ANSWERS TO SELECTED EXCERCISES, CH 4-5

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1 ANSWERS O SELECED EXCERCISES, CH 4-5 Exercise 4.4 a. Peak detection is, in many cases, a simple process it's usually just a matter of identifying regions of your signal that rise above a certain threshold, and then finding the highest point within those regions. he script below is a slight improvement upon that approach. IMPORAN! Make sure the data file ActionPotential.mat is in your directory. % Exercise 4.4 % Peak Detection clear; close all; % parameters threshold = -0.4; % AU % load data and reshape data = load('actionpotential.mat'); data.ap = data.ap.'; % pick out points above threshold gt_thresh_mask = data.ap > threshold; % pick out points preceded by a rising slope rising_mask = diff([0 data.ap]) > 0; % pick out points followed by a falling slope falling_mask = diff([data.ap 0]) <= 0; % combine to pick out peaks peak_mask = gt_thresh_mask & rising_mask & falling_mask; % generate peaks vector peak_times = find(peak_mask); peaks = data.ap(peak_times); % plot figure(1) figure plot(data.ap) hold all scatter(peak_times, peaks) xlabel('ime (samples)') ylabel('membrane voltage (a.u.)') title('intracellular activity, with peak detection') It's worth pointing out, though, that in many measurement systems, peak detection can become a great deal more complicated. For example, it's common in extracellular electrophysiology setups to obtain more than one unit on a single electrode. What would happen if both units were to fire very closely together? A simple peak-detection script such as the one above might only count that as a single event.

2 ANSWERS O SELECED EXCERCISES, CH 4-5 Similarly, when using calcium fluorescence as a proxy for neuronal activity, the timescales of the indicator dyes and the recording apparatus are often much longer than the events being recorded sometimes as much as 100ms blurring closelysynced voltage spikes into a single longer-scale change in calcium fluorescence. In these cases, it is necessary to rely on more complex strategies to determine how many events contributed to the measured signal. b.

3 ANSWERS O SELECED EXCERCISES, CH 4-5 Exercise 4.5 a. he SNR of this measurement is SNR = 10 log 10 [ms signal ms noise ] = 10 log 10 [5 50] = 10dB b. o obtain a higher SNR, we have to either boost the signal or attenuate the noise. Signal averaging attenuates noise by taking advantage of the (presumably) random nature of most of the noise in a well-set-up measurement system. First, we determine by how much the noise variance (or mean-square remember, for a zero-mean system, variance and mean-square are the same thing) must decrease to obtain an SNR of 0dB. 0dB = 10 log 10 [5μV ] So ms noise ms noise = 0.05μV, which is a 1000-fold decrease from an original msnoise of 50μV. We know from equation (4.13) that the variance of the noise will decrease inversely with the number of trials we average over, so to obtain a 1000-fold decrease in the noise variance, we need to average over 1000 trials. c. he +/- average should only contain our noise, so its variance will be 0.05μV d. Script % Exercise 4.5 % Signal Averaging % parameters signal_scaling = sqrt(10); % to obtain variance of 5 noise_scaling = sqrt(50); % to obtain variance of 50 num_points = 1000; num_trials = 1000; interval = linspace(0, *pi, num_points); % 1 period % generate sample signal = signal_scaling * sin(interval); noise = noise_scaling * randn(1, num_points); sample_msrmnt = signal + noise; % plot before averaging

4 ANSWERS O SELECED EXCERCISES, CH 4-5 figure(1) plot(interval, sample_msrmnt, interval, signal) xlim([0 *pi]) set(gca, 'Xick', [0 pi/ pi 3*pi/ *pi]) set(gca, 'XickLabel',{'0', '\pi/', '\pi', '3\pi/', '\pi'}) ylabel('amplitude (a.u.)') title('sample measurement and signal, pre-averaging') % average avg_msrmnt = zeros(1, num_points); plus_minus_avg = zeros(1, num_points); for n = 1:num_trials noise_instance = noise_scaling * randn(1, num_points); msrmnt_instance = noise_instance + signal; avg_msrmnt = avg_msrmnt + msrmnt_instance; plus_minus_avg = plus_minus_avg + msrmnt_instance * (-1)^n; end avg_msrmnt = avg_msrmnt / num_trials; plus_minus_avg = plus_minus_avg / num_trials; % plot after averaging figure() plot(interval, avg_msrmnt, interval, signal) xlim([0 *pi]) set(gca, 'Xick', [0 pi/ pi 3*pi/ *pi]) set(gca, 'XickLabel', {'0', '\pi/', '\pi', '3\pi/', '\pi'}) ylabel('amplitude (a.u.)') title('measurement and signal, post-averaging') e. Our analysis above relied upon the noise source having zero mean. he MALAB rand command samples from a uniform distribution over [0, 1], which has mean 0.5. randn samples from a Gaussian distribution with mean 0 and variance (by default) 1. We could just as well have used rand if we had de-meaned it before adding it to our signal i.e., if we had just shifted all values down by 0.5 (but then our noise would be uniform over [- 0.5, 0.5], instead of Gaussian).

