GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL and COMPUTER ENGINEERING ECE 3084 3084 Sum Su mmer 2014 2014 Problem Set #6
Assigned: 4-July-14 Due Date: 11-July-14
Your homework is due at 3:00 at 3:00 PM on Friday, on Friday, July 11, 11, to Prof. Lanterman in his Van Leer 431 office. If he is not there, you may simply slide your homework under his office door. Refrain from looking at backfiles of homework and exam solutions – i.e., “word” in Georgia Tech parlance – from previous versions of ECE2025, ECE2026, or ECE3084, beyond your own materials assembled while taking those classes and any old material we explicitly provide to you. Try posting questions on piazza. on piazza. PROBLEM 6.1:
Consider a lowpass filter specified by the system function H (s) = where ωn = 2π
ωn2 , s2 + 2ζ 2ζ ωn s + ω + ωn2
(1)
× 10000 radians per second, but for the moment, we’re not telling you what ζ is.
(a) The unit step response of this lowpass lowpass filter is given given by y (t) = 1
)sin(ωd t + φ + φ))u(t), − ωωnd exp(−ζ ωnt)sin(ω
where φ = arctan and ω and ω d = ω n
− 1
− ζ 2
1
ζ
formula for ζ in and ω n . ζ 2 is the damped frequency . Find a formula for ζ in terms of ω ω d and ω
(b) Figure Figure 1 shows shows the step response of this lowpass filter (notice (notice that the horizontal horizontal scale is in milliseconds, not seconds.) Estimate ωd by measuring the time between the first and third crossing of the y = 1 line, which correspond correspond to zero crossings crossings of the underlying underlying sinusoid. sinusoid. To 1 help with this, Figure 2 shows a zoomed-in version of the plot. Then use your answer from part (a) to estimate ζ .
1
Notice Notice that a peak-to-peak peak-to-peak measuremen measurement, t, while useful, useful, would would be an approximat approximation; ion; because because of the decay, decay, the distance between the peaks changes slightly from “period” to “period.”
Figure 1: Step response for Problem 1.
Figure 2: Zoomed-in step response for Problem 1.
PROBLEM 6.2:
It is quite useful to estimate the transfer function of a system from its frequency response. Here we consider doing that for a second-order low pass filter that has a resonant peak. This is done by first measuring the magnitude and location of the peak (A pk and ωr ) as well as A0 , the magnitude at D.C. (ω = 0). Since we don’t know the overall gain of each system, the unknown transfer function is of the form C H (s) = 2 . s + 2ζωn s + ωn2 (a) Using the observation in the sentence just after Equation 12.8 as well as Equations 12.13 and 12.18 in the typeset class notes, derive equations for C , ζ and ω n in terms of A 0 , A pk , and ωr . (b) Estimate H (s) for the frequency response shown below on the left using your result from part (a) and measurements of A 0 , A pk , and ω r . (c) How would this procedure be different for a high pass filter? PROBLEM 6.3:
Since a bandpass filter always has a peak at ωn and its magnitude is zero at both ω = 0 and ω a different method must be used to estimate the transfer function, which is of the form, H (s) =
→ ∞,
Cs . s2 + 2ζωn s + ωn2
The location of the peak can be used to estimate ωn , but not ζ . It must be estimated from the bandwidth of the frequency response. Referring to Section 12.3 of Chapter 12 of the typeset class notes, we define bandwidth as the width of the peak at the half-power points; that is, BW = ω U ωL where H ( jω U ) = H ( jω L ) = 1/ 2. The text of Chapter 12 indicates that ω n /BW = Q, which is the same as BW = 2ζωn .
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−
√
(a) Derive an equation for C in terms of ω n , ζ , and the measured peak amplitude. (b) Now estimate H (s) for the frequency response shown below on the right using measurements of ωn , B W and A pk . Bandpass Filter − Problem 6.3(c)
Lowpass Filter − Problem 6.2(b) 4.5
6
4
5 3.5 3
4
| ) 2.5 ω j ( H 2 |
| ) ω j ( 3
1.5
2
H |
1
1 0.5 0 0
500
1000 rad/s
1500
0 0
0.5
1
1.5 rad/s
2
2.5
3
PROBLEM 6.4:
To prepare for this problem, review Section 13.2.1 of the typeset class notes. A third-order, lowpass Butterworth filter with cutoff ωc has three s-plane poles, consisting of one conjugate pair at ωc exp( j2π/3) and a single real pole at ωc . Suppose ω c = 2πf c , where f c = 3 kHz. The transfer function of the continuous-time filter is ωc3 H (s) = . (s p1 )(s p2 )(s p3 )
±
−
−
−
−
Perform every part of this problem using MATLAB, Mathematica, Maple, or a similar tool; there’s really nothing that you want to do “by hand.” Turn in your well-commented code in addition to the requested graphs and answers. (a) Plot the magnitude of the frequency response of this continuous time filter, H ( jω) = H ( j2πf ) , for 0 f 20 kHz. Verify that H ( jω c ) is approximately 0.707.
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(b) Using the “pole-zero mapping” method discussed in class, namely z p = exp(s p T s ), where T s is the sampling period, find the three z-plane poles of the discrete-time filter that approximates this Butterworth filter for a sample rate of f s = 40 kHz. (c) We will define the transfer function of the discrete-time pole-zero mapped system as H pzm (z) = =
(1 (z
− z p −
1
z −1 )(1
−
C z p z −1 )(1 2
Cz 3 z p )(z z p )(z 1
−
2
1
− z p z− ) 3
− z p ) . 3
The first version is the most convenient for multiplying out and turning into a difference equation to actually implement the filter. The second form is probably the most convenient for plotting frequency responses, as we’re doing in this assignment. The constant C is a factor that usually gets stuck there “in practice” to force the frequency response of the continuous-time system being approximated and the discrete-time implementation to match up at some desired frequency. You can take ECE4270 to learn all sort of “principled” ways of choosing C . Or, you can do what people really do in practice, which is to hack it. Find the C that makes H pzm (e j 0 ) = H pzm (1) equal 1; i.e., we’ll set up our discrete-time approximation to have unity gain at “D.C.,” just as the original continuous-time system does. (d) Plot H pzm (e j ωˆ ) , the magnitude of the discrete-time frequency response of the filter found in parts (b) and (c), for π ω ˆ π.
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(e) Plot the magnitude of the frequency response of the original continuous-time filter, H ( j2πf ) for f ranging from 0 to 20 kHz. On the same graph, also plot H eff ( j2πf ) = H pzm (e j 2πf/f ) , which is the magnitude of the frequency response of the discrete-time approximation consisting of a cascade of a continuous-to-discrete converter, the IIR (infinite impulse response) filter found in parts (b) and (c), and a discrete-to-continuous converter. (Remember that because of aliasing, the expression for H eff ( j2πf ) is not meaningful for f larger than half the sample rate of f s = 40 kHz.) Comment on how well the two systems match.
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(f) Repeat part (e), except change the cutoff frequency to f c = 12 kHz. (If you set up your code well, you should be able to do this by just changing one line. You need not repeat the code; you can just show the graph). Comment on how well the two systems match now. What do you conclude about the quality of the match as f c approaches half the sampling frequency?
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