plot of HGw)
Exercise: Problem 6.49− |H(j�)| Exercise: Problem 6.49− This is the approximate to |H(j�)| 1 1 0.9 0.9 0.8 0.8 0.7 0.7 0.6 0.6 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0 0 20 40 60 80 frequency (�) 0 100 0 20 40 60 80 100 frequency (�) (j�)| |H |H(j approx �)| 0.5 0.5 The two are shown on the same plot below. The plot of H(j�) is solid and the plot of Hˆ (j�) is dashed. 3
Exercise: Problem 6.49-Both (Hjo)l and H( jol Magnification of (Hjo)) and h jol at low frequencies Hgo) Hol 09 08 0.7 0.6 3工E3 04 04 03 0.5 frequency (o) From the graphs above, it is visible that for w<1, Hw) approximates hw) well As w increases beyond 1, the two diverge from each other until roughly w= 20. This is due to the fact that the approximation to the fast part of the signal reaches its final value instantaneously, hence it must include all fast frequency components We can also compare the step response of the original signal, s(t) with the step response gnal, s(t)
Exercise: Problem 6.49− Both |H(j�)| and |Happrox(j�| Magnification of |H(j�)| and |Happrox(j�| at low frequencies 1 1 |H(j�)| |Happr (j�)| ox |H( |Ha j�)| pprox(j�)| 0.9 0.9 0.8 0.8 0.7 |H(j �)| and |Happrox(j�)| |H(j �)| and |Happrox(j�)| 0.6 0.5 0.4 0.3 0.7 0.6 0.5 0.2 0.4 0.1 0 0 10 20 30 40 50 frequency (�) 0.3 0 0.5 1 1.5 2 frequency (�) From ˆ the graphs above, it is visible that for � < 1, H(j�) approximates H(j�) well. As � increases beyond 1, the two diverge from each other until roughly � = 20. This is due to the fact that the approximation to the fast part of the signal reaches its final value instantaneously, hence it must include all fast frequency components. We can also compare the step response of the original signal, s(t) with the step response of the approximate signal, sˆ(t). 4
h(t)dt dt (0.9-e-2+0.le-l0)u(t) (10) s(t) (t)dt dt-/(-0.1)6(t)dt (12) (0.9-e-2)a() The two step responses are plotted together below
� t s(t) = h(t)dt (8) �−� t � t = e−t dt − e−10t dt (9) o 0 = (0.9 − e−t + 0.1e−10t )u(t). (10) � t ˆ sˆ(t) = h(t)dt (11) �−� t � t = e−t dt − (−0.1)�(t)dt (12) o 0 = (0.9 − e−t )u(t). (13) The two step responses are plotted together below: 5