ME 375. HW 7 Solutions. Original Homework Assigned 10/12, Due 10/19.

Transcription

1 ME 375. HW 7 Solutions. Original Homework Assigned /2, Due /9. Problem. Palm 8.2 a-b Part (a). T (s) = 5 6s+2 = 5 2 3s+. Here τ = 3 and the multiplicative factor 5/2 shifts the magnitude curve up by 2log5/2 = 7.52dB. The magnitude plot looks like that in Figure 8..6, but shifted up by 7.5 db. Because < (5/2) = degrees, the phase curve is identical to that shown in Figure By hand, one should get plots like this for each part: For the whole transfer function response, by hand and in Matlab (sys82a=tf([5],[6 2]);bode(sys82a)), one gets the following:

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3 9s 8s+4 = 2.25 s 2s+. Part (b). T (s) = Here τ = 2 and the multiplicative factor 9/2 shifts the m curve up by 2log9/2 = 3.64 db. The m plot looks like that in Figure 8..9, but shifted up by 3.64 db. Because < (9/2) = degrees the phase curve is identical to that shown in Figure By hand, one should get plots like this for each part: For the whole transfer function response, by hand and in Matlab (sys82b=tf([9 ],[8 4]);bode(sys82b)), one gets the following:

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5 Problem 2. Part (a). First draw the approximate Bode plot of the following transfer function by hand (using asymptotic approximations): T (s) = 6(4s+7) (s+2) = 2 2s+ 5s+. Here τ = 2 and τ 2 = 5. The multiplicative factor 7/2 shifts the m curve up by 2log7/2 =.88 db. The magnitude plot looks like that in part (a) of Figure 8..2 (because τ < τ 2 ), but shifted up by.88 db. Note that Figure 8..2 applies to the case where K = in (8..3). Because < (7/2) = degrees, the phase curve is given by φ(ω) =< ( + 2ωj) < ( + 5ωj). At low frequencies, φ degrees but negative. At ω = /τ 2 = /5 and at ω = /τ = /2, φ = 23 degrees (this identical value is a coincidence because here τ = /τ 2 ). As ω inf, φ degrees through negative values. By hand, one should get plots like this for each part: For the whole transfer function response by hand should look like:

6 Part (b). Compute the Bode frequency response plot by numerically calculating and plotting the magnitude and phase response of the above transfer function. (Hint: try typing help bode and help tf at the Matlab prompt). In Matlab: sysp2b=tf([84 42],[ 2]);bode(sysp2b). One gets the following: Part (c). The transfer function from part (a) is modified, with the addition of a pole. Plot the Bode diagram by hand using the approximation method. What affect does this pole have on the previous curve? T (s) = 6(4s+7) (s+)(s+2) = 2 2s+ 5s+ s/+. Added to part (a) is a pole. This adds a breakpoint frequency at rad/s. It also changes the constant to 7/2. Other than a constant shift in the magnitude plot across all frequencies, this has no other effect below the breakpoint frequency of. Above that frequency, the magnitude plot from part (a) is changed such that the magnitude decreases by an additional 2 db/decade. The phase angle decreases by an additional 9 degrees above ω =. By hand, one should get plots like this for each part:

7 For the whole transfer function response by hand should look like: Part (d). Numerically compute the Bode frequency response plot of part (c) (Hint: try typing help bode and help tf at the Matlab prompt). In Matlab: sysp2d=tf([84 42],[ 2 2]);bode(sysp2d). One gets the following:

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11 Problem 3. Part (a). First draw the approximate Bode plot of the following transfer function by hand (using asymptotic approximations): T (s) = 8 s(s 2 +s+) = 8 s (s 2 +s+). The second order complex poles term has a breakpoint frequency at the natural frequency ω n =, adding an additional decrease of 4 db/decade, and 8 degrees. The integrator term decreases at 2 db/decade. By hand, one should get plots like this for each part: For the whole transfer function response by hand should look like: Part (b). Compute the Bode frequency response plot by numerically calculating and plotting the magnitude and phase response of the above transfer function. (Hint: try typing help bode and help tf at the Matlab prompt).

12 In Matlab: sysp3b=tf([8],[ ]);bode(sysp3b). One gets the following:

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