5 ANSWERS O SELECED EXCERCISES, CH 4-5 Exercise 5.1 a. We begin by re-expressing our product of sines and cosines using Equations (A5.1-4). Recall that sines and cosines integrated over a one or several period(s) evaluate to zero, e.g.: cos(x) dx = 0 so when n m, both of the cosine integrals above equal zero. Our first requirement for orthogonality is satisfied. (See also Appendix 5.1) Next, we have to show that the norm-squared of sin(nωt) is nonzero. We start with another trigonometric identity: sin (nωt) dt = 1 cos (nωt) dt As before, the cosine term integrates to zero, and we are left integrating a constant: 1 dt = his satisfies the second condition, so the two functions are orthogonal. b. Normality will be satisfied if the functions' non-zero value equals one instead of /. hus, we must add a multiplicative factor in front of each function (i.e. sqrt(/)) c. We begin the same way re-expressing the product as a sum: sin(nωt) cos(mωt) dt = 1 [ sin((n m)ωt)dt + sin((n + m)ωt)dt] Following our logic in part (a), when n = m, the right-hand sine integral becomes zero. However, this time, so does the left-hand sine integral, as sin(0) = 0. And when n m, both sine integrals are zero, so we've accomplished the first task. In the meantime, we've also shown that the first condition of orthogonality is satisfied. We showed in part (a) that the norm-squared of sin(nωt) is nonzero, so it remains only to show the same for cos(nωt): cos (nωt) dt = 1 + cos (nωt) dt = hus, both functions have nonzero norm-squared, and so the second condition is met. hey are orthogonal.

6 ANSWERS O SELECED EXCERCISES, CH 4-5 Exercise 5. a. For a cosine function (even symmetric), we know that b n =0 n Additionally we have only energy at radial frequency, thus a n =0 n he only nonzero coefficient is a, and its amplitude value is obviously 4. We can also check this with the official equation: a = 4 cos(ωt) cos(ωt) dt = 4 b. By analogous reasoning, we can conclude that the only nonzero coefficient is b, and its value is 6. Also this can be checked by the official equation: = 4 b = 6 sin(ωt) sin(ωt) dt = 6 c. his function is a linear combination of the functions in parts (a) and (b), and so its Fourier series is as well. he only nonzero coefficients are a = 4 and b = 6 = 6

7 ANSWERS O SELECED EXCERCISES, CH 4-5 Exercise 5.4 a. his function has odd symmetry b. We'll take as our period of integration the interval [/4, 5/4]. he function is centered around 0, so a0 = 0. he function is odd, so all other an are also zero. o avoid issues with the discontinuity at 3/4, we break our integral into two parts. Following the same procedure as in the examples in section 5.4, this yields the following for b n : b n = b n = b n = 0 8A (nπ) 8A if n mod 4 = 1 (nπ) if n mod 4 = 3 if n is even

8 ANSWERS O SELECED EXCERCISES, CH 4-5 Exercise 5.6 a. his function is odd. b. his function is even. c. his function is neither. he even component is 5; the odd component is ax.

9 ANSWERS O SELECED EXCERCISES, CH 4-5 Exercise 5.7 a. Script % Exercise 5.7 % Fourier Series, Square Wave % (modified from pr5_1.m) clear; close all; % parameters num_harmonics = 8; harm_num = 1::( * num_harmonics - 1); interval = 0:0.001:1; freq = 10; % Hz % generate harmonics harmonics = zeros(num_harmonics, length(interval)); for n = 1:num_harmonics freq_ratio = harm_num(n); harmonics(n,:) = (4 / pi) *... (1 / freq_ratio) *... sin( * pi * freq * freq_ratio * interval); end total = sum(harmonics, 1); % plot harmonics (offsets are for display purposes) figure; hold; for n = 1:num_harmonics offset = harm_num(n) - 1; plot(interval, (harmonics(n,:) + offset)) end plot(interval, total - 3, 'k'); axis('off'); title('approximation of a Rectangular Wave in the ime Domain') % plot frequency equivalent figure; hold; for n = 1:num_harmonics freq_ratio = harm_num(n); stem(freq_ratio, 4 / (freq_ratio * pi)); end title('frequency Domain Representation') xlabel('frequency') ylabel('amplitude') axis([0 (harm_num(end) + 1) 0 1.5]);

10 b. Plots ANSWERS O SELECED EXCERCISES, CH 4-5 c. he overall fit is better, but the fast oscillations due to the so-called Gibbs phenomenon is more visible.