{"id":245588,"date":"2025-07-06T09:18:36","date_gmt":"2025-07-06T09:18:36","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245588"},"modified":"2025-07-06T09:18:38","modified_gmt":"2025-07-06T09:18:38","slug":"consider-the-pulse-function-signal-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/06\/consider-the-pulse-function-signal-below\/","title":{"rendered":"Consider the pulse function signal below"},"content":{"rendered":"\n
Problem (8) fft, fftshift, abs functions in MATLAB Consider the pulse function signal below: x(t) = stp_fn(t+1) – stp_fn(t-1) Generate a time variable from -4 to 4 [sec] in 0.001 [sec] step size. Draw x(t) with respect to time using plot. Convert the signal into the “Frequency Domain” signal (amplitude only): – Use fft to convert x(t) into the frequency domain (find syntax using doc fft). – Use fftshift to correct the frequency axis. X(f) (you can also represent this in angular freq. X(@)) – Use abs to find the absolute value (amplitude) of the signals in frequency (find syntax using doc abs). X() (or in angular freq. |X()|) – Plot the amplitude with respect to frequency. |X(f)| vs f (2-D graph)<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
To solve this problem, we will break it into steps and explain each part.<\/p>\n\n\n\n
Problem Breakdown:<\/h3>\n\n\n\n
You are given the signal:x(t)=stp_fn(t+1)\u2212stp_fn(t\u22121)x(t) = \\text{stp\\_fn}(t+1) – \\text{stp\\_fn}(t-1)x(t)=stp_fn(t+1)\u2212stp_fn(t\u22121)<\/p>\n\n\n\n
where stp_fn\\text{stp\\_fn}stp_fn is a step function. A step function is typically represented by:stp_fn(t)={0,t<01,t\u22650\\text{stp\\_fn}(t) = \\begin{cases} 0, & t < 0 \\\\ 1, & t \\geq 0 \\end{cases}stp_fn(t)={0,1,\u200bt<0t\u22650\u200b<\/p>\n\n\n\n
So, in this case:x(t)=step(t+1)\u2212step(t\u22121)x(t) = \\text{step}(t+1) – \\text{step}(t-1)x(t)=step(t+1)\u2212step(t\u22121)<\/p>\n\n\n\n
This can be interpreted as a rectangular pulse that starts at t=\u22121t = -1t=\u22121 and ends at t=1t = 1t=1.<\/p>\n\n\n\n
Step 1: Generating the Time Vector<\/h3>\n\n\n\n
We are asked to generate the time vector from -4 to 4 seconds with a step size of 0.001 seconds.<\/p>\n\n\n\n
matlabCopyEditt = -4:0.001:4; % Time vector from -4 to 4 with step size of 0.001 seconds\n<\/code><\/pre>\n\n\n\n
Step 2: Defining the Signal x(t)x(t)x(t)<\/h3>\n\n\n\n
Now, we can define x(t)x(t)x(t) based on the step function. The step_fn<\/code> function can be created using the heaviside<\/code> function in MATLAB.<\/p>\n\n\n\n
Step 3: Plotting the Time Domain Signal<\/h3>\n\n\n\n
We plot the time-domain signal x(t)x(t)x(t) with respect to time.<\/p>\n\n\n\n
matlabCopyEditfigure;\nplot(t, x);\nxlabel('Time (seconds)');\nylabel('Amplitude');\ntitle('Time Domain Signal x(t)');\ngrid on;\n<\/code><\/pre>\n\n\n\n
Step 4: Convert the Signal to the Frequency Domain<\/h3>\n\n\n\n
To convert the time-domain signal into the frequency domain, we use the fft<\/code> function. The fft<\/code> function computes the Fast Fourier Transform of the signal.<\/p>\n\n\n\n
matlabCopyEditX_f = fft(x); % Apply FFT to the signal\n<\/code><\/pre>\n\n\n\n
Step 5: Correct the Frequency Axis Using fftshift<\/code><\/h3>\n\n\n\n
The fft<\/code> function produces a frequency spectrum with negative frequencies on the left and positive frequencies on the right. We use fftshift<\/code> to center the zero frequency in the middle of the plot.<\/p>\n\n\n\n
matlabCopyEditX_f_shifted = fftshift(X_f); % Shift the zero frequency to the center\n<\/code><\/pre>\n\n\n\n
Step 6: Find the Absolute Value of the Frequency Domain Signal<\/h3>\n\n\n\n
To get the amplitude of the frequency-domain signal, we take the absolute value of X(f)X(f)X(f) using the abs<\/code> function.<\/p>\n\n\n\n
