Consider the square wave shown in the picture below (with a frequency of = 1 rad/s). atos 2π Using Fourier analysis, this square wave can be decomposed into the following set of harmonics: Phase (rad.) 0 -0.2 If this signal is input into an ideal low pass filter with cutoff frequency of 5 rad/s (i.e. a perfect filter with gain = 1 for <5 rad/s and gain=0 for ≥ 5 rad/s) and a phase shift given below, then write the output signal. Explain the reasoning behind your logic. Some potentially useful numbers are provided next to the plot below. -0.4 -0.6 -0.8 _y(t) = 1⁄2 + -1+ -1.2- -1.4 -1.6 102 2 (2k-1) 10-1 = sin((2k – 1)t) = 10⁰ Frequency (rad s¹) 1 2 +=sin(t) + sin(3t) +... 3T 10¹ p(0) = 0 rad. p(0.1) = -0.1 rad. p(1) = -0.79 rad. (3) = -1.25 rad. (5) = -1.37 rad.
Consider the square wave shown in the picture below (with a frequency of = 1 rad/s). atos 2π Using Fourier analysis, this square wave can be decomposed into the following set of harmonics: Phase (rad.) 0 -0.2 If this signal is input into an ideal low pass filter with cutoff frequency of 5 rad/s (i.e. a perfect filter with gain = 1 for <5 rad/s and gain=0 for ≥ 5 rad/s) and a phase shift given below, then write the output signal. Explain the reasoning behind your logic. Some potentially useful numbers are provided next to the plot below. -0.4 -0.6 -0.8 _y(t) = 1⁄2 + -1+ -1.2- -1.4 -1.6 102 2 (2k-1) 10-1 = sin((2k – 1)t) = 10⁰ Frequency (rad s¹) 1 2 +=sin(t) + sin(3t) +... 3T 10¹ p(0) = 0 rad. p(0.1) = -0.1 rad. p(1) = -0.79 rad. (3) = -1.25 rad. (5) = -1.37 rad.
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![**Square Wave and Fourier Analysis**
**Square Wave Representation:**
The image depicts a square wave with a frequency of \(\omega = 1 \, \text{rad/s}\). The wave oscillates between values 1 and -1 at regular intervals, starting from 0, passing through \(\pi\), \(2\pi\), and so on.
**Fourier Series Decomposition:**
Using Fourier analysis, the square wave can be decomposed into an infinite series of harmonics:
\[
y(t) = \frac{1}{2} + \sum_{k=1}^{\infty} \frac{2}{(2k-1)\pi} \sin((2k-1)t) = \frac{1}{2} + \frac{2}{\pi} \sin(t) + \frac{2}{3\pi} \sin(3t) + \ldots
\]
**Low Pass Filter Analysis:**
The signal can be input into an ideal low pass filter with a cutoff frequency of 5 rad/s:
- Gain = 1 for \(\omega < 5\) rad/s
- Gain = 0 for \(\omega \geq 5\) rad/s
Given this filter, only components with frequencies \(\omega < 5\) rad/s will pass through without attenuation.
**Phase Shift Information:**
The phase shift for different frequencies \(\omega\) is given by:
- \(\varphi(0) = 0 \, \text{rad}\)
- \(\varphi(0.1) = -0.1 \, \text{rad}\)
- \(\varphi(1) = -0.79 \, \text{rad}\)
- \(\varphi(3) = -1.25 \, \text{rad}\)
- \(\varphi(5) = -1.37 \, \text{rad}\)
**Graph Description:**
The graph illustrates the phase response (\(\text{Phase (rad)}\)) of the low pass filter against the frequency (\(\text{Frequency (rad s}^{-1}\text{)}\)). The phase decreases with increasing frequency, indicating the phase shift applied by the filter to different harmonic components.
**Output Signal:**
Since the filter allows frequencies below 5 rad/s to pass, the output signal will comprise harmonics with \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F099a9854-7cc3-46b9-a37c-ec0486b4a12a%2F663b99fd-c04a-4020-84bc-37af71dc3299%2Faxd92yz_processed.png&w=3840&q=75)
Transcribed Image Text:**Square Wave and Fourier Analysis**
**Square Wave Representation:**
The image depicts a square wave with a frequency of \(\omega = 1 \, \text{rad/s}\). The wave oscillates between values 1 and -1 at regular intervals, starting from 0, passing through \(\pi\), \(2\pi\), and so on.
**Fourier Series Decomposition:**
Using Fourier analysis, the square wave can be decomposed into an infinite series of harmonics:
\[
y(t) = \frac{1}{2} + \sum_{k=1}^{\infty} \frac{2}{(2k-1)\pi} \sin((2k-1)t) = \frac{1}{2} + \frac{2}{\pi} \sin(t) + \frac{2}{3\pi} \sin(3t) + \ldots
\]
**Low Pass Filter Analysis:**
The signal can be input into an ideal low pass filter with a cutoff frequency of 5 rad/s:
- Gain = 1 for \(\omega < 5\) rad/s
- Gain = 0 for \(\omega \geq 5\) rad/s
Given this filter, only components with frequencies \(\omega < 5\) rad/s will pass through without attenuation.
**Phase Shift Information:**
The phase shift for different frequencies \(\omega\) is given by:
- \(\varphi(0) = 0 \, \text{rad}\)
- \(\varphi(0.1) = -0.1 \, \text{rad}\)
- \(\varphi(1) = -0.79 \, \text{rad}\)
- \(\varphi(3) = -1.25 \, \text{rad}\)
- \(\varphi(5) = -1.37 \, \text{rad}\)
**Graph Description:**
The graph illustrates the phase response (\(\text{Phase (rad)}\)) of the low pass filter against the frequency (\(\text{Frequency (rad s}^{-1}\text{)}\)). The phase decreases with increasing frequency, indicating the phase shift applied by the filter to different harmonic components.
**Output Signal:**
Since the filter allows frequencies below 5 rad/s to pass, the output signal will comprise harmonics with \
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