matlabCopyEditX_f_abs = abs(X_f_shifted); % Find the amplitude of the signal in the frequency domain\n<\/code><\/pre>\n\n\n\n
Step 7: Generate the Frequency Vector<\/h3>\n\n\n\n
The fft<\/code> function assumes the time signal is periodic, so the frequency vector needs to correspond to the length of the signal. The frequency range is given by:f=\u2212N\/2\u0394t,\u2212N\/2+1\u0394t,…,N\/2\u22121\u0394tf = \\frac{-N\/2}{\\Delta t}, \\frac{-N\/2+1}{\\Delta t}, …, \\frac{N\/2-1}{\\Delta t}f=\u0394t\u2212N\/2\u200b,\u0394t\u2212N\/2+1\u200b,…,\u0394tN\/2\u22121\u200b<\/p>\n\n\n\n
where NNN is the number of points in XfX_fXf\u200b and \u0394t\\Delta t\u0394t is the time step.<\/p>\n\n\n\n
matlabCopyEditfs = 1 \/ 0.001; % Sampling frequency\nn = length(t); % Number of points in the signal\nf = (-n\/2:n\/2-1)*(fs\/n); % Frequency vector\n<\/code><\/pre>\n\n\n\n
Step 8: Plot the Amplitude Spectrum \u2223X(f)\u2223|X(f)|\u2223X(f)\u2223<\/h3>\n\n\n\n
Finally, we plot the magnitude of the Fourier Transform \u2223X(f)\u2223|X(f)|\u2223X(f)\u2223 with respect to frequency fff.<\/p>\n\n\n\n
matlabCopyEdit% Step 1: Generate time vector\nt = -4:0.001:4;\n\n% Step 2: Define the signal x(t)\nx = heaviside(t + 1) - heaviside(t - 1);\n\n% Step 3: Plot time domain signal\nfigure;\nplot(t, x);\nxlabel('Time (seconds)');\nylabel('Amplitude');\ntitle('Time Domain Signal x(t)');\ngrid on;\n\n% Step 4: Apply FFT to convert to frequency domain\nX_f = fft(x);\n\n% Step 5: Correct the frequency axis with fftshift\nX_f_shifted = fftshift(X_f);\n\n% Step 6: Compute the absolute value of X(f)\nX_f_abs = abs(X_f_shifted);\n\n% Step 7: Generate the frequency vector\nfs = 1 \/ 0.001; % Sampling frequency\nn = length(t); % Number of points\nf = (-n\/2:n\/2-1)*(fs\/n); % Frequency vector\n\n% Step 8: Plot the frequency domain amplitude\nfigure;\nplot(f, X_f_abs);\nxlabel('Frequency (Hz)');\nylabel('Amplitude');\ntitle('Frequency Domain Amplitude |X(f)|');\ngrid on;\n<\/code><\/pre>\n\n\n\n
Explanation of the Process<\/h3>\n\n\n\n\n
Time Domain Signal<\/strong>: The signal x(t)=step(t+1)\u2212step(t\u22121)x(t) = \\text{step}(t+1) – \\text{step}(t-1)x(t)=step(t+1)\u2212step(t\u22121) is a rectangular pulse between -1 and 1 seconds.<\/li>\n\n\n\n
FFT<\/strong>: The fft<\/code> function computes the Fourier Transform of the signal, converting it from the time domain to the frequency domain.<\/li>\n\n\n\n
FFT Shift<\/strong>: The fftshift<\/code> function is used to shift the zero frequency to the center of the frequency axis.<\/li>\n\n\n\n
Amplitude Spectrum<\/strong>: The abs<\/code> function is used to compute the magnitude of the frequency components of the signal.<\/li>\n\n\n\n
Plotting<\/strong>: We plot both the time-domain and frequency-domain representations of the signal for visual inspection.<\/li>\n<\/ol>\n\n\n\n
This process shows how to convert a time-domain signal to its frequency-domain representation using FFT in MATLAB.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Problem (8) fft, fftshift, abs functions in MATLAB Consider the pulse function signal below: x(t) = stp_fn(t+1) – stp_fn(t-1) Generate a time variable from -4 to 4 [sec] in 0.001 [sec] step size. Draw x(t) with respect to time using plot. Convert the signal into the “Frequency Domain” signal (amplitude only): – Use fft to […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-245588","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245588","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=245588"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245588\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245588"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245588"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245588"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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